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1 |
Theoretical Foundations
|
For an oscillator with charge $q$, its energy operator without an external field is
\begin{equation*}
H_{0}=\frac{p^{2}}{2 m}+\frac{1}{2} m \omega^{2} x^{2}
\end{equation*}
If a uniform electric field $\mathscr{E}$ is applied, causing an additional force on the oscillator $f= q \mathscr{E}$, the total energy operator becomes
\begin{equation*}
H=\frac{p^{2}}{2 m}+\frac{1}{2} m \omega^{2} x^{2}-q \mathscr{E} x
\end{equation*}
Find the expression for the new energy levels $E_{n}$.
|
[
"E_{n} =(n+\\frac{1}{2}) \\hbar \\omega-\\frac{q^{2} \\mathscr{E}^{2}}{2 m \\omega^{2}}"
] |
Expression
|
In $H_{0}$ and $H$, $p$ is the momentum operator,
$p=-\mathrm{i} \hbar \frac{\mathrm{~d}}{\mathrm{~d} x}$
The potential energy term in equation (2) can be rewritten as
$\frac{1}{2} m \omega^{2} x^{2}-q \mathscr{E} x=\frac{1}{2} m \omega^{2}[(x-x_{0})^{2}-x_{0}^{2}]$
where
\begin{equation*}
x_{0}=\frac{q \mathscr{E}}{m \omega^{2}} \tag{3}
\end{equation*}
By performing a coordinate shift, let
\begin{equation*}
x^{\prime}=x-x_{0} \tag{4}
\end{equation*}
Because
\begin{equation*}
p=-\mathrm{i} \hbar \frac{\mathrm{~d}}{\mathrm{~d} x}=-\mathrm{i} \hbar \frac{\mathrm{~d}}{\mathrm{~d} x}=p^{\prime} \tag{5}
\end{equation*}
$H$ can be expressed as
\begin{equation*}
H=\frac{p^{\prime 2}}{2 m}+\frac{1}{2} m \omega^{2} x^{\prime 2}-\frac{1}{2} m \omega^{2} x_{0}^{2} \tag{6}
\end{equation*}
Comparing equations (1) and (6), it is evident that the difference between $H$ and $H_{0}$ is that the variable changes from $x$ to $x^{\prime}$, with an added constant term $(-\frac{1}{2} m \omega^{2} x_{0}^{2})$. Hence, we get
\begin{gather*}
E_{n}=E_{n}^{(0)}-\frac{1}{2} m \omega^{2} x_{0}^{2} \tag{7}\\
\varphi_{n}(x)=\psi_{n}(x^{\prime})=\psi_{n}(x-x_{0}) \tag{8}
\end{gather*}
It is well-known that the energy levels of the free oscillator are
\begin{equation*}
E_{n}^{(0)}=(n+\frac{1}{2}) \hbar \omega, \quad n=0,1,2, \cdots \tag{9}
\end{equation*}
Thus,
\begin{align*}
E_{n} & =(n+\frac{1}{2}) \hbar \omega-\frac{1}{2} m \omega^{2} x_{0}^{2} \\
& =(n+\frac{1}{2}) \hbar \omega-\frac{q^{2} \mathscr{E}^{2}}{2 m \omega^{2}} \tag{10}
\end{align*}
Introducing the coordinate shift operator
\begin{equation*}
D_{x}(x_{0})=\mathrm{e}^{-\mathrm{i} x_{0} p / \hbar}=\mathrm{e}^{-x_{0} \frac{d}{d x}} \tag{11}
\end{equation*}
Its effect on the wave function is
\begin{equation*}
D_{x}(x_{0}) \psi(x)=\psi(x-x_{0}) \tag{11'}
\end{equation*}
Then the eigenfunctions of $H$ and $H_{0}$ can be related through the shift operator:
\begin{equation*}
\varphi_{n}(x)=\psi_{n}(x-x_{0})=D_{x}(x_{0}) \psi_{n}(x) \tag{12}
\end{equation*}
Conversely,
\begin{equation*}
\psi_{n}(x)=\varphi_{n}(x+x_{0})=D_{x}(-x_{0}) \varphi_{n}(x) \tag{$\prime$}
\end{equation*}
Solution two, using the raising and lowering operators of the oscillator
\begin{equation*}
a=\sqrt{\frac{m \omega}{2 \hbar}}(x+\frac{\mathrm{i}}{m \omega} p), \quad a^{+}=\sqrt{\frac{m \omega}{2 \hbar}}(x-\frac{\mathrm{i}}{m \omega} p) \tag{11}
\end{equation*}
Expressing $H_{0}$ and $H$ as
\begin{gather*}
H_{0}=(a^{+} a+\frac{1}{2}) \hbar \omega \tag{14}\\
H=(a^{+} a+\frac{1}{2}) \hbar \omega-q \mathscr{E} \sqrt{\frac{\hbar}{2 m \omega}}(a+a^{+}) \tag{15}
\end{gather*}
Introducing
\begin{equation*}
x_{0}=q \varepsilon / m \omega^{2}
\end{equation*}
Then
\begin{align*}
H & =\hbar \omega[a^{+} a+\frac{1}{2}-x_{0} \sqrt{\frac{m \omega}{2 \hbar}}(a+a^{+})] \\
& =\hbar \omega[(a^{+}-x_{0} \sqrt{\frac{m \omega}{2 \hbar}})(a-x_{0} \sqrt{\frac{m \omega}{2 \hbar}})+\frac{1}{2}-\frac{m \omega x_{0}^{2}}{2 \hbar}] \\
& =\hbar \omega[(a^{+}-\alpha_{0})(a-\alpha_{0})+\frac{1}{2}]-\frac{1}{2} m \omega^{2} x_{0}^{2} \tag{16}
\end{align*}
where
\begin{equation*}
\alpha_{0}=x_{0} \sqrt{m \omega / 2 \hbar} \tag{17}
\end{equation*}
Comparing equations (14), (16), the difference between $H$ and $H_{0}$ is $a \rightarrow a-\alpha_{0}, a^{+} \rightarrow a^{+}-\alpha_{0}$, and an added constant term $(-\frac{1}{2} m \omega^{2} x_{0}^{2})$.
Starting from the fundamental commutation relation
\begin{equation*}
[a, a^{+}]=1 \tag{18}
\end{equation*}
It was proved that the energy level formula (9) and the recursion relations between eigenstates
\begin{equation*}
a \psi_{n}=\sqrt{n} \psi_{n-1}, \quad a^{-} \psi_{n}=\sqrt{n+1} \psi_{n+1} \tag{19}
\end{equation*}
And the ground state wave function satisfies
\begin{equation*}
a \psi_{0}=0 \tag{20}
\end{equation*}
As
$[a-\alpha_{0}, a^{+}-\alpha_{0}]=[a, a^{+}]=1$
So the same reasoning logically leads to similar conclusions for the eigenvalues and eigenfunctions of $H$, simply substituting $(a-\alpha_{0})$ for $a$ in the entire derivation, $(a^{+}-\alpha_{0})$ for $a^{+}$. The eigenvalue of $H$ is clearly equation (10). The recursion relations and ground state equation for the eigenfunctions are
\begin{gather*}
(a-\alpha_{0}) \varphi_{n}(x)=\sqrt{n} \varphi_{n-1}(x) \tag{21}\\
(a^{+}-\alpha_{0}) \varphi_{n}(x)=\sqrt{n+1} \varphi_{n+1}(x) \\
(a-\alpha_{0}) \varphi_{0}(x)=0 \tag{22}
\end{gather*}
$a \rightarrow(a-\alpha_{0})$ (and $a^{+} \rightarrow a^{+}-\alpha_{0})$ are equivalent to $x \rightarrow(x-x_{0})$, thus replacing $x$ with $(x-x_{0})$ in $\psi_{n}(x)$ gives
\begin{equation*}
\varphi_{n}(x)=\psi_{n}(x-x_{0})=D_{x}(x_{0}) \psi_{n}(x) \tag{12}
\end{equation*}
$\varphi_{n}$ and $\varphi_{0}$ can be related through the operator $(a^{+}-\alpha_{0})$:
\begin{equation*}
\varphi_{n}(x)=\frac{1}{\sqrt{n!}}(a^{+}-\alpha_{0})^{n} \varphi_{0}(x) \tag{23}
\end{equation*}
|
{"$q$": "charge of the oscillator", "$H_{0}$": "energy operator without external field", "$p$": "momentum operator", "$m$": "mass", "$\\omega$": "angular frequency", "$x$": "position", "$\\mathscr{E}$": "uniform electric field", "$f$": "force on the oscillator", "$H$": "total energy operator", "$E_{n}$": "new energy levels", "$\\hbar$": "reduced Planck's constant", "$x_{0}$": "displacement caused by the field", "$x^{\\prime}$": "shifted coordinate", "$p^{\\prime}$": "shifted momentum operator", "$\\varphi_{n}$": "eigenfunction of the total energy operator", "$\\psi_{n}$": "eigenfunction of the initial operator", "$a$": "lowering operator", "$a^{+}$": "raising operator", "$\\alpha_{0}$": "constant related to displacement"}
|
Schrödinger equation one-dimensional motion
| |
2 |
Theoretical Foundations
|
A particle of mass $m$ is in the ground state of a one-dimensional harmonic oscillator potential
\begin{equation*}
V_{1}(x)=\frac{1}{2} k x^{2}, \quad k>0
\end{equation*}
When the spring constant $k$ suddenly changes to $2k$, the potential then becomes
\begin{equation*}
V_{2}(x)=k x^{2}
\end{equation*}
Immediately measure the energy of the particle, and find the expression for the probability of the particle being in the ground state of the new potential $V_{2}$.
|
[
"\\frac{2^{5 / 4}}{1+\\sqrt{2}}"
] |
Expression
|
(a) The wave function of the particle $\psi(x, t)$ should satisfy the time-dependent Schrödinger equation
\begin{equation*}
\mathrm{i} \hbar \frac{\partial}{\partial t} \psi=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}} \psi+V \psi \tag{3}
\end{equation*}
When $V$ undergoes a sudden change (from $V_{1} \rightarrow V_{2}$) but with a finite change quantity, $\psi$ remains a continuous function of $t$, implying that $\psi$ does not change when $V$ changes abruptly.
Denote $\psi_{0}(x)$ and $\phi_{0}(x)$ as the ground state wave functions of the potential $V_{1}$ and $V_{2}$, respectively. After the potential suddenly changes from $V_{1}$ to $V_{2}$, the wave function of the particle remains $\psi_{0}$. The probability of measuring the particle in the state $\phi_{0}$ is $|\langle\psi_{0} \mid \phi_{0}\rangle|^{2}$.
Rewrite $V_{1}$ and $V_{2}$ in standard form:
\begin{gather*}
V_{1}(x)=\frac{1}{2} k x^{2}=\frac{1}{2} m \omega_{1}^{2} x^{2} \tag{$\prime$}\\
V_{2}(x)=k x^{2}=\frac{1}{2} m \omega_{2}^{2} x^{2} \tag{$\prime$}
\end{gather*}
It is clear that
\begin{equation*}
\omega_{2}=\sqrt{2} \omega_{1} \tag{4}
\end{equation*}
$\psi_{0}$ and $\phi_{0}$ can be expressed as in formulas (3) and (5) from problem 3.2, namely
\begin{align}
\psi_{0}(x) &= \left( \frac{\alpha}{\sqrt{\pi}} \right)^{\frac{1}{2}} \mathrm{e}^{-\alpha^2 x^2 / 2}, &
\alpha^{2} &= m \omega_{1} / \hbar \notag \\
\phi_{0}(x) &= \left( \frac{\beta}{\sqrt{\pi}} \right)^{\frac{1}{2}} \mathrm{e}^{-\beta^2 x^2 / 2}, &
\beta^{2} &= m \omega_{2} / \hbar \tag{6}
\end{align}
where
\begin{equation*}
\beta^{2} / \alpha^{2}=\omega_{2} / \omega_{1}=\sqrt{2} \tag{7}
\end{equation*}
Thus
\begin{align*}
\langle\psi_{0} \mid \phi_{0}\rangle & =\sqrt{\frac{\alpha \beta}{\pi}} \int_{-\infty}^{+\infty} \mathrm{e}^{-\frac{1}{2}(\alpha^{2}+\beta^{2}) x^{2}} \mathrm{~d} x=(\frac{2 \alpha \beta}{\alpha^{2}+\beta^{2}})^{\frac{1}{2}} \\
|\langle\psi_{0} \mid \phi_{0}\rangle|^{2} & =\frac{2 \alpha \beta}{\alpha^{2}+\beta^{2}}=\frac{2 \beta / \alpha}{1+\beta^{2} / \alpha^{2}}=\frac{2^{5 / 4}}{1+\sqrt{2}}=0.9852 \tag{8}
\end{align*}
This is the required probability.
(b) Consider the time when the potential changes for the first time $(V_{1} \rightarrow V_{2})$ as $t=0$, then the wave function is
\begin{equation*}
\psi(x, 0)=\psi_{0}(x) \tag{9}
\end{equation*}
Let $\phi_{n}(x)$ denote the energy eigenstates of the potential $V_{2}$, corresponding to energy levels
$E_{n}=(n+\frac{1}{2}) \hbar \omega_{2}$
Expand $\psi_{0}$ as a linear combination of $\phi_{n}$,
\begin{equation*}
\psi_{0}(x)=\sum_{n} C_{n} \phi_{n}(x), \quad(n \text { can only take even values }) \tag{10}
\end{equation*}
For $0<t<\tau$, in Schrödinger equation (3) $V=V_{2}(x)$, its solution is
\begin{align*}
\psi(x, t) & =\sum_{n} C_{n} \phi_{n}(x) \mathrm{e}^{-\mathrm{i} E_{n} / / \hbar} \\
& =\mathrm{e}^{-i \omega_{2} t \cdot 2} \sum_{n} C_{n} \phi_{n}(x) \mathrm{e}^{-i \omega_{\omega_{2}} t} \tag{11}
\end{align*}
To have $\psi(x, \tau)=A \psi_{0}(x)$, it must hold that
\begin{equation*}
\mathrm{e}^{-\mathrm{in} \omega_{2} \tau}=1, \quad n=0,2,4, \cdots \tag{12}
\end{equation*}
That is
\begin{equation*}
\mathrm{e}^{\mathrm{i} \omega_{2} \tau}= \pm 1 \tag{12'}
\end{equation*}
The $\tau$ that satisfies this condition is
\begin{equation*}
\tau=l \pi / \omega_{2}=l \pi \sqrt{\frac{m}{2 k}}, \quad l=1,2,3, \cdots \tag{13}
\end{equation*}
When $t=\tau$, after the potential changes from $V_{2}$ back to $V_{1}$, the particle remains in the state $\psi_{0}$, with energy $E=\hbar \omega_{1} / 2$.
|
{"$m$": "mass of the particle", "$k$": "spring constant", "$x$": "position", "$\\omega_{1}$": "angular frequency for potential $V_{1}$", "$\\omega_{2}$": "angular frequency for potential $V_{2}$", "$\\psi_{0}$": "ground state wave function of $V_{1}$", "$\\phi_{0}$": "ground state wave function of $V_{2}$", "$\\hbar$": "reduced Planck's constant", "$\\alpha$": "parameter related to $\\omega_{1}$", "$\\beta$": "parameter related to $\\omega_{2}$", "$t$": "time", "$\\tau$": "specific time when potential changes", "$E_{n}$": "energy level corresponding to state $\\phi_{n}$", "$n$": "quantum number"}
|
Schrödinger equation one-dimensional motion
| |
3 |
Theoretical Foundations
|
A harmonic oscillator with charge $q$ is in a free vibration state at $t<0$ and $t>\tau$, with the total energy operator given by
\begin{equation*}
H_{0}=\frac{1}{2 m} p^{2}+\frac{1}{2} m \omega^{2} x^{2}
\end{equation*}
The energy eigenstates are denoted by $\psi_{n}$, and the energy levels $E_{n}^{(0)}=(n+\frac{1}{2}) \hbar \omega$ . When $0 \leqslant t \leqslant \tau$, a uniform electric field $\mathscr{E}$ is applied, and the total energy operator becomes
\begin{equation*}
H^{\prime}=\frac{1}{2 m} p^{2}+\frac{1}{2} m \omega^{2} x^{2}-q \mathscr{\varepsilon} x
\end{equation*}
Assuming the oscillator is in the ground state $\psi_{0}$ at $t \leqslant 0$, find the expression for the probability $P_n$ that the system is in the energy eigenstate $\psi_{n}$ at $t>\tau$.
|
[
"P_n = \\frac{1}{n!}(2 \\alpha_{0} \\sin \\frac{\\omega \\tau}{2})^{2 n} \\mathrm{e}^{-(2 \\alpha_{0} \\sin \\frac{\\omega \\tau}{2})^{2}}"
] |
Expression
|
At $t>\tau$, the external electric field has vanished, and the wavefunction satisfies the Schrödinger equation
\begin{equation*}
\mathrm{i} \hbar \frac{\partial}{\partial t} \psi(x, t)=H_{0} \psi(x, t) \tag{3}
\end{equation*}
The general solution is
\begin{equation*}
\psi(x, t)=\sum_{n} f_{n} \psi_{n}(x) \mathrm{e}^{-\mathrm{i} E_{n}^{(0)}(t-\tau) / \hbar} \tag{4}
\end{equation*}
where the components of the $\psi_{n}$ states are
\begin{equation*}
|\langle\psi_{n} \mid \psi\rangle|^{2}=|f_{n}|^{2} \tag{5}
\end{equation*}
Each coefficient $f_{n}$ depends on the wavefunction at $t=\tau$
\begin{equation*}
\psi(x, \tau)=\sum_{n} f_{n} \psi_{n}(x) \tag{6}
\end{equation*}
So the key is to find $\psi(x, \tau)$.
In the interval $0 \leqslant t \leqslant \tau$, the Schrödinger equation is
\begin{equation*}
\mathrm{i} \hbar \frac{\partial}{\partial t} \psi(x, t)=H \psi(x, t) \tag{7}
\end{equation*}
The solution is
\begin{equation*}
\psi(x, t)=\sum_{n} C_{n} \varphi_{n}(x) \mathrm{e}^{-\mathrm{i} \mathrm{E}_{n} t / \hbar} \tag{8}
\end{equation*}
where $C_{n}$ depends on the initial wavefunction
\begin{equation*}
\psi(x, 0)=\psi_{0}(x)=\sum_{n} C_{n} \varphi_{n}(x) \tag{9}
\end{equation*}
It has been proven that
\begin{equation*}
\psi_{0}(x)=\varphi_{0}(x+x_{0})=D_{x}(-x_{0}) \varphi_{0}(x) \tag{10}
\end{equation*}
where $x_{0}=q \mathscr{E} / m \omega^{2}$. If we express the displacement operator $D_{x}(-x_{0})$ in terms of the ladder operators,
\begin{equation*}
D_{x}(-x_{0})=\mathrm{e}^{\mathrm{i} x_{0} p / \hbar}=\mathrm{e}^{-a_{0}(a^{+}-a)} \tag{11}
\end{equation*}
Utilizing Glauber's formula
$\mathrm{e}^{A+B}=\mathrm{e}^{A} \mathrm{e}^{B} \mathrm{e}^{-\frac{1}{2}[A, B]}$
We obtain
\begin{align*}
D_{x}(-x_{0}) & =\mathrm{e}^{-\frac{1}{2} \alpha_{0}^{2}} \mathrm{e}^{-\alpha_{0} a^{+}} \mathrm{e}^{\alpha_{0} a} \\
& =\mathrm{e}^{-\frac{1}{2} \alpha_{0}^{2}} \mathrm{e}^{-\alpha_{0}(a^{+}-\alpha_{0})} \mathrm{e}^{\alpha_{0}(a-\alpha_{0})} \tag{$\prime$}
\end{align*}
where $\alpha_{0}=x_{0} \sqrt{m \omega / 2 \hbar}$. Substituting equation ( $11^{\prime}$ ) into equation (10), we obtain
\begin{equation*}
\psi_{0}(x)=\mathrm{e}^{-\frac{1}{2} \alpha_{0}^{2}} \mathrm{e}^{-\alpha_{0}(a^{+}-\alpha_{0})} \mathrm{e}^{\alpha_{0}(a-\alpha_{0})} \varphi_{0}(x) \tag{12}
\end{equation*}
It has been shown in item 3.4
\begin{gather*}
(a-\alpha_{0}) \varphi_{0}=0 \tag{13}\\
(a^{+}-\alpha_{0})^{n} \varphi_{0}=\sqrt{n!} \varphi_{n} \tag{14}
\end{gather*}
Thus,
\begin{align*}
& \mathrm{e}^{\alpha_{0}(a-\alpha_{0})} \varphi_{0}=\sum_{n=0}^{\infty} \frac{\alpha_{0}^{n}}{n!}(a-\alpha_{0})^{n} \varphi_{0}=\varphi_{0} \tag{$\prime$}\\
& \begin{aligned}
\psi_{0}(x) & =\mathrm{e}^{-\frac{1}{2} \alpha_{0}^{2}} \mathrm{e}^{-\alpha_{0}(a^{+}-\alpha_{0})} \varphi_{0} \\
& =\mathrm{e}^{-\frac{1}{2} \alpha_{0}^{2}} \sum_{n} \frac{(-\alpha_{0})^{n}}{n!}(a^{+}-\alpha_{0})^{n} \varphi_{0} \\
& =\mathrm{e}^{-\frac{1}{2} \alpha_{0}^{2}} \sum_{n} \frac{(-\alpha_{0})^{n}}{\sqrt{n!}} \varphi_{n}(x)
\end{aligned}
\end{align*}
namely, $\psi_{0}(x)$ is a coherent state wavefunction composed of the basis vectors $\varphi_{n}$. Comparing equations (9) and (15), we obtain
\begin{equation*}
C_{n}=\frac{(-\alpha_{0})^{n}}{\sqrt{n!}} \mathrm{e}^{-\frac{1}{2} \alpha_{0}^{2}} \tag{16}
\end{equation*}
Substituting equation (16) into equation (8), and using the energy formula derived in item 3.4
$E_{n}=(n+\frac{1}{2}) \hbar \omega-\frac{1}{2} m \omega^{2} x_{0}^{2}$
we obtain
\begin{equation*}
\psi(x, \tau)=\mathrm{e}^{\mathrm{i} \delta} \mathrm{e}^{-\frac{1}{2} a_{0}^{2}} \sum_{n} \frac{[-\alpha(\tau)]^{n}}{\sqrt{n!}} \varphi_{n}(x) \tag{17}
\end{equation*}
where
\begin{gather*}
\alpha(\tau)=\alpha_{0} \mathrm{e}^{-\mathrm{i} \omega \tau} \tag{18}\\
\delta=\frac{m \omega^{2} x_{0}^{2} \tau}{2 \hbar}-\frac{\omega \tau}{2} \tag{19}
\end{gather*}
$\psi(x, \tau)$ is also a coherent state wavefunction composed of the $\varphi_{n}$ states.
Using equation (14), equation (17) can be expressed as
\begin{equation*}
\psi(x, \tau)=\mathrm{e}^{\mathrm{i} \delta} \mathrm{e}^{-\frac{1}{2} \alpha_{0}^{2}} \mathrm{e}^{-\alpha(\tau)(a^{+}-\alpha_{0})} \varphi_{0}(x) \tag{20}
\end{equation*}
From equation (12 ${ }^{\prime}$ ), it follows that
$\mathrm{e}^{-\frac{1}{2} \alpha_{0}^{2}} \varphi_{0}=\mathrm{e}^{\alpha_{0}(a^{+}-\alpha_{0})} \psi_{0}$
Substituting into equation (20), we obtain
\begin{equation*}
\psi(x, \tau)=\mathrm{e}^{\mathrm{i} \delta} \mathrm{e}^{[\alpha_{0}-\alpha(\tau)](a^{+}-a_{0})} \psi_{0}=\mathrm{e}^{\wp \delta} \mathrm{e}^{-\alpha_{0}^{2}+\alpha_{0} \alpha(\tau)} \mathrm{e}^{[\alpha_{0}-\alpha(\tau)] a^{+}} \psi_{0} \tag{21}
\end{equation*}
To express $\psi(x, \tau)$ as a linear superposition of the $\psi_{n}$, we use the formula
\begin{equation*}
(a^{+})^{n} \psi_{0}=\sqrt{n!} \psi_{n} \tag{22}
\end{equation*}
Thus, we have
\begin{equation*}
\psi(x, \tau)=\mathrm{e}^{\mathrm{i} \delta} \mathrm{e}^{\alpha_{0} \alpha(\tau)-\alpha_{0}^{2}} \sum_{n} \frac{[\alpha_{0}-\alpha(\tau)]^{n}}{\sqrt{n!}} \psi_{n}(x) \tag{23}
\end{equation*}
where the component fraction of the $\psi_{n}$ state is
\begin{align*}
|\langle\psi_{n} \mid \psi(x, \tau)\rangle|^{2} & =|\mathrm{e}^{\alpha_{0} \alpha(\tau)-\alpha_{0}^{2}}|^{2} \frac{|\alpha_{0}-\alpha(\tau)|^{2 n}}{n!} \\
& =\frac{1}{n!}(2 \alpha_{0} \sin \frac{\omega \tau}{2})^{2 n} \mathrm{e}^{-(2 \alpha_{0} \sin \frac{\omega \tau}{2})^{2}} \tag{24}
\end{align*}
It can be easily verified that
\begin{equation*}
\sum_{n}|\langle\psi_{n} \mid \psi(x, \tau)\rangle|^{2}=1 \tag{25}
\end{equation*}
This is a specific manifestation of the conservation of total probability.
By changing the external electric field duration $\tau$, each time
$\tau=k 2 \pi / \omega, \quad k=1,2,3, \cdots$
In $\psi(x, \tau)$, the components of each excited state $(n \geqslant 1)$ become 0, and the ground state $(\psi_{0})$ component is 1. Each time
$\tau=(2 k+1) \pi / \omega, \quad k=0,1,2, \cdots$
the ground state component in $\psi(x, \tau)$ reaches a minimum value of $\mathrm{e}^{-4 \alpha_{0}^{2}}, \psi_{n}$ state components are
$$|f_{n}|^{2}=\frac{(4 \alpha_{0}^{2})^{n}}{n!} \mathrm{e}^{-4 \alpha_{0}^{2}}$$
|
{"$q$": "charge", "$t$": "time", "$\\tau$": "duration period for electric field", "$H_{0}$": "initial energy operator", "$m$": "mass", "$p$": "momentum", "$\\omega$": "angular frequency", "$x$": "position", "$\\psi_{n}$": "energy eigenstate", "$E_{n}^{(0)}$": "energy level without electric field", "$\\hbar$": "reduced Planck's constant", "$\\mathscr{E}$": "electric field", "$H^{\\prime}$": "energy operator with electric field", "$\\mathscr{\\varepsilon}$": "electric field", "$\\psi_{0}$": "ground state wavefunction", "$P_n$": "probability of the system being in the energy eigenstate $\\psi_{n}$", "$f_{n}$": "coefficients for the wavefunction at a specific time", "$E_{n}$": "energy level with electric field", "$C_{n}$": "initial coefficients in wavefunction expansion", "$\\varphi_{n}$": "basis vectors", "$x_{0}$": "displacement due to electric field", "$D_{x}$": "displacement operator", "$a$": "annihilation operator", "$a^{+}$": "creation operator", "$\\alpha_{0}$": "dimensionless displacement parameter", "$\\alpha(\\tau)$": "time-dependent displacement parameter", "$\\delta$": "phase factor", "$\\wp$": "phase factor"}
|
Schrödinger equation one-dimensional motion
| |
4 |
Theoretical Foundations
|
Calculate the result of the commutator $[\boldsymbol{p}, \frac{1}{r}]$.
|
[
"\\mathrm{i} \\hbar \\frac{\\boldsymbol{r}}{r^{3}}"
] |
Expression
|
Using the commutator
\begin{equation*}
[\boldsymbol{p}, F(\boldsymbol{r})]=-\mathrm{i} \hbar \nabla F \tag{1}
\end{equation*}
We obtain
\begin{equation*}
[p, \frac{1}{r}]=-\mathrm{i} \hbar(\nabla \frac{1}{r})=\mathrm{i} \hbar \frac{r}{r^{3}} \tag{2}
\end{equation*}
Using the formula (see question 4.2)
$[\boldsymbol{A} \cdot \boldsymbol{B}, F]=[\boldsymbol{A}, F] \cdot \boldsymbol{B}+\boldsymbol{A} \cdot[\boldsymbol{B}, F]$
We obtain
\[
\begin{aligned}
{[\boldsymbol{p}^{2}, \frac{1}{r}] } & =[\boldsymbol{p}, \frac{1}{r}] \cdot \boldsymbol{p}+\boldsymbol{p} \cdot[\boldsymbol{p}, \frac{1}{r}]=\mathrm{i} \hbar(\frac{1}{r^{3}} \boldsymbol{r} \cdot \boldsymbol{p}+\boldsymbol{p} \cdot \frac{\boldsymbol{r}}{r^{3}}) \\
& =2 \mathrm{i} \hbar \frac{1}{r^{3}} \boldsymbol{r} \cdot \boldsymbol{p}+\hbar^{2}(\nabla \cdot \frac{\boldsymbol{r}}{r^{3}})
\end{aligned}
\]
However, since
$\nabla \cdot \frac{\boldsymbol{r}}{r^{3}}=\frac{1}{r^{3}} \nabla \cdot \boldsymbol{r}+\boldsymbol{r} \cdot(\nabla \frac{1}{r^{3}})=\frac{3}{r^{3}}-\boldsymbol{r} \cdot \frac{3 \boldsymbol{r}}{r^{5}}=0$
So
\begin{equation*}
[\boldsymbol{p}^{2}, \frac{1}{r}]=2 \mathrm{i} \hbar \frac{1}{r^{3}} \boldsymbol{r} \cdot \boldsymbol{p}=2 \hbar^{2} \frac{1}{r^{2}} \frac{\partial}{\partial r} \tag{3}
\end{equation*}
Subsequently, using equation (1), we get
\begin{align*}
{[\boldsymbol{p}, r^{2}] } & =-i \hbar \nabla(r^{2})=-2 i \hbar \boldsymbol{r} \tag{4}\\
{[\boldsymbol{p}^{2}, r^{2}] } & =[\boldsymbol{p}, r^{2}] \cdot \boldsymbol{p}+\boldsymbol{p} \cdot[\boldsymbol{p}, r^{2}]=-2 i \hbar(\boldsymbol{r} \cdot \boldsymbol{p}+\boldsymbol{p} \cdot \boldsymbol{r}) \\
& =-4 i \hbar \boldsymbol{r} \cdot \boldsymbol{p}-2 \hbar^{2} \nabla \cdot \boldsymbol{r}=-4 i \hbar \boldsymbol{r} \cdot \boldsymbol{p}-6 \hbar^{2} \\
& =-4 \hbar^{2} r \frac{\partial}{\partial r}-6 \hbar^{2} \tag{5}
\end{align*}
Finally, using the commutator
\begin{equation*}
[\boldsymbol{p}^{2}, \boldsymbol{r}]=-\mathrm{i} \hbar \frac{\partial p^{2}}{\partial \boldsymbol{p}}=-2 \mathrm{i} \hbar \boldsymbol{p} \tag{6}
\end{equation*}
And equation (3), we get
\begin{align*}
{[\boldsymbol{p}^{2}, \frac{\boldsymbol{r}}{r}] } & =[\boldsymbol{p}^{2}, \boldsymbol{r}] \frac{1}{r}+\boldsymbol{r}[\boldsymbol{p}^{2}, \frac{1}{r}] \\
& =-2 \mathrm{i} \hbar \boldsymbol{p} \frac{1}{r}+2 \mathrm{i} \hbar \boldsymbol{r} \frac{1}{r^{3}}(\boldsymbol{r} \cdot \boldsymbol{p}) \\
& =-2 \mathrm{i} \hbar[\frac{1}{r} \boldsymbol{p}-\mathrm{i} \hbar(\nabla \frac{1}{r})]+2 \mathrm{i} \hbar \frac{\boldsymbol{r}}{r^{3}}(\boldsymbol{r} \cdot \boldsymbol{p}) \\
& =2 \hbar^{2} \frac{\boldsymbol{r}}{r^{3}}+2 \mathrm{i} \hbar[\frac{\boldsymbol{r}}{r^{3}}(\boldsymbol{r} \cdot \boldsymbol{p})-\frac{1}{r} \boldsymbol{p}] \\
& =2 \hbar^{2}(\frac{\boldsymbol{r}}{r^{3}}+\boldsymbol{r} \frac{1}{r^{2}} \frac{\partial}{\partial r}-\frac{1}{r} \nabla) \tag{7}
\end{align*}
|
{"$\\boldsymbol{p}$": "momentum operator", "$\\hbar$": "reduced Planck's constant", "$\\boldsymbol{r}$": "position vector", "$r$": "magnitude of the position vector $\\boldsymbol{r}$"}
|
Schrödinger equation one-dimensional motion
| |
5 |
Theoretical Foundations
|
For the hydrogen-like ion (nuclear charge $Z e$) with the $(H, l^{2}, l_{z})$ common eigenstate $\psi_{n l m}$, it is known that the various $\langle r^{\lambda}\rangle$ satisfy the following recursion relation (Kramers' formula):
\begin{equation*}
\frac{\lambda+1}{n^{2}}\langle r^{\lambda}\rangle-(2 \lambda+1) \frac{a_{0}}{Z}\langle r^{\lambda-1}\rangle+\frac{\lambda}{4}[(2 l+1)^{2}-\lambda^{2}] \frac{a_{0}^{2}}{Z^{2}}\langle r^{\lambda-2}\rangle=0
\end{equation*}
It is known that $\langle r^{0}\rangle=1$. Use this formula to calculate the expression for $\langle r\rangle_{n l m}$.
|
[
"\\langle r\\rangle_{n l m}=\\frac{1}{2}[3 n^{2}-l(l+1)] \\frac{a_{0}}{Z}"
] |
Expression
|
The spherical coordinate expression of $\psi_{n l m}$ is
\begin{equation*}
\psi_{n l m}=R_{n l}(r) \mathrm{Y}_{l m}(\theta, \varphi)=\frac{1}{r} u_{n l}(r) \mathrm{Y}_{l m}(\theta, \varphi) \tag{2}
\end{equation*}
The expectation value of $r^{\lambda}$ is
\begin{equation*}
\langle r^{\lambda}\rangle_{n l m}=\int r^{\lambda}|\psi_{n l m}|^{2} \mathrm{~d}^{3} x=\int_{0}^{\infty} r^{\lambda}(u_{n l})^{2} \mathrm{~d} r \tag{3}
\end{equation*}
$u_{n l}$ satisfies the radial equation
\begin{equation*}
-\frac{\hbar^{2}}{2 \mu} u^{\prime \prime}+[l(l+1) \frac{\hbar^{2}}{2 \mu r^{2}}-\frac{Z e^{2}}{r}] u=E_{n} u \tag{4}
\end{equation*}
Because
\begin{equation*}
E_{n}=-\frac{Z^{2} e^{2}}{2 n^{2} a_{0}}, \quad a_{0}=\frac{\hbar^{2}}{\mu e^{2}} \tag{5}
\end{equation*}
Equation (4) can be rewritten as
\begin{equation*}
u^{\prime \prime}+[\frac{2 Z}{a_{0} r}-\frac{l(l+1)}{r^{2}}-(\frac{Z}{n a_{0}})^{2}] u=0 \tag{$\prime$}
\end{equation*}
Multiply each term of equation ($4^{\prime}$) by $r^{\lambda} u$ and integrate $\int_{0}^{\infty} \cdots \mathrm{d} r$, the last three terms clearly yield the expectation values of $r^{\lambda-1}, ~ r^{\lambda-2}$ and $r^{\lambda}$, while
The first term gives
\begin{align*}
\int_{0}^{\infty} r^{\lambda} u u^{\prime \prime} \mathrm{d} r & =r^{\lambda} u u^{\prime}|_{0} ^{\infty}-\int_{0}^{\infty}(r^{\lambda} u^{\prime}+\lambda r^{\lambda-1} u) u^{\prime} \mathrm{d} r \\
& =(r^{\lambda} u u^{\prime}-\frac{\lambda}{2} r^{\lambda-1} u^{2})|_{0} ^{\infty}+\frac{\lambda(\lambda-1)}{2}\langle r^{\lambda-2}\rangle-\int_{0}^{\infty} r^{\lambda}(u^{\prime})^{2} \mathrm{~d} r \tag{6}
\end{align*}
If the value of $\lambda$ ensures that
\begin{equation*}
r^{\lambda} u u^{\prime}|_{0} ^{\infty}=0,\quad r^{\lambda-1} u^{2}|_{0} ^{\infty}=0 \tag{7}
\end{equation*}
We obtain the following preliminary result:
\begin{equation*}
[\frac{\lambda(\lambda-1)}{2}-l(l+1)]\langle r^{\lambda-2}\rangle+\frac{2 Z}{a_{0}}\langle r^{\lambda-1}\rangle-(\frac{Z}{n a_{0}})^{2}\langle r^{\lambda}\rangle=\int_{0}^{\infty} r^{\lambda}(u^{\prime})^{2} \mathrm{~d} r \tag{8}
\end{equation*}
Moreover, multiply each term of equation ($4^{\prime}$) by $2 r^{\lambda+1} u^{\prime}$ and integrate, successively obtaining
\[
\begin{aligned}
& \int_{0}^{\infty} 2 r^{\lambda+1} u^{\prime} u^{\prime \prime} \mathrm{d} r=r^{\lambda+1}(u^{\prime})^{2}|_{0} ^{\infty}-\int_{0}^{\infty}(\lambda+1) r^{\lambda}(u^{\prime})^{2} \mathrm{~d} r \\
& \int_{0}^{\infty} 2 r^{\lambda+1} u^{\prime} u \mathrm{~d} r=r^{\lambda+1} u^{2}|_{0} ^{\infty}-(\lambda+1)\langle r^{\lambda}\rangle \\
& \int_{0}^{\infty} 2 r^{\lambda+1} u^{\prime} \frac{u}{r} \mathrm{~d} r=r^{\lambda} u^{2}|_{0} ^{\infty}-\lambda\langle r^{\lambda-1}\rangle \\
& \int_{0}^{\infty} 2 r^{\lambda+1} u^{\prime} \frac{u}{r^{2}} \mathrm{~d} r=r^{\lambda-1} u^{2}|_{0} ^{\infty}-(\lambda-1)\langle r^{\lambda-2}\rangle
\end{aligned}
\]
Under the conditions guaranteed by equation (7), all first terms in the above equations are zero, thus
\begin{equation*}
(\lambda-1) l(l+1)\langle r^{\lambda-2}\rangle-2 \lambda \frac{Z}{a_{0}}\langle r^{\lambda-1}\rangle+(\lambda+1)(\frac{Z}{n a_{0}})^{2}\langle r^{\lambda}\rangle=(\lambda+1) \int_{0}^{\infty} r^{\lambda}(u^{\prime})^{2} \mathrm{~d} r \tag{9}
\end{equation*}
Combine equations (8) and (9), eliminating the integrals on the right to obtain equation (1). The condition for the validity of equation (1) is equation (7). Given ${ }^{(1)}$
\begin{align*}
& r \rightarrow 0, \quad u \sim r^{l+1} \\
& r \rightarrow \infty, \quad u \sim r^{n} \mathrm{e}^{-Z r / n a_{0}} \tag{10}
\end{align*}
It is evident that the necessary and sufficient condition for equation (7) to hold is
\begin{equation*}
\lambda>-(2 l+1) \tag{11}
\end{equation*}
In equation (1), taking $\lambda=0$ and noting $\langle r^{0}\rangle=1$ immediately yields
\begin{equation*}
\langle\frac{1}{r}\rangle_{n l m}=\frac{Z}{n^{2} a_{0}} \tag{12}
\end{equation*}
\footnotetext{
(1) Refer to Zeng Jin-Yan. Quantum Mechanics Volume I. Beijing: Science Press, 1997. §6.3.
}
This result has been obtained using the virial theorem.
Sequentially taking $\lambda=1, 2$, leads to
\begin{gather*}
\langle r\rangle_{n l m}=\frac{1}{2}[3 n^{2}-l(l+1)] \frac{a_{0}}{Z} \tag{13}\\
\langle r^{2}\rangle_{n l m}=\frac{n^{2}}{2}[1+5 n^{2}-3 l(l+1)](\frac{a_{0}}{Z})^{2} \tag{14}
\end{gather*}
For example
\[
\begin{array}{rc}
1 \mathrm{~s} \text { state (ground state), } & \langle r\rangle_{100}=\frac{3}{2},\langle r^{2}\rangle_{100}=3 \\
2 \mathrm{~s} \text { state, } & \langle r\rangle_{200}=6,\langle r^{2}\rangle_{200}=42 \tag{15}\\
2 \mathrm{p} \text { state, } & \langle r\rangle_{21 m}=5,\langle r^{2}\rangle_{21 m}=30
\end{array}
\]
In equation (15), $\langle r\rangle$ is measured in units of $a_{0} / Z$ and $\langle r^{2}\rangle$ is measured in units of $a_{0}^{2} / Z^{2}$.
Note that in the context of this problem, equation (1) cannot be used to calculate $\langle r^{-2}\rangle$, but if the result from the previous problem on $\langle r^{-2}\rangle$ is used, then by substituting $\lambda=-1$ into equation (1), $\langle r^{-3}\rangle$ can be calculated, with results consistent with the previous problem. Furthermore, by taking $\lambda=-2(l \geqslant 1)$, it is possible to calculate
\begin{equation*}
\langle r^{-4}\rangle=(\frac{Z}{a_{0}})^{4} \frac{3 n^{2}-l(l+1)}{2 n^{5}(l-\frac{1}{2}) l(l+\frac{1}{2})(l+1)(l+\frac{3}{2})} \tag{16}
\end{equation*}
Calculating other expectations $\langle r^{\lambda}\rangle$ can follow this analogy ${ }^{(1)}$.
|
{"$Z$": "nuclear charge divided by elementary charge", "$e$": "elementary charge", "$H$": "hydrogen atom", "$l^{2}$": "squared orbital angular momentum", "$l_{z}$": "z component of angular momentum", "$\\psi_{n l m}$": "wavefunction in the common eigenstate basis", "$n$": "principal quantum number", "$a_{0}$": "Bohr radius", "$r$": "radial distance", "$\\lambda$": "exponent of the radial distance", "$E_{n}$": "energy of the level n", "$\\hbar$": "reduced Planck constant", "$\\mu$": "reduced mass", "$u_{n l}$": "radial component of the wavefunction", "$R_{n l}$": "radial wavefunction", "$\\mathrm{Y}_{l m}$": "spherical harmonics", "$\\theta$": "polar angle", "$\\varphi$": "azimuthal angle", "$l$": "orbital quantum number", "$m$": "magnetic quantum number", "$n^{2}$": "principal quantum number squared"}
|
Schrödinger equation one-dimensional motion
| |
6 |
Theoretical Foundations
|
Three-dimensional isotropic harmonic oscillator, the total energy operator is
\begin{equation*}
H=\frac{\boldsymbol{p}^{2}}{2 \mu}+\frac{1}{2} \mu \omega^{2} r^{2}=-\frac{\hbar^{2}}{2 \mu} \nabla^{2}+\frac{1}{2} \mu \omega^{2} r^{2}
\end{equation*}
For the common eigenstates of $(H, l^{2}, l_{z})$
\begin{equation*}
\psi_{n_{r} l m}=R_{n_{r} l}(r) \mathrm{Y}_{l m}(\theta, \varphi)=\frac{1}{r} u_{n_{r} l}(r) \mathrm{Y}_{l m}(\theta, \varphi)
\end{equation*}
Calculate the expression for $\langle r^{-2}\rangle_{n_{r} l m}$.
|
[
"\\langle\\frac{1}{r^{2}}\\rangle_{n, l m}=\\frac{1}{l+\\frac{1}{2}} \\frac{\\mu \\omega}{\\hbar}"
] |
Expression
|
The energy levels of the three-dimensional isotropic harmonic oscillator are
\begin{equation*}
E_{n, l m}=E_{N}=\left(N+\frac{3}{2}\right) \hbar \omega, \quad N=l+2 n_{r} . \tag{3}
\end{equation*}
For $\psi_{n_{r}, l m}, H$ is equivalent to
\footnotetext{
(1) Refer to H. A. Kramers. Quantum Mechanics. Amsterdam: North-Holland, 1958. § 59.
}
\begin{equation*}
H \rightarrow-\frac{\hbar^{2}}{2 \mu} \frac{1}{r} \frac{\partial^{2}}{\partial r^{2}} r+l(l+1) \frac{\hbar^{2}}{2 \mu r^{2}}+\frac{1}{2} \mu \omega^{2} r^{2} \tag{4}
\end{equation*}
According to the Hellmann theorem, there should be
\begin{equation*}
\frac{\partial E_{n_{l} l m}}{\partial l}=\langle\frac{\partial H}{\partial l}\rangle_{n_{r} l m}=(l+\frac{1}{2}) \frac{\hbar^{2}}{\mu}\langle\frac{1}{r^{2}}\rangle_{n_{r}, l m} \tag{5}
\end{equation*}
From equation (3) it is evident
\begin{equation*}
\frac{\partial E_{n, l m}}{\partial l}=\frac{\partial E_{N}}{\partial N}=\hbar \omega \tag{6}
\end{equation*}
Substituting into equation (5), we get
\begin{equation*}
\langle\frac{1}{r^{2}}\rangle_{n, l m}=\frac{1}{l+\frac{1}{2}} \frac{\mu \omega}{\hbar}=\frac{\alpha^{2}}{l+\frac{1}{2}}, \quad \alpha=\sqrt{\frac{\mu \omega}{\hbar}} \tag{7}
\end{equation*}
The average value of the centrifugal potential energy is
\begin{equation*}
\langle\frac{l^{2}}{2 \mu r^{2}}\rangle_{n_{r}, m}=\frac{l(l+1)}{2 l+1} \hbar \omega \tag{8}
\end{equation*}
Note that the average value of the centrifugal potential energy is directly determined by the angular quantum number $l$ and is independent of the principal quantum number $N$. Among the states with the same energy level $E_{N}$, the centrifugal potential energy is highest in the state $l=N$.
\begin{equation*}
\langle\frac{l^{2}}{2 \mu r^{2}}\rangle_{l=N}=\frac{N(N+1)}{2 N+1} \hbar \omega \tag{9}
\end{equation*}
The corresponding radial kinetic energy is only
\begin{align*}
\langle\frac{p_{r}^{2}}{2 \mu}\rangle_{l=N} & =\langle\frac{\boldsymbol{p}^{2}}{2 \mu}-\frac{l^{2}}{2 \mu r^{2}}\rangle_{l=N}=\frac{E_{N}}{2}-\frac{N(N+1)}{2 N+1} \hbar \omega \\
& =(1+\frac{1}{4 N+2}) \frac{\hbar \omega}{2} \tag{10}
\end{align*}
When $N \gg 1$, it follows
\begin{equation*}
\langle\frac{p_{r}^{2}}{2 \mu}\rangle_{l=N} \approx \frac{1}{2} \hbar \omega \tag{$\prime$}
\end{equation*}
This situation corresponds to the circular orbit in Bohr's quantum theory. The general formula for the average value of radial kinetic energy is
\begin{equation*}
\langle\frac{p_{r}^{2}}{2 \mu}\rangle_{n_{r} l m}=\frac{E_{N}}{2}-\langle\frac{l^{2}}{2 \mu r^{2}}\rangle_{n_{r} l m}=[N+\frac{3}{2}-\frac{l(l+1)}{l+1 / 2}] \frac{\hbar \omega}{2} \tag{11}
\end{equation*}
If $n_{r} \ggg 1$ (quasi-classical case), the approximate treatment can be made as follows:
$$\frac{l(l+1)}{l+\frac{1}{2}} \approx l+\frac{1}{2}$$
Equations (8) and (11) become
\begin{align*}
& \langle\frac{l^{2}}{2 \mu r^{2}}\rangle_{n_{r} l m} \approx(l+\frac{1}{2}) \frac{\hbar \omega}{2}=(\frac{l}{2}+\frac{1}{4}) \hbar \omega \tag{$\prime$}\\
& \langle\frac{p_{r}^{2}}{2 \mu}\rangle_{n_{r} l m} \approx(N+1-l) \frac{\hbar \omega}{2}=(n_{r}+\frac{1}{2}) \hbar \omega \tag{11'}
\end{align*}
|
{"$H$": "Hamiltonian (total energy operator)", "$\\boldsymbol{p}$": "momentum vector", "$\\mu$": "reduced mass", "$\\omega$": "angular frequency", "$r$": "radial coordinate", "$\\hbar$": "reduced Planck's constant", "$l$": "orbital angular momentum quantum number", "$m$": "magnetic quantum number", "$n_{r}$": "radial quantum number", "$\\theta$": "polar angle", "$\\varphi$": "azimuthal angle", "$\\psi_{n_{r} l m}$": "wave function for specified quantum numbers", "$R_{n_{r} l}(r)$": "radial wave function component", "$\\mathrm{Y}_{l m}(\\theta, \\varphi)$": "spherical harmonics", "$u_{n_{r} l}(r)$": "radial wave function in alternative form", "$N$": "total quantum number, related to principal energy level", "$E_{n, l m}$": "energy of the state defined by quantum numbers", "$E_{N}$": "energy associated with total quantum number", "$\\langle\\frac{1}{r^{2}}\\rangle_{n_{r} l m}$": "average inverse-square radial distance"}
|
Schrödinger equation one-dimensional motion
| |
7 |
Theoretical Foundations
|
A particle with mass $\mu$ moves in a "spherical square well" potential, $$V(r)= \begin{cases}0, & r<a \\ V_{0}>0, & r \geqslant a.\end{cases}$$ Consider only the bound state $(0<E<V_{0})$. As $V_{0}$ gradually increases from small to large, find the value of $V_{0} a^{2}$ when the first bound state (with angular quantum number $l=0$ and energy $E \approx V_0$) appears.
|
[
"V_{0} a^{2}=\\frac{\\pi^{2} \\hbar^{2}}{8 \\mu}"
] |
Expression
|
As a central force problem, the bound state wave function can be taken as the common eigenfunction of $(H, l^{2}, l_{z})$, written as
\begin{equation*} \psi=R(r) \mathrm{Y}_{l m}(\theta, \varphi) \tag{2} \end{equation*}
The radial equation is
\begin{equation*} R^{\prime \prime}+\frac{2}{r} R^{\prime}+[\frac{2 \mu}{\hbar^{2}}(E-V)-\frac{l(l+1)}{r^{2}}] R=0 \tag{3} \end{equation*}
When $V_{0}$ gradually increases, and a new bound state $E \approx V_{0}$ appears, the above equation becomes
\begin{align*} & R^{\prime \prime}+\frac{2}{r} R^{\prime}+[\frac{2 \mu V_{0}}{\hbar^{2}}-\frac{l(l+1)}{r^{2}}] R=0, \quad r<a(\text { inside the well }) \tag{4a}\\ & R^{\prime \prime}+\frac{2}{r} R^{\prime}-\frac{l(l+1)}{r^{2}} R=0, \quad r>a(\text { outside the well }) \tag{4b} \end{align*}
The equation inside the well (4a) is precisely the spherical Bessel equation, and the physically allowed solution is the spherical Bessel function
\begin{equation*} R(r)=j_{l}(k_{0} r), \quad k_{0}=\sqrt{2 \mu V_{0}} / \hbar \tag{5a} \end{equation*}
The solution outside the well (4b) is
\begin{equation*} R(r)=C / r^{l+1} \tag{5b} \end{equation*}
(The other solution $r^{l}$ does not satisfy the bound state boundary condition $R \rightarrow 0$ as $r \rightarrow \infty$, so it is discarded.) For the wave function outside the well (5b), it is evident that
\begin{equation*} \frac{\mathrm{d}}{\mathrm{~d} r}[r^{l+1} R(r)]=0, \quad r \geqslant a \tag{6} \end{equation*}
At $r=a$, $R$ and $R^{\prime}$ should both be continuous. Therefore, for the wave function inside the well (5a), as $r \rightarrow a$, it should also satisfy condition (6), that is,
\begin{equation*} \frac{\mathrm{d}}{\mathrm{~d} r}[r^{l+1} j_{l}(k_{0} r)]_{r=a}=0 \tag{$\prime$} \end{equation*}
Using the formula
$\frac{\mathrm{d}}{\mathrm{~d} x}[x^{l+1} j_{l}(x)]=x^{l+1} j_{l-1}(x)$ Equation $(6^{\prime})$
can be transformed into
\begin{equation*} j_{l-1}(k_{0} a)=0 \tag{7} \end{equation*}
This is the condition for a new bound state (angular quantum number $l$, energy level $E_{n l} \approx V_{0}$) to appear. For the first bound state $l=0$, considering that $j_{-1}(k_{0} a)=\frac{\cos k_{0} a}{k_{0} a}$ The condition for the appearance of a new s-state $(l=0)$ energy level $(E \approx V_{0})$ is \begin{equation*} \cos k_{0} a=0, \quad k_{0} a=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \cdots \tag{8} \end{equation*} When the first bound state appears, $k_{0} a=\pi / 2$, which means \begin{equation*} V_{0} a^{2}=\frac{\pi^{2} \hbar^{2}}{8 \mu} \tag{9} \end{equation*} As $V_{0}$ gradually increases, whenever condition (7) is satisfied, a new energy level $E_{n l} \approx V_{0}$ appears. The order of appearance of each energy level can be determined based on the zeros of the spherical Bessel function $j_{l}(x)$:
\[
\begin{aligned}
&1~\mathrm{s},\quad 1~\mathrm{p},\quad 1~\mathrm{d},\quad 2~\mathrm{s},\quad 1~\mathrm{f},\quad 2~\mathrm{p},\quad 1~\mathrm{g},\quad 2~\mathrm{d},\quad 3~\mathrm{s},\\
&1~\mathrm{h},\quad 2~\mathrm{f},\quad 1~\mathrm{i},\quad 3~\mathrm{p},\quad 2~\mathrm{g},\quad 1~\mathrm{k},\quad 3~\mathrm{d},\quad 4~\mathrm{s},\ \cdots
\end{aligned}
\]
The $l$ values corresponding to each spectral notation are:
\[
\begin{array}{ccccccccc}
l & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\text{Letter} & \mathrm{s} & \mathrm{p} & \mathrm{d} & \mathrm{f} & \mathrm{g} & \mathrm{h} & \mathrm{i} & \mathrm{j}
\end{array}
\]
When $V_0$ is large, the total number of bound states can be approximated by “one bound state per $h^3$ volume in phase space.” The maximum momentum for a particle in a bound state inside the well is:
\begin{equation*} p_{0}=\hbar k_{0}=\sqrt{2 \mu V_{0}} \tag{10} \end{equation*}
Therefore, the total phase space volume occupied by the bound states is $\frac{4 \pi}{3} a^{3} \cdot \frac{4 \pi}{3} p_{0}^{3}=\frac{16}{9} \pi^{2}(a \hbar k_{0})^{3}=\Omega$ The total number of bound states is \begin{align*} N & \approx \Omega /(2 \pi \hbar)^{3}=\frac{2 \pi^{2}}{9}(\frac{a k_{0}}{\pi})^{3} \\ & =\frac{2 \pi^{2}}{9}(\frac{a}{\pi \hbar})^{3}(2 \mu V_{0})^{3 / 2} \tag{11} \end{align*} For example, when $a k_{0}=7 \pi / 2$, the highest energy level is 4 s, and counting all the states from 1 s to 4 s (consider the degeneracy of energy levels $E_{n l}$ as $2 l+1$) gives 99, while equation (11) yields $N \approx 94$, which is indeed very close ${ }^{(1)}$.
|
{"$\\mu$": "mass of the particle", "$V_{0}$": "potential outside the well", "$a$": "radius of the spherical square well", "$E$": "energy of the bound state", "$\\hbar$": "reduced Planck's constant"}
|
Schrödinger equation one-dimensional motion
| |
8 |
Theoretical Foundations
|
For the common eigenstate $|l m\rangle$ of $l^{2}$ and $l_{z}$, calculate the expectation value $\overline{l_{n}}$ of $l_{n}=\boldsymbol{n} \cdot \boldsymbol{l}$. Here, $\boldsymbol{n}$ is a unit vector in an arbitrary direction, and its angle with the $z$-axis is $\gamma$.
|
[
"m \\hbar \\cos \\gamma"
] |
Expression
|
Using the basic commutation relation $\boldsymbol{l} \times \boldsymbol{l}=\mathrm{i} \hbar \boldsymbol{l}$, we have
\[
\begin{aligned}
& \mathrm{i} \hbar l_{x} l_{y}=(l_{y} l_{z}-l_{z} l_{y}) l_{y}=l_{y} l_{z} l_{y}-l_{z} l_{y}^{2} y \\
& \mathrm{i} \hbar l_{y} l_{x}=l_{y}(l_{y} l_{z}-l_{z} l_{y})=l_{y}^{2} l_{z}-l_{y} l_{z} l_{y}
\end{aligned}
\]
Calculating the expectation value in the state $|l m\rangle$, since
$\overline{l_{z} l_{y}^{2}}=\overline{l_{y}^{2} l_{z}}=m \hbar \overline{l_{y}^{2}}$
Therefore
$\overline{l_{y} l_{x}}=-\overline{l_{x} l_{y}}$
Thus
\begin{align*}
& \overline{l_{x} l_{y}}-\overline{l_{y} l_{x}}=2 \overline{l_{x} l_{y}}=\mathrm{i} \hbar \overline{l_{z}}=\mathrm{i} \hbar^{2} m \\
& \overline{l_{x} l_{y}}=\mathrm{i}^{2} m / 2, \quad \overline{l_{y} l_{x}}=-\mathrm{i}^{2} m / 2 \tag{1}
\end{align*}
The projection operator of $\boldsymbol{l}$ in the direction of $\boldsymbol{n}$ is
$l_{n}=\boldsymbol{n} \cdot \boldsymbol{l}=l_{x} \cos \alpha+l_{y} \cos \beta+l_{z} \cos \gamma$
For the state $|l m\rangle$, since $\overline{l_{x}}=0, \overline{l_{y}}=0$, we have
\begin{gather*}
\overline{l_{n}}=m \hbar \cos \gamma \tag{2}\\
l_{n}^{2}=l_{x}^{2} \cos ^{2} \alpha+l_{y}^{2} \cos ^{2} \beta+l_{z}^{2} \cos ^{2} \gamma \\
\\
+(l_{x} l_{y}+l_{y} l_{x}) \cos \alpha \cos \beta+\cdots \text { (similar terms with cyclic permutations) }
\end{gather*}
Calculating the expectation value in the state $|l m\rangle$, since
\begin{gather*}
\overline{l_{x} l_{y}}+\overline{l_{y} l_{x}}=0, \quad \overline{l_{y} l_{z}}+\overline{l_{z} l_{y}}=2 m \hbar \overline{l_{y}}=0, \cdots \\
\overline{l_{x}^{2}}=\overline{l_{y}^{2}}=\frac{1}{2}(\overline{l^{2}}-\overline{l_{z}^{2}})=\frac{\hbar^{2}}{2}{l(l+1)-m^{2}} \tag{3}
\end{gather*}
Thus
\begin{align*}
\overline{l_{n}^{2}} & =m^{2} \hbar^{2} \cos ^{2} \gamma+\frac{\hbar^{2}}{2}{l(l+1)-m^{2}}(\cos ^{2} \alpha+\cos ^{2} \beta) \\
& =m^{2} \hbar^{2} \cos ^{2} \gamma+\frac{\hbar^{2}}{2}{l(l+1)-m^{2}}(1-\cos ^{2} \gamma) \\
& =\frac{\hbar^{2}}{2}{l(l+1)(1-\cos ^{2} \gamma)+m^{2}(3 \cos ^{2} \gamma-1)} \tag{4}
\end{align*}
If $\boldsymbol{n}$ is orthogonal to the $z$-axis, that is, $\cos \gamma=0$, then $\overline{l_{n}^{2}}=\overline{l_{x}^{2}}=\overline{l_{y}^{2}}, \overline{l_{n}}=0$.
|
{"$|l m\\rangle$": "common eigenstate of $l^{2}$ and $l_{z}$", "$l_{z}$": "component of angular momentum along the z-axis", "$l_{x}$": "component of angular momentum along the x-axis", "$l_{y}$": "component of angular momentum along the y-axis", "$l_{n}$": "projection of angular momentum in direction of $\\boldsymbol{n}$", "$\\boldsymbol{n}$": "unit vector in an arbitrary direction", "$\\gamma$": "angle between the unit vector $\\boldsymbol{n}$ and z-axis", "$\\overline{l_{n}}$": "expectation value of projection of angular momentum in direction $\\boldsymbol{n}$", "$m$": "magnetic quantum number", "$\\hbar$": "reduced Planck's constant", "$l$": "angular momentum quantum number"}
|
Schrödinger equation one-dimensional motion
| |
9 |
Theoretical Foundations
|
For an electron's spin state $\chi_{\frac{1}{2}}(\sigma_{z}=1)$ (i.e., the state where the Pauli matrix $\sigma_z$ has an eigenvalue of $+1$), if we measure its spin projection in an arbitrary direction $\boldsymbol{n}$, $\sigma_n = \boldsymbol{\sigma} \cdot \boldsymbol{n}$, where $\boldsymbol{n}$ is a unit vector and $n_z$ is its component along the $z$-axis. Find the expression for the probability of measuring $\sigma_n = +1$ (expressed in terms of $n_z$).
|
[
"\\frac{1}{2}(1+n_z)"
] |
Expression
|
One Using the eigenfunctions of $\sigma_{n}$ obtained from the previous problem, it is easy to find
(a) In the spin state $\chi_{\frac{1}{2}}=[\begin{array}{l}1 \\ 0\end{array}]$,
the probability of $\sigma_{n}=1$ is
\begin{equation*}
|\langle\phi_{1} \lvert\, \chi_{\frac{1}{2}}\rangle|^{2}=\cos ^{2} \frac{\theta}{2}=\frac{1}{2}(1+n_{z}) \tag{1}
\end{equation*}
the probability of $\sigma_{n}=-1$ is
\begin{equation*}
|\langle\phi_{-1} \lvert\, \chi_{\frac{1}{2}}\rangle|^{2}=\sin ^{2} \frac{\theta}{2}=\frac{1}{2}(1-n_{z}) \tag{2}
\end{equation*}
(b) In the spin state $\phi_{1}(\sigma_{n}=1)$,
the probability of $\sigma_{z}=1$ is
\begin{equation*}
|\langle\chi_{\frac{1}{2}} \rvert\, \phi_{1}\rangle|^{2}=\frac{1}{2}(1+n_{z}) \tag{3}
\end{equation*}
the probability of $\sigma_{z}=-1$ is
\begin{equation*}
1-\frac{1}{2}(1+n_{z})=\frac{1}{2}(1-n_{z}) \tag{4}
\end{equation*}
\begin{equation*}
\langle\sigma_{z}\rangle=\frac{1}{2}(1+n_{z})-\frac{1}{2}(1-n_{z})=n_{z} \tag{5}
\end{equation*}
Considering
$\sigma_{n}=\sigma_{x} n_{x}+\sigma_{y} n_{y}+\sigma_{z} n_{z}$
The components of $\boldsymbol{\sigma}$ and $\boldsymbol{n}$ have symmetric roles in the construction of $\sigma_{n}$, so using formulas (3), (4), (5), and cyclically permuting $x, ~ y$, $z$, the following can be deduced:
\begin{gather*}
\sigma_{x}= \pm 1 \text { with probability } \frac{1}{2}(1 \pm n_{x}) \tag{6}\\
\langle\sigma_{x}\rangle=n_{x} \tag{7}\\
\sigma_{y}= \pm 1 \text { with probability } \frac{1}{2}(1 \pm n_{y}) \tag{8}\\
\langle\sigma_{y}\rangle=n_{y} \tag{9}
\end{gather*}
By combining equations (5), (7), (9) in vector form as follows:
In the spin state $\phi_{1}(\sigma_{n}=1)$,
\begin{equation*}
\langle\boldsymbol{\sigma}\rangle=\boldsymbol{n} \tag{10}
\end{equation*}
Similarly, it is easy to calculate
In the spin state $\phi_{-1}(\sigma_{n}=-1)$,
\begin{equation*}
\langle\boldsymbol{\sigma}\rangle=-\boldsymbol{n} \tag{11}
\end{equation*}
Solution two (a) In the spin state $\chi_{\frac{1}{2}}$ where $\sigma_{z}=1$, the possible measured values of $\sigma_{n}$ are the eigenvalues $\pm 1$; let the corresponding probabilities be $w_{+}$ and $w_{-}$, then
\begin{equation*}
\langle\sigma_{n}\rangle=w_{+} \times 1+w_{-} \times(-1)=w_{+}-w_{-} \tag{12}
\end{equation*}
Since
\begin{equation*}
\sigma_{n}=\sigma_{x} n_{x}+\sigma_{y} n_{y}+\sigma_{z} n_{z} \tag{13}
\end{equation*}
Considering that in the eigenstate of $\sigma_{z}$ the average value of $\sigma_{x}$ and $\sigma_{y}$ is zero, and the average value of $\sigma_{z}$ is the eigenvalue, hence in the state $\chi_{\frac{1}{2}}$,
\begin{equation*}
\langle\sigma_{n}\rangle=\langle\sigma_{z}\rangle n_{z}=n_{z}=\cos \theta \tag{14}
\end{equation*}
From equations (12), (14), and using $w_{+}+w_{-}=1$, it can be found that
\begin{equation*}
w_{+}=\frac{1}{2}(1+n_{z}), \quad w_{-}=\frac{1}{2}(1-n_{z}) \tag{15}
\end{equation*}
These are the equations (1), (2) in solution one.
(b) In equation (14), $\theta$ is the angle parameter in the $z$-axis and $\boldsymbol{n}$. The choices of the $z$-axis and $\boldsymbol{n}$ are arbitrary, and the original $z$-axis can be taken as the new $\boldsymbol{n}$, while the original $\boldsymbol{n}$ is taken as the new $z$-axis. Thus, it can be known that in the spin state where $\sigma_{n}=1$
The average value of $\sigma_{z}$ remains $\cos \theta$, which is $n_{z}$. By letting $x, ~ y, ~ z$ permute, we obtain
\begin{equation*}
\text { In the spin state } \phi_{1}(\sigma_{n}=1) \text {, }\langle\boldsymbol{\sigma}\rangle=\boldsymbol{n} \tag{10}
\end{equation*}
In the state $\phi_{1}$, the values of each component of $\boldsymbol{\sigma}$ are of course all $\pm 1$, and their probabilities can be written similarly to those in (a), thus
\begin{align*}
\sigma_{x} & = \pm 1 \text { with probability } \frac{1}{2}(1 \pm n_{x}) \tag{6}\\
\sigma_{y} & = \pm 1 \text { with probability } \frac{1}{2}(1 \pm n_{y}) \tag{8}\\
\sigma_{z} & = \pm 1 \text { with probability } \frac{1}{2}(1 \pm n_{z}) \tag{3,4}
\end{align*}
|
{"$\\chi_{\\frac{1}{2}}$": "electron's spin state (spin-up along z-axis)", "$\\sigma_{z}$": "Pauli matrix corresponding to spin measurement along the z-axis", "$\\boldsymbol{n}$": "arbitrary unit vector direction for spin projection", "$n_{z}$": "component of vector \\boldsymbol{n} along the z-axis", "$\\sigma_{n}$": "spin projection in the direction of vector \\boldsymbol{n}", "$\\boldsymbol{\\sigma}$": "Pauli spin operator as a vector", "$\\sigma_{x}$": "Pauli matrix corresponding to spin measurement along the x-axis", "$\\sigma_{y}$": "Pauli matrix corresponding to spin measurement along the y-axis", "$\\phi_{1}$": "spin state with eigenvalue +1 for \\sigma_{n}"}
|
Schrödinger equation one-dimensional motion
| |
10 |
Theoretical Foundations
|
Express the operator $(I+\sigma_{x})^{1 / 2}$ (where $\sigma_x$ is the Pauli matrix) as a linear combination of the $2 \times 2$ identity matrix (denoted as $1$ in the expression) and $\sigma_x$. The operation takes the principal square root.
|
[
"\\frac{1}{\\sqrt{2}}(1+\\sigma_x)"
] |
Expression
|
(a) $(I+\sigma_{x})^{I / 2}$. The eigenvalues of $\sigma_{x}$ are $\pm 1$, and for each eigenvalue, $(I+\sigma_{x})^{1 / 2}$ gives a clear value (principal root is taken), so it can be concluded that $(I+\sigma_{x})^{1 / 2}$ exists, and it is a function of $\sigma_{x}$. According to the argument in problem 6.14, we can set
\begin{equation*}
(I+\sigma_{x})^{1 / 2}=C_{0}+C_{1} \sigma_{x} \tag{1}
\end{equation*}
Squaring this equation, we get
$I+\sigma_{x}=(C_{0}+C_{1} \sigma_{x})^{2}=C_{0}^{2}+C_{1}^{2}+2 C_{0} C_{1} \sigma_{x}$
Therefore
\[
\begin{gathered}
C_{0}^{2}+C_{1}^{2}=1 \\
2 C_{0} C_{1}=1
\end{gathered}
\]
Adding and subtracting the two equations, we get
\[
\begin{aligned}
& (C_{0}+C_{1})^{2}=2 \\
& (C_{0}-C_{1})^{2}=0
\end{aligned}
\]
If we set $(C_{0}+C_{1})$ to take positive value, we can solve
$C_{0}=C_{1}=1 / \sqrt{2}$
Substituting into equation (1), we obtain
\begin{equation*}
(1+\sigma_{z})^{1 / 2}=\frac{1}{\sqrt{2}}(I+\sigma_{x}) \tag{2}
\end{equation*}
It is easy to verify that for any eigenvalue $( \pm 1)$ of $\sigma_{x}$, this equation holds true.
|
{"$\\sigma_x$": "Pauli matrix (x-component)"}
|
Schrödinger equation one-dimensional motion
| |
11 |
Theoretical Foundations
|
For a spin $1/2$ particle, $\langle\boldsymbol{\sigma}\rangle$ is often called the polarization vector, denoted as $\boldsymbol{P}$, which is the spatial orientation of the spin angular momentum. Given the initial spin wave function at $t=0$ (in the $\sigma_{z}$ representation) as:
\chi(0)=[\begin{array}{c}
\cos \delta \mathrm{e}^{-\mathrm{i} \alpha} \\
\sin \delta \mathrm{e}^{\mathrm{i} \alpha}
\end{array}]
where $\delta, ~ \alpha$ are positive real numbers (or 0), $\delta \leqslant \pi / 2, \alpha \leqslant \pi$. Determine the azimuthal angle $\theta_{0}$ (polar angle) of the initial polarization vector $\boldsymbol{P}(t=0)$.
|
[
"2 \\delta"
] |
Expression
|
Any definite spin state is an eigenstate (with eigenvalue 1) of the projection $\sigma_{n}$ of $\boldsymbol{\sigma}$ in some direction $(\theta, \varphi)$, and
\begin{equation*}
\boldsymbol{P}=\langle\boldsymbol{\sigma}\rangle=\boldsymbol{n} \tag{3}
\end{equation*}
where $\boldsymbol{n}$ is the unit vector in the direction $(\theta, \varphi)$, the eigenfunction of $\sigma_{n}=1$ is
\begin{equation}
\phi_{1}(\theta, \varphi) =[\begin{array}{l}
\cos \frac{\theta}{2} \mathrm{e}^{-\mathrm{i} \varphi / 2} \tag{4}\\
\sin \frac{\theta}{2} \mathrm{e}^{\mathrm{i} \varphi / 2}
\end{array}]
\end{equation}
By comparing Equation (1) and (4), we find the azimuthal angle of the initial polarization vector $\boldsymbol{P}(t=0)$
\begin{equation*}
\theta_{0}=2 \delta, \quad \varphi_{0}=2 \alpha \tag{5}
\end{equation*}
|
{"$\\langle\\boldsymbol{\\sigma}\\rangle$": "polarization vector", "$\\boldsymbol{P}$": "polarization vector", "$\\boldsymbol{n}$": "unit vector direction", "$\\sigma_{n}$": "projection of spin operator in direction n", "$\\theta_{0}$": "azimuthal angle of initial polarization vector", "$\\theta$": "azimuthal angle", "$\\varphi$": "polar angle", "$\\phi_{1}$": "eigenfunction of spin operator", "$\\delta$": "angle parameter", "$\\alpha$": "angle parameter"}
|
Schrödinger equation one-dimensional motion
| |
12 |
Theoretical Foundations
|
For a system composed of two spin $1 / 2$ particles, let $s_{1}$ and $s_{2}$ denote their spin angular momentum operators. Calculate and simplify the triple product $s_{1} \cdot (s_{1} \times s_{2})$ (take $\hbar=1$ ).
|
[
"\\mathrm{i} s_{1} \\cdot s_{2}"
] |
Expression
|
The basic relationships are as follows (the single particle formula is only written for particle 1)
(a) $s_{1}^{2}=\frac{3}{4}, \boldsymbol{\sigma}_{1}^{2}=3 ;(s_{1 x})^{2}=\frac{1}{4},(\sigma_{1 x})^{2}=1$, and so on.
(b) $s_{1} \times s_{1}=\mathrm{i} s_{1}, \boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{1}=2 \mathrm{i} \boldsymbol{\sigma}_{1}$
(c) $\sigma_{1 x} \sigma_{1 y}=-\sigma_{1}, \sigma_{1 x}=\mathrm{i} \sigma_{1 z}$, and so on.
(d) $s_{1}$ commutes with $s_{2}$, $s_{1} \cdot s_{2}=s_{2} \cdot s_{1}, s_{1} \times s_{2}=-s_{2} \times s_{1}$
For the triple product, there are the following types:
\begin{equation*}
1^{\circ} \tag{5}
\end{equation*}
\begin{align*}
s_{1} \cdot(s_{1} \times s_{2}) & =(s_{1} \times s_{1}) \cdot s_{2}=\mathrm{i} s_{1} \cdot s_{2} \\
s_{2} \cdot(s_{1} \times s_{2}) & =-s_{2} \cdot(s_{2} \times s_{1}) \\
& =-(s_{2} \times s_{2}) \cdot s_{1}=-\mathrm{i} s_{1} \cdot s_{2} \tag{6}\\
\boldsymbol{\sigma}_{1} \cdot(\boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2}) & =2 \mathrm{i} \boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2} \tag{7}\\
\boldsymbol{\sigma}_{2} \cdot(\boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2}) & =-2 \mathrm{i} \boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2} \tag{8}
\end{align*}
$\mathbf{2}^{\circ} \boldsymbol{\sigma}_{1}(\boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2})$ type. Since $\boldsymbol{\sigma}_{1}$ and $\boldsymbol{\sigma}_{2}$ commute, using the formula proven in question 6.21
\begin{aligned}
\boldsymbol{\sigma}(\boldsymbol{\sigma} \cdot \boldsymbol{A})-\boldsymbol{A} & =\boldsymbol{A}-(\boldsymbol{\sigma} \cdot \boldsymbol{A}) \boldsymbol{\sigma} \\
& =\mathrm{i} \boldsymbol{A} \times \boldsymbol{\sigma}, \quad(\boldsymbol{\sigma}, \boldsymbol{A} \text { commute })
\end{aligned}
Let $\boldsymbol{\sigma}$ and $\boldsymbol{A}$ be equal to $\boldsymbol{\sigma}_{1}$ and $\boldsymbol{\sigma}_{2}$ respectively, and obtain
\begin{gather*}
\boldsymbol{\sigma}_{1}(\boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2})=\boldsymbol{\sigma}_{2}-\mathrm{i} \boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2} \tag{9}\\
\boldsymbol{\sigma}_{2}(\boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2})=\boldsymbol{\sigma}_{1}+\mathrm{i} \boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2} \tag{10}\\
(\boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2}) \boldsymbol{\sigma}_{1}=\boldsymbol{\sigma}_{2}+\mathrm{i} \boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2} \tag{11}\\
(\boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2}) \boldsymbol{\sigma}_{2}=\boldsymbol{\sigma}_{1}-\mathrm{i} \boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2} \tag{12}\\
s_{1}(s_{1} \cdot s_{2})=\frac{1}{4} s_{2}-\frac{\mathrm{i}}{2} s_{1} \times s_{2}, \text { etc. } \tag{13}\\
{[s_{1} \cdot s_{2}, s_{1}]=\mathrm{i} s_{1} \times s_{2}} \tag{14}\\
{[s_{1} \cdot s_{2}, s_{2}]=-\mathrm{i} s_{1} \times s_{2}} \tag{15}
\end{gather*}
$3^{\circ} s_{1} \times(s_{1} \times s_{2})$ type. Using the vector operator formula (see question 4.1)
\boldsymbol{A} \times(\boldsymbol{B} \times \boldsymbol{C})=\overparen{\boldsymbol{A} \cdot(\boldsymbol{B C})}-(\boldsymbol{A} \cdot \boldsymbol{B}) \boldsymbol{C}
(\boldsymbol{A} \times \boldsymbol{B}) \times \boldsymbol{C}=\boldsymbol{A \cdot ( B C )}-\boldsymbol{A}(\boldsymbol{B} \cdot \boldsymbol{C})
Hence, we obtain
\begin{align*}
s_{1} \times(s_{1} \times s_{2}) & =(s_{1} \cdot s_{2}) s_{1}-\frac{3}{4} s_{2}=\frac{\mathrm{i}}{2} s_{1} \times s_{2}-\frac{1}{2} s_{2} \tag{16}\\
(s_{1} \times s_{2}) \times s_{1} & =\frac{3}{4} s_{2}-s_{1}(s_{1} \cdot s_{2})=\frac{1}{2} s_{2}+\frac{\mathrm{i}}{2} s_{1} \times s_{2} \tag{17}\\
s_{2} \times(s_{1} \times s_{2}) & =\frac{1}{2} s_{1}+\frac{\mathrm{i}}{2} s_{1} \times s_{2} \tag{18}\\
(s_{1} \times s_{2}) \times s_{2} & =\frac{\mathrm{i}}{2} s_{1} \times s_{2}-\frac{1}{2} s_{1} \tag{19}\\
\boldsymbol{\sigma}_{1} \times(\boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2}) & =\mathrm{i} \boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2}-2 \boldsymbol{\sigma}_{2} \tag{20}\\
(\boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2}) \times \boldsymbol{\sigma}_{1} & =\mathrm{i} \boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2}+2 \boldsymbol{\sigma}_{2} \tag{21}\\
\boldsymbol{\sigma}_{2} \times(\boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2}) & =\mathrm{i} \boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2}+2 \boldsymbol{\sigma}_{1} \tag{22}\\
(\boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2}) \times \boldsymbol{\sigma}_{2} & =\mathrm{i} \boldsymbol{\sigma}_{1} \times \boldsymbol{\sigma}_{2}-2 \boldsymbol{\sigma}_{1} \tag{23}
\end{align*}
$4^{\circ}$ Total spin $\boldsymbol{S}=s_{1}+s_{2}$,
\begin{align*}
& \boldsymbol{S}^{2}=2 s_{1} \cdot s_{2}+\frac{3}{2} \tag{24}\\
& {[\boldsymbol{S}^{2}, s_{1}]=2[s_{1} \cdot s_{2}, s_{1}]=2 \mathrm{i} s_{1} \times s_{2}} \tag{25}\\
& {[\boldsymbol{S}^{2}, s_{2}]=-2 \mathrm{i} s_{1} \times s_{2}} \tag{26}\\
& \boldsymbol{S} \cdot(s_{1} \times s_{2})=(s_{1} \times s_{2}) \cdot \boldsymbol{S}=0 \tag{27}\\
& \boldsymbol{S}(s_{1} \cdot s_{2})=(s_{1} \cdot s_{2}) \boldsymbol{S}=\frac{1}{4} \boldsymbol{S} \tag{28}\\
& \boldsymbol{S S}^{2}=\boldsymbol{S}^{2} \boldsymbol{S}=2 \boldsymbol{S} \tag{29}
\end{align*}
|
{"$s_{1}$": "spin angular momentum operator for particle 1", "$s_{2}$": "spin angular momentum operator for particle 2", "$\\hbar$": "reduced Planck's constant"}
|
Schrödinger equation one-dimensional motion
| |
13 |
Theoretical Foundations
|
Two localized non-identical particles with spin $1/2$ (ignoring orbital motion) have an interaction energy given by (setting $\hbar=1)$
\begin{equation*}
H=A \boldsymbol{s}_{1} \cdot \boldsymbol{s}_{2}
\end{equation*}
At $t=0$, particle 1 has spin 'up' $(s_{1 z}=1 / 2)$, and particle 2 has spin 'down' $(s_{2 z}=-\frac{1}{2})$. Find the probability that particle 1 has spin 'up' $(s_{1 z}=1 / 2)$ at any time $t>0$.
|
[
"\\cos^{2}(\\frac{A t}{2})"
] |
Expression
|
Start by finding the spin wave function of the system. Since
\begin{equation*}
H=A s_{1} \cdot s_{2}=\frac{A}{2}(\boldsymbol{S}^{2}-\frac{3}{2}) \tag{$\prime$}
\end{equation*}
It is evident that the total spin $\boldsymbol{S}$ is a conserved quantity, so the stationary wave function can be chosen as a common eigenfunction of $\boldsymbol{S}^{2}, ~ S_{z}$. According to different values of the total spin quantum number $S$, the eigenfunctions and energy levels are
\begin{array}{ll}
S=1, & \chi_{1 M_{s}}, \quad E_{1}=A / 4 \tag{2}\\
S=0, & \chi_{00}, \quad E_{0}=-3 A / 4
\end{array}}
At $t=0$, the spin state of the system is
\begin{equation*}
\chi(0)=\alpha(1) \beta(2)=\frac{1}{\sqrt{2}}(\chi_{10}+\chi_{00}) \tag{3}
\end{equation*}
Therefore, the wave function at $t>0$ is
\begin{equation*}
\chi(t)=\frac{1}{\sqrt{2}} \chi_{10} \mathrm{e}^{-\mathrm{i} E_{1} t}+\frac{1}{\sqrt{2}} \chi_{00} \mathrm{e}^{-\mathrm{i} E_{0} t} \tag{4}
\end{equation*}
That is
\begin{align*}
\chi(t) & =\frac{1}{2}[\alpha(1) \beta(2)+\beta(1) \alpha(2)] \mathrm{e}^{-\mathrm{i} A / 4}+\frac{1}{2}[\alpha(1) \beta(2)-\beta(1) \alpha(2)] \mathrm{e}^{3 \mathrm{i} t / 4} \\
& =[\alpha(1) \beta(2) \cos \frac{A t}{2}-\mathrm{i} \beta(1) \alpha(2) \sin \frac{A t}{2}] \mathrm{e}^{\mathrm{i} A / 4} \tag{$4^\prime$}
\end{align*}
From formula ($4^{\prime}$), it can be seen that at time $t$, the probability of particle 1 having spin 'up' [while particle 2 has spin 'down', corresponding to the $\alpha(1) \beta(2)$ term] is $\cos ^{2}(\frac{A t}{2})$.
|
{"$A$": "interaction constant", "$\\boldsymbol{s}_{1}$": "spin vector of particle 1", "$\\boldsymbol{s}_{2}$": "spin vector of particle 2", "$t$": "time", "$s_{1 z}$": "z-component of spin vector of particle 1", "$s_{2 z}$": "z-component of spin vector of particle 2", "$\\hbar$": "reduced Planck's constant", "$\\boldsymbol{S}$": "total spin vector", "$S$": "total spin quantum number", "$S_{z}$": "z-component of total spin vector", "$\\chi_{1 M_{s}}$": "eigenfunction for total spin S=1 with magnetic quantum number $M_s$", "$E_{1}$": "energy level for total spin S=1", "$\\chi_{00}$": "eigenfunction for total spin S=0", "$E_{0}$": "energy level for total spin S=0", "$\\chi(t)$": "wave function at time $t$"}
|
Schrödinger equation one-dimensional motion
| |
14 |
Theoretical Foundations
|
Consider a system consisting of three distinguishable particles each with spin $1/2$, with the Hamiltonian given by
\begin{equation*}
H=A(s_{1} \cdot s_{2}+s_{2} \cdot s_{3}+s_{3} \cdot s_{1}) \quad \text { ( } A \text { is real) }
\end{equation*}
Let $S$ denote the total spin quantum number of the system. Determine the expression for the energy level $E_{3/2}$ when $S=3/2$ (taking $\hbar=1$).
|
[
"E_{3/2} = \\frac{3}{4}A"
] |
Expression
|
The Hamiltonian $H$ can be written as
\begin{equation*}
H=\frac{A}{2}(\boldsymbol{S}_{123}^{2}-3 \times \frac{3}{4}) \tag{$\prime$}
\end{equation*}
Therefore, the energy levels (taking $\hbar=1$) are
\begin{equation*}
E_{S}=\frac{A}{2}[S(S+1)-\frac{9}{4}] \tag{2}
\end{equation*}
A complete set of conserved quantities can be taken as $[S_{123}^{2}, ~(S_{123})_{z}, S_{12}^{2}]$, with eigenvalues
\begin{array}{l}
\boldsymbol{S}_{12}^{2}=S^{\prime}(S^{\prime}+1), \quad S^{\prime}=1,0 \tag{3}\\
\boldsymbol{S}_{123}^{2}=S(S+1), \quad S=\frac{3}{2}, \frac{1}{2}, \frac{1}{2} \\
(S_{123})_{z}=M=S, S-1, \cdots,(-S)
\end{array}}
The possible combinations for the quantum numbers $S^{\prime}, ~ S$ are $S=S^{\prime} \pm \frac{1}{2}>0$, i.e.,
\begin{array}{l}
S=3 / 2, \quad S^{\prime}=1 \tag{4}\\
S=1 / 2, \quad S^{\prime}=1,0
\end{array}}
For each pair $(S, S^{\prime})$, the degeneracy of the energy level is $(2 S+1)$, so
\begin{align*}
& E_{3 / 2}=\frac{3}{4} A, \quad S=\frac{3}{2}, \quad S^{\prime}=1, \quad \text { Degeneracy }=4 \\
& E_{1 / 2}=-\frac{3}{4} A, \quad S=\frac{1}{2}, \quad S^{\prime}=1,0, \quad \text { Degeneracy }=4 \tag{5}
\end{align*}
We now aim to determine the common eigenstates of $[\boldsymbol{S}_{123}^{2},(S_{123})_{z}, \boldsymbol{S}_{12}^{2}]$. The spin "up" and "down" states of the $k$-th particle $(k=1,2,3)$ are denoted as $\alpha(k), ~ \beta(k)$, corresponding to $s_{k z}=\frac{1}{2}, ~-\frac{1}{2}$. As is known, the common eigenstates of $S_{12}^{2}$ and $(S_{12})_{z}$ are given by:
\begin{aligned}
& \chi_{11}(1,2)=\alpha(1)_{\alpha}(2), \quad S^{\prime}=1, \quad(S_{12})_{z}=1 \\
& \chi_{10}(1,2)=\frac{1}{\sqrt{2}}[\alpha(1) \beta(2)+\beta(1)_{\alpha}(2)], \quad S^{\prime}=1, \quad(S_{12})_{z}=0 \\
& \chi_{1,-1}(1,2)=\beta(1) \beta(2), \quad S^{\prime}=1, \quad(S_{12})_{z}=-1 \\
& \chi_{00}(1,2)=\frac{1}{\sqrt{2}}[\alpha(1) \beta(2)-\beta(1) \alpha(2)], \quad S^{\prime}=0, \quad(S_{12})_{z}=0
\end{aligned}
The common eigenstates of $\boldsymbol{S}_{12}^{2}, ~ \boldsymbol{S}_{123}^{2}, ~(S_{123})_{z}$ are denoted as $\chi_{S_{S M}}(1,2,3)$. When all three quantum numbers take their maximum values, the eigenstate is clearly
\begin{equation*}
\chi_{1 \frac{3}{2} \frac{3}{2}}=\chi_{11}(1,2)_{\alpha}(3)=\alpha(1)_{\alpha}(2) \alpha(3) \tag{6a}
\end{equation*}
When $S=3 / 2$, $M$ has 4 possible values; corresponding eigenstates for $M=\frac{1}{2},-\frac{1}{2},-\frac{3}{2}$ can be obtained by repeatedly applying the ladder operator $[(S_{123})_{x}-\mathrm{i}(S_{123})_{y}]$ to $\chi_{1 \frac{3}{2}} \frac{3}{2}$. Since the ladder operator and $\chi_{1 \frac{3}{2} \frac{3}{2}}$ are symmetric under permutation of the particles, each $\chi_{1 \frac{3}{2} M}$ obtained from this is a symmetric function. Based on this observation and considering the values of $M$, the only possible structure of these functions can be immediately written:
\begin{align*}
\chi_{1 \frac{3}{2} \frac{1}{2}} & =\frac{1}{\sqrt{3}}[\alpha(1) \alpha(2) \beta(3)+\beta(1) \alpha(2) \alpha(3)+\alpha(1) \beta(2) \alpha(3)] \tag{6b}\\
\chi_{1 \frac{3}{2},-\frac{1}{2}} & =\frac{1}{\sqrt{3}}[\alpha(1) \beta(2) \beta(3)+\beta(1) \alpha(2) \beta(3)+\beta(1) \beta(2) \alpha(3)] \tag{6c}\\
\chi_{1 \frac{3}{2},-\frac{3}{2}} & =\beta(1) \beta(2) \beta(3) \tag{6d}
\end{align*}
When $S^{\prime}=0$, the part of the system wave function concerning particles $1, ~ 2$ can only be $\chi_{00}$, and considering the values of $M$, it can be immediately concluded that
\begin{align*}
\chi_{0 \frac{1}{2} \frac{1}{2}} & =\chi_{00}(1,2) \alpha(3) \\
& =\frac{1}{\sqrt{2}}[\alpha(1) \beta(2) \alpha(3)-\beta(1) \alpha(2) \alpha(3)] \tag{7a}\\
\chi_{0 \frac{1}{2},-\frac{1}{2}} & =\chi_{00}(1,2) \beta(3) \\
& =\frac{1}{\sqrt{2}}[\alpha(1) \beta(2) \beta(3)-\beta(1) \alpha(2) \beta(3)] \tag{7b}
\end{align*}
Now only $\chi_{1 \frac{1}{2} \frac{1}{2}}$ and $\chi_{1 \frac{1}{2},-\frac{1}{2}}$ remain to be determined. In the construction of $\chi_{1 \frac{1}{2} \frac{1}{2}}$, each term should include two $\alpha$ states and one $\beta$ state. It must be a linear combination of $\chi_{11}(1,2) \beta(3)$ and $\chi_{10}(1,2) \alpha(3)$, and should be orthogonal to $\chi_{1 \frac{3}{2} \frac{1}{2}}$ since they have different $S$ values. Expression (6b) for $\chi_{1 \frac{3}{2} \frac{1}{2}}$ can be written as
\begin{equation*}
\chi_{1 \frac{3}{2} \frac{1}{2}}=\frac{1}{\sqrt{3}}[\chi_{11}(1,2) \beta(3)+\sqrt{2} \chi_{10}(1,2)_{\alpha}(3)] \tag{$\prime$}
\end{equation*}
Thus, the construction of $\chi_{1 \frac{1}{2} \frac{1}{2}}$ can only be
\begin{align*}
\chi_{1 \frac{1}{2} \frac{1}{2}} & =\frac{1}{\sqrt{3}}[\sqrt{2} \chi_{11}(1,2) \beta(3)-\chi_{10}(1,2) \alpha(3)] \\
& =\frac{1}{\sqrt{6}}[2 \alpha(1) \alpha(2) \beta(3)-\alpha(1) \beta(2) \alpha(3)-\beta(1) \alpha(2) \beta(3)] \tag{8a}
\end{align*}
Similarly, expression (6c) can be written as
\begin{equation*}
\chi_{1 \frac{3}{2},-\frac{1}{2}}=\frac{1}{\sqrt{3}}[\sqrt{2} \chi_{10}(1,2) \beta(3)+\chi_{1-1}(1,2)_{\alpha}(3)] \tag{$\prime$}
\end{equation*}
$\chi_{1 \frac{1}{2},-\frac{1}{2}}$ should be formed with the same terms but orthogonal to the above expression, so
$$\chi_{1 \frac{1}{2},-\frac{1}{2}}=\frac{1}{\sqrt{3}}[\chi_{\mathrm{i} 0}(1,2) \beta(3)-\sqrt{2} \chi_{1,-1}(1,2) \alpha(3)]$$
\begin{equation*}
=\frac{1}{\sqrt{6}}[\alpha(1) \beta(2) \beta(3)+\beta(1) \alpha(2) \beta(3)-2 \beta(1) \beta(2) \alpha(3)] \tag{8b}
\end{equation*}
Expressions (6), (7), and (8) above are the complete set of common eigenfunctions for $\boldsymbol{S}_{12}^{2}, ~ \boldsymbol{S}_{123}^{2}, ~(S_{123})_{z}$. It is easy to verify that they are indeed orthogonal to each other. Since each particle has two spin states $\alpha$ and $\beta$, the system of three particles has a total of 8 independent states, thus expressions (6) to (8) form the sought orthogonal and complete set of energy eigenstates.
The reader can easily verify that according to the theory of angular momentum coupling, similar results would be obtained using the C.G. coefficient table.
|
{"$H$": "Hamiltonian", "$A$": "real constant in the Hamiltonian", "$s_{1}$": "spin component of the first particle", "$s_{2}$": "spin component of the second particle", "$s_{3}$": "spin component of the third particle", "$S$": "total spin quantum number of the system", "$E_{3/2}$": "energy level for total spin quantum number 3/2", "$\\hbar$": "reduced Planck's constant", "$\\boldsymbol{S}_{123}$": "total spin operator for particles 1, 2, and 3", "$S_{123}^{2}$": "squared total spin operator for particles 1, 2, and 3", "$(S_{123})_{z}$": "z-component of the total spin operator for particles 1, 2, and 3", "$S_{12}^{2}$": "squared spin operator for particles 1 and 2", "$S^{\\prime}$": "intermediate spin quantum number", "$M$": "magnetic quantum number", "$\\alpha(k)$": "spin up state of the k-th particle", "$\\beta(k)$": "spin down state of the k-th particle", "$s_{k z}$": "z-component of the spin for the k-th particle", "$\\chi_{11}$": "state with specific spin configuration for two particles", "$\\chi_{10}$": "state with specific spin configuration for two particles", "$\\chi_{1,-1}$": "state with specific spin configuration for two particles", "$\\chi_{00}$": "state with specific spin configuration for two particles", "$\\chi_{1 \\frac{3}{2} \\frac{3}{2}}$": "common eigenstate when all quantum numbers are at maximum values", "$\\chi_{1 \\frac{3}{2} \\frac{1}{2}}$": "common eigenstate for total spin 3/2 and magnetic quantum number 1/2", "$\\chi_{1 \\frac{3}{2},-\\frac{1}{2}}$": "common eigenstate for total spin 3/2 and magnetic quantum number -1/2", "$\\chi_{1 \\frac{3}{2},-\\frac{3}{2}}$": "common eigenstate for total spin 3/2 and magnetic quantum number -3/2", "$\\chi_{0 \\frac{1}{2} \\frac{1}{2}}$": "common eigenstate for intermediate spin 0 and magnetic quantum number 1/2", "$\\chi_{0 \\frac{1}{2},-\\frac{1}{2}}$": "common eigenstate for intermediate spin 0 and magnetic quantum number -1/2", "$\\chi_{1 \\frac{1}{2} \\frac{1}{2}}$": "common eigenstate for total spin 1/2 and magnetic quantum number 1/2", "$\\chi_{1 \\frac{1}{2},-\\frac{1}{2}}$": "common eigenstate for total spin 1/2 and magnetic quantum number -1/2"}
|
Schrödinger equation one-dimensional motion
| |
15 |
Theoretical Foundations
|
An electron moves freely in a one-dimensional region $-L/2 \leqslant x \leqslant L/2$, with the wave function satisfying periodic boundary conditions $\psi(x)=\psi(x+L)$. Its unperturbed energy eigenvalue is $E_n^{(0)}$. A perturbation $H^{\prime}=\varepsilon \cos q x$ is applied to this system where $L q=4 \pi N$ ($N$ is a large positive integer). Consider the degenerate state where the electron's momentum is $|p|=q \hbar / 2$ (corresponding to energy $E_N^{(0)}$) without perturbation. Find the expression for the second-order energy correction $E_N^{(2)}$ of these degenerate states caused by the perturbation $H^{\prime}$.
|
[
"-\\frac{\\varepsilon^{2}}{32 E_{N}^{(0)}}"
] |
Expression
|
(a) For a free particle, the common eigenfunction of the energy $(H_{0})$ and momentum $(p)$ satisfying the periodic condition is
\begin{equation*}
\psi_{n}^{(0)}=\frac{1}{\sqrt{L}} \mathrm{e}^{\mathrm{i} 2 \pi x / L}, \quad n=0, \pm 1, \pm 2, \cdots \tag{1}
\end{equation*}
The eigenvalue is
\begin{align*}
p_{n} & =\hbar k_{n}=2 \pi n \hbar / L \\
E_{n}^{(0)} & =\frac{p_{n}^{2}}{2 m}=\frac{(2 \pi n \hbar)^{2}}{2 m L^{2}} \tag{2}
\end{align*}
(b) The perturbation operator is
\begin{equation*}
H^{\prime}=\varepsilon \cos q x=\frac{\varepsilon}{2}(\mathrm{e}^{\mathrm{i} \pi \lambda N_{x} / L}+\mathrm{e}^{-\mathrm{i} 4 \pi N_{x} / L}) \tag{3}
\end{equation*}
For $|p|=q \hbar / 2=2 \pi N \hbar / L$ , the corresponding zeroth-order wave function is
\begin{equation*}
\psi^{(0)}=C_{N} \psi_{N}^{(0)}+C_{-, ~} \psi_{-, ~}^{(0)} \tag{4}
\end{equation*}
It can be easily calculated that
\begin{equation*}
H_{\mathrm{vv}}^{\prime}=H_{-\mathrm{v},-\mathrm{N}}^{\prime}=0, \quad H_{v,-\mathrm{v}}^{\prime}=H_{-\mathrm{N}, \mathrm{v}}^{\prime}=\frac{\varepsilon}{2} \tag{5}
\end{equation*}
Therefore, in the subspace ${\psi_{N}^{(0)}, \psi_{-N}^{(0)}}$, the matrix representation of $H^{\prime}$ is
H^{\prime}=\frac{\varepsilon}{2}[\begin{array}{ll}
0 & 1 \tag{$\prime$}\\
1 & 0
\end{array}]
In this subspace, $\psi^{(0)}$ should satisfy the eigenvalue equation
$H^{\prime} \psi^{(0)}=E_{N}^{(1)} \psi^{(0)}
That is
\begin{equation}
\frac{\varepsilon}{2}\left[\begin{array}{ll}
0 & 1 \tag{6}\\
1 & 0
\end{array}\right]\left[\begin{array}{l}
C_{\mathrm{N}} \\
C_{-\mathrm{N}}
\end{array}\right]=E_{N}^{(1)}\left[\begin{array}{l}
C_{\mathrm{N}} \\
C_{-N}
\end{array}\right]
\end{equation}
It is easy to solve
\begin{align*}
& E_{\mathrm{N}+}^{(1)}=\varepsilon / 2 \tag{7}\\
& \psi^{(0)}=\frac{1}{\sqrt{2}}[\psi_{N}^{(0)}+\psi_{-\mathrm{N}}^{(0)}]=\psi_{N+}^{(0)} \\
& E_{\mathrm{N}-}^{(1)}=-\varepsilon / 2 \\
& \psi^{(0)}=\frac{1}{\sqrt{2}}[\psi_{\mathrm{N}}^{(0)}-\psi_{-\mathrm{N}}^{(0)}]=\psi_{N-}^{(0)}
\end{align*}
The first-order correction to the wave function is
\begin{equation*}
\psi^{(1)}=\sum_{n}^{\prime} \frac{1}{E_{N}^{(0)}-E_{n}^{(0)}}\langle\psi_{n}^{(0)}| H^{\prime}|\psi^{(0)}\rangle \psi_{n}^{(0)} \quad(n \neq \pm N) \tag{8}
\end{equation*}
The matrix elements contributing to $\psi^{(1)}$ are (corresponding to $n= \pm 3 N$)
$$ H_{3, \mathrm{~V} . \mathrm{N}}^{\prime}=H_{-3, \mathrm{~V},-\mathrm{N}}^{\prime}=\varepsilon / 2$$
The corresponding energy difference is
$$ E_{N}^{(0)}-E_{3 N}^{(0)}=-8 E_{N}^{(0)}=-\frac{8(2 \pi N \hbar)^{2}}{2 m L^{2}}$$
Thus, the first-order correction to the wave function is
\begin{align}
E^{(1)} & =E_{N+}^{(1)}=\frac{\varepsilon}{2} \\
\psi^{(1)} & =-\frac{\varepsilon}{2} \frac{2 m L^{2}}{8(2 \pi N \hbar)^{2}} \frac{1}{\sqrt{2}}[\psi_{3 N}^{(0)}+\psi_{-3 N}^{(0)}]
\end{align}
\begin{align*}
E^{(1)} & =E_{N-}^{(1)}=-\frac{\varepsilon}{2} \\
\psi^{(1)} & =-\frac{\varepsilon}{2} \frac{2 m L^{2}}{8(2 \pi N \hbar)^{2}} \frac{1}{\sqrt{2}}[\psi_{3 N}^{(0)}-\psi_{-3 N}^{(0)}] \tag{9}
\end{align*}
(c) The second-order correction to the energy is
\begin{equation*}
E_{N}^{(2)}=\langle\psi^{(0)}| H^{\prime}|\psi^{(1)}\rangle \tag{10}
\end{equation*}
Substituting equations (7) and (9) into the above formula, we obtain
\begin{equation*}
E_{N}^{(2)}=-\frac{\varepsilon^{2}}{32} \frac{2 m L^{2}}{(2 \pi N \hbar)^{2}}=-\frac{\varepsilon^{2}}{32 E_{N}^{(0)}} \tag{11}
\end{equation*}
Combining equations (7) and (11), the conclusion for the energy level up to $\varepsilon^{2}$ order is:
\begin{equation*}
E_{N}=E_{N}^{(0)}+E_{N}^{(1)}+E_{N}^{(2)}=E_{\mathrm{N}}^{(0)} \pm \frac{\varepsilon}{2}-\frac{\varepsilon^{2}}{32 E_{N}^{(0)}} \tag{12}
\end{equation*}
Corresponding to
$$\psi^{(0)}=\frac{1}{\sqrt{2}}[\psi_{N}^{(0)} \pm \psi_{-, .}^{(0)}] $$
|
{"$x$": "position", "$L$": "length of the region", "$\\psi(x)$": "wave function", "$\\psi(x+L)$": "wave function after a period", "$E_n^{(0)}$": "unperturbed energy eigenvalue", "$H^{\\prime}$": "perturbation to the system", "$\\varepsilon$": "perturbation strength", "$q$": "wave number associated with the perturbation", "$N$": "large positive integer", "$p$": "momentum of the electron", "$\\hbar$": "reduced Planck's constant", "$E_N^{(0)}$": "energy corresponding to the momentum state before perturbation", "$E_N^{(2)}$": "second-order energy correction", "$H_{0}$": "Hamiltonian for unperturbed system", "$p_{n}$": "momentum eigenvalue", "$E_{n}^{(0)}$": "energy eigenvalue for state n", "$H_{\\mathrm{vv}}^{\\prime}$": "matrix element of the perturbation", "$H_{-\\mathrm{v},-\\mathrm{N}}^{\\prime}$": "off-diagonal matrix element of perturbation", "$H_{v,-\\mathrm{v}}^{\\prime}$": "off-diagonal matrix element of perturbation", "$H_{-\\mathrm{N},\\mathrm{v}}^{\\prime}$": "off-diagonal matrix element of perturbation", "$E_{N+}^{(1)}$": "first-order energy correction (positive)", "$\\psi^{(0)}$": "zeroth-order wave function", "$C_{N}$": "coefficient for positive momentum state", "$C_{-N}$": "coefficient for negative momentum state", "$\\psi_{N+}^{(0)}$": "positive superposition state", "$E_{N-}^{(1)}$": "first-order energy correction (negative)", "$\\psi_{N-}^{(0)}$": "negative superposition state", "$E_{3 N}^{(0)}$": "energy of 3N state", "$E^{(1)}$": "first-order correction", "$\\psi^{(1)}$": "first-order correction to wave function"}
|
Schrödinger equation one-dimensional motion
| |
16 |
Theoretical Foundations
|
For the $n$-th bound state $\psi_{n}, ~ E_{n}$ of a square well (depth $V_{0}$, width $a$), under the condition $V_{0} \gg E_{n}$, calculate the probability of the particle appearing outside the well.
|
[
"\\frac{2 \\hbar E_n}{a V_0 \\sqrt{2 m V_0}}"
] |
Expression
|
Take the even parity state as an example. The energy eigenvalue equation can be written as
\begin{equation}
\begin{array}{lll}
\psi^{\prime \prime}+k^{2} \psi=0, & |x| \leqslant a / 2 & \text { (inside the well) } \tag{1}\\
\psi^{\prime \prime}-\beta^{2} \psi=0, & |x| \geqslant a / 2 & \text { (outside the well) }
\end{array}
\end{equation}
where
\begin{equation*}
k=\sqrt{2 m E} / \hbar, \quad \beta=\sqrt{2 m(V_{0}-E)} / \hbar \tag{2}
\end{equation*}
Note that under the condition $V_{0} \gg E$, $\beta \gg k$.
The even parity solution of equation (1) is
\begin{array}{ll}
\psi=\cos k x, & |x| \leqslant a / 2 \\
\psi=C \mathrm{e}^{-\beta|x|}, & |x| \geqslant a / 2 \tag{3}
\end{array}
At $x=a / 2$, $\psi$ should be continuous, thus yielding
\begin{equation*}
C=\mathrm{e}^{\beta a / 2} \cos \frac{k a}{2} \tag{4}
\end{equation*}
At $x=a / 2$, $\psi^{\prime}$ should also be continuous, thus yielding
$$ C=\frac{k}{\beta} \mathrm{e}^{\beta a / 2} \sin \frac{k a}{2} $$
Dividing by equation (4), we obtain the energy level formula
\begin{equation*}
\tan \frac{k a}{2}=\frac{\beta}{k} \tag{5}
\end{equation*}
Under the condition $\beta \gg k$, the solution of equation (5) is
\begin{equation*}
k a \approx n \pi, \quad n=1,3,5, \cdots \tag{6}
\end{equation*}
Substituting into equation (2), the energy levels are
\begin{equation*}
E_{n}=\frac{\hbar^{2} k^{2}}{2 m} \approx \frac{1}{2 m}(\frac{n \pi \hbar}{a})^{2} \tag{7}
\end{equation*}
This is precisely the energy level formula for an infinitely deep potential well.
Now calculate the probability of the particle appearing inside and outside the well. From equations (3) and (4), it is easy to find
\begin{align*}
& \int_{\text {outside }}|\psi|^{2} \mathrm{~d} x=2 C^{2} \int_{a / 2}^{\infty} \mathrm{e}^{-2 \beta x} \mathrm{~d} x=\frac{C^{2}}{\beta} \mathrm{e}^{-\beta a}=\frac{1}{\beta} \cos ^{2} \frac{k a}{2} \tag{8}\\
& \int_{\text {inside }}|\psi|^{2} \mathrm{~d} x=2 \int_{0}^{a / 2} \cos ^{2} k x \mathrm{~d} x=\frac{a}{2}(1+\frac{\sin k a}{k a}) \tag{9}
\end{align*}
Considering $k a \approx n \pi(n=1,3,5, \cdots), \sin k a$ and $\cos (k a / 2)$ are both close to zero, it is understood that the probability of the particle appearing outside the well is much smaller than that inside the well. Additionally,
\begin{gather*}
\int_{-\infty}^{+\infty}|\psi|^{2} \mathrm{~d} x \approx \int_{\text {inside }}|\psi|^{2} \mathrm{~d} x \approx \frac{a}{2} \\
\text { Outside probability }=\frac{\int_{\text {outside }}|\psi|^{2} \mathrm{~d} x}{\int_{-\infty}^{+\infty}|\psi|^{2} \mathrm{~d} x} \approx \frac{2}{\beta a} \cos ^{2} \cdot \frac{k a}{2} \tag{10}
\end{gather*}
Using the energy level formula (5), it is easy to obtain
\begin{gather*}
1+\tan ^{2} \frac{k a}{2}=\frac{k^{2}+\beta^{2}}{k^{2}}=\frac{V_{0}}{E} \\
\cos ^{2} \frac{k a}{2}=\frac{E}{V_{0}} \tag{11}
\end{gather*}
Substituting into equation (10), we get
\begin{equation*}
\text { Outside probability }=\frac{2 E}{a \beta V_{0}} \approx \frac{2 \hbar}{a \sqrt{2 m V_{0}}} \frac{E}{V_{0}} \tag{12}
\end{equation*}
Considering $V_{0} \gg E$ and the energy level formula (7), it is easily seen
\begin{equation*}
\sqrt{2 m V_{0}} \gg n \pi \hbar / a \tag{13}
\end{equation*}
Thus,
\begin{equation*}
\text { Outside probability } \ll \frac{2 E}{n \pi V_{0}} \tag{14}
\end{equation*}
|
{"$n$": "quantum number representing the bound state", "$\\psi_{n}$": "wave function for the n-th bound state", "$E_{n}$": "energy of the n-th bound state", "$V_{0}$": "depth of the potential well", "$a$": "width of the potential well", "$k$": "wave number inside the well", "$m$": "mass of the particle", "$E$": "energy", "$\\beta$": "exponential decay constant outside the well", "$C$": "normalization constant for the wave function"}
|
Schrödinger equation one-dimensional motion
| |
17 |
Theoretical Foundations
|
A particle moves freely, and the initial wave function at $t=0$ is given as
$$ \psi(x, 0)=(2 \pi a^{2})^{-1 / 4} \exp [\mathrm{i} k_{0}(x-x_{0})-(\frac{x-x_{0}}{2 a})^{2}], \quad a>0 $$
Find the wave function $\varphi(p)$ in the $p$ representation at $t=0$;
|
[
"\\varphi(p) = (\\frac{2 a^{2}}{\\pi})^{\\frac{1}{4}} \\exp [-\\mathrm{i} \\frac{p x_{0}}{\\hbar}-a^{2}(\\frac{p}{\\hbar}-k_{0})^{2}]"
] |
Expression
|
To simplify, we set $\hbar=1$ during calculations, then add it to the final result. First, consider the shape of the wave packet at $t=0$
\begin{equation*}
|\psi(x, 0)|^{2}=(2 \pi a^{2})^{-1 / 2} \mathrm{e}^{-(x-x_{0})^{2} / 2 a^{2}} \tag{1}
\end{equation*}
which is a Gaussian distribution. According to the mean value formula
\begin{equation}
\overline{f(x)}=\int_{-\infty}^{+\infty}|\psi|^{2} f(x) \mathrm{d} x
\end{equation}
it is easy to calculate that at $t=0$
\begin{align*}
& \bar{x}=x_{0}, \quad \bar{x}^{2}=a^{2}+x_{0}^{2} \\
& \Delta x=(\bar{x}^{2}-\bar{x}^{2})^{1 / 2}=a \tag{2}
\end{align*}
which means at $t=0$, the center of the wave packet is at $x_{0}$, and the width of the wave packet is $a$. Let
\begin{equation*}
\psi(x, 0)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \varphi(k) \mathrm{e}^{\mathrm{i} k x} \mathrm{~d} k \tag{3}
\end{equation*}
where $\varphi(k)$ is the initial wave function in the momentum representation. According to the Fourier transform formula
\begin{align*}
\varphi(k)&=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \psi(x, 0) \mathrm{e}^{-\mathrm{i} k x} \mathrm{~d} x\\
& =(\frac{1}{2 \pi})^{3 / 4} \frac{1}{\sqrt{a}} \mathrm{e}^{-\mathrm{k} x_{0}} \int_{-\infty}^{+\infty} \mathrm{d} x \exp [-(\frac{x-x_{0}}{2 a})^{2}-\mathrm{i}(k-k_{0})(x-x_{0})] \\
& =(\frac{2 a^{2}}{\pi})^{\frac{1}{4}} \exp [-\mathrm{i} k x_{0}-a^{2}(k-k_{0})^{2}] \tag{4}
\end{align*}
The integration formula is used in the calculation (refer to the note at the end of the question). From equation (4)
\begin{equation*}
|\varphi(k)|^{2}=\sqrt{\frac{2}{\pi}} a \mathrm{e}^{-2 a^{2}(k-k_{0})^{2}} \tag{5}
\end{equation*}
which indicates that the probability distribution of momentum is also a Gaussian distribution. According to the mean value formula
\begin{equation}
\overline{f(k)}=\int_{-\infty}^{+\infty}|\varphi(k)|^{2} f(k) \mathrm{d} k
\end{equation}
it is readily calculated that
\begin{equation*}
\bar{k}=k_{0}, \quad \overline{k^{2}}=k_{0}^{2}+\frac{1}{4 a^{2}} \tag{6}
\end{equation*}
Since $p=\hbar k$, thus
\begin{align*}
& \bar{p}=\hbar k_{0}, \quad \bar{p}^{2}=\hbar^{2} k_{0}^{2}+\frac{\hbar^{2}}{4 a^{2}} \tag{7}\\
& \Delta p=(\overline{p^{2}}-\bar{p}^{2})^{1 / 2}=\hbar / 2 a
\end{align*}
These are the characteristics of the momentum distribution at $t=0$. Since it is a free particle, momentum conservation implies that the probability distribution of momentum is also conserved. Therefore, equation (7) applies to any time. From equations (2) and (7), it follows that
\begin{equation*}
\Delta x \cdot \Delta p=\hbar / 2 \quad(t=0 \text { instant }) \tag{8}
\end{equation*}
|
{"$t$": "time", "$\\psi$": "wave function in position representation", "$x$": "position", "$a$": "width of the wave packet", "$k_{0}$": "initial wave number", "$x_{0}$": "initial position", "$\\varphi$": "wave function in momentum representation", "$p$": "momentum", "$\\hbar$": "reduced Planck's constant", "$k$": "wave number"}
|
Schrödinger equation one-dimensional motion
| |
18 |
Theoretical Foundations
|
The particle moves freely, with the initial wave function at $t=0$ given as
$$ \psi(x, 0)=(2 \pi a^{2})^{-1 / 4} \exp [\mathrm{i} k_{0}(x-x_{0})-(\frac{x-x_{0}}{2 a})^{2}], \quad a>0 $$
Find the wave function $\psi(x, t)$ for $t>0$
|
[
"\\psi(x, t) = \\frac{\\exp [\\mathrm{i} k_{0}(x-x_{0})-\\mathrm{i} t k_{0}^{2} / 2 m]}{(2 \\pi)^{1 / 4}(a+\\mathrm{i} t / 2 m a)^{1 / 2}} \\exp [-\\frac{1}{4}(x-x_{0}-\\frac{k_{0} t}{m})^{2} \\frac{1-\\mathrm{i} t / 2 m a^{2}}{a^{2}+(t / 2 m a)^{2}}]"
] |
Expression
|
In equation \begin{equation*}
\psi(x, 0)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \varphi(k) \mathrm{e}^{\mathrm{i} k x} \mathrm{~d} k, \tag{3}
\end{equation*} let
$$ \mathrm{e}^{\mathrm{i} k x} \rightarrow \mathrm{e}^{\mathrm{i}(k x-\omega t)}, \quad \omega=\frac{\hbar k^{2}}{2 m}=\frac{k^{2}}{2 m} \quad(\hbar=1) $$
thus obtaining the wave function $\psi(x, t)$ for $t>0$, i.e.,
\begin{equation*}
\psi(x, t)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \varphi(k) \exp (\mathrm{i} k x-\frac{\mathrm{i} k^{2} t}{2 m}) \mathrm{d} k \tag{9}
\end{equation*}
Substitute equation \begin{align*}
\varphi(k)&= (\frac{2 a^{2}}{\pi})^{\frac{1}{4}} \exp [-\mathrm{i} k x_{0}-a^{2}(k-k_{0})^{2}]
\end{align*} into the above expression and we obtain
\begin{align*}
& \psi(x, t) \\
= & \frac{\exp [\mathrm{i} k_{0}(x-x_{0})-\mathrm{i} t k_{0}^{2} / 2 m]}{(2 \pi)^{1 / 4}(a+\mathrm{i} t / 2 m a)^{1 / 2}} \exp [-\frac{1}{4}(x-x_{0}-\frac{k_{0} t}{m})^{2} \frac{1-\mathrm{i} t / 2 m a^{2}}{a^{2}+(t / 2 m a)^{2}}] \tag{10}\\
\end{align*}
|
{"$t$": "time", "$\\psi$": "wave function", "$x$": "position", "$a$": "positive constant related to wave packet width", "$k_{0}$": "initial wave number", "$x_{0}$": "initial position", "$\\varphi$": "function in wave number space", "$\\omega$": "angular frequency", "$m$": "mass"}
|
Schrödinger equation one-dimensional motion
| |
19 |
Theoretical Foundations
|
Using the raising and lowering operators $a^{+}, ~ a$, find the energy eigenfunctions of the harmonic oscillator (in the $x$ representation), and briefly discuss their mathematical properties.
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[
"\\psi_n(x) = N_n H_n(\\alpha x) e^{-\\frac{1}{2}(\\alpha x)^2}",
"\\psi_n(x) = (\\frac{\\alpha}{\\sqrt{\\pi} 2^n n!})^{1/2} H_n(\\alpha x) e^{-\\frac{1}{2}(\\alpha x)^2}"
] |
Expression
|
Start with the ground state wave function $\psi_{0}(x)$. We have,
\begin{equation*}
a|0\rangle=0 \tag{1}
\end{equation*}
In the $x$ representation this reads as
\begin{equation*}
(\mathrm{i} \hat{p}+m \omega x) \psi_{0}(x)=(\hbar \frac{\mathrm{d}}{\mathrm{~d} x}+m \omega x) \psi_{0}(x)=0 \tag{2}
\end{equation*}
Let $\alpha=\sqrt{m \omega / \hbar}$, the above equation can be written as
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{~d} x} \psi_{0}+\alpha^{2} x \psi_{0}=0 \tag{2'}
\end{equation*}
It is easy to solve
\begin{equation*}
\psi_{0}(x)=N_{0} \mathrm{e}^{-\alpha^{2} x^{2} / 2} \tag{3}
\end{equation*}
where $N_{0}$ is the normalization constant. From the normalization condition
\begin{equation*}
\int_{-\infty}^{+\infty}|\psi_{n}(x)|^{2} \mathrm{~d} x=1 \tag{4}
\end{equation*}
Find (taking as real)
\begin{equation*}
N_{0}=\sqrt{\alpha} / \pi^{1 / 4} \tag{5}
\end{equation*}
Secondly, we have
$$ a^{+}|0\rangle=|1\rangle $$
That is
\begin{equation*}
\psi_{1}(x)=a^{+} \psi_{0}(x)=\frac{1}{\sqrt{2}}(\alpha x-\frac{1}{\alpha} \frac{\mathrm{~d}}{\mathrm{~d} x}) \psi_{0}(x) \tag{6}
\end{equation*}
Substituting equation (3) and (5) into equation (6), it's easy to find
\begin{equation*}
\psi_{1}=\sqrt{2} \alpha x \psi_{0}=\frac{\sqrt{2 \alpha}}{\pi^{1 / 4}} \alpha x \mathrm{e}^{-\alpha^{2} x^{2} / 2} \tag{7}
\end{equation*}
Generally, we have
\begin{equation*}
\psi_{n}=\frac{1}{\sqrt{n}} a^{+} \psi_{n-1}=\frac{1}{\sqrt{2 n}}(\alpha x-\frac{1}{\alpha} \frac{\mathrm{~d}}{\mathrm{~d} x}) \psi_{n-1} \tag{8}
\end{equation*}
Introducing the dimensionless variable
\begin{equation*}
\xi=\alpha x, \tag{9}
\end{equation*}
we thus obtain
\begin{equation*}
\psi_{n}=\frac{1}{\sqrt{2 n}}(\xi-\frac{\mathrm{d}}{\mathrm{~d} \xi}) \psi_{n-1} \tag{10}.
\end{equation*}
Recursively, we get
\begin{equation*}
\psi_{n}=(\frac{1}{2^{n} n!})^{\frac{1}{2}}(\xi-\frac{\mathrm{d}}{\mathrm{~d} \xi})^{n} \psi_{0}=(\frac{\alpha}{\sqrt{\pi} 2^{n} n!})^{\frac{1}{2}}(\xi-\frac{\mathrm{d}}{\mathrm{~d} \xi})^{n} \mathrm{e}^{-\xi^{2} / 2}=N_{n} H_{n}(\xi) \mathrm{e}^{-\xi^{2} / 2} \tag{11}
\end{equation*}
Where
\begin{equation*}
N_{n}=(\frac{\alpha}{\sqrt{\pi} 2^{n} n!})^{\frac{1}{2}} \tag{12}
\end{equation*}
is the normalization constant. Notice
\begin{equation*}
N_{n}=N_{n-1} / \sqrt{2 n} \tag{13}
\end{equation*}
In equation (11),
\begin{equation*}
H_{n}(\xi)=\mathrm{e}^{\xi^{2} / 2}(\xi-\frac{\mathrm{d}}{\mathrm{~d} \xi})^{n} \mathrm{e}^{-\xi^{2} / 2} \tag{1}
\end{equation*}
It is obvious that $H_{n}(\xi)$ is an $n$-th degree polynomial in $\xi$, called the Hermite polynomial.
Below is a brief discussion of the mathematical properties of $H_{n}(\xi)$.
{Parity}
It is obvious from equation (14)
\begin{equation*}
H_{n}(-\xi)=(-1)^{n} H_{n}(\xi) \tag{15}
\end{equation*}
When $n$ is even, $H_{n}(\xi)$ has even parity; when $n$ is odd, $H_{n}(\xi)$ has odd parity.
{Recursion relations}
We also have,
$$ a \psi_{n}=\frac{1}{\sqrt{2}}(\xi+\frac{\mathrm{d}}{\mathrm{~d} \xi}) \psi_{n}=\sqrt{n} \psi_{n-1}$$
Substituting equation (11) into this equation, and using equation (13) to eliminate the normalization constant, we get
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{~d} \xi} H_{n}(\xi)=2 n H_{n-1}(\xi) \tag{16},
\end{equation*}
We also have:
$$ \xi \psi_{n}=\sqrt{\frac{n+1}{2}} \psi_{n+1}+\sqrt{\frac{n}{2}} \psi_{n-1} $$
Substituting equations (11), (13) into this equation, we get
\begin{equation*}
2 \xi H_{n}(\xi)=H_{n+1}(\xi)+2 n H_{n-1}(\xi) \tag{17}
\end{equation*}
{Differential equation}
Combining equations (16) and (17), eliminate $H_{n-1}$, then differentiate once more, and using equation (16) to eliminate $\mathrm{d} H_{n+1} / \mathrm{d} \xi$, we find that $H_{n}$ satisfies the following differential equation:
\begin{equation*}
\frac{\mathrm{d}^{2}}{\mathrm{~d} \xi^{2}} H_{n}(\xi)-2 \xi \frac{\mathrm{~d}}{\mathrm{~d} \xi} H_{n}(\xi)+2 n H_{n}(\xi)=0 \tag{18}
\end{equation*}
This equation is known as the Hermite equation. Equation (14) is the unique polynomial solution to this equation.
$4{ }^{\circ}$ Differential expression
Equation (14) is equivalent to the following:
\begin{equation*}
H_{n}(\xi)=(-1)^{n} e^{\xi^{2}}(\frac{d}{d \xi})^{n} e^{-\xi^{2}} \tag{$\prime$}
\end{equation*}
This can be proved by mathematical induction. First, for $n=0,1$, both equations (14) and ($14'$) give
$$H_{0}(\xi)=1, \quad H_{1}(\xi)=2 \xi $$
Assume the proposition holds for $n=k$, i.e., assume
\mathrm{e}^{\xi^{2}}(-\frac{\mathrm{d}}{\mathrm{~d} \xi})^{k} \mathrm{e}^{-\xi^{2}}=\mathrm{e}^{\xi^{2} / 2}(\xi-\frac{\mathrm{d}}{\mathrm{~d} \xi})^{k} \mathrm{e}^{-\xi^{2} / 2}=H_{k}(\xi)
Then
\begin{gathered}
(-\frac{\mathrm{d}}{\mathrm{~d} \xi})^{k+1} \mathrm{e}^{-\xi^{2}}=-\frac{\mathrm{d}}{\mathrm{~d} \xi}(\mathrm{e}^{-\xi^{2}} H_{k})=(2 \xi H_{k}-H_{k}^{\prime}) \mathrm{e}^{-\xi^{2}} \\
(\xi-\frac{\mathrm{d}}{\mathrm{~d} \xi})^{k+1} \mathrm{e}^{-\xi^{2} / 2}=(\xi-\frac{\mathrm{d}}{\mathrm{~d} \xi})(\mathrm{e}^{-\xi^{2} / 2} H_{k})=(2 \xi H_{k}-H_{k}^{\prime}) \mathrm{e}^{-\xi^{2} / 2}
\end{gathered}
Thus
\mathrm{e}^{\xi^{2}}(-\frac{\mathrm{d}}{\mathrm{~d} \xi})^{k+1} \mathrm{e}^{-\xi^{2}}=2 \xi H_{k}-H_{k}^{\prime}=\mathrm{e}^{\xi^{2} / 2}(\xi-\frac{\mathrm{d}}{\mathrm{~d} \xi})^{k+1} \mathrm{e}^{-\xi^{2} / 2}
The proposition holds for $n=k+1$. Proof complete.
$5^{\circ} H_{n}(0)$ and $\psi_{n}(0)$
When $n=2 k+1$ (odd), from equation (15), we have
H_{2 k+1}(-\xi)=-H_{2 k+1}(\xi)
Let $\xi \rightarrow 0$, we get
\begin{equation*}
H_{2 k+1}(0)=0, \quad k=1,2,3, \cdots \tag{19}
\end{equation*}
When $n=2 k$ (even), in equation (17) let $\xi \rightarrow 0$, and change $n$ to ( $n-1$ ), we get
H_{2 k}(0)=-2(2 k-1) H_{2 k-2}(0)
Using this repeatedly, and noting that $H_{0}=1$, we get
\begin{equation*}
H_{2 k}(0)=(-1)^{k} 2^{k}(2 k-1)!!=(-1)^{k}(2 k)!/ k! \tag{20}
\end{equation*}
Using equations (11), (12) again, we get
\begin{gather*}
\psi_{2 k+1}(0)=0 \tag{21}\\
\psi_{2 k}(0)=(-1)^{k} \frac{\sqrt{\alpha}}{\pi^{1 / 4}} \frac{\sqrt{(2 k)!}}{(2 k)!!} \tag{22}
\end{gather*}
\begin{equation*}
[\psi_{2 k}(0)]^{2}=\frac{\alpha}{\sqrt{\pi}} \frac{(2 k)!}{[(2 k)!!]^{2}}=\frac{\alpha}{\sqrt{\pi}} \frac{(2 k-1)!!}{(2 k)!!} \tag{23}
\end{equation*}
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{"$a$": "lowering operator", "$a^{+}$": "raising operator", "$x$": "position variable", "$m$": "mass of the particle", "$\\omega$": "angular frequency of the oscillator", "$\\hbar$": "reduced Planck's constant", "$\\alpha$": "dimensionless constant related to mass, frequency, and Planck's constant", "$N_{0}$": "normalization constant of the ground state wave function", "$N_{n}$": "normalization constant of the nth energy eigenfunction", "$\\psi_{0}(x)$": "ground state wave function in the position representation", "$\\psi_{n}(x)$": "nth energy eigenfunction in the position representation", "$\\xi$": "dimensionless position variable", "$H_{n}(\\xi)$": "Hermite polynomial of degree n"}
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Schrödinger equation one-dimensional motion
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20 |
Theoretical Foundations
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The emission of recoil-free $\gamma$ radiation by nuclei bound in a lattice is a necessary condition for the Mössbauer effect. The potential acting on the nuclei in the lattice can be approximated as a harmonic oscillator potential
$$ V(x)=\frac{1}{2} M \omega^{2} x^{2} $$
where $M$ is the mass of the nucleus, $x$ is the coordinate of the nucleus's center of mass, and $\omega$ is the vibration frequency. Assume that initially, the nucleus's center of mass motion (harmonic vibration) is in its ground state, and at $t=0$, due to a transition of energy levels within the nucleus, a photon is emitted along the $x$-axis with energy $E_{\gamma}$, and momentum $E_{\gamma} / c$. Since the $\gamma$ radiation occurs suddenly, the only effect on the nucleus's center of mass motion is that its momentum eigenvalue changes from $p$ to $(p-E_{\gamma} / c)$. Determine the probability that the nucleus's center of mass motion remains in the ground state after the photon is emitted. For instance, for a ${ }^{57} \mathrm{Fe}$ nucleus, if $E_{\gamma}=18 \mathrm{keV}, \omega=10^{12} \mathrm{~Hz}$, calculate the probability of this 'recoil-free emission' (i.e., no energy is transferred to the atom).
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[
"P=\\exp (-\\frac{E_{\\gamma}^{2}}{2 \\hbar \\omega M c^{2}})"
] |
Expression
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Due to the harmonic oscillator potential, the center of mass motion of the nucleus is harmonic, initially (for $t<0$) in the ground state $\psi_{0}(x)$. Expanding $\psi_{0}(x)$ using momentum eigenfunctions gives:
\begin{equation*}
\psi_{0}(x)=(2 \pi \hbar)^{-1 / 2} \int_{-\infty}^{+\infty} \varphi(p) \mathrm{e}^{\mathrm{i} p x / \hbar} \mathrm{d} p \tag{1}
\end{equation*}
Due to the emission of $\gamma$ photon, momentum changes (total momentum is conserved!) from $p$ to $p-p_{0},(p_{0}=E_{\gamma} / c)$, which means the wave function $\varphi(p)$ transforms as follows:
\begin{equation*}
\varphi(p) \rightarrow \varphi(p+p_{0}) \tag{2}
\end{equation*}
Therefore, after $\gamma$ emission, the wave function of the nucleus's center of mass motion becomes
\begin{align*}
\psi(x) & =(2 \pi \hbar)^{-1 / 2} \int_{-\infty}^{+\infty} \varphi(p+p_{0}) \mathrm{e}^{\mathrm{i} p x / \hbar} \mathrm{d} p \\
& =(2 \pi \hbar)^{-1 / 2} \int_{-\infty}^{+\infty} \varphi(p^{\prime}) \mathrm{e}^{\mathrm{i}(p^{\prime}-p_{0}) x / \hbar} \mathrm{d} p^{\prime} \\
& =\psi_{0}(x) \mathrm{e}^{-\mathrm{i} p_{0} x / \hbar} \tag{3}
\end{align*}
Here, the exponential factor indicates that the nucleus's center of mass momentum is reduced by $p_{0}$. The probability that the nucleus's center of mass vibration remains in the ground state after photon emission is
\begin{equation*}
P=|\langle\psi_{0} \mid \psi\rangle|^{2}=|\int_{-\infty}^{+\infty} \psi_{0}^{2}(x) \mathrm{e}^{-i p_{0} x / \hbar} \mathrm{d} x|^{2} \tag{4}
\end{equation*}
Using the explicit form of the ground state wave function $\psi_{0}$
\begin{equation*}
\psi_{0}(x)=\frac{\sqrt{\alpha}}{\pi^{1 / 4}} \mathrm{e}^{-\alpha^{2} x^{2} / 2}, \quad \alpha=\sqrt{\frac{M_{\omega}}{\hbar}} \tag{5}
\end{equation*}
It is straightforward to calculate
\begin{align*}
\langle\psi_{0} \mid \psi\rangle & =\frac{\alpha}{\sqrt{\pi}} \int_{-\infty}^{+\infty} \mathrm{e}^{-\alpha^{2} x^{2}-i p_{0} x / \hbar} \mathrm{d} x \\
& =\frac{\alpha}{\sqrt{\pi}} \int_{-\infty}^{+\infty} \cos (\frac{p_{0} x}{\hbar}) \mathrm{e}^{-\alpha^{2} x^{2}} \mathrm{~d} x \\
& =\exp (-\frac{p_{0}^{2}}{4 \hbar^{2} \alpha^{2}})=\exp (-\frac{E_{\gamma}^{2}}{4 \hbar \omega M c^{2}}) \tag{6}
\end{align*}
Substituting into equation (4), we obtain the probability, which is
\begin{equation*}
P=|\langle\psi_{0} \mid \psi\rangle|^{2}=\exp (-\frac{E_{\gamma}^{2}}{2 \hbar \omega M c^{2}}) \tag{7}
\end{equation*}
Calculated numerical values are as follows:
\begin{gathered}
{ }^{57} \mathrm{Fe} \text { nucleus, } M c^{2} \approx 57 m_{\mathrm{p}} c^{2} \approx 57 \times 938 \mathrm{MeV}=5.35 \times 10^{10} \mathrm{eV} \\
E_{\gamma}=18 \mathrm{keV}=1.8 \times 10^{4} \mathrm{eV} \\
\hbar \omega=\hbar c \cdot \frac{\omega}{c} \approx 1.97 \times 10^{-5} \mathrm{eV} \cdot \mathrm{~cm} \times \frac{2 \pi \times 10^{12}}{3 \times 10^{10}} \mathrm{~cm}^{-1} \\
=4.13 \times 10^{-3} \mathrm{eV} \\
P=\exp [-\frac{(1.8 \times 10^{4})^{2}}{2 \times 4.13 \times 10^{-3} \times 5.35 \times 10^{10}}] \\
=\mathrm{e}^{-0.733}=0.48
\end{gathered}
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{"$\\gamma$": "gamma radiation", "$M$": "mass of the nucleus", "$x$": "coordinate of the nucleus's center of mass", "$\\omega$": "vibration frequency", "$t$": "time", "$E_{\\gamma}$": "energy of the emitted photon", "$c$": "speed of light", "$p$": "momentum", "$p_{0}$": "momentum of the photon (E_{\\gamma}/c)", "$\\hbar$": "reduced Planck's constant", "$\\psi_{0}(x)$": "ground state wave function", "$\\alpha$": "related to the oscillator strength (\\sqrt{M_{\\omega}/\\hbar})"}
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Schrödinger equation one-dimensional motion
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21 |
Theoretical Foundations
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Calculate the probability that the nucleus is in various energy eigenstates after $\gamma$ radiation, as in the previous problem.
|
[
"P_n = \\frac{1}{n!}(\\frac{E_{\\gamma}^{2}}{2 \\hbar \\omega M c^{2}})^{n} \\exp (-\\frac{E_{\\gamma}^{2}}{2 \\hbar \\omega M c^{2}})"
] |
Expression
|
From the previous problem, the center-of-mass wave function of the nucleus after $\gamma$ radiation is
\begin{equation*}
\psi(x)=\mathrm{e}^{-\mathrm{i} p_{0} x / \hbar} \psi_{0}(x), \quad p_{0}=E_{\gamma} / c \tag{1}
\end{equation*}
The probability of being in the ${ }^{\prime} n$-th vibrational excited state $\psi_{n}(x)$ is
\begin{equation*}
P_{n}=|\langle\psi_{n} \mid \psi\rangle|^{2}=|\langle\psi_{n} \mid \mathrm{e}^{-\mathrm{i} p_{0} x / \hbar} \psi_{0}\rangle|^{2} \tag{2}
\end{equation*}
Let $\lambda=-p_{0} / \hbar$, calculate $\langle\psi_{n} \mid \mathrm{e}^{\mathrm{i} \lambda x} \psi_{0}\rangle$ where $\mathrm{e}^{\mathrm{i} \lambda x}$ can be viewed as an operator, expressed in terms of creation and annihilation operators, with the following relations:
\begin{align*}
& \mathrm{i} \lambda x=\alpha a^{+}-\alpha^{*} a, \quad \alpha=\mathrm{i} \lambda \sqrt{\frac{\hbar}{2 M \omega}} \tag{3}\\
& \mathrm{e}^{\mathrm{i} \lambda x}=\mathrm{e}^{\alpha a^{+}-\alpha^{*} a}=\mathrm{e}^{\alpha a^{+}} \mathrm{e}^{-\alpha^{*} a} \mathrm{e}^{-\frac{1}{2}|\alpha|^{2}} \tag{4}\\
& \mathrm{e}^{\mathrm{i} \lambda x}|\psi_{0}\rangle=\mathrm{e}^{-\frac{1}{2}|\alpha|^{2}} \mathrm{e}^{\alpha a^{+}} \mathrm{e}^{-\alpha^{*} a}|\psi_{0}\rangle=|\psi_{\alpha}\rangle \\
&= \mathrm{e}^{-\frac{1}{2}|\alpha|^{2}} \sum_{n} \frac{\alpha^{n}}{\sqrt{n!}}|\psi_{n}\rangle \tag{5}
\end{align*}
where $\psi_{\alpha}$ is the coherent state of the harmonic oscillator. Utilizing equation (5), we get
\begin{gather*}
\langle\psi_{n} \mid \mathrm{e}^{\mathrm{i} \lambda x} \psi_{0}\rangle=\frac{\alpha^{n}}{\sqrt{n!}} \mathrm{e}^{-\frac{1}{2}|\alpha|^{2}} \tag{6}\\
P_{n}=\frac{|\alpha|^{2 n}}{n!} \mathrm{e}^{-|\alpha|^{2}}=\frac{1}{n!}(\frac{\lambda^{2} \hbar}{2 M \omega})^{n} \mathrm{e}^{-\frac{\lambda^{2} \hbar}{2 M \omega}} \\
= \frac{1}{n!}(\frac{E_{\gamma}^{2}}{2 \hbar \omega M c^{2}})^{n} \exp (-\frac{E_{\gamma}^{2}}{2 \hbar \omega M c^{2}}) \tag{7}
\end{gather*}
It is easy to verify $\sum_{n} P_{n}=1$. For the example of the ${ }^{57} \mathrm{Fe}$ nucleus, the probabilities for the first few states are
$$ P_{0}=0.48, \quad P_{1}=0.35, \quad P_{2}=0.13 $$
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{"$\\gamma$": "gamma radiation energy", "$\\psi$": "wave function", "$p_{0}$": "momentum of the nucleus", "$x$": "position", "$\\hbar$": "reduced Planck constant", "$\\psi_{0}$": "initial wave function", "$P_{n}$": "probability of being in the nth vibrational excited state", "$\\psi_{n}$": "nth vibrational eigenstate", "$\\lambda$": "parameter related to momentum and reduced Planck constant", "$M$": "mass of the nucleus", "$\\omega$": "angular frequency", "$E_{\\gamma}$": "energy of the gamma photon", "$a^{+}$": "creation operator", "$a$": "annihilation operator", "$\\alpha$": "displacement parameter"}
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Schrödinger equation one-dimensional motion
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22 |
Theoretical Foundations
|
Suppose the operator $\hat{H}$ has continuous eigenvalues $\omega$, and its eigenfunctions $u_{\omega}(\boldsymbol{x})$ form an orthonormal complete system, i.e.
\begin{gather*}
\hat{H} u_{\omega}(\boldsymbol{x})=\omega u_{\omega}(\boldsymbol{x}) \tag{1}\\
\int u_{\omega^{*}}^{*}(\boldsymbol{x}) u_{\omega}(\boldsymbol{x}) \mathrm{d}^{3} x=\delta(\omega-\omega^{\prime}) \tag{2}\\
\int u_{\omega}^{*}(\boldsymbol{x}^{\prime}) u_{\omega}(\boldsymbol{x}) \mathrm{d} \omega=\delta(\boldsymbol{x}-\boldsymbol{x}^{\prime}) \tag{3}
\end{gather*}
Solve the equation
\begin{equation*}
(\hat{H}-\omega_{0}) u(\boldsymbol{x})=F(\boldsymbol{x}) \tag{4}
\end{equation*}
where $F(\boldsymbol{x})$ is a known function, and $\omega_{0}$ is a specific eigenvalue of $H$.
If the answer exists in an integral, then find the integrand
|
[
"\\frac{F_{\\omega}}{\\omega-\\omega_{0}} u_{\\omega}(\\boldsymbol{x})"
] |
Expression
|
Since $u_{\omega}(\boldsymbol{x})$ is a complete system, the solution of equation (4) can always be expressed as
\begin{equation*}\nu(x)=\int C_{\omega} u_{\omega}(x) \mathrm{d} \omega \tag{5}
\end{equation*}
Substitute into equation (4), we obtain
\begin{equation*}
\int(\omega-\omega_{0}) C_{\omega} u_{\omega}(\boldsymbol{x}) \mathrm{d} \omega=F(\boldsymbol{x}) \tag{6}
\end{equation*}
Expand $F(\boldsymbol{x})$ as
\begin{align*}
F(\boldsymbol{x}) & =\int F(\boldsymbol{x}^{\prime}) \delta(\boldsymbol{x}-\boldsymbol{x}^{\prime}) \mathrm{d}^{3} x^{\prime} \\
& =\iint F(\boldsymbol{x}^{\prime}) u_{\omega}^{*}(\boldsymbol{x}^{\prime}) u_{\omega}(\boldsymbol{x}) \mathrm{d}^{3} x^{\prime} \mathrm{d} \omega \\
& =\int F_{\omega} u_{\omega}(\boldsymbol{x}) \mathrm{d} \omega \tag{7}
\end{align*}
where
\begin{equation*}
F_{\omega}=\int F(\boldsymbol{x}^{\prime}) u_{\omega}^{*}(\boldsymbol{x}^{\prime}) \mathrm{d}^{3} x^{\prime}=\int F(\boldsymbol{x}) u_{\omega}^{*}(\boldsymbol{x}) \mathrm{d}^{3} x \tag{8}
\end{equation*}
Substitute equation (7) into equation (6), we obtain
\begin{equation*}
C_{\omega}=F_{\omega} /(\omega-\omega_{0}) \tag{9}
\end{equation*}
Substitute back into equation (5), the solution to equation (4) is
\begin{equation*}\nu(\boldsymbol{x})=\int \frac{F_{\omega}}{\omega-\omega_{0}} u_{\omega}(\boldsymbol{x}) \mathrm{d} \omega \tag{10}
\end{equation*}
|
{"$\\hat{H}$": "operator", "$\\omega$": "continuous eigenvalue", "$u_{\\omega}(\\boldsymbol{x})$": "eigenfunction corresponding to eigenvalue $\\omega$", "$F(\\boldsymbol{x})$": "known function", "$\\omega_{0}$": "specific eigenvalue of $H$", "$F_{\\omega}$": "coefficient related to function $F(\\boldsymbol{x})$"}
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Schrödinger equation one-dimensional motion
| |
23 |
Theoretical Foundations
|
For a hydrogen-like ion (nuclear charge $Z e$ ), an electron is in the bound state $\psi_{n l m}$, calculate $\langle r^{\lambda}\rangle, \lambda=-3．
|
[
"\\langle\\frac{1}{r^{3}}\\rangle_{n l m}=\\frac{1}{n^{3} l(l+\\frac{1}{2})(l+1)}(\\frac{Z}{a_{0}})^{3}"
] |
Expression
|
Known energy levels of a hydrogen-like ion are
\begin{equation*}
E_{n l m}=E_{n}=-\frac{Z^{2} e^{2}}{2 n^{2} a_{0}}, \quad n=n_{r}+l+1 \tag{1}
\end{equation*}
where $a_{0}=\hbar^{2} / \mu e^{2}$ is the Bohr radius. According to the virial theorem,
\begin{equation*}
\langle\frac{p^{2}}{2 \mu}\rangle_{n l m}=\langle\frac{r}{2} \frac{\mathrm{~d} V}{\mathrm{~d} r}\rangle_{n l m}=\frac{Z e^{2}}{2}\langle\frac{1}{r}\rangle_{n l m}=-\frac{1}{2}\langle V\rangle_{n l m} \tag{2}
\end{equation*}
so
E_{n}=\frac{1}{2}\langle V\rangle_{n l m}=-\frac{Z e^{2}}{2}\langle\frac{1}{r}\rangle_{n l m}
\begin{equation*}
\langle\frac{1}{r}\rangle_{n l m}=-\frac{2 E_{n}}{Z e^{2}}=\frac{Z}{n^{2} a_{0}} . \quad n=1,2,3, \cdots \tag{3}
\end{equation*}
Further, the spherical coordinate representation of $\psi_{n l m}$ is
\begin{equation*}
\psi_{n l m}(r, \theta, \varphi)=R_{n l}(r) Y_{l m}(\theta, \varphi) \tag{4}
\end{equation*}
It is a common eigenfunction of $(H, l^{2}, l_{z})$, satisfying the energy eigenvalue equation
\begin{equation*}
-\frac{\hbar^{2}}{2 \mu} \frac{1}{r} \frac{\partial^{2}}{\partial r^{2}} r \psi_{n l m}+[\frac{l(l+1) \hbar^{2}}{2 \mu r^{2}}-\frac{Z e^{2}}{r}] \psi_{n l m}=E_{n} \psi_{n l m} \tag{5}
\end{equation*}
The total energy operator is equivalent to
\begin{equation*}
H \rightarrow-\frac{\hbar^{2}}{2 \mu} \frac{1}{r} \frac{\partial^{2}}{\partial r^{2}} r+l(l+1) \frac{\hbar^{2}}{2 \mu r^{2}}-\frac{Z e^{2}}{r} \tag{6}
\end{equation*}
Considering $l$ as a parametric variable, differentiate equation (6) with respect to $l$, using the Hellmann theorem (refer to Chapter 8), we get
\begin{equation*}
\frac{\partial E_{n}}{\partial l}=\langle\frac{\partial H}{\partial l}\rangle_{n l m}=(l+\frac{1}{2}) \frac{\hbar^{2}}{\mu}\langle\frac{1}{r^{2}}\rangle_{n l m} \tag{7}
\end{equation*}
Since $n=n_{r}+l+1$, we find
\begin{equation*}
\frac{\partial E_{n}}{\partial l}=\frac{\partial E_{n}}{\partial n}=\frac{Z^{2} e^{2}}{n^{3} a_{0}} \tag{8}
\end{equation*}
Substituting into equation (7) and using $a_{0}=\hbar^{2} / \mu e^{2}$, we obtain
\begin{equation*}
\langle\frac{1}{r^{2}}\rangle_{n l m}=\frac{1}{(l+\frac{1}{2}) n^{3}} \frac{Z^{2}}{a_{0}^{2}} \tag{9}
\end{equation*}
Finally, compute $(r^{-3})$．
For the s state $(l=0), r \rightarrow 0$, $\psi \rightarrow C$ (constant), therefore
\begin{equation*}
\langle r^{-3}\rangle_{n 00} \rightarrow \infty \tag{10}
\end{equation*}
When $l \neq 0$, using formula (7b) in problem (5.7), we get
\begin{equation*}
\langle\frac{1}{r^{3}}\rangle_{n l m}=\frac{Z}{l(l+1) a_{0}}\langle\frac{1}{r^{2}}\rangle_{n l m} \tag{11}
\end{equation*}
Thus
\begin{equation*}
\langle\frac{1}{r^{3}}\rangle_{n l m}=\frac{1}{n^{3} l(l+\frac{1}{2})(l+1)}(\frac{Z}{a_{0}})^{3} \tag{12}
\end{equation*}
As $l \rightarrow 0$, the right side of the equation $\rightarrow \infty$, so this formula is actually applicable for all $l$ values.
Discussion: Since both the total energy operator and radial equation are independent of the magnetic quantum number $m$, $\langle r^{\lambda}\rangle$ is independent of $m$. However, $\langle r^{-1}\rangle$ is also independent of the angular quantum number $l$, depending only on the principal quantum number $n.$ $\langle r^{-2}\rangle$ and $\langle r^{-3}\rangle$ depend on both $n$ and $l$, meaning for states with the same energy level but different "orbital shapes" (different $l$), $\langle r^{-2}\rangle$ or $\langle r^{-3}\rangle$ have different values.
Using formula (9), it can be easily obtained that the average value of the centrifugal potential energy in the state $\psi_{n l m}$ is
\begin{equation*}
\langle\frac{l^{2}}{2 \mu r^{2}}\rangle_{n l m}=\frac{l(l+1) Z^{2} e^{2}}{(2 l+1) n^{3} a_{0}}=-\frac{l(l+1)}{(l+\frac{1}{2}) n} E_{n} \tag{13}
\end{equation*}
Since $(-E_{n})$ is the average kinetic energy, the proportion of centrifugal potential energy within kinetic energy is $l(l+1) /(l+\frac{1}{2}) n$. When $n$ is fixed, as $l$ increases, this proportion grows. When $l$ takes the maximum value $(l=n-1)$, this proportion is $(n-1) /$ $(n-\frac{1}{2})$, and the radial kinetic energy occupies only $1 /(2 n-1)$ of the kinetic energy in this case. Therefore, if $n \gg 1,(n, n-1, m)$ implies small radial kinetic energy, corresponding to circular orbits in Bohr's quantum theory.
|
{"$Z$": "nuclear charge", "$e$": "elementary charge", "$n$": "principal quantum number", "$l$": "angular quantum number", "$m$": "magnetic quantum number", "$E_{n}$": "energy of the state with principal quantum number n", "$a_{0}$": "Bohr radius", "$\\mu$": "reduced mass", "$\\hbar$": "reduced Planck's constant", "$r$": "radial distance", "$\\psi_{n l m}$": "wavefunction of the state with quantum numbers n, l, m", "$R_{n l}$": "radial part of the wavefunction corresponding to n and l", "$Y_{l m}$": "spherical harmonics corresponding to l and m", "$H$": "total energy operator", "$l_{z}$": "z-component of angular momentum"}
|
Schrödinger equation one-dimensional motion
| |
24 |
Theoretical Foundations
|
The potential acting on the valence electron (outermost electron) of a monovalent atom by the atomic nucleus (atomic nucleus and inner electrons) can be approximately expressed as
\begin{equation*}
V(r)=-\frac{e^{2}}{r}-\lambda \frac{e^{2} a_{0}}{r^{2}}, \quad 0<\lambda \ll 1 \tag{1}
\end{equation*}
where $a_{0}$ is the Bohr radius. Find the energy level of the valence electron and compare it with the energy level of the hydrogen atom.
|
[
"E_{n l}=-\\frac{e^{2}}{2(n^{\\prime})^{2} a_{0}}"
] |
Expression
|
Take the complete set of conserved quantities as $(H, l^{2}, l_{z})$, whose common eigenfunctions are
\begin{equation*}
\psi(r, \theta, \varphi)=R(r) \mathrm{Y}_{l m}(\theta, \varphi)=\frac{u(r)}{r} \mathrm{Y}_{l m}(\theta, \varphi) \tag{2}
\end{equation*}
$u(r)$ satisfies the radial equation
\begin{equation*}
-\frac{\hbar^{2}}{2 \mu} u^{\prime \prime}+[l(l+1) \frac{\hbar^{2}}{2 \mu r^{2}}-\frac{e^{2}}{r}-\lambda \frac{e^{2} a_{0}}{r^{2}}] u=E u \tag{3}
\end{equation*}
Let
\begin{equation*}
l(l+1)-2 \lambda=l^{\prime}(l^{\prime}+1) \tag{4}
\end{equation*}
Equation (3) can then be transformed into
\begin{equation*}
-\frac{\hbar^{2}}{2 \mu} u^{\prime \prime}+[l^{\prime}(l^{\prime}+1) \frac{\hbar^{2}}{2 \mu r^{2}}-\frac{e^{2}}{r}] u=E u \tag{3'}
\end{equation*}
This is equivalent to the radial equation of the hydrogen atom with $l$ replaced by $l^{\prime}$. Therefore, the solution process for equation ($3^{\prime}$) is entirely analogous to the hydrogen atom problem. The latter's energy levels are
\begin{equation*}
E_{n}=-\frac{e^{2}}{2 n^{2} a_{0}}, \quad n=n_{r}+l+1, \quad n_{r}=0,1,2, \cdots \tag{5}
\end{equation*}
Replacing $l$ with $l^{\prime}$ gives the energy levels of the valence electron:
\begin{equation*}
E_{n l}=-\frac{e^{2}}{2(n^{\prime})^{2} a_{0}}, \quad n^{\prime}=n_{r}+l^{\prime}+1 \tag{6}
\end{equation*}
It is generally assumed that
\begin{equation*}
l^{\prime}=l+\Delta_{l} \tag{7}
\end{equation*}
\begin{equation*}
n^{\prime}=n_{r}+l+\Delta_{l}+1=n+\Delta_{l} \tag{8}
\end{equation*}
$\Delta_{l}$ is referred to as the 'correction number' of the quantum numbers $l$ and $n$. Since $\lambda \ll 1$, equation (4) can be approximated as follows:
\begin{aligned}
l(l+1)-2 \lambda & =l^{\prime}(l^{\prime}+1)=(l+\Delta_{l})(l+\Delta_{l}+1) \\
& =l(l+1)+(2 l+1) \Delta_{l}+(\Delta_{l})^{2}
\end{aligned}
Neglecting $(\Delta_{l})^{2}$, we get
\begin{equation*}
\Delta_{l} \approx-\lambda /(l+\frac{1}{2}) \tag{9}
\end{equation*}
Since $\lambda \ll 1$, $|\Delta_{l}| \ll 1$. Thus, the energy level $E_{n l}$ obtained in this problem has only a small difference from the hydrogen atomic energy level, but the 'degeneracy in $l$' of the energy levels has been removed. Equation (6) is broadly consistent with experimental data on alkali metal spectra; in particular, the correction number $|\Delta_{l}|$ decreases as $l$ increases, which agrees well with experiments.
The exact solution for equation (4) is
\begin{equation*}
l^{\prime}=-\frac{1}{2}+(l+\frac{1}{2})[1-\frac{8 \lambda}{(2 l+1)^{2}}]^{\frac{1}{2}} \tag{10}
\end{equation*}
Expanding the above equation as a binomial series and retaining the $\lambda$ term while neglecting terms of order $\lambda^{2}$ and higher, yields equation (9).
|
{"$V(r)$": "potential function", "$e$": "elementary charge", "$r$": "radial distance", "$\\lambda$": "dimensionless small parameter", "$a_{0}$": "Bohr radius", "$H$": "Hamiltonian operator", "$l$": "angular momentum quantum number", "$l_{z}$": "z-component of angular momentum", "$\\psi$": "wave function", "$R(r)$": "radial wave function component", "$u(r)$": "radial function", "$\\hbar$": "reduced Planck's constant", "$\\mu$": "reduced mass", "$E$": "energy", "$l^{\\prime}$": "modified angular momentum quantum number", "$E_{n}$": "energy level of hydrogen atom", "$n$": "principal quantum number", "$n_{r}$": "radial quantum number", "$E_{n l}$": "energy level of the valence electron", "$n^{\\prime}$": "modified principal quantum number", "$\\Delta_{l}$": "correction number of quantum numbers"}
|
Schrödinger equation one-dimensional motion
| |
25 |
Theoretical Foundations
|
For a particle of mass $\mu$ moving in a spherical shell $\delta$ potential well
\begin{equation*}
V(r)=-V_{0} \delta(r-a), \quad V_{0}>0, a>0 \tag{1}
\end{equation*}
find the minimum value of $V_{0}$ required for bound states to exist.
|
[
"V_{0}=\\frac{\\hbar^{2}}{2 \\mu a}"
] |
Expression
|
The ground state is an s-state $(l=0)$, and the wave function can be expressed as
\begin{equation*}
\psi(r)=u(r) / r \tag{2}
\end{equation*}
$u(r)$ satisfies the radial equation
\begin{equation*}\nu^{\prime \prime}+\frac{2 \mu E}{\hbar^{2}} u+\frac{2 \mu V_{0}}{\hbar^{2}} \delta(r-a) u=0 \tag{3}
\end{equation*}
As $r \rightarrow \infty$, $V(r) \rightarrow 0$, so for a bound state $E<0$. Define
\begin{equation*}
\beta=\sqrt{-2 \mu E} / \hbar \tag{4}
\end{equation*}
Equation (3) can be rewritten as
\begin{equation*}
\nu^{\prime \prime} - \beta^{2} u + \frac{2 \mu V_{0}}{\hbar^{2}} \delta(r-a) u=0 \tag{3^{\prime}}
\end{equation*}
The boundary conditions are
$$r \rightarrow 0, \infty \text {, } u \rightarrow 0$$
Integrating equation ($3^{\prime}$) around $r \sim a$, we obtain the jump condition for $u^{\prime}$ (see Problem 2.1)
\begin{equation*}
\nu^{\prime}(a+0)-u^{\prime}(a-0)=-\frac{2 \mu V_{0}}{\hbar^{2}} u(a) \tag{5}
\end{equation*}
This means
\begin{equation*}
\frac{u^{\prime}}{u}|_{r=a-0} ^{r=a+0}=-\frac{2 \mu V_{0}}{\hbar^{2}} \tag{$\prime$}
\end{equation*}
For $r \neq a$, equation ($3^{\prime}$) becomes
\begin{equation*}\nu^{\prime \prime}-\beta^{2} u=0 \tag{\prime\prime}
\end{equation*}
In the region $r>a$, the solution satisfying the boundary condition at infinity is
\begin{equation*}\nu=C \mathrm{e}^{-\beta r}, \quad r>a \tag{6}
\end{equation*}
Thus
\begin{equation*}
(\frac{u^{\prime}}{u})_{r=a+0}=-\beta \tag{7}
\end{equation*}
If the value of $V_{0}$ is just sufficient to form the first bound state, the energy level must be $E=0^{-}$, at which point $\beta=0$, and equation ($3^{\prime\prime}$) becomes
\begin{equation*}\nu^{\prime \prime}=0 \quad(E \rightarrow 0^{-}) \tag{8}
\end{equation*}
Equation (7) becomes
\begin{equation*}
(\frac{u^{\prime}}{u})_{r=a+0}=0 \quad(E \rightarrow 0^{-}) \tag{7'}
\end{equation*}
When $E \rightarrow 0^{-}$, the solution of equation (8) in the region $r<a$, which satisfies the boundary condition $u(0)=0$, is
\begin{equation*}\nu=A r, \quad r<a \quad(E \rightarrow 0^{-}) \tag{9}
\end{equation*}
Thus
\begin{equation*}
(\frac{u^{\prime}}{u})_{r=a-0}=\frac{1}{a} \quad(E \rightarrow 0^{-}) \tag{10}
\end{equation*}
Substituting equations ($7^{\prime}$) and (10) into equation ($5^{\prime}$) gives
\begin{equation*}
V_{0}=\hbar^{2} / 2 \mu a \tag{11}
\end{equation*}
This is the minimum value of $V_{0}$ required for the existence of bound states, with the corresponding energy level being $E=0^{-}$.
|
{"$\\mu$": "mass of the particle", "$\\delta$": "spherical shell delta potential", "$V_{0}$": "potential well depth", "$a$": "radius of the spherical shell", "$r$": "radial distance", "$E$": "energy of the bound state", "$\\hbar$": "reduced Planck's constant", "$\\beta$": "wavefunction decay parameter", "$C$": "integration constant for the wavefunction", "$A$": "integration constant for the wavefunction"}
|
Schrödinger equation one-dimensional motion
| |
26 |
Theoretical Foundations
|
Simplify $\mathrm{e}^{\mathrm{i} \lambda_{\sigma_{2}}} \sigma_{\alpha} \mathrm{e}^{-\mathrm{i} \lambda \sigma_{z}}, \alpha=x, ~ y, \lambda$ are constants.
|
[
"\\mathrm{e}^{\\mathrm{i} \\lambda \\sigma_{z}} \\sigma_{x} \\mathrm{e}^{-\\mathrm{i} \\lambda \\sigma_{z}} = \\sigma_{x} \\cos 2 \\lambda-\\sigma_{y} \\sin 2 \\lambda"
] |
Expression
|
Solution 1: Utilize the formula:
\begin{equation*}
\mathrm{e}^{\mathrm{i} \lambda \sigma_{z}}=\cos \lambda+\mathrm{i} \sigma_{z} \sin \lambda \tag{1}
\end{equation*}
We get
$$\mathrm{e}^{\mathrm{i} \lambda \sigma_{z} \sigma_{x} \mathrm{e}^{-\mathrm{i} \lambda \sigma_{z}}=(\cos \lambda+\mathrm{i} \sigma_{z} \sin \lambda) \sigma_{x}(\cos \lambda-\mathrm{i} \sigma_{z} \sin \lambda) .}$$
Further using
\begin{equation*}
\sigma_{z}^{2}=1, \quad \sigma_{z} \sigma_{x}=-\sigma_{x} \sigma_{z}=\mathrm{i} \sigma_{y}, \tag{2}
\end{equation*}
we obtain
\begin{equation*}
\mathrm{e}^{\mathrm{i} \lambda_{\sigma_{z}} \sigma_{y} \mathrm{e}^{-\mathrm{i} \alpha_{z}}}=\sigma_{x} \sin 2 \lambda+\sigma_{y} \cos 2 \lambda \tag{4}
\end{equation*}
Solution 2: let
\begin{equation*}
\mathrm{e}^{\mathrm{i} \lambda \sigma_{z} \sigma_{x} \mathrm{e}^{-\mathrm{i} \lambda \sigma_{z}}=C_{0} I+C_{1} \sigma_{x}+C_{2} \sigma_{y}+C_{3} \sigma_{z} .} \tag{5}
\end{equation*}
Applying the above expression to the eigenstate $\chi_{\frac{1}{2}}$ of $\sigma_{z}$, using
\begin{equation*}
\sigma_{z} \chi_{\frac{1}{2}}=\chi_{\frac{1}{2}}, \quad \sigma_{x} \chi_{\frac{1}{2}}=\chi_{-\frac{1}{2}}, \quad \sigma_{y} \chi_{\frac{1}{2}}=\mathrm{i} \chi_{-\frac{1}{2}} \tag{6}
\end{equation*}
we get
$$ \mathrm{e}^{-2 \mathrm{a}} \chi_{-\frac{1}{2}}=(C_{0}+C_{3}) \chi_{\frac{1}{2}}+(C_{1}+\mathrm{i} C_{2}) \chi_{-\frac{1}{2}}, $$
Since $\chi_{\frac{1}{2}}$ and $\chi_{-\frac{1}{2}}$ are linearly independent, comparing the coefficients on both sides, we have
\begin{align*}
& C_{0}+C_{3}=0 \tag{7}\\
& C_{1}+\mathrm{i}_{2}=\mathrm{e}^{-2 \mathrm{u}}
\end{align*}
Now apply expression (5) to the eigenstate $\chi_{-\frac{1}{2}}$ of $\sigma_{z}$, using
\begin{equation*}
\sigma_{z} \chi_{-\frac{1}{2}}=-\chi_{-\frac{1}{2}}, \quad \sigma_{x} \chi_{-\frac{1}{2}}=\chi_{\frac{1}{2}}, \quad \sigma_{y} \chi_{-\frac{1}{2}}=-\mathrm{i} \chi_{\frac{1}{2}} \tag{8}
\end{equation*}
which gives
$$ \mathrm{e}^{2 i \mathrm{i}} \chi_{\frac{1}{2}}=(C_{1}-\mathrm{i} C_{2}) \chi_{\frac{1}{2}}+(C_{0}-C_{3}) \chi_{-\frac{1}{2}}$$
Comparing the coefficients, we have
\begin{gather*}
C_{0}-C_{3}=0 \\
C_{1}-\mathrm{i} C_{2}=\mathrm{e}^{2 \mathrm{i}} \tag{9}
\end{gather*}
From equations (7) and (9), it is easy to solve for
\begin{equation*}
C_{0}=C_{3}=0, \quad C_{1}=\cos 2 \lambda, \quad C_{2}=-\sin 2 \lambda \tag{10}
\end{equation*}
Substitute back into expression (5), thus we have
\begin{equation*}
\mathrm{e}^{\mathrm{i} \lambda \sigma_{z}} \sigma_{x} \mathrm{e}^{-\mathrm{i} \lambda \sigma_{z}}=\sigma_{x} \cos 2 \lambda-\sigma_{y} \sin 2 \lambda \tag{3}
\end{equation*}
Similarly, it can be proven that
\begin{equation*}
\mathrm{e}^{\mathrm{i} \lambda \sigma_{z}} \sigma_{y} \mathrm{e}^{-\mathrm{i} \lambda \sigma_{z}}=\sigma_{x} \sin 2 \lambda+\sigma_{y} \cos 2 \lambda \tag{4}
\end{equation*}
Solution 3: Treat $\lambda$ as a parameter, let
\begin{align*}
& f(\lambda)=\mathrm{e}^{\mathrm{i} \lambda \sigma_{z}} \sigma_{x} \mathrm{e}^{-\mathrm{i} \lambda \sigma_{z}} \tag{11}\\
& g(\lambda)=\mathrm{e}^{\mathrm{i} \lambda \sigma_{z}} \sigma_{y} \mathrm{e}^{-\mathrm{i} \lambda \sigma_{z}}. \tag{12}
\end{align*}
Note that
\begin{equation*}
f(0)=\sigma_{x}, \quad g(0)=\sigma_{y} \tag{13}
\end{equation*}
Differentiate equations (11) and (12) with respect to $\lambda$, we have
\begin{align*}
& f^{\prime}(\lambda)=\mathrm{e}^{\mathrm{i} \lambda \sigma_{z}} \mathrm{i}(\sigma_{z} \sigma_{x}-\sigma_{x} \sigma_{z}) \mathrm{e}^{-\mathrm{i} \lambda \sigma_{z}}=-2 g(\lambda) \tag{14}\\
& g^{\prime}(\lambda)=\mathrm{e}^{\mathrm{i} \lambda \sigma_{z}} \mathrm{i}(\sigma_{z} \sigma_{y}-\sigma_{y} \sigma_{z}) \mathrm{e}^{-\mathrm{i} \lambda \sigma_{z}}=2 f(\lambda)
\end{align*}
In equation (14), multiply the second equation by i, then add it to the first, yielding
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{~d} \lambda}[f(\lambda)+\mathrm{i} g(\lambda)]=2 \mathrm{i}[f(\lambda)+\mathrm{i} g(\lambda)] \tag{$\prime$}
\end{equation*}
As a first-order differential equation, the solution is evidently
\begin{equation*}
f(\lambda)+\mathrm{i} g(\lambda)=[f(0)+\mathrm{i} g(0)] \mathrm{e}^{2 i \lambda}=(\sigma_{x}+\mathrm{i} \sigma_{y}) \mathrm{e}^{2 i \lambda} \tag{15}
\end{equation*}
In equation (14), multiply the second equation by (-i), then add it to the first, yielding
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{~d} \lambda}[f(\lambda)-\mathrm{i} g(\lambda)]=-2 \mathrm{i}[f(\lambda)-\mathrm{i} g(\lambda)] \tag{$\prime\prime$}
\end{equation*}
The solution is
\begin{equation*}
f(\lambda)-\mathrm{i} g(\lambda)=[f(0)-\mathrm{i} g(0)] \mathrm{e}^{-2 \mathrm{i}}=(\sigma_{x}-\mathrm{i} \sigma_{y}) \mathrm{e}^{-2 \mathrm{i} \lambda} \tag{16}
\end{equation*}
Adding and subtracting equations (15) and (16), we obtain
\begin{align*}
& f(\lambda)=\sigma_{x} \cos 2 \lambda-\sigma_{y} \sin 2 \lambda \tag{17}\\
& g(\lambda)=\sigma_{x} \sin 2 \lambda+\sigma_{y} \cos 2 \lambda
\end{align*}
This is exactly the equations (3) and (4).
(Note) If $\lambda$ is a real number, then both $f(\lambda)$ and $g(\lambda)$ are Hermitian operators. At this time, equation (16) is the conjugate equation of equation (15), and equation (17) can be derived directly from equation (15).
|
{"$\\lambda_{\\sigma_{2}}$": "constant associated with the operator $\\sigma_{2}$", "$\\sigma_{\\alpha}$": "Pauli matrix corresponding to the variable $\\alpha$", "$\\alpha$": "variable representing $x$ or $y$, specifically used for components", "$x$": "constant representing one of the spatial axes", "$y$": "constant representing one of the spatial axes", "$\\lambda$": "constant parameter used in the exponential expressions", "$\\sigma_{z}$": "Pauli matrix representing the spin in the z-direction", "$\\sigma_{x}$": "Pauli matrix representing the spin in the x-direction", "$\\sigma_{y}$": "Pauli matrix representing the spin in the y-direction", "$C_{0}$": "constant coefficient corresponding to the identity matrix component", "$C_{1}$": "constant coefficient corresponding to the $\\sigma_{x}$ component", "$C_{2}$": "constant coefficient corresponding to the $\\sigma_{y}$ component", "$C_{3}$": "constant coefficient corresponding to the $\\sigma_{z}$ component", "$\\chi_{\\frac{1}{2}}$": "eigenstate of the operator $\\sigma_{z}$ with eigenvalue +1/2", "$\\chi_{-\\frac{1}{2}}$": "eigenstate of the operator $\\sigma_{z}$ with eigenvalue -1/2"}
|
Schrödinger equation one-dimensional motion
| |
27 |
Theoretical Foundations
|
A particle with spin $\hbar / 2$ and magnetic moment $\boldsymbol{\mu}=\mu_{0} \boldsymbol{\sigma}$ is placed in a uniform magnetic field $\boldsymbol{B}$ that points in an arbitrary direction $(\theta, \varphi)$ (where $n_3 = \cos \theta$ is the projection of the direction vector of the magnetic field on the $z$ axis). Neglecting orbital motion, assume at $t=0$ the particle is polarized along the positive $z$ direction, namely $\sigma_{z}=1,\langle\boldsymbol{\sigma}\rangle_{t=0}=$ $\boldsymbol{e}_{3}$. Find the expression for the $z$ component of the polarization direction of the particle for $t>0$, i.e., $\langle\sigma_{z}\rangle_{t}$.
|
[
"n_{3}^{2}+(1-n_{3}^{2}) \\cos 2 \\omega t"
] |
Expression
|
First, determine the spin wave function, then compute $\langle\boldsymbol{\sigma}\rangle$. Using $\boldsymbol{n}$ to denote the unit vector in the $(\theta, \varphi)$ direction, $\boldsymbol{e}_{1}, ~ \boldsymbol{e}_{2}, ~ \boldsymbol{e}_{3}$ denote the unit vectors in the $x, ~ y, ~ z$ directions, respectively:
\begin{align*}
\boldsymbol{n} & =n_{1} \boldsymbol{e}_{1}+n_{2} \boldsymbol{e}_{2}+n_{3} \boldsymbol{e}_{3} \\
& =\sin \theta \cos \varphi \boldsymbol{e}_{1}+\sin \theta \sin \varphi \boldsymbol{e}_{2}+\cos \theta \boldsymbol{e}_{3} \tag{1}
\end{align*}
The Hamiltonian associated with the spin motion is
\begin{equation*}
H=-\boldsymbol{\mu} \cdot \boldsymbol{B}=-\mu_{0} B \boldsymbol{\sigma} \cdot \boldsymbol{n}=-\mu_{0} B_{\sigma_{n}} \tag{2}
\end{equation*}
In the $\sigma_{z}$ representation, the matrix representation of $\sigma_{n}$ is
\sigma_{n}=\sigma_{x} n_{1}+\sigma_{y} n_{2}+\sigma_{z} n_{3}=[\begin{array}{cc}
n_{3} & n_{1}-\mathrm{i} n_{2} \tag{3}\\
n_{1}+\mathrm{i} n_{2} & -n_{3}
\end{array}]
Assume the spin wave function of the particle is
\chi(t)=[\begin{array}{l}
a(t) \tag{4}\\
b(t)
\end{array}]
$\chi(t)$ satisfies the Schrödinger equation
\begin{equation*}
\mathrm{i} \hbar \frac{\mathrm{~d}}{\mathrm{~d} t} \chi(t)=H \chi(t)=-\mu_{0} B \sigma_{n} \chi(t) \tag{5}
\end{equation*}
The eigenvalues of $\sigma_{n}$ are $\pm 1$, the eigenvalues of $H$ (i.e., stationary energy levels) are
\begin{gather*}
E=\mp \mu_{0} B=\mp \hbar \omega \\
\omega=\mu_{0} B / \hbar \tag{6}
\end{gather*}
Since $\omega$ and $\mu_{0}$ have the same sign, if $\mu_{0}<0$ then $\omega<0$. The eigenfunctions of $\sigma_{n}$ and $H$ are
\begin{gather*}
\phi_{1}=\frac{1}{\sqrt{2(1+n_{3})}}[\begin{array}{c}
1+n_{3} \\
n_{1}+\mathrm{i} n_{2}
\end{array}] \quad(\sigma_{n}=1, E=-\hbar \omega) \\
\phi_{-1}=\frac{1}{\sqrt{2(1+n_{3})}}[\begin{array}{c}
n_{1}-\mathrm{i} n_{2} \\
-1-n_{3}
\end{array}] \quad(\sigma_{n}=-1, E=\hbar \omega) \tag{7}
\end{gather*}
The general solution of equation (5) is
\begin{equation*}
\chi(t)=C_{1} \phi_{1} \mathrm{e}^{\mathrm{i} \omega t}+C_{-1} \phi_{-1} \mathrm{e}^{-\mathrm{i} \omega t} \tag{8}
\end{equation*}
$C_{1}, ~ C_{-1}$ are determined by initial conditions:
\begin{equation*}
C_{1}=\phi_{1}^{+} \chi(0), \quad C_{-1}=\phi_{-1}^{+} \chi(0) \tag{9}
\end{equation*}
The initial wave function for this problem is the eigenfunction of $\sigma_{z}=1$, that is
\chi(0)=\chi_{\frac{1}{2}}=[\begin{array}{l}
1 \tag{10}\\
0
\end{array}]
Thus
\begin{equation*}
C_{1}=\sqrt{\frac{1+n_{3}}{2}}, \quad C_{-1}=\frac{n_{1}+\mathrm{i} n_{2}}{\sqrt{2(1+n_{3})}} \tag{11}
\end{equation*}
Substituting into equation (8), we get
\begin{align*}
\chi(t) & =\frac{1}{\sqrt{2(1+n_{3})}}[\begin{array}{l}
(1+n_{3}) C_{1} \mathrm{e}^{\mathrm{i} \omega t}+(n_{1}-\mathrm{i} n_{2}) C_{-1} \mathrm{e}^{-\mathrm{i} \omega t} \\
(n_{1}+\mathrm{i} n_{2}) C_{1} \mathrm{e}^{\mathrm{i} \omega t}-(1+n_{3}) C_{-1} \mathrm{e}^{-\mathrm{i} \omega t}
\end{array}] \\
& =[\begin{array}{l}
\cos \omega t+\mathrm{i} n_{3} \sin \omega t \\
(\mathrm{i} n_{1}-n_{2}) \sin \omega t
\end{array}]=[\begin{array}{l}
a(t) \\
b(t)
\end{array}] \tag{12}
\end{align*}
The expectation values of $\boldsymbol{\sigma}$ in the $\chi(t)$ state are
\begin{align*}
\langle\sigma_{x}\rangle & =\chi^{+} \sigma_{x} \chi=[a^{*} b^{*}][\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}][\begin{array}{l}
a \\
b
\end{array}]=a^{*} b+b^{*} a \\
& =n_{1} n_{3}(1-\cos 2 \omega t)-n_{2} \sin 2 \omega t \tag{13a}\\
\langle\sigma_{y}\rangle & =\chi^{+} \sigma_{y} \chi=\mathrm{i}(b^{*} a-a^{*} b) \\
& =n_{2} n_{3}(1-\cos 2 \omega t)+n_{1} \sin 2 \omega t \tag{1;b}\\
\langle\sigma_{z}\rangle & =\chi^{-} \sigma_{*} \chi=a^{*} a-b^{*} b \\
& =n_{3}^{2}+(1-n_{3}^{2}) \cos 2 \omega t \tag{13c}
\end{align*}
If the magnetic field $\boldsymbol{B}$ points in the positive $x$ direction, then
n_{1}=1, \quad n_{2}=n_{3}=0
Then equation (13) becomes
\langle\sigma_{x}\rangle=0,\langle\sigma_{y}\rangle=\sin 2 \omega t, \quad\langle\sigma_{z}\rangle=\cos 2 \omega t
This is precisely the result obtained in the previous problem (note that $\omega$ in the previous problem corresponds to $-\omega$ in this problem).
|
{"$\\hbar$": "reduced Planck's constant", "$\\boldsymbol{\\mu}$": "magnetic moment", "$\\mu_{0}$": "proportionality constant for magnetic moment", "$\\boldsymbol{\\sigma}$": "Pauli spin matrices", "$\\boldsymbol{B}$": "magnetic field", "$\\theta$": "polar angle defining direction", "$\\varphi$": "azimuthal angle defining direction", "$n_3$": "projection of direction vector on the z-axis", "$t$": "time", "$\\sigma_{z}$": "Pauli spin matrix in z direction", "$\\langle\\boldsymbol{\\sigma}\\rangle$": "expectation value of Pauli matrices", "$\\langle\\sigma_{z}\\rangle_{t}$": "expectation value of z-component of spin at time t", "$\\boldsymbol{n}$": "unit vector in specified direction", "$\\boldsymbol{e}_{1}$": "unit vector in x direction", "$\\boldsymbol{e}_{2}$": "unit vector in y direction", "$\\boldsymbol{e}_{3}$": "unit vector in z direction", "$n_{1}$": "x-component of unit vector", "$n_{2}$": "y-component of unit vector", "$H$": "Hamiltonian", "$\\sigma_{n}$": "component of Pauli spin matrix in direction n", "$\\chi(t)$": "spin wave function", "$E$": "energy", "$\\omega$": "angular frequency", "$C_{1}$": "coefficient for initial condition related to positive eigenvalue", "$C_{-1}$": "coefficient for initial condition related to negative eigenvalue", "$\\chi(0)$": "initial wave function", "$\\phi_{1}$": "eigenfunction corresponding to eigenvalue 1", "$\\phi_{-1}$": "eigenfunction corresponding to eigenvalue -1", "$a(t)$": "component of wave function", "$b(t)$": "component of wave function", "$\\langle\\sigma_{x}\\rangle$": "expectation value of x-component of spin", "$\\langle\\sigma_{y}\\rangle$": "expectation value of y-component of spin"}
|
Schrödinger equation one-dimensional motion
| |
28 |
Theoretical Foundations
|
In a system with a spin of $\hbar / 2$, the magnetic moment $\boldsymbol{\mu}=\mu_{0} \boldsymbol{\sigma}$ is placed in a uniform magnetic field $\boldsymbol{B}_{0}$ directed along the positive $z$ direction for $t<0$. At $t \geqslant 0$, an additional rotating magnetic field $\boldsymbol{B}_{1}(t)$, perpendicular to the $z$ axis, is applied:
$$ \boldsymbol{B}_{1}(t)=B_{1} \cos 2 \omega_{0} t e_{1}-B_{1} \sin 2 \omega_{0} t e_{2},$$
where $\omega_{0}=\mu_{0} B_{0} / \hbar$. It is known that for $t \leqslant 0$, the system is in the eigenstate $\chi_{\frac{1}{2}}$ with $s_{z}=\hbar / 2$. Find the expression for the time $\Delta t$ it takes for the system's spin to first reverse from $s_z = \hbar/2$ (along the positive $z$ direction) to $s_z = -\hbar/2$ (along the negative $z$ direction) starting from $t=0$. Express this in terms of $\omega_1 = \mu_0 B_1 / \hbar$ and relevant constants.
|
[
"\\pi \\hbar / 2 \\mu_{0} B_{1}"
] |
Expression
|
The Hamiltonian related to the spin motion of the system is
\begin{equation*}
H=-\mu \cdot[\boldsymbol{B}_{0}+\boldsymbol{B}_{1}(t)], \quad t \geqslant 0 \tag{1}
\end{equation*}
In the $s_{z}$ representation, the matrix form of $H$ is
\begin{align}
H & =-\mu_{0} B_{1}(\sigma_{x} \cos 2 \omega_{0} t-\sigma_{y} \sin 2 \omega_{0} t)-\mu_{0} B_{0} \sigma_{z} \\
& =-\mu_{0} \begin{pmatrix}
B_{0} & B_{1} \mathrm{e}^{2 i \omega_{0} t} \\
B_{1} \mathrm{e}^{-2 i \omega_{0} t} & -B_{0}
\end{pmatrix}
\tag{\prime}
\end{align}
Assuming the wave function for $t \geqslant 0$ is
$$\chi(t)=\begin{pmatrix}
a(t) \tag{2}\\
b(t)
\end{pmatrix}.$$
Substituting into the Schrödinger equation
\begin{equation*}
\mathrm{i} \hbar \frac{\mathrm{~d}}{\mathrm{~d} t} \chi(t)=H \chi(t) \tag{3},
\end{equation*}
we have
\begin{gather*}
\frac{\mathrm{d}}{\mathrm{~d} t} a(t)=\mathrm{i} \omega_{0} a(t)+\mathrm{i} \omega_{1} b(t) \mathrm{e}^{2 \mathrm{i} \omega_{0} t} \\
\frac{\mathrm{~d}}{\mathrm{~d} t} b(t)=-\mathrm{i} \omega_{0} b(t)+\mathrm{i} \omega_{1} a(t) \mathrm{e}^{-2 \mathrm{i} \omega_{0} t}. \tag{4}
\end{gather*}
where
\begin{equation*}
\omega_{0}=\frac{\mu_{0} B_{0}}{\hbar}, \quad \omega_{1}=\frac{\mu_{0} B_{1}}{\hbar} \tag{5}
\end{equation*}
Let
\begin{equation*}
a(t)=c_{1}(t) \mathrm{e}^{\mathrm{i} \omega_{0} t}, \quad b(t)=c_{2}(t) \mathrm{e}^{-\mathrm{i} \omega_{0} t} \tag{6}
\end{equation*}
Substituting into equation (4), yields equations for $c_{1}$ and $c_{2}$
\begin{align*}
& \frac{\mathrm{d}}{\mathrm{~d} t} c_{1}(t)=\mathrm{i} \omega_{1} c_{2}(t) \\
& \frac{\mathrm{d}}{\mathrm{~d} t} c_{2}(t)=\mathrm{i} \omega_{1} c_{1}(t) \tag{7}
\end{align*}
The initial conditions are
\begin{equation*}
\chi(0)=\begin{pmatrix}{l}
a(0) \\
b(0)
\end{pmatrix}=\begin{pmatrix}
1 \\
0
\end{pmatrix}=\chi_{\frac{1}{2}}, \quad \text { i.e., } \quad \left\{\begin{array}{l}
c_{1}(0)=1 \\
c_{2}(0)=0
\end{array}\right.
\tag{8}
\end{equation*}
By adding and subtracting equations in (7), we obtain
\begin{gather*}
\frac{\mathrm{d}}{\mathrm{~d} t}(c_{1}+c_{2})=\mathrm{i} \omega_{1}(c_{1}+c_{2}) \\
\frac{\mathrm{d}}{\mathrm{~d} t}(c_{1}-c_{2})=-\mathrm{i} \omega_{1}(c_{1}-c_{2}) \tag{9}
\end{gather*}
The solution is
\begin{align*}
c_{1}(t)+c_{2}(t) & =[c_{1}(0)+c_{2}(0)] \mathrm{e}^{\mathrm{i} \omega_{1} t} = \mathrm{e}^{\mathrm{i} \omega_{1} t}\\
c_{1}(t)-c_{2}(t) &= (c_{1}(0)-c_{2}(0)) \mathrm{e}^{-\mathrm{i} \omega_{1} t}=\mathrm{e}^{-\mathrm{i} \omega_{1} t} .
\end{align*}
Adding and subtracting these, we find
\begin{equation*}
c_{1}(t)=\cos \omega_{1} t, \quad c_{2}(t)=\mathrm{i} \sin \omega_{1} t \tag{11}
\end{equation*}
Substituting into equation (6), we obtain
\begin{equation*}
a(t)=\cos \omega_{1} t \mathrm{e}^{\mathrm{i} \omega_{0} t}, \quad b(t)=\mathrm{i} \sin \omega_{1} t \mathrm{e}^{-\mathrm{i} \omega_{0} t} \tag{12}
\end{equation*}
Substituting into equation (2), we find
\begin{equation}
\begin{split}
\chi(t)&=\begin{pmatrix}
\cos \omega_{1} t \mathrm{e}^{\mathrm{i} \omega_{0} t} \\
\mathrm{i} \sin \omega_{1} t \mathrm{e}^{-\mathrm{i} \omega_{0} t}
\end{pmatrix} \\
&=\cos \omega_{1} t \mathrm{e}^{\mathrm{i} \omega_{0} t} \chi_{\frac{1}{2}}+\mathrm{i} \sin \omega_{1} t \mathrm{e}^{-\mathrm{i} \omega_{0} t} \chi_{-\frac{1}{2}} \tag{13}
\end{split}
\end{equation}
Clearly
\begin{gathered}
t=0, \quad \chi=\chi_{\frac{1}{2}}=[\begin{array}{l}
1 \\
0
\end{array}] \\
s_{z}=\frac{\hbar}{2}, \quad\langle s\rangle=\frac{\hbar}{2} e_{3} \\
t=\frac{\pi}{2 \omega_{1}}, \quad \chi=\mathrm{ie}^{-\mathrm{i} \omega_{0} t} \chi_{-\frac{1}{2}}=\mathrm{ie}^{-\mathrm{i} \omega_{0} t}[\begin{array}{l}
0 \\
1
\end{array}] \\
s_{\tilde{z}}=-\frac{\hbar}{2}, \quad\langle\boldsymbol{s}\rangle=-\frac{\hbar}{2} e_{3} \\
t=\frac{\pi}{\omega_{1}}, \quad \chi=-\mathrm{e}^{\mathrm{i} \omega_{0} t} \chi_{\frac{1}{2}}=-\mathrm{e}^{\mathrm{i} \omega_{0} t}[\begin{array}{l}
1 \\
0
\end{array}] \\
s_{z}=\frac{\hbar}{2}, \quad\langle\boldsymbol{s}\rangle=\frac{\hbar}{2} e_{3}
\end{gathered}
In other words, the system's spin direction changes once every $\Delta t=\pi / 2 \omega_{1}=\pi \hbar / 2 \mu_{0} B_{1}$ . The spin state of the system undergoes periodic oscillation between $\chi_{\frac{1}{2}}$ and $\chi_{-\frac{1}{2}}$, with a period $T=2 \Delta t=\pi \hbar / \mu_{0} B_{1}$.
This problem illustrates the basic principle of magnetic resonance.
|
{"$\\hbar$": "reduced Planck's constant", "$\\mu_{0}$": "magnetic permeability constant", "$\\boldsymbol{\\mu}$": "magnetic moment vector", "$\\boldsymbol{\\sigma}$": "Pauli matrices vector", "$\\boldsymbol{B}_{0}$": "uniform magnetic field in positive z direction", "$t$": "time", "$\\boldsymbol{B}_{1}(t)$": "rotating magnetic field", "$B_{1}$": "magnitude of the rotating magnetic field", "$\\omega_{0}$": "precession angular frequency due to $B_{0}$", "$\\chi_{\\frac{1}{2}}$": "eigenstate with spin $s_{z} = \\hbar/2$", "$s_{z}$": "spin component along z axis", "$\\Delta t$": "time for the spin to reverse", "$\\omega_{1}$": "precession angular frequency due to $B_{1}$", "$c_{1}(t)$": "time-dependent coefficient in the wave function", "$c_{2}(t)$": "time-dependent coefficient in the wave function"}
|
Schrödinger equation one-dimensional motion
| |
29 |
Theoretical Foundations
|
The magnetic moment (operator) of an electron is
\begin{equation*}
\boldsymbol{\mu}=\boldsymbol{\mu}_{l}+\boldsymbol{\mu}_{s}=-\frac{e}{2 m_{\mathrm{e}} c}(\boldsymbol{l}+2 \boldsymbol{s}) \tag{1}
\end{equation*}
Try to calculate the expectation value of $\mu_{z}$ for the $|l j m_{j}\rangle$ state.
|
[
"\\langle l j m_{j}| \\mu_{z}|l j m_{j}\\rangle=-(1+\\frac{j(j+1)-l(l+1)+3 / 4}{2 j(j+1)}) m_{j}"
] |
Expression
|
If we use the Bohr magneton $\mu_{\mathrm{B}}=e \hbar / 2 m_{\mathrm{e}} c$ as the unit of magnetic moment, then the magnetic moment operator of an electron can be written as (here $\hbar=1$ is taken)
\begin{equation*}
\boldsymbol{\mu}=-(\boldsymbol{l}+2 \boldsymbol{s})=-(\boldsymbol{j}+\boldsymbol{s})=-(\boldsymbol{j}+\frac{1}{2} \boldsymbol{\sigma}) \tag{2}
\end{equation*}
Thus, we have
\begin{equation*}
\langle l j m_{j}| \mu_{z}|l j m_{j}\rangle=-g m_{j} \tag{3}
\end{equation*}
where
\begin{equation*}
g=1+\frac{j(j+1)-l(l+1)+3 / 4}{2 j(j+1)} \quad \text { (Landè } g \text { factor) } \tag{4}
\end{equation*}
The average value of $\mu_{z}$ for $m_{j}=j$ (its maximum value) is usually taken as the definition of the magnetic moment observable, denoted as $\mu$. For an electron,
\begin{equation*}
\mu=\langle l j j| \mu_{z}|l j j\rangle=-g j \tag{5}
\end{equation*}
that is
\mu= \begin{cases}-(j+\frac{1}{2}), & j=l+\frac{1}{2} \tag{$\prime$}\\ -j(2 j+1) /(2 j+2), & j=l-\frac{1}{2}\end{cases}
|
{"$e$": "electron charge", "$m_{\\mathrm{e}}$": "electron mass", "$c$": "speed of light in vacuum", "$l$": "orbital angular momentum", "$s$": "spin angular momentum", "$\\mu_{z}$": "z-component of the magnetic moment", "$\\mu_{\\mathrm{B}}$": "Bohr magneton", "$\\hbar$": "reduced Planck's constant", "$j$": "total angular momentum quantum number", "$m_{j}$": "magnetic quantum number related to total angular momentum", "$\\sigma$": "Pauli matrices", "$g$": "Landè g factor", "$\\mu$": "magnetic moment"}
|
Schrödinger equation one-dimensional motion
| |
30 |
Theoretical Foundations
|
A system composed of two spin-$1 / 2$ particles is placed in a uniform magnetic field, with the magnetic field direction as the $z$-axis. The Hamiltonian of the system related to spin is given by
\begin{equation*}
H=a \sigma_{1 z}+b \sigma_{2 z}+c_{0} \boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2} \tag{1}
\end{equation*}
where $a, ~ b$ terms arise from the interaction between the magnetic field and the particles' intrinsic magnetic moments, and the $c_{0}$ term arises from the interaction between the two particles. $a, ~ b, ~ c_{0}$ are real constants. If the system is in the common eigenstate $\chi_{11}=\alpha(1) \alpha(2)$ of the total spin operators $(\boldsymbol{S}^{2}, S_z)$, find the energy level of the system in this case.
|
[
"c_{0} + a + b"
] |
Expression
|
We will solve using matrix methods in spin state vector space. The basis vectors can be chosen as the common eigenstates of $(\sigma_{1 z}, \sigma_{2 z})$
$$\alpha(1) \alpha(2), \quad \alpha(1) \beta(2), \quad \beta(1) \alpha(2), \quad \beta(1) \beta(2) $$
or as the common eigenstates $\chi_{S M_{s}}$ of the total spin operators $(\boldsymbol{S}^{2}, S_{z})$. For this problem, considering the diagonalization of $\boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2}$, it is more convenient to use $\chi_{S M_{S}}$ as the basis vectors. For convenience, the basis vector order is as follows:
\begin{array}{l}
\chi_{1}=\chi_{11}=\alpha(1) \alpha(2) \tag{2}\\
\chi_{2}=\chi_{1-1}=\beta(1) \beta(2) \\
\chi_{3}=\chi_{10}=\frac{1}{\sqrt{2}}[\alpha(1) \beta(2)+\beta(1) \alpha(2)] \\
\chi_{4}=\chi_{00}=\frac{1}{\sqrt{2}}[\alpha(1) \beta(2)-\beta(1) \alpha(2)]
\end{array}}
The Hamiltonian operator can be rewritten as
\begin{gather*}
H=c_{0} \boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2}+c_{1}(\sigma_{1 z}+\sigma_{2 z})+c_{2}(\sigma_{1 z}-\sigma_{2 z}) \tag{1'}\\
c_{1}=\frac{1}{2}(a+b), \quad c_{2}=\frac{1}{2}(a-b) \tag{3}
\end{gather*}
All four basis vectors are common eigenstates of $(\boldsymbol{S}^{2}, S_{z})$, and also common eigenstates of $\boldsymbol{\sigma}_{1} \cdot \boldsymbol{\sigma}_{2}$ and $(\sigma_{1 z}+\sigma_{2 z})$. It is easy to see that $\chi_{1}$ and $\chi_{2}$ are also eigenstates of $(\sigma_{1 z}-\sigma_{2 z})$, thus $\chi_{1}$ and $\chi_{2}$ are already eigenstates of $H$.
\begin{align*}
& H \chi_{1}=(c_{0}+2 c_{1}) \chi_{1}=(c_{0}+a+b) \chi_{1} \tag{4}\\
& H \chi_{2}=(c_{0}-2 c_{1}) \chi_{2}=(c_{0}-a-b) \chi_{2} \tag{5}
\end{align*}
Thus, we obtain two energy levels of the system: $E=c_{0} \pm 2 c_{1}$.
It is easy to calculate the action of $(\sigma_{1 z}-\sigma_{2 z})$ on the basis vectors, which is
\begin{array}{cl}
(\sigma_{1 z}-\sigma_{2 z}) \chi_{1}=0, & (\sigma_{1 z}-\sigma_{2 z}) \chi_{2}=0 \\
(\sigma_{1 z}-\sigma_{2 z}) \chi_{3}=2 \chi_{4}, & (\sigma_{1 z}-\sigma_{2 z}) \chi_{4}=2 \chi_{3} \tag{7}
\end{array}
Therefore, in the subspace ${\chi_{3}, \chi_{4}}$, the matrix elements of $(\sigma_{1 z}-\sigma_{2 z})$ are
\begin{array}{l}
(\sigma_{1 z}-\sigma_{2 z})_{33}=(\sigma_{1 z}-\sigma_{2 z})_{44}=0 \tag{7'}\\
(\sigma_{1 z}-\sigma_{2 z})_{34}=(\sigma_{1 z}-\sigma_{2 z})_{43}=2
\end{array}
All matrix elements of $(\sigma_{1 z}+\sigma_{2 z})$ are zero. The matrix representation of $H$ is
\begin{equation}
H=\begin{pmatrix}
c_{0} & 2 c_{2} \tag{8}\\
2 c_{2} & -3 c_{0}
\end{pmatrix}
\end{equation}
Assume the energy eigenstate is
\begin{equation*}
\chi=f_{3} \chi_{3}+f_{4} \chi_{4} \tag{9}
\end{equation*}
Substitute into the energy eigenvalue equation
\begin{equation*}
H \chi=E \chi \tag{10}
\end{equation*}
resulting in
[\begin{array}{cc}
c_{0}-E & 2 c_{2} \tag{$\prime$}\\
2 c_{2} & -3 c_{0}-E
\end{array}][\begin{array}{l}
f_{1} \\
f_{2}
\end{array}]=0
The energy level $E$ is determined by:
\begin{equation*}
\operatorname{det}(H-E)=0 \tag{11}
\end{equation*}
namely
|\begin{array}{cc}
c_{0}-E & 2 c_{2} \tag{11'}\\
2 c_{2} & -3 c_{0}-E
\end{array}|=(E-c_{0})(E+3 c_{0})-4 c_{2}^{2}=0
Solving gives
E=-c_{0} \pm 2 \sqrt{c_{0}^{2}+c_{2}^{2}}
Conclusion: This problem has four energy levels (excluding accidental degeneracy), which are
\begin{equation*}
E=c_{0} \pm 2 c_{1}, \quad-c_{0} \pm 2 \sqrt{c_{0}^{2}+c_{2}^{2}} \tag{12}
\end{equation*}
The energy eigenstates of the first two energy levels are $\chi_{1}$ and $\chi_{2}$, respectively, and the energy eigenstates of the last two energy levels are linear combinations of $\chi_{3}$ and $\chi_{4}$.
|
{"$H$": "Hamiltonian of the system", "$a$": "real constant related to interaction with magnetic field", "$b$": "real constant related to interaction with magnetic field", "$c_{0}$": "real constant related to interaction between particles", "$\\sigma_{1 z}$": "spin operator for the first particle along the z-axis", "$\\sigma_{2 z}$": "spin operator for the second particle along the z-axis", "$\\boldsymbol{\\sigma}_{1}$": "spin operator vector for the first particle", "$\\boldsymbol{\\sigma}_{2}$": "spin operator vector for the second particle", "$\\chi_{11}$": "eigenstate of total spin operators for maximum projection", "$\\chi_{1-1}$": "eigenstate of total spin operators for minimum projection", "$\\chi_{10}$": "eigenstate of total spin operators with intermediate projection", "$\\chi_{00}$": "eigenstate of total spin operators for zero projection", "$c_{1}$": "transformed constant as a function of a and b", "$c_{2}$": "transformed constant as a function of a and b"}
|
Schrödinger equation one-dimensional motion
| |
31 |
Theoretical Foundations
|
Consider a system composed of three non-identical spin $1/2$ particles, with the Hamiltonian given by
\begin{equation*}
H=A \boldsymbol{s}_{1} \cdot \boldsymbol{s}_{2}+B(\boldsymbol{s}_{1}+\boldsymbol{s}_{2}) \cdot \boldsymbol{s}_{3}, \tag{1}
\end{equation*}
where $A, ~ B$ are real constants. Let $\boldsymbol{S}_{12}=\boldsymbol{s}_{1}+\boldsymbol{s}_{2}$ and $\boldsymbol{S}_{123}=\boldsymbol{S}_{12}+\boldsymbol{s}_{3}$. Try to express the Hamiltonian $H$ as a function of $\boldsymbol{S}_{12}^{2}$ and $\boldsymbol{S}_{123}^{2}$. (Take $\hbar=1$ )
|
[
"H = \\frac{1}{2}(A-B) \\boldsymbol{S}_{12}^{2}+\\frac{B}{2} \\boldsymbol{S}_{123}^{2}-\\frac{3}{8}(2 A+B)"
] |
Expression
|
The sum of the spins of particles 1 and 2 is denoted as $\boldsymbol{S}_{12}$, and the total spin is denoted as $\boldsymbol{S}_{123}$, that is
\begin{equation*}
\boldsymbol{S}_{12}=\boldsymbol{s}_{1}+\boldsymbol{s}_{2}, \quad \boldsymbol{S}_{123}=\boldsymbol{s}_{1}+\boldsymbol{s}_{2}+\boldsymbol{s}_{3}=\boldsymbol{S}_{12}+\boldsymbol{s}_{3} \tag{2}
\end{equation*}
Evidently, $\boldsymbol{S}_{12}$ and $\boldsymbol{S}_{123}$ both possess the properties of angular momentum, satisfying the commutation relation
\begin{array}{ll}
\boldsymbol{S}_{12} \times \boldsymbol{S}_{12}=\mathrm{i} \boldsymbol{S}_{12}, & {[\boldsymbol{S}_{12}^{2}, \boldsymbol{S}_{12}]=0} \\
\boldsymbol{S}_{123} \times \boldsymbol{S}_{123}=\mathrm{i} \boldsymbol{S}_{123}, & {[\boldsymbol{S}_{123}^{2}, \boldsymbol{S}_{123}]=0} \tag{4}
\end{array}
$\boldsymbol{s}_{1}, ~ \boldsymbol{s}_{2}, ~ \boldsymbol{s}_{3}$ commute with each other, and
\begin{equation*}
s_{1}^{2}=s_{2}^{2}=s_{3}^{2}=\frac{3}{4} \tag{5}
\end{equation*}
Therefore
\begin{align*}
& \boldsymbol{S}_{12}^{2}=\boldsymbol{s}_{1}^{2}+\boldsymbol{s}_{2}^{2}+2 \boldsymbol{s}_{1} \cdot \boldsymbol{s}_{2}=\frac{3}{2}+2 \boldsymbol{s}_{1} \cdot \boldsymbol{s}_{2} \tag{6}\\
& \boldsymbol{S}_{123}^{2}=\frac{9}{4}+2(\boldsymbol{s}_{1} \cdot \boldsymbol{s}_{2}+\boldsymbol{s}_{2} \cdot \boldsymbol{s}_{3}+\boldsymbol{s}_{3} \cdot \boldsymbol{s}_{1}) \tag{7}
\end{align*}
Based on this, $H$ can be written as
\begin{align*}
H & =(A-B) \boldsymbol{s}_{1} \cdot \boldsymbol{s}_{2}+B(\boldsymbol{s}_{1} \cdot \boldsymbol{s}_{2}+\boldsymbol{s}_{2} \cdot \boldsymbol{s}_{3}+\boldsymbol{s}_{3} \cdot \boldsymbol{s}_{1}) \\
& =\frac{1}{2}(A-B) \boldsymbol{S}_{12}^{2}+\frac{B}{2} \boldsymbol{S}_{123}^{2}-\frac{3}{8}(2 A+B) \tag{$\prime$}
\end{align*}
|
{"$H$": "Hamiltonian", "$A$": "real constant related to interaction", "$B$": "real constant related to interaction", "$\\boldsymbol{s}_{1}$": "spin vector of particle 1", "$\\boldsymbol{s}_{2}$": "spin vector of particle 2", "$\\boldsymbol{s}_{3}$": "spin vector of particle 3", "$\\boldsymbol{S}_{12}$": "combined spin vector of particles 1 and 2", "$\\boldsymbol{S}_{123}$": "combined spin vector of particles 1, 2, and 3"}
|
Schrödinger equation one-dimensional motion
| |
32 |
Theoretical Foundations
|
Same as the previous question, for any value, find
\begin{equation*}
d_{j m}^{j}(\lambda)=\langle j j| \mathrm{e}^{-\mathrm{i} \lambda J_{y}}|j m\rangle \tag{1}
\end{equation*}
|
[
"d_{j m}^{j}(\\lambda)=(-1)^{j-m}[\\frac{(2 j)!}{(j+m)!(j-m)!}]^{\\frac{1}{2}}(\\cos \\frac{\\lambda}{2})^{j+m}(\\sin \\frac{\\lambda}{2})^{j-m}"
] |
Expression
|
According to the general theory of angular momentum,
\begin{array}{rl}
J_{+}|j m\rangle & =(J_{x}+i J_{y})|j m\rangle \tag{2}\\
J_{-}|j m\rangle & =a_{j m}|j m+1\rangle \\
J_{x}-i J_{y})|j m\rangle & =a_{j,-m}|j m-1\rangle
\end{array}}
where
\begin{equation*}
a_{j m}=\sqrt{(j-m)(j+m+1)} \tag{3}
\end{equation*}
When $m=j$
\begin{equation*}
J_{+}|j j\rangle=0, \quad\langle j j| J_{-}=0 \tag{4}
\end{equation*}
(The second equation is the conjugate of the first equation.) Therefore
\begin{equation*}
\langle j j| J_{-} \mathrm{e}^{-\mathrm{i} \lambda J_{y}}|j m\rangle=0. \tag{5}
\end{equation*}
We also have:
\begin{align*}
J_{-} \mathrm{e}^{-\mathrm{i} \mathrm{i} J_{y}} & =\mathrm{e}^{-\mathrm{i} \lambda J_{y}}(J_{z} \sin \lambda+J_{x} \cos \lambda-\mathrm{i} J_{y}) \\
& =\mathrm{e}^{-\mathrm{i} \lambda J_{y}}[J_{z} \sin \lambda+\frac{1}{2}(\cos \lambda-1) J_{+}+\frac{1}{2}(\cos \lambda+1) J_{-}] \tag{6}
\end{align*}
Substitute into equation (5), and use equation (2) to obtain
\begin{gather*}
m \sin \lambda d_{j m}^{j}(\lambda)+\frac{1}{2}(\cos \lambda-1) a_{j m} d_{j m+1}^{j}(\lambda) \\
\quad+\frac{1}{2}(\cos \lambda+1) a_{j,-m} d_{j m-1}^{j}(\lambda)=0 \tag{7}
\end{gather*}
Additionally, we can also obtain
\begin{equation*}
J_{z} \mathrm{e}^{-\mathrm{i} \lambda J_{y}}=\mathrm{e}^{-\mathrm{i} \lambda J_{y}}(J_{z} \cos \lambda-J_{x} \sin \lambda)=\mathrm{e}^{-\mathrm{i} \lambda J_{y}}[J_{z} \cos \lambda-\frac{1}{2} \sin \lambda(J_{+}+J_{-})] \tag{8}
\end{equation*}
Multiply the above equation from the left with $\langle j|$ and from the right with $|j m\rangle$, yielding
\begin{equation*}
(j-m \cos \lambda) d_{j m}^{j}(\lambda)+\frac{1}{2} \sin \lambda[a_{j m} d_{j m+1}^{j}(\lambda)+a_{j,-m} d_{j m-1}^{j}(\lambda)]=0 \tag{9}
\end{equation*}
Combine equations (7) and (9), eliminate $d_{j m-1}^{j}(\lambda)$, and obtain a simpler recursive relation:
\begin{equation*}
\sin \lambda \cdot a_{j m} d_{j m+1}^{j}(\lambda)=-(j-m)(1+\cos \lambda) d_{j m}^{j}(\lambda) \tag{10}
\end{equation*}
That is
\begin{equation*}
d_{j m}^{j}(\lambda)=-\frac{\sin \frac{\lambda}{2}}{\cos \frac{\lambda}{2}}(\frac{j+m+1}{j-m})^{\frac{1}{2}} d_{j m+1}^{j}(\lambda) \tag{$10^\prime$}
\end{equation*}
Thus, once $d_{j j}^{j}$ is found, all $d_{j m}^{j}$ can be recursively derived. Let's first find $d_{j j}^{j}$.
\begin{equation*}
d_{j j}^{j}(\lambda)=\langle j j| \mathrm{e}^{-\mathrm{i} \lambda J_{y}}|j j\rangle \tag{11}
\end{equation*}
Note that $d_{j j}^{j}(0)=1$, differentiating the above equation with respect to $\lambda$, we get
\begin{align*}
\frac{\mathrm{d}}{\mathrm{~d} \lambda d_{j j}^{j}(\lambda)} & =\langle j j| \mathrm{e}^{-\mathrm{i} \lambda J_{y}}(-\mathrm{i} J_{y})|j j\rangle \\
& =\frac{1}{2}\langle j j| \mathrm{e}^{-\mathrm{i} \lambda J_{y}}(J_{-}-J_{+})|j j\rangle \\
& =\frac{1}{2} a_{j-j} d_{j j-1}^{j}(\lambda) \tag{12}
\end{align*}
While equation (7) takes $m=j$, noticing $a_{j j}=0$, we get
\begin{equation*}
\frac{1}{2}(\cos \lambda+1) a_{j,-j} d_{j j-1}^{j}(\lambda)=-j \sin \lambda \cdot d_{j j}^{j}(\lambda) \tag{13}
\end{equation*}
Substituting into equation (12), we get
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{~d} \lambda} d_{j j}^{j}(\lambda)=-j \frac{\sin \lambda}{1+\cos \lambda} d_{j j}^{j}(\lambda) \tag{14}
\end{equation*}
Integrate, and using $d_{j j}^{j}(0)=1$, we obtain
\begin{equation*}
d_{j j}^{j}(\lambda)=(\cos \frac{\lambda}{2})^{2 j} \tag{15}
\end{equation*}
This formula can also be derived using the model in another format. Using equation (15) and the recursive relation ( $10^{\prime}$), we can sequentially obtain
\begin{aligned}
& d_{j j-1}^{j}(\lambda)=-(2 j)^{1 / 2}(\cos \frac{\lambda}{2})^{2 j-1} \sin \frac{\lambda}{2} \\
& d_{j j-2}^{j}(\lambda)=[\frac{2 j(2 j-1)}{2!}]^{\frac{1}{2}}(\cos \frac{\lambda}{2})^{2 j-2}(\sin \frac{\lambda}{2})^{2} \\
& \vdots
\end{aligned}
\begin{align*}
& d_{j m}^{j}(\lambda)=(-1)^{j-m}[\frac{(2 j)!}{(j+m)!(j-m)!}]^{\frac{1}{2}}(\cos \frac{\lambda}{2})^{j+m}(\sin \frac{\lambda}{2})^{j-m} \tag{16}\\
& \vdots \\
& d_{j-j}^{j}(\lambda)=(-1)^{2 j}(\sin \frac{\lambda}{2})^{2 j}
\end{align*}
Discussion $1^{\circ}$ When $\lambda \rightarrow-\lambda$, clearly
d_{j m}^{j_{m}}(-\lambda)=(-1)^{\jmath^{-m}} d_{j m}^{\prime}(\lambda)
This result matches the general property of $d_{m^{\prime} m}^{\prime}$.
$\mathbf{2}^{\circ}$ When $\lambda=\pi$, the only non-zero matrix element is clearly
\begin{equation*}
d_{j-\jmath}^{j}(\pi)=(-1)^{2 j} \tag{18}
\end{equation*}
This is because
\begin{align*}
& \mathrm{e}^{-i \pi J_{y}}|j m\rangle=(-1)^{j-m}|j-m\rangle \tag{19}\\
& \mathrm{e}^{-i \pi J_{y}}|j-j\rangle=(-1)^{2 j}|j j\rangle
\end{align*}
[The meaning of operator $\mathrm{e}^{-\mathrm{i} \pi_{y}}$ is to rotate the system by $180^{\circ}$ about the y-axis, so the state $|j m\rangle$ becomes the state $|j-m\rangle$, and equation (19) specifies the relative phase factor for the relationship between $|j m\rangle$ and $|j-m\rangle$.]
|
{"$d_{j m}^{j}(\\lambda)$": "matrix element in the context of angular momentum theory, parameterized by angular momentum quantum numbers and an angle", "$j$": "total angular momentum quantum number", "$m$": "magnetic quantum number", "$\\lambda$": "rotation angle related to angular momentum", "$J_{y}$": "y-component of angular momentum operator", "$J_{+}$": "raising operator for angular momentum", "$J_{-}$": "lowering operator for angular momentum", "$a_{j m}$": "coefficient related to angular momentum quantum numbers j and m", "$J_{z}$": "z-component of angular momentum operator", "$J_{x}$": "x-component of angular momentum operator"}
|
Schrödinger equation one-dimensional motion
| |
33 |
Theoretical Foundations
|
Let $\boldsymbol{J}_{1}$ and $\boldsymbol{J}_{2}$ be angular momenta corresponding to different degrees of freedom, then their sum $\boldsymbol{J}=\boldsymbol{J}_{1}+\boldsymbol{J}_{2}$ is also an angular momentum. Try to compute the expectation values of $J_{1 z}$ for the common eigenstate $|j_{1} j_{2} j m\rangle$ of $(\boldsymbol{J}_{1}^{2}, \boldsymbol{J}_{2}^{2}, \boldsymbol{J}^{2}, J_{z})$. (Take $\hbar=1$)
|
[
"m \\frac{j(j+1)+j_{1}(j_{1}+1)-j_{2}(j_{2}+1)}{2 j(j+1)}"
] |
Expression
|
$J_{1}, ~ J_{2}$ satisfy the fundamental commutation relations of angular momentum operators
\begin{equation*}
\boldsymbol{J}_{1} \times \boldsymbol{J}_{1}=\mathrm{i} \boldsymbol{J}_{1}, \quad \boldsymbol{J}_{2} \times \boldsymbol{J}_{2}=\mathrm{i} \boldsymbol{J}_{2} \tag{1}
\end{equation*}
$\boldsymbol{J}_{1}, ~ J_{2}$ belong to different degrees of freedom and commute with each other, so
\begin{align*}
& {[J_{x}, J_{1 x}]=[J_{1 x}+J_{2 x}, J_{1 x}]=0} \tag{2}\\
& {[J_{x}, J_{1 y}]=[J_{1 x}+J_{2 x}, J_{1 y}]=\mathrm{i} J_{1 z}}
\end{align*}
$\boldsymbol{J}$ and $\boldsymbol{J}_{2}$ have similar relationships. In summary, $\boldsymbol{J}_{1}$ or $\boldsymbol{J}_{2}$ and $\boldsymbol{J}$ satisfy all the relations between the vector operator $\boldsymbol{A}$ and $\boldsymbol{J}$ from the previous problem. Moreover,
\begin{align*}
& \boldsymbol{J} \cdot \boldsymbol{J}_{1}=\boldsymbol{J}_{1}^{2}+\boldsymbol{J}_{2} \cdot \boldsymbol{J}_{1}=\frac{1}{2}(\boldsymbol{J}^{2}+\boldsymbol{J}_{1}^{2}-\boldsymbol{J}_{2}^{2}) \tag{3}\\
& \boldsymbol{J} \cdot \boldsymbol{J}_{2}=\boldsymbol{J}_{2}^{2}+\boldsymbol{J}_{1} \cdot \boldsymbol{J}_{2}=\frac{1}{2}(\boldsymbol{J}^{2}+\boldsymbol{J}_{2}^{2}-\boldsymbol{J}_{1}^{2}) \tag{4}
\end{align*}
Using equation (5) from the previous problem, we have
\begin{align*}
& j(j+1)\langle\boldsymbol{J}_{1}\rangle_{j_{1} j_{2} j m}=\frac{1}{2}[j(j+1)+j_{1}(j_{1}+1)-j_{2}(j_{2}+1)]\langle\boldsymbol{J}\rangle_{j_{1} j_{2} j m} \tag{5}\\
& j(j+1)\langle\boldsymbol{J}_{2}\rangle_{j_{1} j_{2} j m}=\frac{1}{2}[j(j+1)+j_{2}(j_{2}+1)-j_{1}(j_{1}+1)]\langle\boldsymbol{J}\rangle_{j_{1} j_{2} j m} \tag{6}
\end{align*}
Since in the state $|J_{z}=m\rangle$
$$ \langle J_{x}\rangle=\langle J_{y}\rangle=0, \quad\langle J_{z}\rangle=m $$
Therefore, equations (5) and (6) yield
$$ \langle J_{1 x}\rangle_{j_{1} j_{2} j m}=\langle J_{1 y}\rangle_{j_{1} j_{2} j m}=\langle J_{2 x}\rangle_{j_{1} j_{2} j m}=\langle J_{2 y}\rangle_{j_{1} j_{2} j m}=0 $$
\begin{align*}
\langle J_{1 z}\rangle_{j_{1} j_{2} j m} & =m \frac{j(j+1)+j_{1}(j_{1}+1)-j_{2}(j_{2}+1)}{2 j(j+1)} \tag{7}\\
\langle J_{2 z}\rangle_{j_{1} j_{2} j m} & =m \frac{j(j+1)+j_{2}(j_{2}+1)-j_{1}(j_{1}+1)}{2 j(j+1)} \\
& =m-\langle J_{1 z}\rangle_{j_{1} j_{2}>m}
\end{align*}
|
{"$\\boldsymbol{J}_{1}$": "angular momentum for degree of freedom 1", "$\\boldsymbol{J}_{2}$": "angular momentum for degree of freedom 2", "$\\boldsymbol{J}$": "total angular momentum", "$J_{1 z}$": "z-component of angular momentum for degree of freedom 1", "$J_{2 z}$": "z-component of angular momentum for degree of freedom 2", "$J_{x}$": "x-component of total angular momentum", "$J_{1 x}$": "x-component of angular momentum for degree of freedom 1", "$J_{2 x}$": "x-component of angular momentum for degree of freedom 2", "$J_{y}$": "y-component of total angular momentum", "$J_{1 y}$": "y-component of angular momentum for degree of freedom 1", "$J_{2 y}$": "y-component of angular momentum for degree of freedom 2", "$m$": "magnetic quantum number", "$j_{1}$": "angular momentum quantum number for degree of freedom 1", "$j_{2}$": "angular momentum quantum number for degree of freedom 2", "$j$": "total angular momentum quantum number"}
|
Schrödinger equation one-dimensional motion
| |
34 |
Theoretical Foundations
|
Two angular momenta $\boldsymbol{J}_{1}$ and $\boldsymbol{J}_{2}$, of equal magnitude but belonging to different degrees of freedom, couple to form a total angular momentum $\boldsymbol{J}=\boldsymbol{J}_{1}+\boldsymbol{J}_{2}$, with $\hbar=1$, and assume $\boldsymbol{J}_{1}^{2}=\boldsymbol{J}_{2}^{2}=j(j+1)$. In the state where the total angular momentum quantum number $J=0$, what is the probability when $J_{1z}$ takes the value $m$ (with $J_{2z}$ simultaneously taking the value $-m$)?
|
[
"1/(2j+1)"
] |
Expression
|
The eigenstate of $(\boldsymbol{J}_{1}^{2}, J_{1 z})$ is denoted by $|j m_{1}\rangle_{1}$, and the eigenstate of $(\boldsymbol{J}_{2}^{2}, J_{2 z})$ is denoted by $|j m_{2}\rangle_{2}$. The common eigenstate of $(\boldsymbol{J}_{1}^{2}, \boldsymbol{J}_{2}^{2}$, $\mathbf{J}^{2}, J_{z})$ is denoted by $|j j J M\rangle$, where $M$ is the eigenvalue of $J_{z}$.
$|j j J M\rangle$ can also be abbreviated as $|J M\rangle$.
When $J=0$, $M=0, m_{1}=-m_{2}=m$, so the state under discussion can be expressed as
\begin{equation*}
|j j 00\rangle=\sum_{m} C_{m}|j m\rangle_{1}|j-m\rangle_{2} \tag{3}
\end{equation*}
$C_{m}$ is the C.G. coefficient $\langle j_{1} m_{1} j_{2} m_{2} \mid J M\rangle$ when $j_{1}=j_{2}=j, J=M=0, m_{1}=-m_{2}=m$. $|C_{m}|^{2}$ is the probability that $J_{1}$ takes the value $m$ (with $J_{2 z}$ taking the value $-m$ at the same time). Below, we solve for $C_{m}$.
Since $\boldsymbol{J}^{2}|j j 00\rangle=0$, and $\boldsymbol{J}^{2}$ is positive definite, it must be
\begin{equation*}
\boldsymbol{J}|j j 00\rangle=0 \tag{4}
\end{equation*}
Thus,
\begin{equation*}
(J_{x}+\mathrm{i} J_{y})|j j 00\rangle=0 \tag{5}
\end{equation*}
Where
\begin{equation*}
J_{x}+\mathrm{i} J_{y}=(J_{1 x}+\mathrm{i} J_{1 y})+(J_{2 x}+\mathrm{i} J_{2 y})=J_{1+}+J_{2+} \tag{6}
\end{equation*}
According to the basic formula for angular momentum ladder operators,
\begin{align*}
J_{1+}|j m\rangle_{1} & =(J_{1 x}+\mathrm{i} J_{1 y})|j m\rangle_{1}=a_{j m}|j m+1\rangle_{1} \\
a_{j m} & =\sqrt{(j-m)(j+m+1)} \tag{7a}\\
J_{2+}|j,-m\rangle_{2} & =(J_{2 x}+\mathrm{i} J_{2 y})|j,-m\rangle_{2}=a_{j,-m}|j, 1-m\rangle_{2} \\
a_{j,-m} & =\sqrt{(j+m)(j-m+1)}=a_{,, m-1} \tag{7b}
\end{align*}
Substituting expressions (5) to (7) into (3), we obtain
\begin{equation*}
\sum_{m} C_{m}[a_{j m}|j, m+1\rangle_{1}|j,-m\rangle_{2}+a_{J, m-1}|j m\rangle_{1}|j, 1-m\rangle_{2}]=0 \tag{8}
\end{equation*}
Since
\sum_{m} C_{m} a_{j, m-1}|j m\rangle_{1}|j, 1-m\rangle_{2} \xrightarrow{m \rightarrow m+1} \sum_{m} C_{m+1} a_{j m}|j m+1\rangle_{1}|j,-m\rangle_{2}
Therefore, equation (8) becomes
\begin{equation*}
\sum_{m}(C_{m}+C_{m+1}) a_{j m}|j m+1\rangle_{1}|j,-m\rangle_{2}=0 \tag{$\prime$}
\end{equation*}
Since each basis vector is linearly independent, it must be
\begin{equation*}
(C_{m}+C_{m+1}) a_{j m}=0 \tag{9}
\end{equation*}
Which implies
\begin{equation*}
C_{m}=-C_{m+1}, \quad m=j-1, j-2, \cdots,(-j) \tag{\prime}
\end{equation*}
In equation (3), $|j j 00\rangle, ~|j m\rangle_{1}, ~|j,-m\rangle_{2}$ etc. are all orthonormalized, and $m$ has a total of $(2 j+1)$ possible values. According to the normalization condition
\begin{equation*}
\sum_{m}|C_{m}|^{2}=1 \tag{10}
\end{equation*}
And equation ($9^{\prime}$), we immediately have $|C_{m}|^{2}=1 /(2 j+1)$. If we take
C_{j}=1 / \sqrt{2 j+1}
We get
\begin{equation*}
C_{m}=(-1)^{j-m} C_{j}=(-1)^{j-m} \frac{1}{\sqrt{2 j+1}} \tag{11}
\end{equation*}
Substituting into equation (3), we obtain
\begin{equation*}
|j j 00\rangle=\frac{1}{\sqrt{2 j+1}} \sum_{m}(-1)^{1^{-m}}|j m\rangle_{1}|j,-m\rangle_{2} \tag{12}
\end{equation*}
Clearly, under the premise $J_{1 z}=-J_{2 z}$, the probabilities that $J_{1 z}$ and $J_{2 z}$ take each eigenvalue $(j, j-1, \cdots,-j)$ are equal, both being $1 /(2 j+1)$.
|
{"$\\boldsymbol{J}_{1}$": "angular momentum 1", "$\\boldsymbol{J}_{2}$": "angular momentum 2", "$\\boldsymbol{J}$": "total angular momentum", "$\\hbar$": "reduced Planck's constant", "$j$": "quantum number for angular momentum magnitude", "$J$": "total angular momentum quantum number", "$J_{1z}$": "z-component of angular momentum 1", "$m$": "eigenvalue of $J_{1z}$", "$J_{2z}$": "z-component of angular momentum 2", "$m_1$": "eigenvalue of $J_{1z}$", "$m_2$": "eigenvalue of $J_{2z}$", "$M$": "eigenvalue of total angular momentum $J_{z}$", "$C_{m}$": "Clebsch-Gordan coefficient for state $m$", "$C_{m+1}$": "Clebsch-Gordan coefficient for state $m+1$", "$a_{j m}$": "ladder operator coefficient for state $m$", "$a_{j,-m}$": "ladder operator coefficient for state $-m$", "$J_{x}$": "x-component of total angular momentum", "$J_{y}$": "y-component of total angular momentum", "$J_{1x}$": "x-component of angular momentum 1", "$J_{1y}$": "y-component of angular momentum 1", "$J_{2x}$": "x-component of angular momentum 2", "$J_{2y}$": "y-component of angular momentum 2", "$J_{1+}$": "ladder operator for angular momentum 1", "$J_{2+}$": "ladder operator for angular momentum 2"}
|
Schrödinger equation one-dimensional motion
| |
35 |
Theoretical Foundations
|
A particle with mass $\mu$ and charge $q$ moves in a magnetic field $\boldsymbol{B}=\nabla \times \boldsymbol{A}$, where the Hamiltonian is $H = \frac{1}{2} \mu \boldsymbol{v}^{2}$, with $\boldsymbol{v}$ as the velocity operator. Calculate $\mathrm{d} \boldsymbol{v} / \mathrm{d} t$.
|
[
"\\frac{q}{2 \\mu c}(\\boldsymbol{v} \\times \\boldsymbol{B}-\\boldsymbol{B} \\times \\boldsymbol{v})"
] |
Expression
|
The Hamiltonian operator can be expressed as
\begin{equation*}
H=\frac{1}{2 \mu}(\boldsymbol{p}-\frac{q}{c} \boldsymbol{A})^{2}=\frac{1}{2} \mu \boldsymbol{v}^{2} \tag{2}
\end{equation*}
Using the commutation relations of $\boldsymbol{v}$ and $v^{2}$, it can be easily demonstrated that
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{~d} t} \boldsymbol{v}=\frac{1}{\mathrm{i} \hbar}[\boldsymbol{v}, H]=\frac{q}{2 \mu c}(\boldsymbol{v} \times \boldsymbol{B}-\boldsymbol{B} \times \boldsymbol{v}) \tag{3}
\end{equation*}
|
{"$\\mu$": "mass of the particle", "$q$": "charge of the particle", "$\\boldsymbol{B}$": "magnetic field", "$\\boldsymbol{A}$": "magnetic vector potential", "$H$": "Hamiltonian", "$\\boldsymbol{v}$": "velocity operator", "$c$": "speed of light in vacuum"}
|
Schrödinger equation one-dimensional motion
| |
36 |
Theoretical Foundations
|
A particle with mass $\mu$ and charge $q$ moves in a magnetic field $\boldsymbol{B}=\nabla \times \boldsymbol{A}$, where the Hamiltonian is $H = \frac{1}{2} \mu \boldsymbol{v}^{2}$, with $\boldsymbol{v}$ as the velocity operator. Let $\boldsymbol{L}$ be the the angular momentum operator. Calculate $\mathrm{d} \boldsymbol{L} / \mathrm{d} t$.
|
[
"\\frac{q}{2 c}[\\boldsymbol{r} \\times(\\boldsymbol{v} \\times \\boldsymbol{B})+(\\boldsymbol{B} \\times \\boldsymbol{v}) \\times \\boldsymbol{r}]"
] |
Expression
|
The Hamiltonian operator can be expressed as
\begin{equation*}
H=\frac{1}{2 \mu}(\boldsymbol{p}-\frac{q}{c} \boldsymbol{A})^{2}=\frac{1}{2} \mu \boldsymbol{v}^{2} \tag{2}
\end{equation*}
Using the commutation relations of $\boldsymbol{v}$ and $v^{2}$, it can be easily demonstrated that
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{~d} t} \boldsymbol{v}=\frac{1}{\mathrm{i} \hbar}[\boldsymbol{v}, H]=\frac{q}{2 \mu c}(\boldsymbol{v} \times \boldsymbol{B}-\boldsymbol{B} \times \boldsymbol{v}) \tag{3}
\end{equation*}
In classical electrodynamics, the Lorentz force is
$$ \boldsymbol{f}=\frac{q}{c} \boldsymbol{v} \times \boldsymbol{B}. $$
The equation of motion is
\begin{equation*}
\mu \frac{\mathrm{d}}{\mathrm{~d} t} \boldsymbol{v}=\boldsymbol{f}=\frac{q}{c} \boldsymbol{v} \times \boldsymbol{B} \tag{$\prime$}
\end{equation*}
Equation (3) is the quantum mechanical extension of the classical equation of motion ($3^{\prime}$). In equation (3),
\begin{equation*}
\boldsymbol{v} \times \boldsymbol{B}=\frac{1}{\mu}(\boldsymbol{p}-\frac{q}{c} \boldsymbol{A}) \times \boldsymbol{B}=-\boldsymbol{B} \times \boldsymbol{v}-\frac{\mathrm{i} \hbar}{\mu} \nabla \times \boldsymbol{B} \tag{4}
\end{equation*}
From electrodynamics,
$$\nabla \times \boldsymbol{B}=\frac{4 \pi}{c} \boldsymbol{j}$$
$\boldsymbol{j}$ is the source current that generates the magnetic field. Thus, if in the space where the particle moves, the source current $\boldsymbol{j}=0$, then
\begin{equation*}
\frac{q^{v}}{c} \boldsymbol{v} \times \boldsymbol{B}=-\frac{q}{c} \boldsymbol{B} \times \boldsymbol{v}=\boldsymbol{f} \tag{5}
\end{equation*}
Equation (3) can still be rewritten in the form of equation ($3^{\prime}$). For uniform fields, it is obvious that equations (3) and ($3^{\prime}$) are equivalent.
The mechanical angular momentum can also be expressed as
\begin{equation*}
\boldsymbol{L}=\mu \boldsymbol{r} \times \boldsymbol{v}=\frac{1}{2} \mu(\boldsymbol{r} \times \boldsymbol{v}-\boldsymbol{v} \times \boldsymbol{r}) \tag{$\prime$}
\end{equation*}
(Note that in $\boldsymbol{r} \times \boldsymbol{v}$, the relevant components of $\boldsymbol{r}$ and $\boldsymbol{v}$ are commutative.) The time derivative of $\boldsymbol{L}$ is
$$\frac{\mathrm{d} \boldsymbol{L}}{\mathrm{~d} t}=\frac{\mu}{2}(\frac{\mathrm{~d} \boldsymbol{r}}{\mathrm{~d} t} \times \boldsymbol{v}+\boldsymbol{r} \times \frac{\mathrm{d} \boldsymbol{v}}{\mathrm{~d} t}-\frac{\mathrm{d} \boldsymbol{v}}{\mathrm{~d} t} \times \boldsymbol{r}-\boldsymbol{v} \times \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{~d} t}).$$
Using equation (3) and $\mathrm{d} \boldsymbol{r} / \mathrm{d} t=\boldsymbol{v}$, we obtain
\begin{equation*}
\frac{\mathrm{d} \boldsymbol{L}}{\mathrm{~d} t}=\frac{q}{4 c}[\boldsymbol{r} \times(\boldsymbol{v} \times \boldsymbol{B})+(\boldsymbol{B} \times \boldsymbol{v}) \times \boldsymbol{r}-\boldsymbol{r} \times(\boldsymbol{B} \times \boldsymbol{v})-(\boldsymbol{v} \times \boldsymbol{B}) \times \boldsymbol{r}] \tag{6}
\end{equation*}
Under the conditions where equation (5) holds, equation (6) can be simplified to
\begin{equation*}
\frac{\mathrm{d} \boldsymbol{L}}{\mathrm{~d} t}=\frac{q}{2 c}[\boldsymbol{r} \times(\boldsymbol{v} \times \boldsymbol{B})+(\boldsymbol{B} \times \boldsymbol{v}) \times \boldsymbol{r}]=\frac{1}{2}(\boldsymbol{r} \times \boldsymbol{f}-\boldsymbol{f} \times \boldsymbol{r}) \tag{7}
\end{equation*}
Note that in $\boldsymbol{r} \times \boldsymbol{f}$, the relevant components of $\boldsymbol{r}$ and $\boldsymbol{f}$ are non-commutative, and $\boldsymbol{r} \times \boldsymbol{f} \neq-\boldsymbol{f} \times \boldsymbol{r}$, hence the right side of equation (7) is not equivalent to $\boldsymbol{r} \times \boldsymbol{f}$.
|
{"$\\mu$": "mass of the particle", "$q$": "charge of the particle", "$\\boldsymbol{B}$": "magnetic field", "$\\boldsymbol{A}$": "vector potential", "$H$": "Hamiltonian", "$\\boldsymbol{v}$": "velocity operator", "$\\boldsymbol{L}$": "angular momentum operator", "$\\boldsymbol{f}$": "Lorentz force", "$\\boldsymbol{r}$": "position vector", "$\\boldsymbol{p}$": "momentum operator", "$\\boldsymbol{j}$": "source current generating the magnetic field", "$c$": "speed of light in vacuum", "$\\hbar$": "reduced Planck's constant"}
|
Schrödinger equation one-dimensional motion
| |
37 |
Theoretical Foundations
|
A three-dimensional isotropic oscillator with mass $\mu$, charge $q$, and natural frequency $\omega_{0}$ is placed in a uniform external magnetic field $\boldsymbol{B}$. Find the formula for energy levels.
|
[
"E_{n_{1} n_{2} m}=(n_{1}+\\frac{1}{2}) \\hbar \\omega_{0}+(2 n_{2}+1+|m|) \\hbar \\omega-m \\hbar \\omega_{\\mathrm{L}}"
] |
Expression
|
Compared to the previous two questions, an additional harmonic oscillator potential $\frac{1}{2} \mu \omega_{0}^{2}(x^{2}+y^{2}+z^{2})$ should be added to the Hamiltonian in this question, thus
\begin{align*}
H & =\frac{1}{2 \mu} \boldsymbol{p}^{2}+\frac{1}{2} \mu \omega_{0}^{2}(x^{2}+y^{2}+z^{2})+\frac{q^{2} B^{2}}{8 \mu c^{2}}(x^{2}+y^{2})-\frac{q B}{2 \mu c} l_{z} \\
& =H_{1}+H_{2}-\omega_{\mathrm{L}} l_{z} \tag{1}
\end{align*}
where
\begin{align*}
& H_{1}=\frac{1}{2 \mu} p_{z}^{2}+\frac{1}{2} \mu \omega_{0}^{2} z^{2} \\
& H_{2}=\frac{1}{2 \mu}(p_{x}^{2}+p_{y}^{2})+\frac{1}{2} \mu(\omega_{0}^{2}+\omega_{\mathrm{L}}^{2})(x^{2}+y^{2}) \tag{2}\\
& \omega_{\mathrm{L}}=\frac{q B}{2 \mu c}
\end{align*}
$H_{1}, ~ H_{2}, ~ l_{z}$ are mutually commuting conserved quantities and can be chosen as a complete set of commuting observables. $H_{1}$ corresponds to a one-dimensional harmonic oscillator energy operator, with eigenvalues given by
\begin{equation*}
E_{n_{1}}=(n_{1}+\frac{1}{2}) \hbar \omega_{0}, \quad n_{1}=0,1,2, \cdots \tag{3}
\end{equation*}
$\mathrm{H}_{2}$ corresponds to a two-dimensional isotropic oscillator total energy operator, with eigenvalues given by
\begin{align*}
& E_{n_{2} m}=(2 n_{2}+1+|m|) \hbar \omega \tag{4}\\
& n_{2}=0,1,2, \cdots, \quad m=0, \pm 1, \pm 2, \cdots
\end{align*}
where
\begin{equation*}
\omega=\sqrt{\omega_{0}^{2}+\omega_{L}^{2}} \tag{5}
\end{equation*}
The eigenvalue of $l_{z}$ is $m \hbar$. Therefore, the energy levels for this problem are
\begin{equation*}
E_{n_{1} n_{2} m}=(n_{1}+\frac{1}{2}) \hbar \omega_{0}+(2 n_{2}+1+|m|) \hbar \omega-m \hbar \omega_{\mathrm{L}} \tag{6}
\end{equation*}
Note: $\omega_{\mathrm{L}}$ can be positive or negative depending on the sign of charge $q$; the magnetic quantum number $m$ can also be positive or negative. Therefore, the sign of $q$ does not affect the overall energy spectrum. For precision, only the case of $q>0(\omega_{\mathrm{L}}>0)$ is discussed below.
The relationship among the three frequencies $\omega_{0}, ~ \omega_{L}, ~ \omega$ is given by:
\begin{equation*}
0<(\omega-\omega_{\mathrm{L}})<\omega_{0} \tag{7}
\end{equation*}
The dependence of the energy levels on the quantum numbers (listed in the order of their magnitude of change) is
\begin{array}{lll}
m \geqslant 0, & m \text { increases by } 1, & E \text { increases by } \hbar(\omega-\omega_{\mathrm{L}}) \\
& n_{1} \text { increases by } 1, & E \text { increases by } \hbar \omega_{0} \\
m \leqslant 0, & |m| \text { increases by } 1, & E \text { increases by } \hbar(\omega+\omega_{\mathrm{L}}) \\
& n_{2} \text { increases by } 1, & E \text { increases by } 2 \hbar \omega
\end{array}
For the ground state $E_{0}$, obviously, $n_{1}, ~ n_{2}, ~|m|$ all take their minimum values,
\begin{equation*}
E_{0}=E_{000}=\frac{1}{2} \hbar \omega_{0}+\hbar \omega \tag{8}
\end{equation*}
For the first excited state $E_{1}, n_{1}=n_{2}=0, m=1$,
\begin{equation*}
E_{1}=E_{001}=\frac{1}{2} \hbar \omega_{0}+2 \hbar \omega-\hbar \omega_{\mathrm{L}}=E_{0}+\hbar(\omega-\omega_{\mathrm{L}}) \tag{9}
\end{equation*}
|
{"$\\mu$": "mass of the oscillator", "$q$": "charge", "$\\omega_{0}$": "natural frequency of the oscillator", "$\\boldsymbol{B}$": "external magnetic field", "$H$": "Hamiltonian operator of the system", "$\\boldsymbol{p}$": "momentum vector", "$c$": "speed of light", "$l_{z}$": "z-component of angular momentum", "$H_{1}$": "one-dimensional harmonic oscillator Hamiltonian", "$H_{2}$": "two-dimensional isotropic oscillator Hamiltonian", "$p_{z}$": "momentum along the z-axis", "$p_{x}$": "momentum along the x-axis", "$p_{y}$": "momentum along the y-axis", "$\\omega_{\\mathrm{L}}$": "Larmor frequency", "$n_{1}$": "quantum number for z-axis oscillator", "$n_{2}$": "quantum number for two-dimensional oscillator", "$m$": "magnetic quantum number", "$\\hbar$": "reduced Planck's constant", "$\\omega$": "derived angular frequency"}
|
Schrödinger equation one-dimensional motion
| |
38 |
Theoretical Foundations
|
A particle with mass $\mu$ and charge $q$ moves in a uniform electric field $\mathscr{E}$ (along the x-axis) and a uniform magnetic field $\boldsymbol{B}$ (along the z-axis) that are perpendicular to each other. If the momentum of the particle in the y-direction is $p_y$ and in the z-direction is $p_z$, find the energy level expression of the system.
|
[
"E = (n+\\frac{1}{2}) \\frac{\\hbar B|q|}{\\mu c}-\\frac{c^{2} \\mathscr{E}^{2} \\mu}{2 B^{2}}-\\frac{c \\mathscr{E}}{B} p_{y}+\\frac{1}{2 \\mu} p_{z}^{2}"
] |
Expression
|
With the electric field direction as the $x$ axis and the magnetic field direction as the $z$ axis, then
\begin{equation*}
\mathscr{L}=(\mathscr{E}, 0,0), \quad \boldsymbol{B}=(0,0, B) \tag{1}
\end{equation*}
Taking the scalar and vector potentials of the electromagnetic field as
\begin{equation*}
\phi=-\mathscr{E} x, \quad \boldsymbol{A}=(0, B x, 0) \tag{2}
\end{equation*}
Satisfying the relation
$$\mathscr{E}=-\nabla \phi, \quad \boldsymbol{B}=\nabla \times \boldsymbol{A}.$$
The Hamiltonian of the particle is
\begin{equation*}
H=\frac{1}{2 \mu}[p_{x}^{2}+(p_{y}-\frac{q B}{c} x)^{2}+p_{z}^{2}]-q \mathscr{E} x \tag{3}
\end{equation*}
With the constants of motion set as $(H, p_{y}, p_{z})$, their common eigenfunction can be written as
\begin{equation*}
\psi(x, y, z)=\psi(x) \mathrm{e}^{\mathrm{i}(p_{y} y+p_{z} z) / \hbar} \tag{4}
\end{equation*}
where $p_{y}$ and $p_{z}$ are eigenvalues and can be any real numbers.
$\psi(x, y, z)$ satisfies the energy eigen-equation
$$ H \psi(x, y, z)=E \psi(x, y, z) $$
Thus, $\psi(x)$ satisfies the equation
\begin{equation*}
\frac{1}{2 \mu}[p_{x}^{2}+(p_{y}-\frac{q B}{c} x)^{2}+p_{z}^{2}] \psi-q \mathscr{E} x \psi=E \psi \tag{5}
\end{equation*}
That is, for $\psi(x)$, $H$ is equivalent to the following:
\begin{align*}
H & \Rightarrow-\frac{\hbar^{2}}{2 \mu} \frac{\partial^{2}}{\partial x^{2}}+\frac{q^{2} B^{2}}{2 \mu c^{2}} x^{2}-(q \mathscr{E}+\frac{q B}{\mu c} p_{y}) x+\frac{1}{2 \mu}(p_{y}^{2}+p_{x}^{2}) \\
& =-\frac{\hbar^{2}}{2 \mu} \frac{\partial^{2}}{\partial x^{2}}+\frac{q^{2} B^{2}}{2 \mu c^{2}}(x-x_{0})^{2}-\frac{q^{2} B^{2}}{2 \mu c^{2}} x_{0}^{2}+\frac{1}{2 \mu}(p_{y}^{2}+p_{z}^{2}) \tag{6}
\end{align*}
where
\begin{equation*}
x_{0}=\frac{\mu c^{2}}{q^{2} B^{2}}(q \mathscr{E}+\frac{q B}{\mu c} p_{y})=\frac{\mu c}{q B}(\frac{c \mathscr{E}}{B}+\frac{p_{y}}{\mu}) \tag{7}
\end{equation*}
Equation (6) corresponds to a one-dimensional harmonic oscillator energy operator
$$-\frac{\hbar^{2}}{2 \mu} \frac{\partial^{2}}{\partial x^{2}}+\frac{1}{2} \mu \omega^{2}(x-x_{0})^{2}, \quad \omega=\frac{|q| B}{\mu c} $$
Plus two constant terms. Therefore, the energy level of this problem is
\begin{align*}
E & =(n+\frac{1}{2}) \hbar \omega-\frac{q^{2} B^{2}}{2 \mu c^{2}} x_{0}^{2}+\frac{1}{2 \mu}(p_{y}^{2}+p_{z}^{2}) \\
& =(n+\frac{1}{2}) \frac{\hbar B|q|}{\mu c}-\frac{c^{2} \mathscr{E}^{2} \mu}{2 B^{2}}-\frac{c \mathscr{E}}{B} p_{y}+\frac{1}{2 \mu} p_{z}^{2} \tag{8}
\end{align*}
where $p_{y}, ~ p_{z}$ are any real numbers, $n=0,1,2, \cdots$.
In equation (4), $\psi(x)$ is the one-dimensional harmonic oscillator energy eigenfunction with $(x-x_{0})$ as the variable, i.e.,
\begin{equation*}
\psi(x)=\psi_{n}(x-x_{0})=H_{n}(\xi) \mathrm{e}^{-\frac{1}{2} \xi^{2}} \tag{9}
\end{equation*}
$H_{n}(\xi)$ is the Hermite polynomial, $\xi=(\frac{|q| B}{\hbar c})^{\frac{1}{2}}(x-x_{0})$.
|
{"$\\mu$": "mass of the particle", "$q$": "charge of the particle", "$\\mathscr{E}$": "electric field", "$\\boldsymbol{B}$": "magnetic field", "$p_y$": "momentum in the y-direction", "$p_z$": "momentum in the z-direction", "$B$": "magnitude of the magnetic field", "$c$": "speed of light", "$H$": "Hamiltonian", "$\\hbar$": "reduced Planck's constant", "$x_0$": "displacement term related to particle's motion", "$n$": "quantum number"}
|
Schrödinger equation one-dimensional motion
| |
39 |
Theoretical Foundations
|
A particle moves in one dimension. When the total energy operator is
\begin{equation*}
H_{0}=\frac{p^{2}}{2 m}+V(x) \tag{1}
\end{equation*}
the energy level is $E_{n}^{(0)}$. If the total energy operator becomes
\begin{equation*}
H=H_{0}+\frac{\lambda p}{m} \tag{2}
\end{equation*}
find the energy level $E_{n}$.
|
[
"E_{n}=E_{n}^{(0)}-\\frac{\\lambda^{2}}{2 m}"
] |
Expression
|
First, treat $\lambda$ as a parameter, then
\begin{equation*}
\frac{\partial H}{\partial \lambda}=\frac{p}{m} \tag{3}
\end{equation*}
According to the Hellmann theorem, we have
\begin{equation*}
\frac{\partial E_{n}}{\partial \lambda}=\langle\frac{\partial H}{\partial \lambda}\rangle_{n}=\frac{1}{m}\langle p\rangle_{n} \tag{4}
\end{equation*}
However, since
\begin{equation*}
\frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{1}{\mathrm{i} \hbar}[x, H]=\frac{1}{\mathrm{i} \hbar m}[x, \frac{p^{2}}{2}+\lambda p]=\frac{1}{m}(p+\lambda) \tag{5}
\end{equation*}
For any bound state,
$$ \langle\frac{\mathrm{d} x}{\mathrm{~d} t}\rangle_{n}=\frac{1}{\mathrm{i} \hbar}\langle x H-H x\rangle_{n}=0, $$
therefore
\begin{equation*}
\langle p\rangle_{n}=-\lambda \tag{6}
\end{equation*}
Substitute into equation (4), obtaining
\begin{equation*}
\frac{\partial E_{n}}{\partial \lambda}=-\frac{\lambda}{m} \tag{7}
\end{equation*}
Integrate to get
\begin{equation*}
E_{n}=-\frac{\lambda^{2}}{2 m}+C \tag{8}
\end{equation*}
$C$ is the integration constant. Since when $\lambda=0$, $H=H_{0}, E_{n}=E_{n}^{(0)}$, so $C=E_{n}^{(0)}$. Substitute into equation (8), obtaining
\begin{equation*}
E_{n}=E_{n}^{(0)}-\frac{\lambda^{2}}{2 m} \tag{9}
\end{equation*}
Solution Two: Write $H$ as
\begin{equation*}
H=\frac{p^{2}}{2 m}+\frac{\lambda p}{m}+V(x)=\frac{P^{2}}{2 m}+V(x)-\frac{\lambda^{2}}{2 m} \tag{10}
\end{equation*}
where
\begin{equation*}
P=p+\lambda \tag{11}
\end{equation*}
In the momentum representation,
\begin{equation*}
x=\mathrm{i} \hbar \frac{\partial}{\partial p}=\mathrm{i} \hbar \frac{\partial}{\partial P} \tag{12}
\end{equation*}
Therefore,
\begin{align*}
H_{0} & =\frac{p^{2}}{2 m}+V(i \hbar \frac{\partial}{\partial p}) \tag{13}\\
H & =\frac{P^{2}}{2 m}+V(i \hbar \frac{\partial}{\partial P})-\frac{\lambda^{2}}{2 m} \tag{14}
\end{align*}
The difference between $H$ and $H_{0}$, aside from the constant term, is just replacing $p$ with $P$, which does not affect the energy levels. Therefore,
\begin{equation*}
E_{n}=E_{n}^{(0)}-\frac{\lambda^{2}}{2 m} \tag{15}
\end{equation*}
|
{"$H_{0}$": "initial total energy operator", "$p$": "momentum", "$m$": "mass", "$V(x)$": "potential energy as a function of position", "$E_{n}^{(0)}$": "initial energy level", "$H$": "total energy operator including perturbation", "$\\lambda$": "perturbation parameter", "$E_{n}$": "energy level after perturbation", "$\\hbar$": "reduced Planck's constant", "$x$": "position", "$C$": "integration constant", "$P$": "modified momentum", "$i$": "imaginary unit"}
|
Schrödinger equation one-dimensional motion
| |
40 |
Theoretical Foundations
|
A particle with mass $\mu$ moves in a central force field,
\begin{equation*}
V(r)=\lambda r^{\nu}, \quad-2<\nu, \quad \nu / \lambda>0 \tag{1}
\end{equation*}
Use the Hellmann theorem and the virial theorem to analyze the dependence of the energy level structure on $\hbar, ~ \lambda, ~ \mu$.
|
[
"E=C \\lambda^{2 /(2+\\nu)}(\\frac{\\hbar^{2}}{2 \\mu})^{\\nu /(2+\\nu)}"
] |
Expression
|
The energy operator is
\begin{equation*}
H=T+V=-\frac{\hbar^{2}}{2 \mu} \nabla^{2}+\lambda r^{\nu} \tag{2}
\end{equation*}
Let $\beta=\hbar^{2} / 2 \mu, \lambda$ and $\beta$ be independent parameters. It is evident that
\begin{equation*}
\beta \frac{\partial H}{\partial \beta}=T, \quad \lambda \frac{\partial H}{\partial \lambda}=V \tag{3}
\end{equation*}
According to the Hellmann theorem, for any bound state,
\begin{align*}
& \langle T\rangle=\frac{\beta \partial E}{\partial \beta} \tag{4}\\
& \langle V\rangle=\frac{\lambda \partial E}{\partial \lambda} \tag{5}
\end{align*}
Adding the two equations yields
\begin{equation*}
\beta \frac{\partial E}{\partial \beta}+\lambda \frac{\partial E}{\partial \lambda}=\langle T+V\rangle=E \tag{6}
\end{equation*}
And from the virial theorem, we have
$$ \langle T\rangle=\frac{\nu}{2}\langle V\rangle$$
That is
\begin{equation*}
\beta \frac{\partial E}{\partial \beta}=\frac{\nu}{2} \lambda \frac{\partial E}{\partial \lambda} \tag{7}
\end{equation*}
Substituting equation (7) into equation (6), we get
\begin{align*}
& (1+\frac{\nu}{2}) \lambda \frac{\partial E}{\partial \lambda}=E \tag{8}\\
& (1+\frac{2}{\nu}) \beta \frac{\partial E}{\partial \beta}=E \tag{9}
\end{align*}
Integrating equation (8), we obtain the construction relationship between $E$ and $\lambda$
\begin{equation*}
E=C_{1} \lambda^{2 /(2+\iota)} \tag{10}
\end{equation*}
$C_{1}$ is the "integration constant" and is independent of $\lambda$. Integrating equation (9), we obtain the construction relationship between $E$ and $\beta$
\begin{equation*}
E=C_{2} \beta^{\nu /(2 \downarrow \imath)} \tag{11}
\end{equation*}
$C_{2}$ is independent of $\nu$. Comparing equations (10) and (11), it follows that
\begin{equation*}
E=C \lambda^{2 /(2+\tau)} \beta^{\nu /(2+\nu)}=C \lambda^{2 /(2+\tau)}(\frac{\hbar^{2}}{2 \mu})^{\nu /(2+\nu)} \tag{12}
\end{equation*}
$C$ is independent of $\lambda, ~ \beta$, and is a dimensionless pure number (related to $\nu$ and quantum numbers). It is easy to verify that the above expression is the only possible energy construction that is dimensionally correct.
From equation (12), it is evident (note $\nu>-2$) that as the interaction strength $|\lambda|$ increases, $|E|$ increases, and the energy level spacing increases. If $\lambda$ is independent of the particle's mass, then when $\nu>0, ~ \mu$ increases, $|E|$ decreases; when $\nu<0, ~ \mu$ increases, $|E|$ increases.
If $\lambda$ is independent of $\mu$, from equations (2) and (4) it can also be seen that
\begin{equation*}
\mu \frac{\partial E}{\partial \mu}=-\langle T\rangle<0 \tag{13}
\end{equation*}
That is, an increase in the particle's mass always leads to a decrease in the algebraic value of the energy levels.
From equation (12) it can also be seen that if $\lambda \propto \mu^{\nu / 2}$, then $E$ is independent of $\mu$. A famous example of this situation is the harmonic oscillator, i.e.,
\begin{align*}
& V(\boldsymbol{r})=\lambda r^{2}=\frac{1}{2} \mu \omega^{2} r^{2} \\
& (\nu=2, \quad \lambda=\frac{1}{2} \mu \omega^{2})
\end{align*}
The energy levels are
$$ E_{N}=(N+\frac{3}{2}) \hbar \omega, \quad N=0,1,2, \cdots $$
If $\omega$ remains constant, $E_{N}$ is independent of $\mu$.
|
{"$\\mu$": "mass of the particle", "$\\nu$": "exponent in the potential function", "$\\lambda$": "scaling factor of the potential", "$\\hbar$": "reduced Planck's constant", "$\\beta$": "parameter related to kinetic energy and mass", "$E$": "energy", "$T$": "kinetic energy", "$V$": "potential energy", "$C$": "dimensionless integration constant related to energy", "$C_{1}$": "integration constant independent of $\\lambda$", "$C_{2}$": "integration constant independent of $\\nu$", "$\\omega$": "angular frequency", "$E_{N}$": "energy level for quantum number $N$", "$N$": "quantum number"}
|
Schrödinger equation one-dimensional motion
| |
41 |
Theoretical Foundations
|
A particle moves in a potential field
\begin{equation*}
V(x)=V_{0}|x / a|^{\nu}, \quad V_{0}, a>0 \tag{1}.
\end{equation*}
Find the dependence of energy levels on parameters as $\nu \rightarrow \infty$.
|
[
"E_{n}=\\frac{n^{2} \\pi^{2} \\hbar^{2}}{8 \\mu a^{2}}"
] |
Expression
|
The total energy operator is
\begin{equation*}
H=T+V=-\frac{\hbar^{2}}{2 \mu} \frac{\mathrm{~d}^{2}}{\mathrm{~d} x^{2}}+V_{0}|x / a|^{\nu} \tag{2}
\end{equation*}
From dimensional analysis, if $x_{0}$ represents the characteristic length, we have
\begin{equation*}
E \sim \frac{\hbar^{2}}{\mu x_{0}^{2}} \sim \frac{V_{0} x_{0}^{\nu}}{a^{\nu}} \tag{3}
\end{equation*}
It is not difficult to solve for
\begin{equation*}
x_{0} \sim(\frac{\hbar^{2} a^{\nu}}{\mu V_{0}})^{\frac{1}{\nu+2}} \xrightarrow{\nu \rightarrow \infty} a \tag{4}
\end{equation*}
\begin{equation*}
E \sim(\frac{\hbar^{2}}{\mu})^{\frac{1}{\nu+2}} V_{0}^{2 /(\nu+2)} a^{-2 \nu /(\nu+2)} \xrightarrow{\nu \rightarrow \infty} \frac{\hbar^{2}}{\mu a^{2}} \tag{5}
\end{equation*}
The construction of the characteristic length and energy levels is independent of $V_{0}$,
In fact, it is easy to see from Equation (1)
$$ \nu \rightarrow \infty, \quad V(x) \rightarrow \begin{cases}0, & |x|<a \tag{6}\\ \infty, & |x|>a\end{cases} $$
This is precisely an infinitely deep potential well of width $2 a$, with energy levels
\begin{equation*}
E_{n}=\frac{n^{2} \pi^{2} \hbar^{2}}{8 \mu a^{2}} . \quad n=1,2,3, \cdots \tag{7}
\end{equation*}
When $\nu$ is finite, according to the virial theorem, for any bound state, there is a relationship between the average kinetic energy and the average potential energy
\begin{equation*}
\frac{2}{\nu}\langle T\rangle=\langle V\rangle \tag{8}
\end{equation*}
Thus, as $u \rightarrow \infty$, it follows
\begin{equation*}
\langle V\rangle \rightarrow 0, \quad\langle T\rangle \rightarrow E \tag{9}
\end{equation*}
This conclusion is also consistent with the infinite potential well problem.
|
{"$V$": "potential energy", "$x$": "position", "$V_{0}$": "scaling factor of the potential", "$a$": "characteristic length scale of the potential", "$\\nu$": "exponent indicating the power to which the scaled position is raised in the potential", "$H$": "total energy operator", "$T$": "kinetic energy", "$\\hbar$": "reduced Planck's constant", "$\\mu$": "reduced mass", "$E$": "energy", "$x_{0}$": "characteristic length", "$E_{n}$": "energy levels", "$n$": "quantum number representing different energy levels"}
|
Schrödinger equation one-dimensional motion
| |
42 |
Theoretical Foundations
|
A particle with mass $m$ moves in a uniform force field $V(x)=F x(F>0)$, with the motion constrained to the range $x \geqslant 0$. Find its ground state energy level.
|
[
"E_{1}=1.8558(\\frac{\\hbar^{2} F^{2}}{m})^{1 / 3}"
] |
Expression
|
The total energy operator is
\begin{equation*}
H=T+V=\frac{p^{2}}{2 m}+F x \tag{1}
\end{equation*}
In momentum representation, the operator for $x$ is given by
\begin{equation*}
\hat{x}=\mathrm{i} \hbar \frac{\mathrm{~d}}{\mathrm{~d} p} \tag{2}
\end{equation*}
The operator for $H$ is given by
\begin{equation*}
\hat{H}=\frac{p^{2}}{2 m}+\mathrm{i} \hbar F \frac{\mathrm{~d}}{\mathrm{~d} p} \tag{3}
\end{equation*}
The time-independent Schrödinger equation is
\begin{equation*}
\frac{p^{2}}{2 m} \varphi(p)+\mathrm{i} \hbar F \frac{\mathrm{~d}}{\mathrm{~d} p} \varphi(p)=E \varphi(p) \tag{4}
\end{equation*}
Here $\varphi(p)$ is the wave function in the momentum representation. Equation (4) is a very simple first-order differential equation, whose solution is
\begin{equation*}
\varphi(p)=A \exp [\frac{\mathrm{i}}{\hbar F}(\frac{p^{3}}{6 m}-E p)] \tag{5}
\end{equation*}
$A$ is the normalization constant.
Transforming to the $x$ representation, the wave function is
\begin{align*}
\psi(x) & =(2 \pi \hbar)^{-1 / 2} \int_{-\infty}^{+\infty} \varphi(p) \mathrm{e}^{\mathrm{i} p x / \hbar} \mathrm{d} p \\
& =A(2 \pi \hbar)^{-1 / 2} \int_{-\infty}^{+\infty} \exp \frac{\mathrm{i}}{\hbar}[\frac{p^{3}}{6 m F}+(x-\frac{E}{F}) p] \mathrm{d} p \\
& =2 A(2 \pi \hbar)^{-1 / 2} \int_{0}^{\infty} \cos [\frac{p^{3}}{6 \hbar m F}+\frac{p}{\hbar}(x-\frac{E}{F})] \mathrm{d} p \\
& =\frac{C}{\sqrt{\pi}} \int_{0}^{\infty} \cos (\frac{u^{3}}{3}+u \xi) \mathrm{d} u \tag{6}
\end{align*}
Where
\begin{align*}
& u=p(2 \hbar m F)^{-1 / 3} \tag{7}\\
& \xi=(x-\frac{E}{F})(\frac{2 m F}{\hbar^{2}})^{1 / 3}=\frac{x}{l}-\lambda \tag{8}\\
& l=(\frac{\hbar^{2}}{2 m F})^{1 / 3} \tag{9}\\
& \lambda=(\frac{2 m}{\hbar^{2} F^{2}})^{1 / 3} E=\frac{2 m E}{\hbar^{2}} l^{2} \tag{10}
\end{align*}
$l$ is the characteristic length of this problem.
Except for the normalization constant $C$, the right-hand side of equation (6) resembles the Airy function with $\xi$ as the variable.
When $\xi>0(x>E / F$ , i.e., the classically forbidden region), the Airy function behaves like the modified Bessel function of the second kind, i.e.,
\begin{equation*}
\psi(x)=\sqrt{\xi} K \frac{1}{3}(\frac{2}{3} \xi^{3 / 2}) \xrightarrow{\xi \rightarrow \infty} \sqrt{\xi}(\frac{3 \pi}{4 \xi^{3 / 2}})^{1 / 2} \mathrm{e}^{-\frac{2}{3} \xi^{3 / 2}} \tag{11}
\end{equation*}
When $\xi<0$ ($x<E / F$, i.e., the classically allowed region), the Airy function behaves like the Bessel function of the first kind, i.e.,
\begin{equation*}
\psi(x)=\sqrt{|\xi|}[J_{\frac{1}{3}}(\frac{2}{3}|\xi|^{3 / 2})+J_{-\frac{1}{3}}(\frac{2}{3}|\xi|^{3 / 2})] \tag{12}
\end{equation*}
The energy level $E$ is determined by the boundary condition $\psi(0)=0$, which implies
\begin{equation*}
J_{\frac{1}{3}}(\frac{2}{3} \lambda^{3 / 2})+J_{-\frac{1}{3}}(\frac{2}{3} \lambda^{3 / 2})=0 \tag{13}
\end{equation*}
Solutions can be found in Bessel function tables, yielding
\lambda=2.3381,4.0880,5.5206,6.7867,7.9441, \cdots
The relationship between the energy eigenvalue and $\lambda$ is given by equation (10). The ground state energy level is
\begin{equation*}
E_{1}=1.8558(\frac{\hbar^{2} F^{2}}{m})^{1 / 3} \tag{14}
\end{equation*}
|
{"$m$": "mass of the particle", "$V$": "potential energy", "$x$": "position", "$F$": "force", "$H$": "total energy operator", "$p$": "momentum", "$E$": "energy", "$\\hbar$": "reduced Planck's constant", "$\\varphi$": "wave function in the momentum representation", "$A$": "normalization constant", "$\\psi$": "wave function in the position representation", "$C$": "constant related to normalization", "$u$": "substitution variable", "$\\xi$": "scaled position variable", "$l$": "characteristic length", "$\\lambda$": "dimensionless parameter related to energy"}
|
Schrödinger equation one-dimensional motion
| |
43 |
Theoretical Foundations
|
Calculate the transmission probability of a particle (energy $E>0$) through a $\delta$ potential barrier $V(x)=V_{0} \delta(x)$ in the momentum representation.
|
[
"\\frac{1}{1+(m V_{0} / \\hbar^{2} k)^{2}}"
] |
Expression
|
The stationary Schrödinger equation in the $x$ representation is
\begin{equation*}
\psi^{\prime \prime}+k^{2} \psi-\frac{2 m V_{0}}{\hbar^{2}} \delta(x) \psi=0, \quad k=\sqrt{2 m E} / \hbar \tag{1}
\end{equation*}
Let
\begin{equation*}
\psi(x)=(2 \pi \hbar)^{-1 / 2} \int_{-\infty}^{+\infty} \mathrm{d} p \varphi(p) \mathrm{e}^{\mathrm{i} p x / \hbar} \tag{2}
\end{equation*}
where $\varphi(p)$ is the wave function in the momentum representation, which should satisfy the equation
\begin{equation*}
\frac{p^{2}}{2 m} \varphi(p)+\int_{-\infty}^{+\infty} V_{p p^{\prime}} \varphi(p^{\prime}) \mathrm{d} p^{\prime}=E \varphi(p) \tag{3}
\end{equation*}
where
\begin{equation*}
V_{p p^{\prime}}=\frac{1}{2 \pi \hbar} \int_{-\infty}^{+\infty} \mathrm{d} x V_{0} \delta(x) \mathrm{e}^{\mathrm{i}(p^{\prime}-p) x / \hbar}=\frac{V_{0}}{2 \pi \hbar} \tag{4}
\end{equation*}
Thus, using equations (4) and (2), we can solve
\begin{equation*}
\int_{-\infty}^{+\infty} V_{p p^{\prime}} \varphi(p^{\prime}) \mathrm{d} p^{\prime}=\frac{V_{0}}{2 \pi \hbar} \int_{-\infty}^{+\infty} \varphi(p^{\prime}) \mathrm{d} p^{\prime}=\frac{V_{0}}{\sqrt{2 \pi \hbar}} \psi(0) \tag{5}
\end{equation*}
Substituting equations (4) and (5) into equation (3), we obtain
\begin{equation*}
(p^{2}-\hbar^{2} k^{2}) \varphi(p)+\frac{2 m V_{0}}{\sqrt{2 \pi \hbar}} \psi(0)=0 \tag{6}
\end{equation*}
According to the fundamental formula of the $\delta$ function
\begin{equation*}
(\xi-\xi_{0}) \delta(\xi-\xi_{0})=0 \tag{7}
\end{equation*}
The general solution of equation (6) is
\begin{equation*}
\varphi(p)=C_{1} \delta(p-\hbar k)+C_{2} \delta(p+\hbar k)-\frac{2 m V_{0}}{\sqrt{2 \pi \hbar}} \frac{\psi(0)}{(p^{2}-\hbar^{2} k^{2})} \tag{8}
\end{equation*}
where $C_{1}, ~ C_{2}, ~ \psi(0)$ are to be determined. Substituting equation (8) into equation (2), we get the wave function in the $x$ representation
\begin{equation*}
\psi(x)=\frac{C_{1}}{\sqrt{2 \pi \hbar}} \mathrm{e}^{\mathrm{i} k x}+\frac{C_{2}}{\sqrt{2 \pi \hbar}} \mathrm{e}^{-\mathrm{i} k x}-\frac{2 m V_{0}}{2 \pi \hbar} \psi(0) \int_{-\infty}^{+\infty} \frac{\mathrm{d} p}{p^{2}-\hbar^{2} k^{2}} \mathrm{e}^{\mathrm{i} p x / \hbar} \tag{9}
\end{equation*}
The last integral in equation (9) should be taken as the principal value, which can be calculated using the contour integral method in the complex $p$ plane, resulting in
\int_{-\infty}^{+\infty} \frac{\mathrm{d} p}{p^{2}-\hbar^{2} k^{2}} \mathrm{e}^{\mathrm{i} k x / \hbar}= \begin{cases}\frac{\mathrm{i} \pi}{2 \hbar k}(\mathrm{e}^{\mathrm{i} k x}-\mathrm{e}^{-\mathrm{i} k x}), & x>0 \tag{10}\\ \frac{i \pi}{2 \hbar k}(\mathrm{e}^{-\mathrm{i} k x}-\mathrm{e}^{\mathrm{i} k x}), & x<0\end{cases}
When $x \rightarrow 0$, the right side of equation (10) becomes 0, and from equation (9) we get
\begin{equation*}
\psi(0)=(C_{1}+C_{2}) / \sqrt{2 \pi \hbar} \tag{11}
\end{equation*}
Substituting equations (10) and (11) into equation (9), we get
\sqrt{2 \pi \hbar} \psi(x)=C_{1} \mathrm{e}^{\mathrm{i} k x}+C_{2} \mathrm{e}^{-\mathrm{i} k x}-\frac{\mathrm{i} m V_{0}}{2 \hbar^{2} k}(C_{1}+C_{2})(\mathrm{e}^{\mathrm{i} k x}-\mathrm{e}^{-\mathrm{i} k x}), \quad x>0
Given that the incident wave is $\mathrm{e}^{\mathrm{i} k x}$ (i.e., the incident momentum $p=\hbar k$), in the region $x>0$ there should only be the transmitted wave, i.e., the $\mathrm{e}^{\mathrm{i} k x}$ term, and no $\mathrm{e}^{-\mathrm{i} k}$ term. Therefore, $C_{1}, ~ C_{2}$ must satisfy the following relation:
\begin{equation*}
C_{2}=-\mathrm{i} \frac{m V_{0}}{2 \hbar^{2} k}(C_{1}+C_{2}) \tag{12}
\end{equation*}
Thus,
\begin{equation*}
\psi(x)=\frac{C_{1}+C_{2}}{\sqrt{2 \pi \hbar}} \mathrm{e}^{\mathrm{i} k x}, \quad x>0 \tag{13}
\end{equation*}
Similarly, we can get
\begin{equation*}
\psi(x)=\frac{1}{\sqrt{2 \pi \hbar}}[(C_{1}-C_{2}) \mathrm{e}^{i k x}+2 C_{2} \mathrm{e}^{-\mathrm{i} k x}], \quad x<0 \tag{14}
\end{equation*}
where the $\mathrm{e}^{\mathrm{ik} x}$ term is the incident wave, and the $\mathrm{e}^{-\mathrm{i} k x}$ term is the reflected wave. If the incident wave amplitude is set to 1, then equation (14) can be written as
\begin{equation*}
\psi(x)=\mathrm{e}^{\mathrm{i} k \tau}+R \mathrm{e}^{-\mathrm{i} k x} \tag{14'}
\end{equation*}
Then we should take
\begin{equation*}
C_{1}-C_{2}=\sqrt{2 \pi \hbar} \tag{15}
\end{equation*}
From equations (13) and (14), it is easy to see that
$$\text { Transmission coefficient }=|\frac{C_{1}+C_{2}}{C_{1}-C_{2}}|^{2}=\frac{1}{|1-\frac{2 C_{2}}{C_{1}+C_{2}}|^{2}}$$
Using equation (12), we obtain
\begin{equation*}
\text { Transmission coefficient }=\frac{1}{|1+\mathrm{i} m V_{0} / \hbar^{2} k|^{2}}=\frac{1}{1+(m V_{0} / \hbar^{2} k)^{2}} \tag{16}
\end{equation*}
|
{"$E$": "energy of the particle", "$m$": "mass of the particle", "$V_{0}$": "strength of the delta potential barrier", "$p$": "momentum", "$\\hbar$": "reduced Planck's constant", "$k$": "wave number related to the particle's energy", "$C_{1}$": "coefficient related to the incident wave", "$C_{2}$": "coefficient related to the reflected wave"}
|
Schrödinger equation one-dimensional motion
| |
44 |
Theoretical Foundations
|
The energy operator of a one-dimensional harmonic oscillator is
\begin{equation*}
H=\frac{p_{x}^{2}}{2 \mu}+\frac{1}{2} \mu \omega^{2} x^{2} \tag{1}
\end{equation*}
Try to derive its energy level expression using the Heisenberg equation of motion for operators and the fundamental commutation relation.
|
[
"E_{n}=(n+\\frac{1}{2}) \\hbar \\omega"
] |
Expression
|
Using the Heisenberg equation of motion, we obtain
\begin{equation*}
\frac{\mathrm{d} p}{\mathrm{~d} t}=\frac{1}{\mathrm{i} \hbar}[p, H]=\frac{\mu \omega^{2}}{2 \mathrm{i} \hbar}[p, x^{2}]=-\mu \omega^{2} x \tag{2}
\end{equation*}
Take the matrix element in the energy representation, yielding
\begin{equation*}
\mathrm{i} \omega_{k n} p_{k n}=-\mu \omega^{2} x_{k n} \tag{3}
\end{equation*}
From the previous result,
\begin{equation*}
p_{k n}=\mathrm{i} \omega_{k n} \mu x_{k n} \tag{4}
\end{equation*}
Combining equations (3) and (4), we obtain
\begin{equation*}
(\omega^{2}-\omega_{k n}^{2}) x_{k n}=0 \tag{5}
\end{equation*}
where $k, ~ n$ can be understood as quantum state indices. From equation (5) it is evident
\begin{array}{ll}
\text { If } \omega_{k n} \neq \pm \omega, & \text { then } x_{k n}=0 \tag{6}\\
\text { If } x_{k n} \neq 0, \quad \text { then } \omega_{k n}= \pm \omega
\end{array}
Since
\begin{equation*}
(x^{2})_{k k}=\sum_{n} x_{k n} x_{n k}=\sum_{n}|x_{n k}|^{2}>0 \tag{7}
\end{equation*}
For any chosen energy level $E_{k}$, there must exist some $n$ such that $x_{n k} \neq 0$, then from equation (6), the energy level difference $E_{n}-$ $E_{k}= \pm \hbar \omega$, that is, given any energy level, there must exist another energy level differing by $\hbar \omega$. Therefore, all energy levels are
$$ E=E_{0}, \quad E_{0}+\hbar \omega, \quad E_{0}+2 \hbar \omega, \cdots $$
That is
\begin{equation*}
E_{n}=E_{0}+n \hbar \omega, \quad n=0,1,2, \cdots \tag{8}
\end{equation*}
To find the ground state energy $E_{0}$, the equation proven in the previous problem can be used
\begin{equation*}
\sum_{n}(E_{n}-E_{k})|x_{n k}|^{2}=\hbar^{2} / 2 \mu \tag{9}
\end{equation*}
Taking $k$ as the ground state, from equations (6), (7), and (9), we get
$$\frac{\hbar^{2}}{2 \mu}=\hbar \omega \sum_{n}|x_{n 0}|^{2}=\hbar \omega(x^{2})_{00}$$
Thus, the average potential energy of the ground state is
\begin{equation*}
\frac{1}{2} \mu \omega^{2}(x^{2})_{00}=\frac{1}{4} \hbar \omega \tag{10}
\end{equation*}
According to the virial theorem, we have
\langle T\rangle_{0}=\langle V\rangle_{0}=\frac{1}{2} E_{0}
Comparing with equation (10), we obtain
\begin{equation*}
E_{0}=\frac{1}{2} \hbar \omega \tag{11}
\end{equation*}
Substituting into equation (8), we get the energy level formula
\begin{equation*}
E_{n}=(n+\frac{1}{2}) \hbar \omega, \quad n=0,1,2, \cdots \tag{$\prime$}
\end{equation*}
Considering equation (6), equation (9) gives
\begin{equation*}
|x_{k+1, k}|^{2}-|x_{k-1, k}|^{2}=\frac{\hbar}{2 \mu \omega} \tag{12}
\end{equation*}
By appropriately choosing the phase factor $(\mathrm{e}^{\mathrm{i}})$ of each energy eigenfunction, all $x_{n k}$ can be made non-negative real numbers, and equation (12) can be rewritten as (changing $k$ to $n$)
\begin{equation*}
(x_{n+1, n})^{2}-(x_{n, n-1})^{2}=\frac{\hbar}{2 \mu \omega} \tag{$\prime$}
\end{equation*}
When $n=0$, the equation yields
\begin{equation*}
(x_{10})^{2}=\frac{\hbar}{2 \mu \omega}, \quad x_{10}=\sqrt{\frac{\hbar}{2 \mu \omega}} \tag{13}
\end{equation*}
By repeatedly using equation (12＇), we get
\begin{equation*}
x_{n+1, n}=\sqrt{\frac{n+1}{2} \frac{\hbar}{\mu \omega}}, \quad n=0,1,2, \cdots \tag{14}
\end{equation*}
By also using equation (4), we get
\begin{equation*}
p_{n+1, n}=\mathrm{i} \omega \mu x_{n+1, n} \tag{15}
\end{equation*}
Note: The matrix elements of $x$ are real numbers, and the matrix elements of $p$ are purely imaginary numbers. Therefore
\begin{align*}
& x_{n, n+1}=(x_{n+1, n})^{*}=x_{n+1, n} \tag{16}\\
& p_{n, n+1}=(p_{n+1, n})^{*}=-p_{n+1, n} \tag{17}
\end{align*}
|
{"$H$": "Hamiltonian or energy operator", "$p_{x}$": "momentum in the x-direction", "$\\mu$": "reduced mass", "$\\omega$": "angular frequency", "$x$": "position", "$\\hbar$": "reduced Planck's constant", "$p$": "momentum operator", "$t$": "time", "$k$": "quantum state index", "$n$": "quantum state index", "$E_{k}$": "energy of state k", "$E_{n}$": "energy of state n", "$E_{0}$": "ground state energy", "$x_{k n}$": "position matrix element between states k and n", "$\\omega_{k n}$": "angular frequency difference between states k and n", "$x_{n 0}$": "position matrix element between state n and ground state", "$(x^{2})_{00}$": "expectation value of position squared in the ground state", "$\\langle T \\rangle_{0}$": "average kinetic energy in the ground state", "$\\langle V \\rangle_{0}$": "average potential energy in the ground state", "$|x_{k+1, k}|$": "magnitude of position matrix element between states k+1 and k", "$|x_{k-1, k}|$": "magnitude of position matrix element between states k-1 and k", "$x_{n+1, n}$": "position matrix element between states n+1 and n", "$x_{10}$": "position matrix element between first excited state and ground state", "$p_{n+1, n}$": "momentum matrix element between states n+1 and n"}
|
Schrödinger equation one-dimensional motion
| |
45 |
Theoretical Foundations
|
The Hamiltonian of a one-dimensional harmonic oscillator is $H=\frac{p^{2}}{2 m}+\frac{1}{2} m \omega^{2} x^{2}$. Compute $[x(t_1),p(t_2)]$ in the Heisenberg picture.
|
[
"\\mathrm{i} \\hbar \\cos \\omega(t_{2}-t_{1})"
] |
Expression
|
It is easy to obtain
\begin{align*}
{[x(t_{1}), x(t_{2})] } & =[x, p] \frac{1}{m \omega}(\cos \omega t_{1} \sin \omega t_{2}-\sin \omega t_{1} \cos \omega t_{2}) \\
& =\frac{i \hbar}{m \omega} \sin \omega(t_{2}-t_{1}). \tag{1}
\end{align*}
Similarly, it can be obtained
\begin{align*}
& {[p(t_{1}), p(t_{2})]=\mathrm{i} m \omega \hbar \sin \omega(t_{2}-t_{1})} \tag{2}\\
& {[x(t_{1}), p(t_{2})]=\mathrm{i} \hbar \cos \omega(t_{2}-t_{1})} \tag{3}
\end{align*}
In equation (3), when $t_{1}=t_{2}=t$, we obtain
\begin{equation*}
[x(t), p(t)]=\mathrm{i} \hbar \tag{4}
\end{equation*}
This commutation relation corresponds to $[x, p]=\mathrm{i} \hbar$ in the Schrödinger picture, which applies not only to the harmonic oscillator problem but also to any other problem. As for equations (1), (2), and (3), they are specific to the harmonic oscillator.
|
{"$H$": "Hamiltonian", "$p$": "momentum", "$m$": "mass", "$\\omega$": "angular frequency", "$x$": "position", "$t_1$": "time point 1", "$t_2$": "time point 2", "$\\mathrm{i}$": "imaginary unit", "$\\hbar$": "reduced Planck's constant"}
|
Schrödinger equation one-dimensional motion
| |
46 |
Theoretical Foundations
|
Two localized non-identical particles with spin $1 / 2$ (ignoring orbital motion) have an interaction energy given by
$$H=A \boldsymbol{s}_{1} \cdot \boldsymbol{s}_{2} $$
where $\boldsymbol{s}_{1}, \boldsymbol{s}_{2}$ are spin operators (the eigenvalues of their $z$ components $s_{iz}$ being $\pm 1/2$), and $A$ is a constant related to energy.
At time $t=0$, particle 1 has spin "up" (i.e., the measured value of $s_{1z}$ is $1/2$), and particle 2 has spin "down" (i.e., the measured value of $s_{2z}$ is $-1/2$). Determine the probability that at any time $t>0$, particle 1 has spin "up" (i.e., measuring $s_{1z}$ yields $1/2$) in the Heisenberg picture.
|
[
"w(t)=\\cos ^{2} \\frac{A t}{2}"
] |
Expression
|
Solution 1: Starting from the Heisenberg equation of motion for the spin operators. It is easily observed from the construction of $H$ that the total spin $\boldsymbol{S}$ commutes with $H$ and is a conserved quantity, hence the values and corresponding probabilities of the components of total $\boldsymbol{S}$ remain constant, and the distribution probability of $\boldsymbol{S}^{2}$ also does not change. At $t=0$, the spin state of the system $\alpha(1) \beta(2)$ is a common eigenstate of $s_{1 z}$ and $s_{2 z}$, and thus also an eigenstate of total $S_{z}$ with eigenvalue $S_{z}=0$. As $S_{z}$ is a conserved quantity, it can only have the eigenvalue 0 at any time point, and cannot take other eigenvalues. Hence, the probability that both particle 1 and 2 have spin "up" $(s_{1 z}=s_{2 z}=\frac{1}{2}, S_{z}=1)$ is 0. At $t=0$, the total spin quantum numbers $S=1, ~ 0$, each with probabilities $1 / 2$. Since $\boldsymbol{S}^{2}$ is conserved, this probability does not change over time. The above conclusions apply to parts (b) and (c).
In order to calculate $\langle\boldsymbol{s}_{1}\rangle$ and $\langle\boldsymbol{s}_{2}\rangle$, the Heisenberg equation of motion can be solved.
\begin{gather*}
\frac{\mathrm{d}}{\mathrm{~d} t} \boldsymbol{s}_{1}=\frac{1}{\mathrm{i} \hbar}[\boldsymbol{s}_{1}, H]=\mathrm{i} A[\boldsymbol{s}_{1} \cdot \boldsymbol{s}_{2}, \boldsymbol{s}_{1}]=-\boldsymbol{A} \boldsymbol{s}_{1} \times \boldsymbol{s}_{2} \tag{1}\\
\frac{\mathrm{~d}}{\mathrm{~d} t} \boldsymbol{s}_{2}=\frac{1}{\mathrm{i} \hbar}[\boldsymbol{s}_{2}, H]=\mathrm{i} A[\boldsymbol{s}_{1} \cdot \boldsymbol{s}_{2}, \boldsymbol{s}_{2}]=\boldsymbol{A} \boldsymbol{s}_{1} \times \boldsymbol{s}_{2} \tag{2}
\end{gather*}
Adding and subtracting the two equations gives
\begin{align*}
& \frac{\mathrm{d}}{\mathrm{~d} t}(\boldsymbol{s}_{1}+\boldsymbol{s}_{2})=\frac{\mathrm{d}}{\mathrm{~d} t} \boldsymbol{S}=0 \tag{3}\\
& \frac{\mathrm{~d}}{\mathrm{~d} t}(\boldsymbol{s}_{1}-\boldsymbol{s}_{2})=-2 A \boldsymbol{s}_{1} \times \boldsymbol{s}_{2} \tag{4}
\end{align*}
Multiplying equation (2) by $s_{1}$ and equation (1) by $s_{2}$ gives
\begin{align*}
\frac{\mathrm{d}}{\mathrm{~d} t}(\boldsymbol{s}_{1} \times \boldsymbol{s}_{2}) & =A[s_{1} \times(s_{1} \times s_{2})-(s_{1} \times s_{2}) \times s_{2}] \\
& =\frac{A}{2}(s_{1}-s_{2}) \tag{5}
\end{align*}
Multiply equation (5) by $2i$, and add it to equation (4) to get
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{~d} t}(s_{1}-s_{2}+2 \mathrm{i} s_{1} \times s_{2})=\mathrm{i} A(s_{1}-s_{2}+2 \mathrm{i} s_{1} \times s_{2}) \tag{6}
\end{equation*}
Take the average value and integrate with respect to $t$, then
\begin{equation*}
\langle\boldsymbol{s}_{1}-\boldsymbol{s}_{2}+2 \mathrm{i} \boldsymbol{s}_{1} \times \boldsymbol{s}_{2}\rangle_{t}=\langle\boldsymbol{s}_{1}-\boldsymbol{s}_{2}+2 \mathrm{i} \mathbf{s}_{1} \times \boldsymbol{s}_{2}\rangle_{t=0} \mathrm{e}^{\mathrm{i} A t} \tag{7}
\end{equation*}
The initial condition of the problem is
\begin{equation*}
\chi(t=0)=\alpha(1) \beta(2) \tag{8}
\end{equation*}
Therefore,
\begin{array}{l}
\langle\boldsymbol{s}_{1}\rangle_{t=0}=(0,0, \frac{1}{2}), \quad\langle\boldsymbol{s}_{2}\rangle_{t=0}=(0,0,-\frac{1}{2}) \tag{$\prime$}\\
\langle\boldsymbol{s}_{1}+\boldsymbol{s}_{2}\rangle_{t=0}=0, \quad\langle\boldsymbol{s}_{1}-\boldsymbol{s}_{2}\rangle_{t=0}=(0,0,1)=\boldsymbol{e}_{3} \\
\langle\boldsymbol{s}_{1} \times \boldsymbol{s}_{2}\rangle_{t=0}=\langle\boldsymbol{s}_{1}\rangle_{t=0} \times\langle\boldsymbol{s}_{2}\rangle_{t=0}=0
\end{array}}
Substituting into equation (7), we get
\begin{equation*}
\langle\boldsymbol{s}_{1}-\boldsymbol{s}_{2}\rangle_{t}+2 \mathrm{i}\langle\boldsymbol{s}_{1} \times \boldsymbol{s}_{2}\rangle_{t}=\mathrm{e}^{\mathrm{i} A t} \boldsymbol{e}_{3}=(\cos A t+\mathrm{i} \sin A t) \boldsymbol{e}_{3} \tag{9}
\end{equation*}
Since $s_{1}, s_{2}, s_{1} \times s_{2}$ are all Hermitian operators, their averages are real numbers, so
\begin{gather*}
\langle\boldsymbol{s}_{1}-\boldsymbol{s}_{2}\rangle_{t}=\boldsymbol{e}_{3} \cos A t \tag{10}\\
\langle\boldsymbol{s}_{1} \times \boldsymbol{s}_{2}\rangle_{t}=\frac{1}{2} \boldsymbol{e}_{3} \sin A t \tag{11}
\end{gather*}
Equation (3) shows that $\boldsymbol{S}$ is conserved, and from the initial condition ($8^{\prime}$):
\begin{equation*}
\langle\boldsymbol{s}_{1}+\boldsymbol{s}_{2}\rangle_{t}=\langle\boldsymbol{s}_{1}+\boldsymbol{s}_{2}\rangle_{t=0}=0 \tag{12}
\end{equation*}
Adding and subtracting equations (10) and (12), we get
\begin{equation*}
\langle\boldsymbol{s}_{1}\rangle_{t}=-\langle\boldsymbol{s}_{2}\rangle_{t}=\frac{1}{2} e_{3} \cos A t \tag{11}
\end{equation*}
That is:
\begin{align*}
& \langle s_{1 x}\rangle_{t}=\langle s_{1 y}\rangle_{t}=\langle s_{2 x}\rangle_{t}=\langle s_{2 y}\rangle_{t}=0 \\
& \langle s_{1 z}\rangle_{t}=-\langle s_{2 z}\rangle_{t}=\frac{1}{2} \cos A t \tag{13'}
\end{align*}
This result is consistent with equation (5) obtained in Exercise 6.47. Equation (11) corresponds to
\begin{align*}
& \langle s_{1 x} s_{2 y}-s_{1 y} s_{2 x}\rangle_{t}=\frac{1}{2} \sin A t \\
& \langle s_{1 y} s_{2 z}-s_{1 z} s_{2 y}\rangle_{t}=0 \tag{11'}\\
& \langle s_{1 z} s_{2 x}-s_{1 x} s_{2 z}\rangle_{t}=0
\end{align*}
Readers can easily use equation ($4^{\prime}$) obtained in Exercise 6.47 to verify this conclusion, although it is not easy to see this result there.
Assume that at $t>0$, the probability of particle 1 having spin "up" $(s_{1 z}=\frac{1}{2})$ is $w(t)$, then
\langle s_{1 z}\rangle_{t}=\frac{1}{2} \cos A t=\frac{1}{2} w(t)-\frac{1}{2}[1-w(t)]=w(t)-\frac{1}{2}
Thus,
\begin{equation*}
w(t)=\frac{1}{2}(1+\cos A t)=\cos ^{2} \frac{A t}{2} \tag{1}
\end{equation*}
Solution 2: Using the Heisenberg picture of the average value formula
\begin{equation*}
\langle s_{1}\rangle_{t}=\langle\chi(0)| s_{1}(t)|\chi(0)\rangle=\langle\chi(0)| U^{+}(t) s_{1} U(t)|\chi(0)\rangle \tag{15}
\end{equation*}
where
\begin{array}{c}\nu(t)=\mathrm{e}^{-\mathrm{i} H t}=\mathrm{e}^{-\mathrm{i} A s_{1} \cdot s_{2}} \tag{16}\\
U^{+}(t)=\mathrm{e}^{\mathrm{i} H t}=\mathrm{e}^{\mathrm{i} t s_{1} \cdot s_{2}}
\end{array}}
$s_{1}(t)$ can be expressed as (let $\lambda=A t)$
\begin{align*}
\boldsymbol{s}_{1}(t) & =U^{+}(t) \boldsymbol{s}_{1} U(t) \\
& =\frac{1}{2}(\boldsymbol{s}_{1}+\boldsymbol{s}_{2})+\frac{1}{2}(\boldsymbol{s}_{1}-\boldsymbol{s}_{2}) \cos A t-\boldsymbol{s}_{1} \times \boldsymbol{s}_{2} \sin A t \tag{17}
\end{align*}
Substituting into equation (15), and using the initial averages
\begin{align*}
\langle\boldsymbol{s}_{1}\rangle_{t=0} & =\langle\chi(0)| \boldsymbol{s}_{1}|\chi(0)\rangle=\langle\alpha(1)| \boldsymbol{s}_{1}|\alpha(1)\rangle \\
& =(0,0, \frac{1}{2}) \\
\langle\boldsymbol{s}_{2}\rangle_{t=0} & =\langle\chi(0)| \boldsymbol{s}_{2}|\chi(0)\rangle=\langle\beta(2)| \boldsymbol{s}_{2}|\beta(2)\rangle \\
& =(0,0,-\frac{1}{2}) \\
& \langle\boldsymbol{s}_{1} \times \boldsymbol{s}_{2}\rangle_{t=0}=\langle\boldsymbol{s}_{1}\rangle_{t=0} \times\langle\boldsymbol{s}_{2}\rangle_{t=0}=0 \tag{18}
\end{align*}
we get
\begin{align*}
\langle\boldsymbol{s}_{1}\rangle_{t} & =\frac{1}{2}\langle\boldsymbol{s}_{1}+\boldsymbol{s}_{2}\rangle_{t=0}+\frac{1}{2}\langle\boldsymbol{s}_{1}-\boldsymbol{s}_{2}\rangle_{t=0} \cos A t-\langle\boldsymbol{s}_{1} \times \boldsymbol{s}_{2}\rangle_{t=0} \sin A t \\
& =\frac{1}{2} \boldsymbol{e}_{3} \cos A t \tag{19}
\end{align*}
The total spin $\boldsymbol{S}=\boldsymbol{s}_{1}+\boldsymbol{s}_{2}$ commutes with both $\boldsymbol{s}_{1} \cdot \boldsymbol{s}_{2}$ and thus also with $U$ and $U^{+}$, so
\begin{gather*}
\boldsymbol{S}(t)=\boldsymbol{S}=\boldsymbol{s}_{1}+\boldsymbol{s}_{2} \tag{20}\\
\langle\boldsymbol{S}\rangle_{t}=\langle\boldsymbol{s}_{1}+\boldsymbol{s}_{2}\rangle_{t}=\langle\boldsymbol{s}_{1}+\boldsymbol{s}_{2}\rangle_{t=0}=0 \tag{21}
\end{gather*}
So
\begin{equation*}
\langle\boldsymbol{s}_{2}\rangle_{t}=\langle\boldsymbol{S}\rangle_{t}-\langle\boldsymbol{s}_{1}\rangle_{t}=-\langle\boldsymbol{s}_{1}\rangle_{t}=-\frac{1}{2} \boldsymbol{e}_{3} \cos A t \tag{22}
\end{equation*}
All these results match those of Solution 1. The method to calculate the probability is the same as Solution 1 and thus the result is also the same, omitted here.
|
{"$A$": "constant related to energy", "$t$": "time", "$\\boldsymbol{s}_{1}$": "spin operator for particle 1", "$\\boldsymbol{s}_{2}$": "spin operator for particle 2", "$s_{1z}$": "z-component of spin operator for particle 1", "$s_{2z}$": "z-component of spin operator for particle 2", "$\\boldsymbol{S}$": "total spin operator"}
|
Schrödinger equation one-dimensional motion
| |
47 |
Theoretical Foundations
|
For a spin $1/2$ particle, $\langle\boldsymbol{\sigma}\rangle$ is often called the polarization vector, denoted as $\boldsymbol{P}$. It is also the spatial orientation of the spin angular momentum. Assume the particle is localized and subject to a magnetic field $\boldsymbol{B}(t)$ along the $z$ direction with time-varying intensity, with the potential given by
$$H=-\mu_{0} \boldsymbol{\sigma} \cdot \boldsymbol{B}(t)=-\mu_{0} \sigma_{z} B(t),$$
In the Heisenberg picture, find the time evolution of the polarization vector, i.e., find $P_x(t)=\langle{\sigma}_x\rangle_{t}$. Let $\boldsymbol{P}(t=0)$ point in the direction $(\theta_{0}, \varphi_{0})$, where $\theta_{0}=2 \delta, \varphi_{0}=2 \alpha$.
|
[
"\\langle\\sigma_{x}\\rangle_{t}=\\sin 2 \\delta \\cos \\varphi(t)"
] |
Expression
|
According to the Heisenberg equation of motion.
$$\frac{\mathrm{d}}{\mathrm{~d} t} \sigma_{z}=\frac{1}{\mathrm{i} \hbar}[\sigma_{z}, H]=0,$$
Therefore,
\begin{equation*}
\langle\sigma_{z}\rangle_{t}=\langle\sigma_{z}\rangle_{t=0}=\cos \theta_{0}=\cos 2 \delta \tag{1}
\end{equation*}
Moreover,
\begin{align*}
& \frac{\mathrm{d}}{\mathrm{~d} t} \sigma_{x}=\frac{1}{\mathrm{i} \hbar}[\sigma_{x}, H]=-\frac{\mu_{0}}{\mathrm{i} \hbar} B(t)[\sigma_{x}, \sigma_{x}]=\frac{2 \mu_{0}}{\hbar} B(t)_{\sigma_{y}} \tag{2}\\
& \frac{\mathrm{~d}}{\mathrm{~d} t} \sigma_{y}=\frac{1}{\mathrm{i} \hbar}[\sigma_{y}, H]=-\frac{2 \mu_{0}}{\hbar} B(t) \sigma_{x}
\end{align*}
Combining the two equations, it is easy to obtain
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{~d} t}(\sigma_{x}+\mathrm{i} \sigma_{y})=-\mathrm{i} \frac{2 \mu_{0}}{\hbar} B(t)(\sigma_{x}+\mathrm{i} \sigma_{y}) \tag{3}
\end{equation*}
Taking the average value and integrating over $t$, we have
\begin{equation*}
\langle\sigma_{x}+\mathrm{i} \sigma_{y}\rangle_{t}=\langle\sigma_{x}+\mathrm{i} \sigma_{y}\rangle_{t=0} \exp [-\mathrm{i} \frac{2 \mu_{0}}{\hbar} \int_{0}^{t} B(\tau) \mathrm{d} \tau] \tag{4}
\end{equation*}
Since $\langle\sigma_{x}\rangle$, $\langle\sigma_{y}\rangle$ are real numbers, separate the real and imaginary parts in the above equation to get
\begin{align*}
& \langle\sigma_{x}\rangle_{t}=\langle\sigma_{x}\rangle_{t=0} \cos [\frac{2 \mu_{0}}{\hbar} \int_{0}^{t} B(\tau) \mathrm{d} \tau]+\langle\sigma_{y}\rangle_{t=0} \sin [\frac{2 \mu_{0}}{\hbar} \int_{0}^{t} B(\tau) \mathrm{d} \tau] \tag{5}\\
& \langle\sigma_{y}\rangle_{t}=\langle\sigma_{y}\rangle_{t=0} \cos [\frac{2 \mu_{0}}{\hbar} \int_{0}^{t} B(\tau) \mathrm{d} \tau]-\langle\sigma_{x}\rangle_{t=0} \sin [\frac{2 \mu_{0}}{\hbar} \int_{0}^{t} B(\tau) \mathrm{d} \tau]
\end{align*}
The initial average values are
\begin{equation*}
\langle\sigma_{x}\rangle_{t=0}=\sin 2 \delta \cos 2 \alpha, \quad\langle\sigma_{y}\rangle_{t=0}=\sin 2 \delta \sin 2 \alpha \tag{6}
\end{equation*}
Substituting into equation (5) and defining $\varphi(t)$, we get
\begin{gather*}
\langle\sigma_{x}\rangle_{t}=\sin 2 \delta \cos \varphi(t), \quad\langle\sigma_{y}\rangle_{t}=\sin 2 \delta \sin \varphi(t) \tag{7}\\
\boldsymbol{P}(t)=\langle\boldsymbol{\sigma}\rangle_{t}=(\sin 2 \delta \cos \varphi(t), \sin 2 \delta \sin \varphi(t), \cos 2 \delta) \tag{8}
\end{gather*}
|
{"$\\sigma_{x}$": "Pauli spin matrix for x direction", "$\\sigma_{y}$": "Pauli spin matrix for y direction", "$\\sigma_{z}$": "Pauli spin matrix for z direction", "$\\langle\\sigma_{x}\\rangle_{t}$": "average value of sigma_x at time t", "$\\langle\\sigma_{y}\\rangle_{t}$": "average value of sigma_y at time t", "$\\langle\\sigma_{z}\\rangle_{t}$": "average value of sigma_z at time t", "$\\langle\\sigma_{x}\\rangle_{t=0}$": "initial average value of sigma_x", "$\\langle\\sigma_{y}\\rangle_{t=0}$": "initial average value of sigma_y", "$\\langle\\sigma_{z}\\rangle_{t=0}$": "initial average value of sigma_z", "$\\langle\\boldsymbol{\\sigma}\\rangle$": "polarization vector", "$\\boldsymbol{P}$": "polarization vector denoted in problem", "$\\boldsymbol{B}(t)$": "magnetic field at time t", "$\\theta_{0}$": "initial polar angle", "$\\varphi_{0}$": "initial azimuthal angle", "$\\delta$": "parameter related to initial polar angle", "$\\alpha$": "parameter related to initial azimuthal angle", "$\\mu_{0}$": "magnetic moment", "$\\hbar$": "reduced Planck constant"}
|
Schrödinger equation one-dimensional motion
| |
48 |
Others
|
Evaluate the function
$$ \langle 0| \phi(x) \phi(y)|0\rangle=D(x-y)=\int \frac{d^{3} p}{(2 \pi)^{3}} \frac{1}{2 E_{\mathbf{p}}} e^{-i p \cdot(x-y)} $$
for $(x-y)$ spacelike so that $(x-y)^{2}=-r^{2}$, explicitly in terms of Bessel functions.
\footnotetext{
$\ddagger$ With some additional work you can show that there are actually six conserved charges in the case of two complex fields, and $n(2 n-1)$ in the case of $n$ fields, corresponding to the generators of the rotation group in four and $2 n$ dimensions, respectively. The extra symmetries often do not survive when nonlinear interactions of the fields are included.
}
|
[
"D(x-y) = \\frac{m}{4 \\pi^{2} r} K_{1}(m r)"
] |
Expression
|
We evaluate the correlation function of a scalar field at two points,
\begin{equation*}
D(x-y)=\langle 0| \phi(x) \phi(y)|0\rangle \tag{2.28}
\end{equation*}
with $x-y$ being spacelike. Since any spacelike interval $x-y$ can be transformed to a form such that $x^{0}-y^{0}=0$, thus we will simply take:
\begin{equation*}
x^{0}-y^{0}=0, \quad \text { and } \quad|\mathbf{x}-\mathbf{y}|^{2}=r^{2}>0 . \tag{2.29}
\end{equation*}
Now:
\begin{align*}
D(x-y) & =\int \frac{\mathrm{d}^{3} p}{(2 \pi)^{3}} \frac{1}{2 E_{p}} e^{-i p \cdot(x-y)}=\int \frac{\mathrm{d}^{3} p}{(2 \pi)^{3}} \frac{1}{2 \sqrt{m^{2}+p^{2}}} e^{i \mathbf{p} \cdot(\mathbf{x}-\mathbf{y})} \\
& =\frac{1}{(2 \pi)^{3}} \int_{0}^{2 \pi} \mathrm{~d} \varphi \int_{-1}^{1} \mathrm{~d} \cos \theta \int_{0}^{\infty} \mathrm{d} p \frac{p^{2}}{2 \sqrt{m^{2}+p^{2}}} e^{i p r \cos \theta} \\
& =\frac{-\mathrm{i}}{2(2 \pi)^{2} r} \int_{-\infty}^{\infty} \mathrm{d} p \frac{p e^{\mathrm{i} p r}}{\sqrt{m^{2}+p^{2}}} \tag{2.30}
\end{align*}
Now we make the path deformation on $p$-complex plane, as is shown in Figure 2.3 of Peskin \& Daniel V. Schroeder. Then the integral becomes,
\begin{equation*}
D(x-y)=\frac{1}{4 \pi^{2} r} \int_{m}^{\infty} \mathrm{d} \rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^{2}-m^{2}}}=\frac{m}{4 \pi^{2} r} \mathrm{~K}_{1}(m r) . \tag{2.31}
\end{equation*}
|
{"$x$": "coordinate x", "$y$": "coordinate y", "$D(x-y)$": "correlation function of the scalar field", "$p$": "momentum", "$E_{\\mathbf{p}}$": "energy corresponding to momentum p", "$m$": "mass", "$\\phi(x)$": "scalar field at coordinate x", "$\\phi(y)$": "scalar field at coordinate y", "$r$": "spacelike interval", "$\\theta$": "polar angle", "$\\varphi$": "azimuthal angle", "$\\rho$": "dummy variable for integration", "$K_{1}$": "modified Bessel function of the second kind of order 1"}
|
The Klein-Gordon Field
|
The spacelike correlation function
|
49 |
Others
|
This problem concerns the discrete symmetries $P, C$, and $T$. Let $\phi(x)$ be a complex-valued Klein-Gordon field. The current associated with this field is $J^{\mu}=i(\phi^{*} \partial^{\mu} \phi-\partial^{\mu} \phi^{*} \phi)$. The Parity operator $P$ acts on the annihilation operator $a_{\mathbf{p}}$ as $P a_{\mathbf{p}} P=a_{-\mathbf{p}}$, and on the field as $P \phi(t, \mathbf{x}) P =\phi(t,-\mathbf{x})$. Determine the transformation property of the current $J^{\mu}(t, \mathbf{x})$ under the Parity transformation $P$. Express your answer in terms of $J^{\mu}(t,-\mathbf{x})$ and the factor $(-1)^{s(\mu)}$, where $s(\mu)=0$ for $\mu=0$ and $s(\mu)=1$ for $\mu=1,2,3$.
|
[
"P J^{\\mu}(t, \\mathbf{x}) P =(-1)^{s(\\mu)} J^{\\mu}(t,-\\mathbf{x})"
] |
Expression
|
Now we work out the $C, P$ and $T$ transformation properties of a scalar field $\phi$. Our starting point is
P a_{\mathbf{p}} P=a_{-\mathbf{p}}, \quad T a_{\mathbf{p}} T=a_{-\mathbf{p}}, \quad C a_{\mathbf{p}} C=b_{\mathbf{p}}
Then, for a complex scalar field
\begin{equation*}
\phi(x)=\int \frac{\mathrm{d}^{3} k}{(2 \pi)^{3}} \frac{1}{\sqrt{2 k^{0}}}[a_{\mathbf{k}} e^{-\mathrm{i} k \cdot x}+b_{\mathbf{k}}^{\dagger} e^{\mathrm{i} k \cdot x}], \tag{3.63}
\end{equation*}
we have
\begin{align*}
& P \phi(t, \mathbf{x}) P=\int \frac{\mathrm{d}^{3} k}{(2 \pi)^{3}} \frac{1}{\sqrt{2 k^{0}}}[a_{-\mathbf{k}} e^{-\mathrm{i}(k^{0} t-\mathbf{k} \cdot \mathbf{x})}+b_{-\mathbf{k}}^{\dagger} e^{\mathrm{i}(k^{0} t-\mathbf{k} \cdot \mathbf{x})}]=\phi(t,-\mathbf{x}) . \tag{3.64a}\\
& T \phi(t, \mathbf{x}) T=\int \frac{\mathrm{d}^{3} k}{(2 \pi)^{3}} \frac{1}{\sqrt{2 k^{0}}}[a_{-\mathbf{k}} e^{\mathrm{i}(k^{0} t-\mathbf{k} \cdot \mathbf{x})}+b_{-\mathbf{k}}^{\dagger} e^{-\mathrm{i}(k^{0} t-\mathbf{k} \cdot \mathbf{x})}]=\phi(-t, \mathbf{x}) . \tag{3.64b}\\
& C \phi(t, \mathbf{x}) C=\int \frac{\mathrm{d}^{3} k}{(2 \pi)^{3}} \frac{1}{\sqrt{2 k^{0}}}[b_{\mathbf{k}} e^{-\mathrm{i}(k^{0} t-\mathbf{k} \cdot \mathbf{x})}+a_{\mathbf{k}}^{\dagger} e^{\mathrm{i}(k^{0} t-\mathbf{k} \cdot \mathbf{x})}]=\phi^{*}(t, \mathbf{x}) . \tag{3.64c}
\end{align*}
As a consequence, we can deduce the $C, P$, and $T$ transformation properties of the current $J^{\mu}=\mathrm{i}(\phi^{*} \partial^{\mu} \phi-(\partial^{\mu} \phi^{*}) \phi)$, as follows:
\begin{align*}
P J^{\mu}(t, \mathbf{x}) P & =(-1)^{s(\mu)} \mathrm{i}[\phi^{*}(t,-\mathbf{x}) \partial^{\mu} \phi(t,-\mathbf{x})-(\partial^{\mu} \phi^{*}(t,-\mathbf{x})) \phi(t,-\mathbf{x})] \\
& =(-1)^{s(\mu)} J^{\mu}(t,-\mathbf{x}), \tag{3.65a}
\end{align*}
where $s(\mu)$ is the label for space-time indices that equals to 0 when $\mu=0$ and 1 when $\mu=1,2,3$. In the similar way, we have
\begin{align*}
& T J^{\mu}(t, \mathbf{x}) T=(-1)^{s(\mu)} J^{\mu}(-t, \mathbf{x}) \tag{3.65b}\\
& C J^{\mu}(t, \mathbf{x}) C=-J^{\mu}(t, \mathbf{x}) \tag{3.65c}
\end{align*}
One should be careful when playing with $T$ - it is antihermitian rather than hermitian, and anticommutes, rather than commutes, with $\sqrt{-1}$.
|
{"$\\phi(x)$": "complex-valued Klein-Gordon field", "$J^{\\mu}$": "current associated with the Klein-Gordon field", "$P$": "Parity operator", "$a_{\\mathbf{p}}$": "annihilation operator for momentum $\\mathbf{p}$", "$a_{-\\mathbf{p}}$": "annihilation operator for momentum $-\\mathbf{p}$", "$s(\\mu)$": "space-time index label function, 0 if $\\mu=0$ and 1 if $\\mu=1,2,3$"}
|
The Dirac Field
|
The discrete symmetries $P, C$ and $T$
|
50 |
Others
|
Let us return to the problem of the creation of Klein-Gordon particles by a classical source. This process can be described by the Hamiltonian
$$ H=H_{0}+\int d^{3} x(-j(t, \mathbf{x}) \phi(x))$$
where $H_{0}$ is the free Klein-Gordon Hamiltonian, $\phi(x)$ is the Klein-Gordon field, and $j(x)$ is a c-number scalar function. We found that, if the system is in the vacuum state before the source is turned on, the source will create a mean number of particles
$$ \langle N\rangle=\int \frac{d^{3} p}{(2 \pi)^{3}} \frac{1}{2 E_{\mathbf{p}}}|\tilde{\jmath}(p)|^{2} .$$
In this problem we will verify that statement, and extract more detailed information, by using a perturbation expansion in the strength of the source. Compute the probability that the source creates one particle of momentum $k$ to all orders in the source $j$ by summing the perturbation series.
|
[
"P(k) = |\\tilde{j}(k)|^{2} e^{-|\\tilde{j}(k)|}"
] |
Expression
|
The probability that the source creates one particle with momentum $\mathbf{k}$ is given by,
$$P(\mathbf{k})=|\langle\mathbf{k}| T \exp\{i \int \mathrm{~d}^{4} x j(x) \phi_{I}(x)\}| 0\rangle|^{2}. $$
Expanding the amplitude to the first order in \( j \), we get:
\begin{equation}
\begin{split}
P(k) &= \left| \langle \mathbf{k} | 0 \rangle + i \int \mathrm{d}^4x \, j(x) \int \frac{\mathrm{d}^3p}{(2\pi)^3} \frac{e^{i p \cdot x}}{\sqrt{2E_p}} \langle \mathbf{k} | a_p^\dagger | 0 \rangle + O(j^2) \right|^2 \\
&= \left| i \int \frac{\mathrm{d}^3p}{(2\pi)^3} \frac{\tilde{j}(p)}{\sqrt{2E_p}} \sqrt{2E_p (2\pi)^3} \, \delta(\mathbf{p} - \mathbf{k}) \right|^2 = |\tilde{j}(k)|^2 + O(j^3).
\end{split}
\end{equation}
If we go on to work out all the terms in the perturbation expansion, we get:
\begin{equation*}
P(k)=|\sum_{n} \frac{i(2 n+1)(2 n+1)(2 n-1) \cdots 3 \cdot 1}{(2 n+1)!} \tilde{j}^{n+1}(k)|^{2}=|\tilde{j}(k)|^{2} e^{-|\tilde{j}(k)|}
\end{equation*}
|
{"$\\mathbf{k}$": "momentum of the created particle", "$j$": "source function in spacetime", "$k$": "momentum", "$x$": "spacetime coordinate", "$\\phi_{I}(x)$": "interaction field at point x", "$0$": "vacuum state", "$P(k)$": "probability of creating a particle with momentum k", "$\\tilde{j}(k)$": "Fourier transform of the source function at momentum k", "$E_p$": "energy associated with momentum p", "$\\mathbf{p}$": "momentum variable in integration", "$a_p^\\dagger$": "creation operator for momentum p", "$n$": "term index in perturbation series"}
|
Interacting Fields and Feynman Diagrams
|
Scalar field with a classical source
|
51 |
Others
|
Decay of a scalar particle. Consider the following Lagrangian, involving two real scalar fields $\Phi$ and $\phi$ :
$$\mathcal{L}=\frac{1}{2}(\partial_{\mu} \Phi)^{2}-\frac{1}{2} M^{2} \Phi^{2}+\frac{1}{2}(\partial_{\mu} \phi)^{2}-\frac{1}{2} m^{2} \phi^{2}-\mu \Phi \phi \phi .$$
The last term is an interaction that allows a $\Phi$ particle to decay into two $\phi$ 's, provided that $M>2 m$. Assuming that this condition is met, calculate the lifetime of the $\Phi$ to lowest order in $\mu$.
|
[
"\\tau = \\frac{8\\pi M}{\\mu^2} (1 - \\frac{4m^2}{M^2})^{-1/2}"
] |
Expression
|
This problem is based on the following Lagrangian,
\begin{equation*}
\mathcal{L}=\frac{1}{2}(\partial_{\mu} \Phi)^{2}-\frac{1}{2} M^{2} \Phi^{2}+\frac{1}{2}(\partial_{\mu} \phi)^{2}-\frac{1}{2} m^{2} \phi^{2}-\mu \Phi \phi \phi . \tag{4.17}
\end{equation*}
When $M>2 m$, a $\Phi$ particle can decay into two $\phi$ particles. We want to calculate the lifetime of the $\Phi$ particle to lowest order in $\mu$.
The two-body decay rate is given in (4.86) of $\mathrm{P} \& \mathrm{~S}$,
\begin{equation*}
\int \mathrm{d} \Gamma=\frac{1}{2 M} \int \frac{\mathrm{~d}^{3} p_{1} \mathrm{~d}^{3} p_{2}}{(2 \pi)^{6}} \frac{1}{4 E_{\mathbf{p}_{1}} E_{\mathbf{p}_{2}}}|\mathcal{M}(\Phi(0) \rightarrow \phi(p_{1}) \phi(p_{2}))|^{2}(2 \pi)^{4} \delta^{(4)}(p_{\Phi}-p_{1}-p_{2}) . \tag{4.18}
\end{equation*}
To lowest order in $\mu$, the amplitude $\mathcal{M}$ is given by,
\begin{equation*}
i \mathcal{M}=-2 i \mu \tag{4.19}
\end{equation*}
The delta function in our case reads,
\begin{equation*}
\delta^{(4)}(p_{\Phi}-p_{1}-p_{2})=\delta(M-E_{\mathbf{p}_{1}}-E_{\mathbf{p}_{2}}) \delta^{(3)}(\mathbf{p}_{1}+\mathbf{p}_{2}), \tag{4.20}
\end{equation*}
thus,
\begin{equation*}
\Gamma=\frac{1}{2} \cdot \frac{2 \mu^{2}}{M} \int \frac{\mathrm{~d}^{3} p_{1} \mathrm{~d}^{3} p_{2}}{(2 \pi)^{6}} \frac{1}{4 E_{\mathbf{p}_{1}} E_{\mathbf{p}_{2}}}(2 \pi)^{4} \delta(M-E_{\mathbf{p}_{1}}-E_{\mathbf{p}_{2}}) \delta^{(3)}(\mathbf{p}_{1}+\mathbf{p}_{2}), \tag{4.21}
\end{equation*}
where an additional factor of $1 / 2$ takes account of two identical $\phi$ 's in final state. Furthermore, there are two mass-shell constraints,
\begin{equation*}
m^{2}+\mathbf{p}_{i}^{2}=E_{\mathbf{p}_{i}}^{2} . \quad(i=1,2) \tag{4.22}
\end{equation*}
Hence,
\begin{equation*}
\Gamma=\frac{\mu^{2}}{M} \int \frac{\mathrm{~d}^{3} p_{1}}{(2 \pi)^{3}} \frac{1}{4 E_{\mathbf{p}_{1}}^{2}}(2 \pi) \delta(M-2 E_{\mathbf{p}_{1}})=\frac{\mu^{2}}{8 \pi M}(1-\frac{4 m^{2}}{M^{2}})^{1 / 2} \tag{4.23}
\end{equation*}
Then the lifetime $\tau$ of $\Phi$ is,
\begin{equation*}
\tau=\Gamma^{-1}=\frac{8 \pi M}{\mu^{2}}(1-\frac{4 m^{2}}{M^{2}})^{-1 / 2} \tag{4.24}
\end{equation*}
|
{"$\\Phi$": "real scalar field (parent particle)", "$\\phi$": "real scalar field (decay particles)", "$M$": "mass of the $\\Phi$ particle", "$m$": "mass of the $\\phi$ particle", "$\\mu$": "coupling constant", "$\\mathcal{M}$": "decay amplitude", "$\\mathcal{L}$": "Lagrangian density", "$E_{\\mathbf{p}_{1}}$": "energy of the first $\\phi$ particle", "$E_{\\mathbf{p}_{2}}$": "energy of the second $\\phi$ particle", "$p_{\\Phi}$": "momentum of the $\\Phi$ particle", "$p_{1}$": "momentum of the first $\\phi$ particle", "$p_{2}$": "momentum of the second $\\phi$ particle", "$\\Gamma$": "decay rate", "$\\tau$": "lifetime of the $\\Phi$ particle"}
|
Interacting Fields and Feynman Diagrams
|
Decay of a scalar particle
|
52 |
Others
|
Consider a theory with two real scalar fields, $\Phi$ with mass $M$ and $\phi$ with mass $m$, described by the Lagrangian $\mathcal{L}=\frac{1}{2}(\partial_{\mu} \Phi)^{2}-\frac{1}{2} M^{2} \Phi^{2}+\frac{1}{2}(\partial_{\mu} \phi)^{2}-\frac{1}{2} m^{2} \phi^{2}-\mu \Phi \phi \phi$. If $M > 2m$, the $\Phi$ particle can decay into two $\phi$ particles. To the lowest order in the coupling constant $\mu$, the decay rate $\Gamma$ for this process is given by $\Gamma = \frac{\mu^2}{8\pi M} \sqrt{1 - \frac{4m^2}{M^2}}$. Calculate the value of the product $\Gamma M / \mu^2$ if $M=3m$.
|
[
"\\frac{\\sqrt{5}}{24\\pi}"
] |
Expression
|
This problem is based on the following Lagrangian,
\begin{equation*}
\mathcal{L}=\frac{1}{2}(\partial_{\mu} \Phi)^{2}-\frac{1}{2} M^{2} \Phi^{2}+\frac{1}{2}(\partial_{\mu} \phi)^{2}-\frac{1}{2} m^{2} \phi^{2}-\mu \Phi \phi \phi .
\end{equation*}
When $M>2 m$, a $\Phi$ particle can decay into two $\phi$ particles. We want to calculate the lifetime of the $\Phi$ particle to lowest order in $\mu$.
The two-body decay rate is given in (4.86) of $\mathrm{P} \& \mathrm{~S}$,
\begin{equation*}
\int \mathrm{d} \Gamma=\frac{1}{2 M} \int \frac{\mathrm{~d}^{3} p_{1} \mathrm{~d}^{3} p_{2}}{(2 \pi)^{6}} \frac{1}{4 E_{\mathbf{p}_{1}} E_{\mathbf{p}_{2}}}|\mathcal{M}(\Phi(0) \rightarrow \phi(p_{1}) \phi(p_{2}))|^{2}(2 \pi)^{4} \delta^{(4)}(p_{\Phi}-p_{1}-p_{2}) .
\end{equation*}
To lowest order in $\mu$, the amplitude $\mathcal{M}$ is given by,
\begin{equation*}
i \mathcal{M}=-2 i \mu
\end{equation*}
The delta function in our case reads,
\begin{equation*}
\delta^{(4)}(p_{\Phi}-p_{1}-p_{2})=\delta(M-E_{\mathbf{p}_{1}}-E_{\mathbf{p}_{2}}) \delta^{(3)}(\mathbf{p}_{1}+\mathbf{p}_{2}),
\end{equation*}
thus,
\begin{equation*}
\Gamma=\frac{1}{2} \cdot \frac{2 \mu^{2}}{M} \int \frac{\mathrm{~d}^{3} p_{1} \mathrm{~d}^{3} p_{2}}{(2 \pi)^{6}} \frac{1}{4 E_{\mathbf{p}_{1}} E_{\mathbf{p}_{2}}}(2 \pi)^{4} \delta(M-E_{\mathbf{p}_{1}}-E_{\mathbf{p}_{2}}) \delta^{(3)}(\mathbf{p}_{1}+\mathbf{p}_{2}),
\end{equation*}
where an additional factor of $1 / 2$ takes account of two identical $\phi$ 's in final state. Furthermore, there are two mass-shell constraints,
\begin{equation*}
m^{2}+\mathbf{p}_{i}^{2}=E_{\mathbf{p}_{i}}^{2} . \quad(i=1,2)
\end{equation*}
Hence,
\begin{equation*}
\Gamma=\frac{\mu^{2}}{M} \int \frac{\mathrm{~d}^{3} p_{1}}{(2 \pi)^{3}} \frac{1}{4 E_{\mathbf{p}_{1}}^{2}}(2 \pi) \delta(M-2 E_{\mathbf{p}_{1}})=\frac{\mu^{2}}{8 \pi M}(1-\frac{4 m^{2}}{M^{2}})^{1 / 2}
\end{equation*}
Then the lifetime $\tau$ of $\Phi$ is,
\begin{equation*}
\tau=\Gamma^{-1}=\frac{8 \pi M}{\mu^{2}}(1-\frac{4 m^{2}}{M^{2}})^{-1 / 2}.
\end{equation*}
We thus have:
\begin{equation}
\Gamma M / \mu^2 = \frac{1}{8\pi}\sqrt{1-\frac{4m^2}{M^2}} = \frac{1}{8\pi}\sqrt{1-\frac{4}{9}} = \frac{\sqrt{5}}{24\pi}
\end{equation}
|
{"$\\Phi$": "real scalar field with mass $M$", "$M$": "mass of the real scalar field $\\Phi$", "$\\phi$": "real scalar field with mass $m$", "$m$": "mass of the real scalar field $\\phi$", "$\\mu$": "coupling constant", "$\\Gamma$": "decay rate of $\\Phi$ particle"}
|
Interacting Fields and Feynman Diagrams
|
Decay of a scalar particle
|
53 |
Others
|
Equivalent photon approximation. Consider the process in which electrons of very high energy scatter from a target. In leading order in $\alpha$, the electron is connected to the target by one photon propagator. If the initial and final energies of the electron are $E$ and $E^{\prime}$, the photon will carry momentum $q$ such that $q^{2} \approx-2 E E^{\prime}(1-\cos \theta)$. In the limit of forward scattering, whatever the energy loss, the photon momentum approaches $q^{2}=0$; thus the reaction is highly peaked in the forward direction. It is tempting to guess that, in this limit, the virtual photon becomes a real photon. Let us investigate in what sense that is true. Working in the frame where $p=(E, 0,0, E)$, compute explicitly the quantity $\bar{u}_+(p^{\prime}) \boldsymbol{\gamma} \cdot \boldsymbol{\epsilon}_{1} u_+(p)$ using massless electrons, where $u_+(p)$ and $u_+(p^{\prime})$ are spinors of positive helicity, and $\boldsymbol{\epsilon}_{1}$ is a unit vector parallel to the plane of scattering. This quantity is needed only for scattering near the forward direction, and you need only provide the term of order $\theta$. Note that for $\boldsymbol{\epsilon}_{1}$ (in the plane of scattering), the small $\hat{3}$ component of $\boldsymbol{\epsilon}_{1}$ also gives a term of order $\theta$ which must be taken into account.
|
[
"-\\sqrt{E E^{\\prime}} \\frac{E+E^{\\prime}}{|E-E^{\\prime}|} \\theta"
] |
Expression
|
It is easy to find that
\epsilon_{1}^{\mu}=N(0, p^{\prime} \cos \theta-p, 0,-p^{\prime} \sin \theta), \quad \epsilon_{2}^{\mu}=(0,0,1,0),
where $N=(E^{2}+E^{\prime 2}-2 E E^{\prime} \cos \theta)^{-1 / 2}$ is the normalization constant. Then, for the righthanded electron with spinor $u_{+}(p)=\sqrt{2 E}(0,0,1,0)^{T}$ and left-handed electron with $u_{-}(p)=$ $\sqrt{2 E}(0,1,0,0)^{T}$, it is straightforward to show that
\begin{equation*}\nu_{+}(p^{\prime})=\sqrt{2 E^{\prime}}(0,0, \cos \frac{\theta}{2}, \sin \frac{\theta}{2})^{T}, \quad u_{-}(p^{\prime})=\sqrt{2 E^{\prime}}(-\sin \frac{\theta}{2}, \cos \frac{\theta}{2}, 0,0) \tag{6.15}
\end{equation*}
and,
\begin{align*}
& \bar{u}_{ \pm}(p^{\prime}) \boldsymbol{\gamma} \cdot \boldsymbol{\epsilon}_{1} u_{ \pm}(p) \simeq-\sqrt{E E^{\prime}} \frac{E+E^{\prime}}{|E-E^{\prime}|} \theta, \tag{6.16}\\
& \bar{u}_{ \pm}(p^{\prime}) \boldsymbol{\gamma} \cdot \boldsymbol{\epsilon}_{2} u_{ \pm}(p) \simeq \pm \mathrm{i} \sqrt{E E^{\prime}} \theta \tag{6.17}\\
& \bar{u}_{ \pm}(p^{\prime}) \boldsymbol{\gamma} \cdot \boldsymbol{\epsilon}_{1} u_{\mp}(p)=\bar{u}_{ \pm}(p^{\prime}) \boldsymbol{\gamma} \cdot \boldsymbol{\epsilon}_{2} u_{\mp}(p)=0 . \tag{6.18}
\end{align*}
That is to say, we have,
\begin{equation*}
C_{ \pm}=-\sqrt{E E^{\prime}} \frac{E+E^{\prime}}{|E-E^{\prime}|} \theta, \quad \quad D_{ \pm}= \pm \mathrm{i} \sqrt{E E^{\prime}} \theta \tag{6.19}
\end{equation*}
|
{"$p$": "momentum vector of the initial particle", "$E$": "energy of the initial particle", "$p^{\\prime}$": "momentum vector of the scattered particle", "$E^{\\prime}$": "energy of the scattered particle", "$\\bar{u}_+(p^{\\prime})$": "conjugate spinor of positive helicity for scattered particle", "$u_+(p)$": "spinor of positive helicity for initial particle", "$\\boldsymbol{\\gamma}$": "gamma matrices in the Dirac equation", "$\\boldsymbol{\\epsilon}_{1}$": "unit vector in the plane of scattering", "$\\boldsymbol{\\epsilon}_{2}$": "unit vector perpendicular to the plane of scattering", "$\\theta$": "scattering angle", "$N$": "normalization constant", "$\\epsilon_{1}^{\\mu}$": "components of the unit vector parallel to the scattering plane", "$\\epsilon_{2}^{\\mu}$": "components of the unit vector perpendicular to the scattering plane", "$C_{ \\pm}$": "coefficient related to the scattering amplitude", "$D_{ \\pm}$": "coefficient related to the imaginary part of the scattering amplitude"}
|
Radiative Corrections: Introduction
|
Equivalent photon approximation
|
54 |
Others
|
Exotic contributions to $\boldsymbol{g} \mathbf{- 2}$. Any particle that couples to the electron can produce a correction to the electron-photon form factors and, in particular, a correction to $g-2$. Because the electron $g-2$ agrees with QED to high accuracy, these corrections allow us to constrain the properties of hypothetical new particles. The unified theory of weak and electromagnetic interactions contains a scalar particle $h$ called the Higgs boson, which couples to the electron according to
$$ H_{\mathrm{int}}=\int d^{3} x \frac{\lambda}{\sqrt{2}} h \bar{\psi} \psi $$
Compute the contribution of a virtual Higgs boson to the electron $(g-2)$, in terms of $\lambda$ and the mass $m_{h}$ of the Higgs boson.
|
[
"\\frac{\\lambda^{2} m^2}{8 \\pi^{2} m_h^2}[\\log (\\frac{m_h^2}{m^2})-\\frac{7}{6}]"
] |
Expression
|
The 1-loop vertex correction from Higgs boson is,
\begin{align*}
\bar{u}(p^{\prime}) \delta \Gamma^{\mu} u(p) & =(\frac{\mathrm{i} \lambda}{\sqrt{2}})^{2} \int \frac{\mathrm{~d}^{d} k}{(2 \pi)^{d}} \frac{\mathrm{i}}{(k-p)^{2}-m_{h}^{2}} \bar{u}(p^{\prime}) \frac{\mathrm{i}}{\not k+q-m} \gamma^{\mu} \frac{\mathrm{i}}{\not k-m} u(p) \\
& =\frac{\mathrm{i} \lambda^{2}}{2} \int_{0}^{1} \mathrm{~d} x \int_{0}^{1-x} \mathrm{~d} y \int \frac{\mathrm{~d}^{d} k^{\prime}}{(2 \pi)^{d}} \frac{2 \bar{u}(p^{\prime}) N^{\mu} u(p)}{(k^{\prime 2}-\Delta)^{3}}, \tag{6.26}
\end{align*}
with
\begin{align*}
& N^{\mu}=(\not k+q+m) \gamma^{\mu}(k+m), \tag{6.27}\\
& k^{\prime}=k-x p+y q, \tag{6.28}\\
& \Delta=(1-x) m^{2}+x m_{h}^{2}-x(1-x) p^{2}-y(1-y) q^{2}+2 x y p \cdot q . \tag{6.29}
\end{align*}
To put this correction into the following form,
\begin{equation*}
\Gamma^{\mu}=\gamma^{\mu} F_{1}(q)+\frac{i \sigma^{\mu \nu} q_{\nu}}{2 m} F_{2}(q) \tag{6.30}
\end{equation*}
we first rewrite $N^{\mu}$ as,
\begin{equation*}
N^{\mu}=A \gamma^{\mu}+B(p^{\prime}+p)^{\mu}+C(p^{\prime}-p)^{\mu} \tag{6.31}
\end{equation*}
where term proportional to $(p^{\prime}-p)$ can be thrown away by Ward identity $q_{\mu} \Gamma^{\mu}(q)=0$. This can be done by gamma matrix calculations, leading to the following result,
\begin{equation*}
N^{\mu}=[(\frac{2}{d}-1) k^{\prime 2}+(3+2 x-x^{2}) m^{2}+(y-x y-y^{2}) q^{2}] \gamma^{\mu}+(x^{2}-1) m(p^{\prime}+p)^{\mu} . \tag{6.32}
\end{equation*}
Then, using Gordon identity, we find,
\begin{equation*}
N^{\mu}=[(\frac{2}{d}-1) k^{\prime 2}+(x+1)^{2} m^{2}+(y-y^{2}-x y) q^{2}] \gamma^{\mu}+\frac{\mathrm{i} \sigma^{\mu \nu} q_{\nu}}{2 m} \cdot 2 m^{2}(1-x^{2}) \tag{6.33}
\end{equation*}
Comparing this with (6.30), we see that
\begin{align*}
\delta F_{2}(q=0) & =2 \mathrm{i} \lambda^{2} m^{2} \int_{0}^{1} \mathrm{~d} x \int_{0}^{1-x} \mathrm{~d} y \int \frac{\mathrm{~d}^{4} k^{\prime}}{(2 \pi)^{4}} \frac{1-x^{2}}{(k^{\prime 2}-\Delta)^{3}} \\
& =\frac{\lambda^{2}}{(4 \pi)^{2}} \int_{0}^{1} \mathrm{~d} x \frac{(1-x)^{2}(1+x)}{(1-x)^{2}+x(m_{h} / m)^{2}} . \tag{6.34}
\end{align*}
To carry out the integration over $x$, we use the approximation that $m_{h} \gg m$. Then,
\begin{align*}
\delta F_{2}(q=0) & \simeq \frac{\lambda^{2}}{(4 \pi)^{2}} \int_{0}^{1} \mathrm{~d} x[\frac{1}{1+x(m_{h} / m)^{2}}-\frac{1+x-x^{2}}{(m_{h} / m)^{2}}] \\
& \simeq \frac{\lambda^{2}}{(4 \pi)^{2}(m_{h} / m)^{2}}[\log (m_{h}^{2} / m^{2})-\frac{7}{6}] . \tag{6.35}
\end{align*}
|
{"$h$": "Higgs boson", "$m_{h}$": "mass of the Higgs boson", "$\\lambda$": "coupling constant", "$m$": "mass of the electron"}
|
Radiative Corrections: Introduction
|
Exotic contributions to $g-2$
|
55 |
Others
|
Exotic contributions to $\boldsymbol{g} \mathbf{- 2}$. Any particle that couples to the electron can produce a correction to the electron-photon form factors and, in particular, a correction to $g-2$. Because the electron $g-2$ agrees with QED to high accuracy, these corrections allow us to constrain the properties of hypothetical new particles. Some more complex versions of this theory contain a pseudoscalar particle called the axion, which couples to the electron according to
$$ H_{\mathrm{int}}=\int d^{3} x \frac{i \lambda}{\sqrt{2}} a \bar{\psi} \gamma^{5} \psi $$
The axion may be as light as the electron, or lighter, and may couple more strongly than the Higgs boson. Compute the contribution of a virtual axion to the $g-2$ of the electron, and work out the excluded values of $\lambda$ and $m_{a}$.
|
[
"-\\frac{\\lambda^{2}}{(4 \\pi)^{2}(m_{a}^{2} / m^{2})}[\\log (m_{a}^{2} / m^{2})-\\frac{11}{6}]"
] |
Expression
|
The 1-loop correction from the axion is given by,
\begin{align*}
\bar{u}(p^{\prime}) \delta \Gamma^{\mu} u(p) & =(\frac{-\lambda}{\sqrt{2}})^{2} \int \frac{\mathrm{~d}^{d} k}{(2 \pi)^{d}} \frac{\mathrm{i}}{(k-p)^{2}-m_{h}^{2}} \bar{u}(p^{\prime}) \gamma^{5} \frac{\mathrm{i}}{\not k+\not q-m} \gamma^{\mu} \frac{\mathrm{i}}{\nVdash-m} \gamma^{5} u(p) \\
& =-\frac{\mathrm{i} \lambda^{2}}{2} \int_{0}^{1} \mathrm{~d} x \int_{0}^{1-x} \frac{\mathrm{~d} y \int \frac{\mathrm{~d}^{d} k^{\prime}}{(2 \pi)^{d}} \frac{2 \bar{u}(p^{\prime}) N^{\mu} u(p)}{(k^{\prime 2}-\Delta)^{3}},}{} \tag{6.37}
\end{align*}
in which $k^{\prime}$ and $\Delta$ are still defined as in (a) except the replacement $m_{h} \rightarrow m_{a}$, while $N^{\mu}$ is now given by,
\begin{equation*}
N^{\mu}=\gamma^{5}(\not k+q q+m) \gamma^{\mu}(\not k+m) \gamma^{5}=-(\nless q+q-m) \gamma^{\mu}(\not k-m) . \tag{6.38}
\end{equation*}
Repeating the same derivation as was done in (a), we get,
\begin{equation*}
N^{\mu}=[-(\frac{2}{d}-1) k^{\prime 2}-(1-x-y) y q^{2}+(1-x)^{2} m^{2}] \gamma^{\mu}-(1-x)^{2} m(p^{\prime}+p)^{2} . \tag{6.39}
\end{equation*}
Again, using Gordon identity, we get,
\begin{align*}
\delta F_{2}(q=0) & =-2 \mathrm{i} \lambda^{2} m^{2} \int_{0}^{1} \mathrm{~d} x \int_{0}^{1-x} \mathrm{~d} y \int \frac{\mathrm{~d}^{4} k^{\prime}}{(2 \pi)^{4}} \frac{(1-x)^{2}}{(k^{\prime 2}-\Delta)^{3}} \\
& =-\frac{\lambda^{2}}{(4 \pi)^{2}} \int_{0}^{1} \mathrm{~d} x \frac{(1-x)^{3}}{(1-x)^{2}+x m_{a}^{2} / m^{2}} \\
& \simeq-\frac{\lambda^{2}}{(4 \pi)^{2}} \int_{0}^{1} \mathrm{~d} x[\frac{1}{1+x m_{a}^{2} / m^{2}}-\frac{3-3 x+x^{2}}{m_{a}^{2} / m^{2}}] \\
& =-\frac{\lambda^{2}}{(4 \pi)^{2}(m_{a}^{2} / m^{2})}[\log (m_{a}^{2} / m^{2})-\frac{11}{6}] . \tag{6.40}
\end{align*}
For order-of-magnitude estimation, it's easy to see that $\lambda m / m_{a} \gtrsim 10^{-5}$ is excluded.
|
{"$\\lambda$": "axion-electron coupling constant", "$a$": "axion field", "$\\psi$": "electron field", "$\\gamma^{5}$": "gamma matrix in Dirac theory", "$m$": "electron mass", "$m_{a}$": "axion mass", "$p$": "initial momentum of the electron", "$p^{\\prime}$": "final momentum of the electron", "$q$": "momentum transfer in the process", "$k$": "loop momentum", "$k^{\\prime}$": "shifted loop momentum", "$\\Delta$": "parameter related to the shifted loop momentum"}
|
Radiative Corrections: Introduction
|
Exotic contributions to $g-2$
|
56 |
Others
|
Although we have discussed QED radiative corrections at length in the last two chapters, so far we have made no attempt to compute a full radiatively corrected cross section. The reason is of course that such calculations are quite lengthy. Nevertheless it would be dishonest to pretend that one understands radiative corrections after computing only isolated effects as we have done. This "final project" is an attempt to remedy this situation. The project is the computation of one of the simplest, but most important, radiatively corrected cross sections.
Strongly interacting particles-pions, kaons, and protons-are produced in $e^{+} e^{-}$annihilation when the virtual photon creates a pair of quarks. If one ignores the effects of the strong interactions, it is easy to calculate the total cross section for quark pair production. In this final project, we will analyze the first corrections to this formula due to the strong interactions.
Let us represent the strong interactions by the following simple model: Introduce a new massless vector particle, the gluon, which couples universally to quarks:
$$\Delta H=\int d^{3} x g \bar{\psi}_{f i} \gamma^{\mu} \psi_{f i} B_{\mu} $$
Here $f$ labels the type ("flavor") of the quark ( $u, d, s, c$, etc.) and $i=1,2,3$ labels the color. The strong coupling constant $g$ is independent of flavor and color. The electromagnetic coupling of quarks depends on the flavor, since the $u$ and $c$ quarks have charge $Q_{f}=+2 / 3$ while the $d$ and $s$ quarks have charge $Q_{f}=-1 / 3$. By analogy to $\alpha$, let us define
\alpha_{g}=\frac{g^{2}}{4 \pi}
In this exercise, we will compute the radiative corrections to quark pair production proportional to $\alpha_{g}$.
This model of the strong interactions of quarks does not quite agree with the currently accepted theory of the strong interactions, quantum chromodynamics (QCD). However, all of the results that we will derive here are also
correct in QCD with the replacement
$$ \alpha_{g} \rightarrow \frac{4}{3} \alpha_{s} . $$
Throughout this exercise, you may ignore the masses of quarks. You may also ignore the mass of the electron, and average over electron and positron polarizations. To control infrared divergences, it will be necessary to assume that the gluons have a small nonzero mass $\mu$, which can be taken to zero only at the end of the calculation. However (as we discussed in Problem 5.5), it is consistent to sum over polarization states of this massive boson by the replacement:
\sum \epsilon^{\mu} \epsilon^{\nu *} \rightarrow-g^{\mu \nu}
this also implies that we may use the propagator
\widehat{B^{\mu} B^{\nu}}=\frac{-i g^{\mu \nu}}{k^{2}-\mu^{2}+i \epsilon} Draw the Feynman diagrams for the process $e^{+} e^{-} \rightarrow \bar{q} q g$, to leading order in $\alpha$ and $\alpha_{g}$, and compute the differential cross section. You may throw away the information concerning the correlation between the initial beam axis and the directions of the final particles. This is conveniently done as follows: The usual trace tricks for evaluating the square of the matrix element give for this process a result of the structure
\int d \Pi_{3} \frac{1}{4} \sum|\mathcal{M}|^{2}=L_{\mu \nu} \int d \Pi_{3} H^{\mu \nu}
where $L_{\mu \nu}$ represents the electron trace and $H^{\mu \nu}$ represents the quark trace. If we integrate over all parameters of the final state except $x_{1}$ and $x_{2}$, which are scalars, the only preferred 4 -vector characterizing the final state is $q^{\mu}$. On the other hand, $H_{\mu \nu}$ satisfies
q^{\mu} H_{\mu \nu}=H_{\mu \nu} q^{\nu}=0
Why is this true? (There is an argument based on general principles; however, you might find it a useful check on your calculation to verify this property explicitly.) Since, after integrating over final-state vectors, $\int H^{\mu \nu}$ depends only on $q^{\mu}$ and scalars, it can only have the form
$$ \int d \Pi_{3} H^{\mu \nu}=(g^{\mu \nu}-\frac{q^{\mu} q^{\nu}}{q^{2}}) \cdot H $$
where $H$ is a scalar. With this information, show that
$$ L_{\mu \nu} \int d \Pi_{3} H^{\mu \nu}=\frac{1}{3}(g^{\mu \nu} L_{\mu \nu}) \cdot \int d \Pi_{3}(g^{\rho \sigma} H_{\rho \sigma})$$
Using this trick, derive the differential cross section
$$ \frac{d \sigma}{d x_{1} d x_{2}}(e^{+} e^{-} \rightarrow \bar{q} q g)=\frac{4 \pi \alpha^{2}}{3 s} \cdot 3 Q_{f}^{2} \cdot \frac{\alpha_{g}}{2 \pi} \frac{x_{1}^{2}+x_{2}^{2}}{(1-x_{1})(1-x_{2})}$$
in the limit $\mu \rightarrow 0$. If we assume that each original final-state particle is realized physically as a jet of strongly interacting particles, this formula gives the probability for observing three-jet events in $e^{+} e^{-}$annihilation and the kinematic distribution of these events. The form of the distribution in the $x_{i}$ is an absolute prediction, and it agrees with experiment. The
normalization of this distribution is a measure of the strong-interaction coupling constant.
|
[
"\\frac{d \\sigma}{d x_{1} d x_{2}}(e^{+} e^{-} \\rightarrow \\bar{q} q g)=\\frac{4 \\pi \\alpha^{2}}{3 s} \\cdot 3 Q_{f}^{2} \\cdot \\frac{\\alpha_{g}}{2 \\pi} \\frac{x_{1}^{2}+x_{2}^{2}}{(1-x_{1})(1-x_{2})}"
] |
Expression
|
Now we calculate the differential cross section for the process $e^{+} e^{-} \rightarrow q \bar{g} g$ to lowest order in $\alpha$ and $\alpha_{g}$. First, the amplitude is
\begin{equation*}
\mathrm{i} \mathcal{M}=Q_{f}(-\mathrm{i} e)^{2}(-\mathrm{i} g) \epsilon_{\nu}^{*}(k_{3}) \bar{u}(k_{1})[\gamma^{\nu} \frac{\mathrm{i}}{\not k_{1}+\not k_{3}} \gamma^{\mu}-\gamma^{\mu} \frac{\mathrm{i}}{\not k_{2}+\not k_{3}} \gamma^{\nu}] v(k_{2}) \frac{-\mathrm{i}}{q^{2}} \bar{v}(p_{2}) \gamma_{\mu} u(p_{1}) . \tag{7.54}
\end{equation*}
Then, the squared amplitude is
\begin{align*}
\frac{1}{4} \sum|\mathrm{i} \mathcal{M}|^{2}= & \frac{Q_{f}^{2} g^{2} e^{4}}{4 s^{2}}(-g_{\nu \sigma}) \operatorname{tr}(\gamma_{\mu} \not p_{1} \gamma_{\rho} \not \not_{2}) \\
& \times \operatorname{tr}[(\gamma^{\nu} \frac{1}{\not k_{1}+\not k_{3}} \gamma^{\mu}-\gamma^{\mu} \frac{1}{\not k_{2}+\not k_{3}} \gamma^{\nu}) \not k_{2}(\gamma^{\rho} \frac{1}{\not k_{1}+\not k_{3}} \gamma^{\sigma}-\gamma^{\sigma} \frac{1}{\not k_{2}+\not k_{3}} \gamma^{\rho}) \not k_{1}] \\
= & \frac{4 Q_{f}^{2} g^{2} e^{4}}{3 s^{2}}(8 p_{1} \cdot p_{2})[\frac{4(k_{1} \cdot k_{2})(k_{1} \cdot k_{2}+q \cdot k_{3})}{(k_{1}+k_{3})^{2}(k_{2}+k_{3})^{2}} \\
&+(\frac{1}{(k_{1}+k_{3})^{4}}+\frac{1}{(k_{2}+k_{3})^{4}})(2(k_{1} \cdot k_{3})(k_{2} \cdot k_{3})-\mu^{2}(k_{1} \cdot k_{2}))] . \tag{7.55}
\end{align*}
We have used the trick described in Peskin's book (P261) when getting through the last equal sign. Now rewrite the quantities of final-state kinematics in terms of $x_{i}$, and set $\mu \rightarrow 0$, we obtain
\begin{align*}
\frac{1}{4} \sum|\mathrm{i} \mathcal{M}|^{2} & =\frac{2 Q_{f}^{2} g^{2} e^{4}}{3 s^{2}}(8 p_{1} \cdot p_{2})[\frac{2(1-x_{3})}{(1-x_{1})(1-x_{2})}+\frac{1-x_{1}}{1-x_{2}}+\frac{1-x_{2}}{1-x_{1}}] \\
& =\frac{8 Q_{f}^{2} g^{2} e^{4}}{3 s} \frac{x_{1}^{2}+x_{2}^{2}}{(1-x_{1})(1-x_{2})} . \tag{7.56}
\end{align*}
Thus the differential cross section, with 3 colors counted, reads
\begin{align*}
\frac{\mathrm{d} \sigma}{\mathrm{~d} x_{1} \mathrm{~d} x_{2}}|_{\mathrm{COM}} & =\frac{1}{2 E_{\mathbf{p}_{1}} 2 E_{\mathbf{p}_{2}}|v_{\mathbf{p}_{1}}-v_{\mathbf{p}_{2}}|} \frac{s}{128 \pi^{3}}(\frac{1}{4} \sum|\mathcal{M}|^{2}) \\
& =\frac{4 \pi \alpha^{2}}{3 s} \cdot 3 Q_{f}^{2} \cdot \frac{\alpha_{g}}{2 \pi} \frac{x_{1}^{2}+x_{2}^{2}}{(1-x_{1})(1-x_{2})} \tag{7.57}
\end{align*}
where we have used the fact that the initial electron and positron are massless, which implies that $2 E_{\mathbf{p}_{1}}=2 E_{\mathbf{p}_{2}}=\sqrt{s}$ and $|v_{\mathbf{p}_{1}}-v_{\mathbf{p}_{2}}|=2$ in COM frame.
|
{"$e$": "electron", "$\\alpha$": "fine-structure constant", "$\\alpha_{g}$": "strong-interaction coupling constant", "$q$": "quark four-momentum", "$L_{\\mu \\nu}$": "electron trace", "$H^{\\mu \\nu}$": "quark trace", "$s$": "center-of-mass energy squared", "$Q_{f}$": "quark charge factor", "$x_{1}$": "final state scalar 1", "$x_{2}$": "final state scalar 2", "$p_{1}$": "electron four-momentum", "$p_{2}$": "positron four-momentum", "$k_{1}$": "quark four-momentum", "$k_{2}$": "anti-quark four-momentum", "$k_{3}$": "gluon four-momentum", "$\\mu$": "small parameter tending to zero"}
|
Final Project I
|
Radiation of Gluon Jets
|
57 |
Others
|
Although we have discussed QED radiative corrections at length in the last two chapters, so far we have made no attempt to compute a full radiatively corrected cross section. The reason is of course that such calculations are quite lengthy. Nevertheless it would be dishonest to pretend that one understands radiative corrections after computing only isolated effects as we have done. This "final project" is an attempt to remedy this situation. The project is the computation of one of the simplest, but most important, radiatively corrected cross sections.
Strongly interacting particles-pions, kaons, and protons-are produced in $e^{+} e^{-}$annihilation when the virtual photon creates a pair of quarks. If one ignores the effects of the strong interactions, it is easy to calculate the total cross section for quark pair production. In this final project, we will analyze the first corrections to this formula due to the strong interactions.
Let us represent the strong interactions by the following simple model: Introduce a new massless vector particle, the gluon, which couples universally to quarks:
$$\Delta H=\int d^{3} x g \bar{\psi}_{f i} \gamma^{\mu} \psi_{f i} B_{\mu}$$
Here $f$ labels the type ("flavor") of the quark ( $u, d, s, c$, etc.) and $i=1,2,3$ labels the color. The strong coupling constant $g$ is independent of flavor and color. The electromagnetic coupling of quarks depends on the flavor, since the $u$ and $c$ quarks have charge $Q_{f}=+2 / 3$ while the $d$ and $s$ quarks have charge $Q_{f}=-1 / 3$. By analogy to $\alpha$, let us define
$$\alpha_{g}=\frac{g^{2}}{4 \pi}$$
In this exercise, we will compute the radiative corrections to quark pair production proportional to $\alpha_{g}$.
This model of the strong interactions of quarks does not quite agree with the currently accepted theory of the strong interactions, quantum chromodynamics (QCD). However, all of the results that we will derive here are also
correct in QCD with the replacement
$$\alpha_{g} \rightarrow \frac{4}{3} \alpha_{s} .$$
Throughout this exercise, you may ignore the masses of quarks. You may also ignore the mass of the electron, and average over electron and positron polarizations. To control infrared divergences, it will be necessary to assume that the gluons have a small nonzero mass $\mu$, which can be taken to zero only at the end of the calculation. However (as we discussed in Problem 5.5), it is consistent to sum over polarization states of this massive boson by the replacement:
\sum \epsilon^{\mu} \epsilon^{\nu *} \rightarrow-g^{\mu \nu}
this also implies that we may use the propagator
\widehat{B^{\mu} B^{\nu}}=\frac{-i g^{\mu \nu}}{k^{2}-\mu^{2}+i \epsilon} Now replace $\mu \neq 0$ in the formula of the differential cross section, and carefully integrate over the region. You may assume $\mu^{2} \ll q^{2}$. In this limit, you will find infrared-divergent terms of order $\log (q^{2} / \mu^{2})$ and also $\log ^{2}(q^{2} / \mu^{2})$, finite terms of order 1 , and terms explicitly suppressed by powers of $(\mu^{2} / q^{2})$. You may drop terms of the last type throughout this calculation. For the moment, collect and evaluate only the infrared-divergent terms.
|
[
"\\frac{4 \\pi \\alpha^{2}}{3 s} \\cdot 3 Q_{f}^{2} \\cdot \\frac{\\alpha_{g}}{2 \\pi}[\\log ^{2} \\frac{\\mu^{2}}{s}+3 \\log \\frac{\\mu^{2}}{s}]"
] |
Expression
|
Now we reevaluate the averaged squared amplitude, with $\mu$ kept nonzero in the formula
\begin{align*}
\frac{1}{4} \sum|\mathrm{i} \mathcal{M}|^{2}= & \frac{4 Q_{f}^{2} g^{2} e^{4}}{3 s^{2}}(8 p_{1} \cdot p_{2})[\frac{4(k_{1} \cdot k_{2})(k_{1} \cdot k_{2}+q \cdot k_{3})}{(k_{1}+k_{3})^{2}(k_{2}+k_{3})^{2}} \\
&+(\frac{1}{(k_{1}+k_{3})^{4}}+\frac{1}{(k_{2}+k_{3})^{4}})(2(k_{1} \cdot k_{3})(k_{2} \cdot k_{3})-\mu^{2}(k_{1} \cdot k_{2}))].
\end{align*}
The result is
\begin{equation*}
\frac{1}{4} \sum|\mathrm{i} \mathcal{M}|^{2}=\frac{8 Q_{f}^{2} g^{2} e^{4}}{3 s} F(x_{1}, x_{2}, \mu^{2} / s) \tag{7.58}
\end{equation*}
where
\begin{align*}
F(x_{1}, x_{2}, \frac{\mu^{2}}{s})= & \frac{2(x_{1}+x_{2}-1+\frac{\mu^{2}}{s})(1+\frac{\mu^{2}}{s})}{(1-x_{1})(1-x_{2})} \\
& +[\frac{1}{(1-x_{1})^{2}}+\frac{1}{(1-x_{2})^{2}}]((1-x_{1})(1-x_{2})-\frac{\mu^{2}}{s}) \tag{7.59}
\end{align*}
The cross section, then, can be got by integrating over $\mathrm{d} x_{1} \mathrm{~d} x_{2}$ :
\begin{align*}
\sigma(e^{+} e^{-} \rightarrow q \bar{q} g) & =\frac{1}{2 E_{\mathbf{p}_{1}} 2 E_{\mathbf{p}_{2}}|v_{\mathbf{p}_{1}}-v_{\mathbf{p}_{2}}|} \frac{s}{128 \pi^{3}} \int \mathrm{~d} x_{1} \mathrm{~d} x_{2}(\frac{1}{4} \sum|\mathcal{M}|^{2}) \\
& =\frac{4 \pi \alpha^{2}}{3 s} \cdot 3 Q_{f}^{2} \cdot \frac{\alpha_{g}}{2 \pi} \int_{0}^{1-\frac{\mu^{2}}{s}} \frac{\mathrm{~d} x_{1} \int_{1-x_{1}-\frac{\mu^{2}}{s}}^{1-\frac{t}{s(1-x_{1})}} \mathrm{d} x_{2} F(x_{1}, x_{2}, \frac{\mu^{2}}{s})}{} \\
& =\frac{4 \pi \alpha^{2}}{3 s} \cdot 3 Q_{f}^{2} \cdot \frac{\alpha_{g}}{2 \pi}[\log ^{2} \frac{\mu^{2}}{s}+3 \log \frac{\mu^{2}}{s}+5-\frac{1}{3} \pi^{2}+\mathcal{O}(\mu^{2})] \tag{7.60}
\end{align*}
|
{"$\\mu$": "mass scale parameter", "$q$": "momentum transfer", "$Q_{f}$": "quark charge factor", "$g$": "strong coupling constant", "$e$": "electric charge", "$s$": "Mandelstam variable, center-of-mass energy squared", "$x_{1}$": "energy fraction variable (1)", "$x_{2}$": "energy fraction variable (2)", "$E_{\\mathbf{p}_{1}}$": "energy of particle 1", "$E_{\\mathbf{p}_{2}}$": "energy of particle 2", "$v_{\\mathbf{p}_{1}}$": "velocity of particle 1", "$v_{\\mathbf{p}_{2}}$": "velocity of particle 2", "$\\alpha$": "fine structure constant", "$\\alpha_{g}$": "strong coupling constant alpha", "$\\mathcal{M}$": "amplitude", "$F(x_{1}, x_{2}, \\frac{\\mu^{2}}{s})$": "dimensionless function related to energy fraction variables", "$t$": "momentum transfer squared"}
|
Final Project I
|
Radiation of Gluon Jets
|
58 |
Others
|
Scalar QED. This problem concerns the theory of a complex scalar field $\phi$ interacting with the electromagnetic field $A^{\mu}$. The Lagrangian is
$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu}^{2}+(D_{\mu} \phi)^{*}(D^{\mu} \phi)-m^{2} \phi^{*} \phi$$
where $D_{\mu}=\partial_{\mu}+i e A_{\mu}$ is the usual gauge-covariant derivative. Compute, to lowest order, the differential cross section for $e^{+} e^{-} \rightarrow \phi \phi^{*}$. Ignore the electron mass (but not the scalar particle's mass), and average over the electron and positron polarizations. Find the asymptotic angular dependence and total cross section. Compare your results to the corresponding formulae for $e^{+} e^{-} \rightarrow \mu^{+} \mu^{-}$.
Compute the contribution of the charged scalar to the photon vacuum polarization, using dimensional regularization. Note that there are two diagrams. To put the answer into the expected form,
$$\Pi^{\mu \nu}(q^{2})=(g^{\mu \nu} q^{2}-q^{\mu} q^{\nu}) \Pi(q^{2})$$
it is useful to add the two diagrams at the beginning, putting both terms over a common denominator before introducing a Feynman parameter. Show that, for $-q^{2} \gg m^{2}$, the charged boson contribution to $\Pi(q^{2})$ is exactly $1 / 4$ that of a virtual electron-positron pair.
|
[
"(\\frac{d\\sigma}{d\\Omega})_{CM} = \\frac{\\alpha^2}{8s} (1 - \\frac{m^2}{E^2})^{3/2} \\sin^2{\\theta}"
] |
Expression
|
Now we calculate the spin-averaged differential cross section for the process $e^{+} e^{-} \rightarrow$ $\phi^{*} \phi$. The scattering amplitude is given by
\begin{equation*}
\mathrm{i} \mathcal{M}=(-\mathrm{i} e)^{2} \bar{v}(k_{2}) \gamma^{\mu} u(k_{1}) \frac{-\mathrm{i}}{s}(p_{1}-p_{2})_{\mu} . \tag{9.3}
\end{equation*}
Then the spin-averaged and squared amplitude is
\frac{1}{4} \sum_{\text {spins }}|\mathcal{M}|^{2}=\frac{e^{4}}{4 s^{2}} \operatorname{tr}[(\not p_{1}-\not p_{2}) \not k_{1}(\not p_{1}-\not p_{2}) \not k_{2}]
\begin{equation*}
=\frac{e^{4}}{4 s^{2}}[8(k_{1} \cdot p_{1}-k_{1} \cdot p_{2})(k_{2} \cdot p_{1}-k_{2} \cdot p_{2})-4(k_{1} \cdot k_{2})(p_{1}-p_{2})^{2}] . \tag{9.4}
\end{equation*}
We may parameterize the momenta as
\begin{array}{ll}
k_{1}=(E, 0,0, E), & p_{1}=(E, p \sin \theta, 0, p \cos \theta), \\
k_{2}=(E, 0,0,-E), & p_{2}=(E,-p \sin \theta, 0,-p \cos \theta),
\end{array}
with $p=\sqrt{E^{2}-m^{2}}$. Then we have
\begin{equation*}
\frac{1}{4} \sum_{\text {spins }}|\mathcal{M}|^{2}=\frac{e^{4} p^{2}}{2 E^{2}} \sin ^{2} \theta \tag{9.5}
\end{equation*}
Thus the differential cross section is:
\begin{equation*}
(\frac{\mathrm{d} \sigma}{\mathrm{~d} \Omega})_{\mathrm{CM}}=\frac{1}{2(2 E)^{2}} \frac{p}{8(2 \pi)^{2} E}(\frac{1}{4} \sum|\mathcal{M}|^{2})=\frac{\alpha^{2}}{8 s}(1-\frac{m^{2}}{E^{2}})^{3 / 2} \sin ^{2} \theta \tag{9.6}
\end{equation*}
|
{"$e^{+}$": "positron", "$e^{-}$": "electron", "$\\phi$": "scalar particle", "$\\phi^{*}$": "conjugate scalar particle", "$m$": "scalar particle mass", "$\\Pi^{\\mu \\nu}$": "photon polarization tensor", "$q$": "momentum transfer", "$\\Pi(q^{2})$": "polarization function", "$s$": "center of mass energy squared", "$\\mathcal{M}$": "scattering amplitude", "$e$": "elementary charge", "$k_{1}$": "momentum of incoming electron", "$k_{2}$": "momentum of incoming positron", "$p_{1}$": "momentum of outgoing scalar particle", "$p_{2}$": "momentum of outgoing conjugate scalar particle", "$E$": "energy of incoming particles", "$p$": "three-momentum magnitude of outgoing particles", "$\\sigma$": "differential cross section", "$\\alpha$": "fine-structure constant", "$\\theta$": "scattering angle"}
|
Functional Methods
|
Scalar QED
|
59 |
Others
|
We discussed the effective potential for an $O(N)$-symmetric $\phi^{4}$ theory in four dimensions. We computed the perturbative corrections to this effective potential, and used the renormalization group to clarify the behavior of the potential for small values of the scalar field mass. After all this work, however, we found that the qualitative dependence of the theory on the mass parameter was unchanged by perturbative corrections. The theory still possessed a second-order phase transition as a function of the mass. The loop corrections affected this picture only in providing some logarithmic corrections to the scaling behavior near the phase transition.
However, loop corrections are not always so innocuous. For some systems, they can change the structure of the phase transition qualitatively. This Final Project treats the simplest example of such a system, the Coleman-Weinberg model. The analysis of this model draws on a broad variety of topics discussed in Part II; it provides a quite nontrivial application of the effective potential formalism and the use of the renormalization group equation. The phenomenon displayed in this exercise reappears in many contexts, from displacive phase transitions in solids to the thermodynamics of the early universe.
This problem makes use of material in starred sections of the book Peskin \& Schroeder, in particular, Sections 11.3, 11.4, and 13.2. Parts (a) and (e), however, depend only on the unstarred material of Part II. We recommend part (e) as excellent practice in the computation of renormalization group functions.
The Coleman-Weinberg model is the quantum electrodynamics of a scalar field in four dimensions, considered for small values of the scalar field mass. The Lagrangian is
$$\mathcal{L}=-\frac{1}{4}(F_{\mu \nu})^{2}+(D_{\mu} \phi)^{\dagger} D^{\mu} \phi-m^{2} \phi^{\dagger} \phi-\frac{\lambda}{6}(\phi^{\dagger} \phi)^{2},$$
where $\phi(x)$ is a complex-valued scalar field and $D_{\mu} \phi=(\partial_{\mu}+i e A_{\mu}) \phi$. Working in Landau gauge ( $\partial^{\mu} A_{\mu}=0$ ), compute the one-loop correction to the effective potential $V(\phi_{\mathrm{cl}})$. Show that it is renormalized by counterterms for $m^{2}$ and $\lambda$. Renormalize by minimal subtraction, introducing a renormalization scale $M$.
|
[
" V_{\\mathrm{eff}}[\\phi_{\\mathrm{cl}}]= m^{2} \\phi_{\\mathrm{cl}}^{2}+\\frac{\\lambda}{6} \\phi_{\\mathrm{cl}}^{4}-\\frac{1}{4(4 \\pi)^{2}}[3(2 e^{2} \\phi_{\\mathrm{cl}}^{2})^{2}(\\log \\frac{M^{2}}{2 e^{2} \\phi_{\\mathrm{cl}}^{2}}+\\frac{3}{2})+(m^{2}+\\lambda \\phi_{\\mathrm{cl}}^{2})^{2}(\\log \\frac{M^{2}}{m^{2}+\\lambda \\phi_{\\mathrm{cl}}^{2}}+\\frac{3}{2})+(m^{2}+\\frac{\\lambda}{3} \\phi_{\\mathrm{cl}}^{2})^{2}(\\log \\frac{M^{2}}{m^{2}+\\frac{\\lambda}{3} \\phi_{\\mathrm{cl}}^{2}}+\\frac{3}{2})]"
] |
Expression
|
Now we calculate the 1-loop effective potential of the model. We know that 1-loop correction of the effective Lagrangian is given by,
\begin{equation*}
\Delta \mathcal{L}=\frac{\mathrm{i}}{2} \log \operatorname{det}[-\frac{\delta^{2} \mathcal{L}}{\delta \varphi \delta \varphi}]_{\varphi=0}+\delta \mathcal{L} \tag{13.26}
\end{equation*}
where $\varphi$ is the fluctuating fields and $\delta \mathcal{L}$ denotes counterterms.
Let the background value of the complex scalar be $\phi_{\mathrm{cl}}$. By the assumption of Poincaré symmetry, $\phi_{\mathrm{cl}}$ must be a constant. For the same reason, the background value of the vector field $A_{\mu}$ must vanish. In addition, we can set $\phi_{\mathrm{cl}}$ to be real without loss of generality. Then we have,
\phi(x)=\phi_{\mathrm{cl}}+\varphi_{1}(x)+\mathrm{i} \varphi_{2}(x),
where $\varphi_{1}(x), \varphi_{2}(x)$, together with $A_{\mu}(x)$, now serve as fluctuating fields. Expanding the Lagrangian around the background fields and keeping terms quadratic in fluctuating fields only, we get,
\begin{align*}
\mathcal{L}= & -\frac{1}{2} F_{\mu \nu} F^{\mu \nu}+|(\partial_{\mu}+\mathrm{i} e A_{\mu})(\phi_{\mathrm{cl}}+\varphi_{1}+\mathrm{i} \varphi_{2})|^{2} \\
& -m^{2}|\phi_{\mathrm{cl}}+\varphi_{1}+\mathrm{i} \varphi_{2}|^{2}-\frac{\lambda}{6}|\phi_{\mathrm{cl}}+\varphi_{1}+\mathrm{i} \varphi_{2}|^{4} \\
= & \frac{1}{2} A_{\mu}[g^{\mu \nu}(\partial^{2}+2 e^{2} \phi_{\mathrm{cl}}^{2})-\partial^{\mu} \partial^{\nu}] A_{\nu}+\frac{1}{2} \varphi_{1}(-\partial^{2}-m^{2}-\lambda \phi_{\mathrm{cl}}^{2}) \varphi_{1} \\
& +\frac{1}{2} \varphi_{2}(-\partial^{2}-m^{2}-\frac{\lambda}{3} \phi_{\mathrm{cl}}^{2}) \varphi_{2}-2 e \phi_{\mathrm{cl}} A_{\mu} \partial^{\mu} \varphi_{2}+\cdots, \tag{13.27}
\end{align*}
where ". . ." denotes terms other than being quadratic in fluctuating fields. Now we impose the Landau gauge condition $\partial_{\mu} A^{\mu}=0$ to the Lagrangian, which removes the off-diagonal term $-2 e \phi_{\mathrm{cl}} A_{\mu} \partial^{\mu} \varphi_{2}$. Then, according to (13.26), the 1-loop effective Lagrangian can be evaluated as,
\begin{align*}
\frac{\mathrm{i}}{2} \log \operatorname{det}[- & \frac{\delta^{2} \mathcal{L}}{\delta \varphi \delta \varphi}]_{\varphi=0}=\frac{\mathrm{i}}{2}[\log \operatorname{det}(-\eta^{\mu \nu}(\partial^{2}+2 e^{2} \phi_{\mathrm{cl}}^{2})+\partial^{\mu} \partial^{\nu}) \\
& +\log \operatorname{det}(\partial^{2}+m^{2}+\lambda \phi_{\mathrm{cl}}^{2})+\log \operatorname{det}(\partial^{2}+m^{2}+\frac{\lambda}{3} \phi_{\mathrm{cl}}^{2})] \\
= & \frac{\mathrm{i}}{2}
\end{align*} \quad \frac{\mathrm{~d}^{d} k}{(2 \pi)^{d}}[\operatorname{tr} \log (-k^{2}+2 e^{2} \phi_{\mathrm{cl}}^{2})^{3} .
In the second equality we use the following identity,
\begin{equation*}
\operatorname{det}(\lambda I+A B)=\lambda^{n-1}(\lambda+B A) \tag{13.29}
\end{equation*}
where $A$ and $B$ are matrices of $n \times 1$ and $1 \times n$, respectively, $\lambda$ is an arbitrary complex number and $I$ is the $n \times n$ identity matrix. In our case, this gives,
\begin{equation*}
\operatorname{det}(-\eta^{\mu \nu}(\partial^{2}+2 e^{2} \phi_{\mathrm{cl}}^{2})+\partial^{\mu} \partial^{\nu})=-2 e^{2} \phi_{\mathrm{cl}}^{2}(\partial^{2}+2 e^{2} \phi_{\mathrm{cl}}^{2})^{3} \tag{13.30}
\end{equation*}
Then the second equality follows up to an irrelevant constant term. The third equality makes use of the trick in (11.72) of P\&S. Then, for $d=4-\epsilon$ and $\epsilon \rightarrow 0$, we have,
\frac{\mathrm{i}}{2} \log \operatorname{det}[-\frac{\delta^{2} \mathcal{L}}{\delta \varphi \delta \varphi}]_{\varphi=0}=\frac{1}{4(4 \pi)^{2}}[3(2 e^{2} \phi_{\mathrm{cl}}^{2})^{2}(\Delta-\log (2 e^{2} \phi_{\mathrm{cl}}^{2}))
\begin{align*}
& +(m^{2}+\lambda \phi_{\mathrm{cl}}^{2})^{2}(\Delta-\log (m^{2}+\lambda \phi_{\mathrm{cl}}^{2})) \\
& +(m^{2}+\frac{\lambda}{3} \phi_{\mathrm{cl}}^{2})^{2}(\Delta-\log (m^{2}+\frac{\lambda}{3} \phi_{\mathrm{cl}}^{2}))], \tag{13.31}
\end{align*}
where we define $\Delta \equiv \frac{2}{\epsilon}-\gamma+\log 4 \pi+\frac{3}{2}$ for brevity.
Now, with $\overline{M S}$ scheme, we can determine the counterterms in (13.26) to be
\begin{equation*}
\delta \mathcal{L}=\frac{-1}{4(4 \pi)^{2}}[\frac{2}{\epsilon}-\gamma+\log 4 \pi-\log M^{2}](3(2 e^{2} \phi_{\mathrm{cl}}^{2})^{2}+(m^{2}+\lambda \phi_{\mathrm{cl}}^{2})^{2}+(m^{2}+\frac{\lambda}{3} \phi_{\mathrm{cl}}^{2})^{2}) . \tag{13.32}
\end{equation*}
where $M$ is the renormalization scale. Now the effective potential follows directly from (13.26), (13.31) and (13.32),
\begin{align*}
& V_{\mathrm{eff}}[\phi_{\mathrm{cl}}]=m^{2} \phi_{\mathrm{cl}}^{2}+\frac{\lambda}{6} \phi_{\mathrm{cl}}^{4}-\frac{1}{4(4 \pi)^{2}}[3(2 e^{2} \phi_{\mathrm{cl}}^{2})^{2}(\log \frac{M^{2}}{2 e^{2} \phi_{\mathrm{cl}}^{2}}+\frac{3}{2}) \\
& \quad+(m^{2}+\lambda \phi_{\mathrm{cl}}^{2})^{2}(\log \frac{M^{2}}{m^{2}+\lambda \phi_{\mathrm{cl}}^{2}}+\frac{3}{2})+(m^{2}+\frac{\lambda}{3} \phi_{\mathrm{cl}}^{2})^{2}(\log \frac{M^{2}}{m^{2}+\frac{\lambda}{3} \phi_{\mathrm{cl}}^{2}}+\frac{3}{2})] . \tag{13.33}
\end{align*}
|
{"$A_{\\mu}$": "vector field", "$\\phi_{\\mathrm{cl}}$": "classical field value", "$m$": "mass parameter", "$\\lambda$": "coupling constant", "$M$": "renormalization scale", "$e$": "electric charge", "$\\gamma$": "Euler-Mascheroni constant", "$\\varphi_{1}$": "fluctuating field component 1", "$\\varphi_{2}$": "fluctuating field component 2", "$d$": "spacetime dimensionality"}
|
Final Project II
|
The Coleman-Weinberg Potential
|
60 |
Others
|
We discussed the effective potential for an $O(N)$-symmetric $\phi^{4}$ theory in four dimensions. We computed the perturbative corrections to this effective potential, and used the renormalization group to clarify the behavior of the potential for small values of the scalar field mass. After all this work, however, we found that the qualitative dependence of the theory on the mass parameter was unchanged by perturbative corrections. The theory still possessed a second-order phase transition as a function of the mass. The loop corrections affected this picture only in providing some logarithmic corrections to the scaling behavior near the phase transition.
However, loop corrections are not always so innocuous. For some systems, they can change the structure of the phase transition qualitatively. This Final Project treats the simplest example of such a system, the ColemanWeinberg model. The analysis of this model draws on a broad variety of topics discussed in Part II; it provides a quite nontrivial application of the effective potential formalism and the use of the renormalization group equation. The phenomenon displayed in this exercise reappears in many contexts, from displacive phase transitions in solids to the thermodynamics of the early universe.
This problem makes use of material in starred sections of the book Peskin \% Schroeder, in particular, Sections 11.3, 11.4, and 13.2. Parts (a) and (e), however, depend only on the unstarred material of Part II. We recommend part (e) as excellent practice in the computation of renormalization group functions.
The Coleman-Weinberg model is the quantum electrodynamics of a scalar field in four dimensions, considered for small values of the scalar field mass. The Lagrangian is
$$\mathcal{L}=-\frac{1}{4}(F_{\mu \nu})^{2}+(D_{\mu} \phi)^{\dagger} D^{\mu} \phi-m^{2} \phi^{\dagger} \phi-\frac{\lambda}{6}(\phi^{\dagger} \phi)^{2},
$$
where $\phi(x)$ is a complex-valued scalar field and $D_{\mu} \phi=(\partial_{\mu}+i e A_{\mu}) \phi$. Construct the renormalization-group-improved effective potential at $\mu^{2}=$ 0 by applying the results of part (e) to the calculation of part (c). Compute $\langle\phi\rangle$ and the mass of the $\sigma$ particle as a function of $\lambda, e^{2}, M$. Compute the ratio $m_{\sigma} / m_{A}$ to leading order in $e^{2}$, for $\lambda \ll e^{2}$.
|
[
"m_{\\sigma}/m_A = \\frac{\\sqrt{6}e}{4\\pi}"
] |
Expression
|
The effective potential obtained in (c) is not a solution to the renormalization group equation, since it is only a first order result in perturbation expansion. However, it is possible to find an effective potential as a solution to the RG equation, with the result in (c) serving as a sort of "initial condition". The effective potential obtained in this way is said to be RG improved.
The Callan-Symansik equation for the effective potential reads
\begin{equation*}
(M \frac{\partial}{\partial M}+\beta_{\lambda} \frac{\partial}{\partial \lambda}+\beta_{e} \frac{\partial}{\partial e}-\gamma_{\phi} \phi_{\mathrm{cl}} \frac{\partial}{\partial \phi_{\mathrm{cl}}}) V_{\mathrm{eff}}(\phi_{\mathrm{cl}}, \lambda, e ; M)=0 . \tag{13.53}
\end{equation*}
The solution to this equation is well known, that is, the dependence of the sliding energy scale $M$ is described totally by running parameters,
\begin{equation*}
V_{\mathrm{eff}}(\phi_{\mathrm{cl}}, \lambda, e ; M)=V_{\mathrm{eff}}(\bar{\phi}_{\mathrm{cl}}(M^{\prime}), \bar{\lambda}(M^{\prime}), \bar{e}(M^{\prime}) ; M^{\prime}), \tag{13.54}
\end{equation*}
where barred quantities satisfy
\begin{equation*}
M \frac{\partial \bar{\lambda}}{\partial M}=\beta_{\lambda}(\bar{\lambda}, \bar{e}), \quad M \frac{\partial \bar{e}}{\partial M}=\beta_{e}(\bar{\lambda}, \bar{e}), \quad M \frac{\partial \bar{\phi}_{\mathrm{cl}}}{\partial M}=-\gamma_{\phi}(\bar{\lambda}, \bar{e}) \bar{\phi}_{\mathrm{cl}} \tag{13.55}
\end{equation*}
The RG-improved effective potential should be such that when expanded in terms of coupling constants $\lambda$ and $e$, it will recover the result in (c) at the given order. For simplicity here we work under the assumption that $\lambda \sim e^{4}$, so that all terms of higher orders of coupling constants than $\lambda$ and $e^{4}$ can be ignored. In this case, the perturbative calculation in (c) gives
\begin{equation*}
V_{\mathrm{eff}}=\frac{\lambda}{6} \phi_{\mathrm{cl}}^{4}+\frac{3 e^{4} \phi_{\mathrm{cl}}^{4}}{(4 \pi)^{2}}(\log \frac{2 e^{2} \phi_{\mathrm{cl}}^{2}}{M^{2}}-\frac{3}{2}) . \tag{13.56}
\end{equation*}
Now we claim that the RG-improved edition of this result reads
\begin{equation*}
V_{\mathrm{eff}}=\frac{\bar{\lambda}}{6} \bar{\phi}_{\mathrm{cl}}^{4}+\frac{3 \bar{e}^{4} \bar{\phi}_{\mathrm{cl}}^{4}}{(4 \pi)^{2}}(\log 2 \bar{e}^{2}-\frac{3}{2}) . \tag{13.57}
\end{equation*}
To see this, we firstly solve the renormalization group equations (13.55),
\begin{align*}
\bar{\lambda}(M^{\prime}) & =\bar{e}^{4}(\frac{\lambda}{e^{4}}+\frac{9}{4 \pi^{2}} \log \frac{M^{\prime}}{M}) \tag{13.58}\\
\bar{e}^{2}(M^{\prime}) & =\frac{e^{2}}{1-(e^{2} / 24 \pi^{2}) \log (M^{\prime} / M)} \tag{13.59}\\
\bar{\phi}_{\mathrm{cl}}(M^{\prime}) & =\phi_{\mathrm{cl}}(\frac{M^{\prime}}{M})^{2 e^{2} /(4 \pi)^{2}} \tag{13.60}
\end{align*}
where the unbarred quantities $\lambda, e$ and $\phi_{\mathrm{cl}}$ are evaluated at scale $M$. Now we substitute these results back into the RG-improved effective potential (13.57) and expand in terms of coupling constants. Then it is straightforward to see that the result recovers (13.56). To see the spontaneous symmetry breaking still occurs, we note that the running coupling $\bar{\lambda}(M^{\prime})$ flows to negative value rapidly for small $M^{\prime}=\phi_{\mathrm{cl}}$, while $\bar{e}(M^{\prime})$ changes mildly along the $\phi_{\mathrm{cl}}$ scale, as can be seen directly from Figure 13.2. Therefore the the coefficient before $\phi_{\mathrm{cl}}^{4}$ is negative for small $\phi_{\mathrm{cl}}$ and positive for large $\phi_{\mathrm{cl}}$. As a consequence, the minimum of this effective potential should be away from $\phi_{\mathrm{cl}}=0$, namely the $U(1)$ symmetry is spontaneously broken.
To find the scalar mass $m_{\sigma}$ in this case (with $\mu=0$ ), we calculate the second derivative of the effective potential $V_{\text {eff }}$ with respect to $\phi_{\mathrm{cl}}$. Since the renormalization scale $M$ can be arbitrarily chosen, we set it to be $M^{2}=2 e^{2}\langle\phi_{\mathrm{cl}}^{2}\rangle$ to simplify the calculation. Then the vanishing of the first derivative of $V_{\mathrm{eff}}$ at $\phi_{\mathrm{cl}}=\langle\phi_{\mathrm{cl}}\rangle$ implies that $\lambda=9 e^{4} / 8 \pi^{2}$. Insert this back to $V_{\text {eff }}$ in (13.56), we find that
\begin{equation*}
V_{\mathrm{eff}}=\frac{3 e^{4} \phi_{\mathrm{cl}}^{4}}{16 \pi^{2}}(\log \frac{\phi_{\mathrm{cl}}^{2}}{\langle\phi_{\mathrm{cl}}^{2}\rangle}-\frac{1}{2}) . \tag{13.61}
\end{equation*}
Then, taking the second derivative of this expression with respect to $\phi_{\mathrm{cl}}$, we get the scalar mass $m_{\sigma}^{2}=3 e^{4}\langle\phi_{\mathrm{cl}}^{2}\rangle / 4 \pi^{2}=3 e^{4} v^{2} / 8 \pi^{2}$. Recall that the gauge boson's mass $m_{A}$ is given by $m_{A}=e^{2} v^{2}$ at the leading order, thus we conclude that $m_{\sigma}^{2} / m_{A}^{2}=3 e^{2} / 8 \pi^{2}$ at the leading order in $e^{2}$.
|
{"$\\mu$": "scale parameter", "$\\phi$": "field value", "$\\sigma$": "scalar particle", "$\\lambda$": "coupling constant", "$e$": "electric charge", "$M$": "energy scale", "$\\langle\\phi\\rangle$": "vacuum expectation value of the field", "$m_{\\sigma}$": "mass of the sigma particle", "$m_{A}$": "mass of the gauge boson A", "$\\beta_{\\lambda}$": "beta function for the coupling constant $\\lambda$", "$\\beta_{e}$": "beta function for the electric charge $e$", "$\\gamma_{\\phi}$": "anomalous dimension of the field", "$\\phi_{\\mathrm{cl}}$": "classical field", "$V_{\\mathrm{eff}}$": "effective potential", "$\\bar{\\lambda}$": "running coupling constant $\\lambda$", "$\\bar{e}$": "running electric charge", "$\\bar{\\phi}_{\\mathrm{cl}}$": "running classical field", "$\\langle\\phi_{\\mathrm{cl}}^{2}\\rangle$": "vacuum expectation value of the squared classical field", "$v$": "vacuum expectation value of the field"}
|
Final Project II
|
The Coleman-Weinberg Potential
|
61 |
Others
|
Deep inelastic scattering from a photon. Consider the problem of deepinelastic scattering of an electron from a photon. This process can actually be measured by analyzing the reaction $e^{+} e^{-} \rightarrow e^{+} e^{-}+X$ in the regime where the positron goes forward, with emission of a collinear photon, which then has a hard reaction with the electron. Let us analyze this process to leading order in QED and to leading-log order in QCD. To predict the photon structure functions, it is reasonable to integrate the renormalization group equations with the initial condition that the parton distribution for photons in the photon is $\delta(x-1)$ at $Q^{2}=(\frac{1}{2} \mathrm{GeV})^{2}$. Take $\Lambda=150 \mathrm{MeV}$. Assume for simplicity that there are four flavors of quarks, $u, d, c$, and $s$, with charges $2 / 3$, $-1 / 3,2 / 3,-1 / 3$, respectively, and that it is always possible to ignore the masses of these quarks. Use the Altarelli-Parisi equations to compute the parton distributions for quarks and antiquarks in the photon, to leading order in QED and to zeroth order in QCD.
|
[
"f_{q}(x, Q)=f_{\\bar{q}}(x, Q)=\\frac{3 Q_{q}^{2} \\alpha}{2 \\pi} \\log \\frac{Q^{2}}{Q_{0}^{2}}[x^{2}+(1-x)^{2}]"
] |
Expression
|
The A-P equation for parton distributions in the photon can be easily written down by using the QED splitting functions listed in (17.121) of Peskin \& Schroeder. Taking account of quarks' electric charge properly, we have,
\begin{align*}
& \frac{\mathrm{d}}{\mathrm{~d} \log Q} f_{q}(x, Q)= \frac{3 Q_{q}^{2} \alpha}{\pi} \int_{x}^{1} \frac{\mathrm{d} z}{z}{P_{e \leftarrow e}(z) f_{q}(\frac{x}{z}, Q)+P_{e \leftarrow \gamma}(z) f_{\gamma}(\frac{x}{z}, Q)}, \tag{18.68}\\
& \frac{\mathrm{d}}{\mathrm{~d} \log Q} f_{\bar{q}}(x, Q)= \frac{3 Q_{q}^{2} \alpha}{\pi} \int_{x}^{1} \frac{\mathrm{d} z}{z}{P_{e \leftarrow e}(z) f_{\bar{q}}(\frac{x}{z}, Q)+P_{e \leftarrow \gamma}(z) f_{\gamma}(\frac{x}{z}, Q)}, \tag{18.69}\\
& \frac{\mathrm{d}}{\mathrm{~d} \log Q} f_{\gamma}(x, Q)= \sum_{q} \frac{3 Q_{q}^{2} \alpha}{\pi} \int_{x}^{1} \frac{\mathrm{d} z}{z}{P_{\gamma \leftarrow e}(z)[f_{q}(\frac{x}{z}, Q)+f_{\bar{q}}(\frac{x}{z}, Q)] \\
&\quad+P_{\gamma \leftarrow \gamma}(z) f_{\gamma}(\frac{x}{z}, Q)}, \tag{18.70}
\end{align*}
where the splitting functions are
\begin{align*}
& P_{e \leftarrow e}(z)=\frac{1+z^{2}}{(1-z)_{+}}+\frac{3}{2} \delta(1-z), \tag{18.71}\\
& P_{\gamma \leftarrow e}(z)=\frac{1+(1-z)^{2}}{z}, \tag{18.72}\\
& P_{e \leftarrow \gamma}(z)=z^{2}+(1-z)^{2} \tag{18.73}\\
& P_{\gamma \leftarrow \gamma}(z)=-\frac{2}{3} \delta(1-z) . \tag{18.74}
\end{align*}
We take $q=u, d, c, s$, and $Q_{u, c}=+2 / 3, Q_{d, s}=-1 / 3$. The factor 3 in the A-P equations above takes account of 3 colors. Since no more leptons appear in final states other than original $e^{+} e^{-}$, they are not included in the photon structure. With the initial condition
$f_{\gamma}(x, Q_{0})=\delta(1-x)$ and $f_{q, \bar{q}}(x, Q_{0})=0$ where $Q_{0}=0.5 \mathrm{GeV}$, these distribution functions can be solved from the equations above to the first order in $\alpha$, to be
\begin{align*}
& f_{q}(x, Q)=f_{\bar{q}}(x, Q)=\frac{3 Q_{q}^{2} \alpha}{2 \pi} \log \frac{Q^{2}}{Q_{0}^{2}}[x^{2}+(1-x)^{2}] \tag{18.75}\\
& f_{\gamma}(x, Q)=(1-\sum_{q} \frac{Q_{q}^{2} \alpha}{\pi} \log \frac{Q^{2}}{Q_{0}^{2}}) \delta(1-x) \tag{18.76}
\end{align*}
|
{"$f_{q}$": "parton distribution for quarks", "$f_{\\bar{q}}$": "parton distribution for antiquarks", "$Q_{q}$": "electric charge of quark", "$\\alpha$": "fine-structure constant (QED coupling constant)", "$Q$": "momentum transfer or energy scale", "$x$": "Bjorken scaling variable", "$f_{\\gamma}$": "parton distribution for the photon", "$Q_{0}$": "initial energy scale", "$u$": "up quark", "$d$": "down quark", "$c$": "charm quark", "$s$": "strange quark"}
|
Operator Products and Effective Vertices
|
Deep inelastic scattering from a photon
|
62 |
Others
|
Neutral-current deep inelastic scattering. In Eq. (17.35) of Peskin \& Schroeder, we wrote formulae for neutrino and antineutrino deep inelastic scattering with $W^{\pm}$ exchange. Neutrinos and antineutrinos can also scatter by exchanging a $Z^{0}$. This process, which leads to a hadronic jet but no observable outgoing lepton, is called the neutral current reaction. Compute $\frac{\mathrm{d}^{2} \sigma}{\mathrm{d} x \mathrm{d} y}(\nu p \rightarrow \nu X)$ for neutral current deep inelastic scattering of neutrinos from protons, accounting for scattering from $u$ and $d$ quarks and antiquarks.
|
[
"\\frac{\\mathrm{d}^{2} \\sigma}{\\mathrm{d} x \\mathrm{d} y}(\\nu p \\rightarrow \\nu X)=\\frac{G_{F}^{2} s x}{4 \\pi} {[(1-\\frac{4}{3} s_{w}^{2})^{2}+\\frac{16}{9} s_{w}^{4}(1-y^{2})] f_{u}(x) + [(1-\\frac{2}{3} s_{w}^{2})^{2}+\\frac{4}{9} s_{w}^{4}(1-y^{2})] f_{d}(x) +[\\frac{16}{9} s_{w}^{4}+(1-\\frac{4}{3} s_{w}^{2})^{2}(1-y^{2})] f_{\\bar{u}}(x) +[\\frac{4}{9} s_{w}^{4}+(1-\\frac{2}{3} s_{w}^{2})^{2}(1-y^{2})] f_{\\bar{d}}(x)}"
] |
Expression
|
In this problem we study the neutral-current deep inelastic scattering. The process is mediated by $Z^{0}$ boson. Assuming $m_{Z}$ is much larger than the energy scale of the scattering process, we can write down the corresponding effective operators, from the neutral-current Feynman rules in electroweak theory,
\begin{align*}
\Delta \mathcal{L}=\frac{g^{2}}{4 m_{W}^{2}}(\bar{\nu} \gamma^{\mu}) P_{L} \nu & {[\bar{u} \gamma_{\mu}((1-\frac{4}{3} s_{w}^{2}) P_{L}-\frac{4}{3} s_{w}^{2} P_{R}) u} \\
& +\bar{d} \gamma_{\mu}((1-\frac{2}{3} s_{w}^{2}) P_{L}-\frac{2}{3} s_{w}^{2} P_{R}) d]+ \text { h.c. } \tag{20.30}
\end{align*}
where $P_{L}=(1-\gamma^{5}) / 2$ and $P_{R}=(1+\gamma^{5}) / 2$ are left- and right-handed projectors, respectively. Compare the effective operator with the charged-operator in (17.31) of Peskin \& Schroeder, we can write down directly the differential cross section for neutrino scattering by modifying (17.35) in Peskin \& Schroeder properly, as
\begin{align*}
\frac{\mathrm{d}^{2} \sigma}{\mathrm{~d} x \mathrm{~d} y}(\nu p \rightarrow \nu X)=\frac{G_{F}^{2} s x}{4 \pi} & {[(1-\frac{4}{3} s_{w}^{2})^{2}+\frac{16}{9} s_{w}^{4}(1-y^{2})] f_{u}(x) \\
+ & {[(1-\frac{2}{3} s_{w}^{2})^{2}+\frac{4}{9} s_{w}^{4}(1-y^{2})] f_{d}(x) } \\
& +[\frac{16}{9} s_{w}^{4}+(1-\frac{4}{3} s_{w}^{2})^{2}(1-y^{2})] f_{\bar{u}}(x) \\
& +[\frac{4}{9} s_{w}^{4}+(1-\frac{2}{3} s_{w}^{2})^{2}(1-y^{2})] f_{\bar{d}}(x)}. \tag{20.31}
\end{align*}
|
{"$W^{\\pm}$": "charged W boson", "$Z^{0}$": "neutral Z boson", "$m_{Z}$": "mass of the Z boson", "$g$": "electroweak coupling constant", "$m_{W}$": "mass of the W boson", "$P_{L}$": "left-handed projector", "$P_{R}$": "right-handed projector", "$G_{F}$": "Fermi coupling constant", "$s$": "center-of-mass energy squared", "$x$": "Bjorken scaling variable", "$y$": "inelasticity", "$s_{w}$": "sine of the Weinberg angle", "$u$": "up quark", "$d$": "down quark", "$\\nu$": "neutrino", "$f_{u}(x)$": "parton distribution function for up quark", "$f_{d}(x)$": "parton distribution function for down quark", "$f_{\\bar{u}}(x)$": "parton distribution function for up antiquark", "$f_{\\bar{d}}(x)$": "parton distribution function for down antiquark"}
|
Gauge Theories with Spontaneous Symmetry Breaking
|
Neutral-current deep inelastic scattering
|
63 |
Others
|
Neutral-current deep inelastic scattering. In Eq. (17.35) of Peskin \& Schroeder, we wrote formulae for neutrino and antineutrino deep inelastic scattering with $W^{ \pm}$exchange. Neutrinos and antineutrinos can also scatter by exchanging a $Z^{0}$. This process, which leads to a hadronic jet but no observable outgoing lepton, is called the neutral current reaction. Compute $\frac{\mathrm{d}^{2} \sigma}{\mathrm{d} x \mathrm{d} y}(\bar{\nu} p \rightarrow \bar{\nu} X)$ for neutral current deep inelastic scattering of antineutrinos from protons, accounting for scattering from $u$ and $d$ quarks and antiquarks.
|
[
"\\frac{\\mathrm{d}^{2} \\sigma}{\\mathrm{d} x \\mathrm{d} y}(\\bar{\\nu} p \\rightarrow \\bar{\\nu} X)=\\frac{G_{F}^{2} s x}{4 \\pi} {[\\frac{16}{9} s_{w}^{4}+(1-\\frac{4}{3} s_{w}^{2})^{2}(1-y^{2})] f_{u}(x) +[\\frac{4}{9} s_{w}^{4}+(1-\\frac{2}{3} s_{w}^{2})^{2}(1-y^{2})] f_{d}(x) +[(1-\\frac{4}{3} s_{w}^{2})^{2}+\\frac{16}{9} s_{w}^{4}(1-y^{2})] f_{\\bar{u}}(x) +[(1-\\frac{2}{3} s_{w}^{2})^{2}+\\frac{4}{9} s_{w}^{4}(1-y^{2})] f_{\\bar{d}}(x)}"
] |
Expression
|
In this problem we study the neutral-current deep inelastic scattering. The process is mediated by $Z^{0}$ boson. Assuming $m_{Z}$ is much larger than the energy scale of the scattering process, we can write down the corresponding effective operators, from the neutral-current Feynman rules in electroweak theory,
\begin{align*}
\Delta \mathcal{L}=\frac{g^{2}}{4 m_{W}^{2}}(\bar{\nu} \gamma^{\mu}) P_{L} \nu & {[\bar{u} \gamma_{\mu}((1-\frac{4}{3} s_{w}^{2}) P_{L}-\frac{4}{3} s_{w}^{2} P_{R}) u} \\
& +\bar{d} \gamma_{\mu}((1-\frac{2}{3} s_{w}^{2}) P_{L}-\frac{2}{3} s_{w}^{2} P_{R}) d]+ \text { h.c. } \tag{20.30}
\end{align*}
where $P_{L}=(1-\gamma^{5}) / 2$ and $P_{R}=(1+\gamma^{5}) / 2$ are left- and right-handed projectors, respectively. Compare the effective operator with the charged-operator in (17.31) of Peskin \& Schroeder, we can write down directly the differential cross section for antineutrino scattering by modifying (17.35) in Peskin \& Schroeder properly, as
\begin{align*}
\frac{\mathrm{d}^{2} \sigma}{\mathrm{~d} x \mathrm{~d} y}(\bar{\nu} p \rightarrow \bar{\nu} X)=\frac{G_{F}^{2} s x}{4 \pi} & {[\frac{16}{9} s_{w}^{4}+(1-\frac{4}{3} s_{w}^{2})^{2}(1-y^{2})] f_{u}(x) \\
& +[\frac{4}{9} s_{w}^{4}+(1-\frac{2}{3} s_{w}^{2})^{2}(1-y^{2})] f_{d}(x) \\
& +[(1-\frac{4}{3} s_{w}^{2})^{2}+\frac{16}{9} s_{w}^{4}(1-y^{2})] f_{\bar{u}}(x) \\
& +[(1-\frac{2}{3} s_{w}^{2})^{2}+\frac{4}{9} s_{w}^{4}(1-y^{2})] f_{\bar{d}}(x)} . \tag{20.32}
\end{align*}
|
{"$W^{\\pm}$": "charged W boson", "$Z^{0}$": "neutral Z boson", "$m_{Z}$": "mass of the Z boson", "$g$": "electroweak coupling constant", "$m_{W}$": "mass of the W boson", "$\\nu$": "neutrino", "$\\bar{\\nu}$": "antineutrino", "$\\bar{u}$": "up antiquark", "$\\bar{d}$": "down antiquark", "$u$": "up quark", "$d$": "down quark", "$s_{w}$": "sine of the Weinberg angle", "$P_{L}$": "left-handed projector", "$P_{R}$": "right-handed projector", "$G_{F}$": "Fermi coupling constant", "$s$": "Mandelstam variable, center-of-mass energy squared", "$x$": "Bjorken scaling variable", "$y$": "inelasticity", "$f_{u}(x)$": "parton distribution function for up quarks", "$f_{d}(x)$": "parton distribution function for down quarks", "$f_{\\bar{u}}(x)$": "parton distribution function for up antiquarks", "$f_{\\bar{d}}(x)$": "parton distribution function for down antiquarks"}
|
Gauge Theories with Spontaneous Symmetry Breaking
|
Neutral-current deep inelastic scattering
|
64 |
Others
|
A model with two Higgs fields. Find the couplings of the physical charged Higgs boson $\phi^{+}$ to the quark mass eigenstates, given the Lagrangian $$\mathcal{L}_{m} = -\lambda_{d}^{ij} \bar{Q}_{L}^{i} \cdot \phi_{1} d_{R}^{j} - \lambda_{u}^{ij} \epsilon^{ab} \bar{Q}_{La}^{i} \phi_{2b}^{\dagger} u_{R}^{j} + \text{h.c.}$$
|
[
"\\frac{\\sqrt{2}}{v} \\left( \\bar{u}_L V_{\\text{CKM}} m_d d_R \\phi^+ \\tan\\beta + \\bar{d}_L V_{\\text{CKM}}^\\dagger m_u u_R \\phi^- \\cot\\beta \\right) + \\text{h.c.}"
] |
Expression
|
Assuming that the Yukawa interactions between quarks and scalars take the following form:
\begin{align}
\mathcal{L}_m &= -\left(\bar{u}_L \quad \bar{d}_L\right)\left[\lambda_d\left(\frac{\pi_1^+}{\sqrt{2}v_1}\right)d_R + \lambda_u\left(\frac{\frac{1}{\sqrt{2}}v_2}{\pi^-}\right)u_R\right] + \text{h.c.}, \tag{20.44}
\end{align}
where we have suppressed flavor indices and neglected neutral scalar components. We focus on charged components only. Using Peskin \& Schroeder's notation, we perform the replacements \( u_L \to U_u u_L \), \( d_L \to U_d d_L \), \( u_R \to W_u u_R \), and \( d_R \to W_d d_R \). With the diagonalization \(\lambda_d = U_d D_d W_d^\dagger\) and \(\lambda_u = U_u D_u W_u^\dagger\) (where \(D_d\) and \(D_u\) are diagonal matrices), we derive:
\begin{align}
\mathcal{L}_m &= -\frac{1}{\sqrt{2}}\left(v_1\bar{d}_L D_d d_R + v_2\bar{u}_L D_u u_R\right) \nonumber \\
&\quad - \bar{u}_L V_{\text{CKM}} D_d d_R \pi_1^+ + \bar{d}_L V_{\text{CKM}}^\dagger D_u u_R \pi_2^- + \text{h.c.}. \tag{20.45}
\end{align}
From the first term, the diagonal quark mass matrices are given by \( m_u = \frac{v_1}{\sqrt{2}}D_u \) and \( m_d = \frac{v_2}{\sqrt{2}}D_d \). Defining \( v = \sqrt{v_1^2 + v_2^2} \), and using the relations \( \pi_1^+ = -\phi^+ \sin\beta + \cdots \), \( \pi_2^+ = \phi^+ \cos\beta + \cdots \), the Yukawa interactions with charged bosons become:
\begin{align}
\mathcal{L}_m &\Rightarrow -\frac{\sqrt{2}}{v_1} \left( \bar{u}_L V_{\text{CKM}} m_d d_R \pi_1^+ + \bar{d}_L V_{\text{CKM}}^\dagger m_u u_R \pi_2^- \right) + \text{h.c.} \nonumber \\
&\Rightarrow \frac{\sqrt{2}}{v} \left( \bar{u}_L V_{\text{CKM}} m_d d_R \phi^+ \tan\beta + \bar{d}_L V_{\text{CKM}}^\dagger m_u u_R \phi^- \cot\beta \right) + \text{h.c.}. \tag{20.46}
\end{align}
|
{"$\\phi^{+}$": "charged Higgs boson", "$\\mathcal{L}_{m}$": "mass term in the Lagrangian", "$\\lambda_{d}^{ij}$": "Yukawa coupling matrix for down-type quarks", "$\\lambda_{u}^{ij}$": "Yukawa coupling matrix for up-type quarks", "$\\bar{Q}_{L}^{i}$": "left-handed quark doublet", "$d_{R}^{j}$": "right-handed down-type quark", "$\\phi_{1}$": "scalar field 1", "$\\epsilon^{ab}$": "antisymmetric tensor", "$\\bar{Q}_{La}^{i}$": "component of the left-handed quark doublet", "$\\phi_{2b}^{\\dagger}$": "conjugate of the component of scalar field 2", "$u_{R}^{j}$": "right-handed up-type quark", "$u_L$": "left-handed up-type quark", "$d_L$": "left-handed down-type quark", "$v_1$": "vacuum expectation value component 1", "$v_2$": "vacuum expectation value component 2", "$v$": "Higgs vacuum expectation value", "$\\pi_1^+$": "Goldstone boson component for charged Higgs interactions", "$\\pi_2^-$": "Goldstone boson component for charged Higgs interactions", "$V_{\\text{CKM}}$": "CKM matrix", "$m_u$": "up-type quark mass matrix", "$m_d$": "down-type quark mass matrix", "$\\tan\\beta$": "ratio of two Higgs doublet vacuum expectation values", "$\\cot\\beta$": "cotangent of the ratio of the two Higgs doublet vacuum expectation values"}
|
Gauge Theories with Spontaneous Symmetry Breaking
|
A model with two Higgs fields
|
65 |
Others
|
We discussed the mystery of the origin of spontaneous symmetry breaking in the weak interactions. The simplest hypothesis is that the $S U(2) \times U(1)$ gauge symmetry of the weak interactions is broken by the expectation value of a two-component scalar field $\phi$. However, since we have almost no experimental information about the mechanism of this symmetry breaking, many other possibilities can be suggested.
Eventually, this problem should be resolved by experimental observation of the particles associated with the symmetry breaking. To form incisive experimental tests, we should compute the properties expected for these particles. We saw in Section 20.2 of Peskin \& Schroeder that, if the symmetry is indeed broken by a single scalar field $\phi$, the symmetry-breaking sector contributes only one new particle, a scalar $h^{0}$ called the Higgs boson. The mass $m_{h}$ of this particle is unknown. However, the couplings of the $h^{0}$ to known fermions and bosons are completely determined by the masses of those particles and the weak interaction coupling constants. Thus, it is possible to compute the amplitudes for production and decay of the $h^{0}$ in some detail. More complicated models of $S U(2) \times U(1)$ symmetry breaking typically contain one or more particles that share some properties with the $h^{0}$. Thus, this study is a useful starting point for the more general analysis of experimental tests of these models.
In this Final Project you will compute the amplitudes for the Higgs boson $h^{0}$ to decay to pairs of quarks, leptons, and gauge bosons. The computations beautifully illustrate the working of perturbation theory for non-Abelian gauge fields. Those decays of the Higgs boson that involve quarks and gluons bring in aspects of QCD. Thus, this exercise reviews all of the important technical methods of Part III. Except for a question raised at the end of part (a), the problem relies only on material from unstarred sections of Part III. The material in Chapter 20 plays an essential role. Material from Chapter 21 enters the analysis only in parts (b) and (f), and the other parts of the problem (except for the final summary in part (h)) do not rely on these. If you have studied Section 19.5, you will have some additional insight into the results of parts (c) and (f), but this insight is not necessary to work the problem.
Consider, then, the minimal form of the Glashow-Weinberg-Salam electroweak theory with one Higgs scalar field $\phi$. The physical Higgs boson $h^{0}$ of
this theory was discussed in Section 20.2, and we listed there the couplings of this particle to quarks, leptons, and gauge bosons. You can now use that information to compute the amplitudes for the various possible decays of the $h^{0}$ as a function of its mass $m_{h}$. You will discover that the decay pattern has a complicated dependence on the mass of the Higgs boson, with different decay modes dominating in different mass ranges. The dependences of the various decay rates on $m_{h}$ illustrate many aspects of the physics of gauge theories that we have discussed in Part III.
In working this exercise, you should consider $m_{h}$ as a free parameter. For the other parameters of weak-interaction theory, you might use the following values: $m_{b}=5 \mathrm{GeV}, m_{t}=175 \mathrm{GeV}, m_{W}=80 \mathrm{GeV}, m_{Z}=91 \mathrm{GeV}, \sin ^{2} \theta_{w}$ $=0.23, \alpha_{s}(m_{Z})=0.12$. Provide the tree-level decay width $\Gamma(h^0 \rightarrow f\bar{f})$ for a Higgs boson $h^0$ decaying into a fermion-antifermion pair $f\bar{f}$ (where $f$ is a quark or lepton of the Standard Model), expressed in terms of the fine-structure constant $\alpha$, the Higgs mass $m_h$, the fermion mass $m_f$, the W boson mass $m_W$, the weak mixing angle $\theta_w$, and the color factor $N_c(f)$ (1 for leptons, 3 for quarks).
|
[
"\\Gamma(h^0 \\rightarrow f\\bar{f}) = (\\frac{\\alpha m_h}{8\\sin^2\\theta_w}) \\cdot \\frac{m_f^2}{m_W^2} (1 - \\frac{4m_f^2}{m_h^2})^{3/2}"
] |
Expression
|
In this final project, we calculate partial widths of various decay channels of the standard model Higgs boson. Although a standard-model-Higgs-like boson has been found at the LHC with mass around 125 GeV , it is still instructive to treat the mass of the Higgs boson as a free parameter in the following calculation.
The main decay modes of Higgs boson include $h^{0} \rightarrow f \bar{f}$ with $f$ the standard model fermions, $h^{0} \rightarrow W^{+} W^{-}, h^{0} \rightarrow Z^{0} Z^{0}, h^{0} \rightarrow g g$ and $h^{0} \rightarrow \gamma \gamma$. The former three processes appear at the tree level, while the leading order contributions to the latter two processes are at one-loop level. We will work out the decay widths of these processes in the following.
In this problem we only consider the two-body final states. The calculation of decay width needs the integral over the phase space of the two-body final states. By momentum conservation and rotational symmetry, we can always parameterize the momenta of two final particles in CM frame to be $p_{1}=(E, 0,0, p)$ and $p_{2}=(E, 0,0,-p)$, where $E=\frac{1}{2} m_{h}$ by energy conservation. Then the amplitude $\mathcal{M}$ will have no angular dependence. Then the phase space integral reads
\begin{equation*}
\int \mathrm{d} \Pi_{2}|\mathcal{M}|^{2}=\frac{1}{4 \pi} \frac{p}{m_{h}}|\mathcal{M}|^{2} . \tag{21.50}
\end{equation*}
Then the decay width is given by
\begin{equation*}
\Gamma=\frac{1}{2 m_{h}} \int \mathrm{~d} \Pi_{2}|\mathcal{M}|^{2}=\frac{1}{8 \pi} \frac{p}{m_{h}^{2}}|\mathcal{M}|^{2} . \tag{21.51}
\end{equation*}
In part (d) of this problem, we will also be dealing with the production of the Higgs boson from two-gluon initial state, thus we also write down the formula here for the cross section of the one-body final state from two identical initial particle. This time, the two ingoing particles have momenta $k_{1}=(E, 0,0, k)$ and $k_{2}=(E, 0,0,-k)$, with $E^{2}=k^{2}+m_{i}^{2}$ and $2 E=m_{f}$ where $m_{i}$ and $m_{f}$ are masses of initial particles and final particle, respectively. The final particle has momentum $p=(m_{f}, 0,0,0)$. Then, the cross section is given by
\begin{align*}
\sigma & =\frac{1}{2 \beta s} \int \frac{\mathrm{~d}^{3} p}{(2 \pi)^{3}} \frac{1}{2 E_{p}}|\mathcal{M}|^{2}(2 \pi)^{4} \delta^{(4)}(p-k_{1}-k_{2}) \\
& =\frac{1}{4 m_{f} \beta s}|\mathcal{M}|^{2}(2 \pi) \delta(2 k-m_{f})=\frac{\pi}{\beta m_{f}^{2}}|\mathcal{M}|^{2} \delta(s-m_{f}^{2}), \tag{21.52}
\end{align*}
where $\beta=\sqrt{1-(4 m_{i} / m_{f})^{2}}$ is the magnitude of the velocity of the initial particle in the center-of-mass frame.
(a) The easiest calculation of above processes is $h^{0} \rightarrow f \bar{f}$, where $f$ represents all quarks and charged leptons. The tree level contribution to this process involves a single Yukawa vertex only. The corresponding amplitude is given by
\begin{equation*}
\mathrm{i} \mathcal{M}(h^{0} \rightarrow f \bar{f})=-\frac{\mathrm{i} m_{f}}{v} \bar{u}^{*}(p_{1}) v(p_{2}) . \tag{21.53}
\end{equation*}
Then it is straightforward to get the squared amplitude with final spins summed to be
\begin{equation*}
\sum|\mathcal{M}(h^{0} \rightarrow f \bar{f})|^{2}=\frac{m_{f}^{2}}{v^{2}} \operatorname{tr}[(\not p_{1}+m_{f})(\not p_{2}-m_{f})]=\frac{2 m_{f}^{2}}{v^{2}}(m_{h}^{2}-4 m_{f}^{2}) \tag{21.54}
\end{equation*}
In CM frame, the final states momenta can be taken to be $p_{1}=(E, 0,0, p)$ and $p_{2}=$ $(E, 0,0,-p)$, with $E=\frac{1}{2} m_{h}$ and $p^{2}=E^{2}-m_{f}^{2}$. Then the decay width is given by
\begin{equation*}
\Gamma(h^{0} \rightarrow f \bar{f})=\frac{1}{8 \pi} \frac{p}{m_{h}^{2}}|\mathcal{M}|^{2}=\frac{m_{h} m_{f}^{2}}{8 v^{2}}(1-\frac{4 m_{f}^{2}}{m_{h}^{2}})^{3 / 2} \tag{21.55}
\end{equation*}
This expression can be expressed in terms of the fine structure constant $\alpha$, the mass of $W$ boson $m_{w}$ and Weinberg angle $\sin \theta_{w}$, as
\begin{equation*}
\Gamma(h^{0} \rightarrow f \bar{f})=\frac{\alpha m_{h}}{8 \sin ^{2} \theta_{w}} \frac{m_{f}^{2}}{m_{W}^{2}}(1-\frac{4 m_{f}^{2}}{m_{h}^{2}})^{3 / 2} \tag{21.56}
\end{equation*}
|
{"$\\Gamma$": "decay width", "$h^0$": "Higgs boson", "$f$": "fermion", "$\\bar{f}$": "antifermion", "$\\alpha$": "fine-structure constant", "$m_h$": "Higgs mass", "$m_f$": "fermion mass", "$m_W$": "W boson mass", "$\\theta_w$": "weak mixing angle", "$N_c(f)$": "color factor"}
|
Final Project III
|
Decays of the Higgs Boson
|
66 |
Others
|
We discussed the mystery of the origin of spontaneous symmetry breaking in the weak interactions. The simplest hypothesis is that the $S U(2) \times U(1)$ gauge symmetry of the weak interactions is broken by the expectation value of a two-component scalar field $\phi$. However, since we have almost no experimental information about the mechanism of this symmetry breaking, many other possibilities can be suggested.
Eventually, this problem should be resolved by experimental observation of the particles associated with the symmetry breaking. To form incisive experimental tests, we should compute the properties expected for these particles. We saw in Section 20.2 that, if the symmetry is indeed broken by a single scalar field $\phi$, the symmetry-breaking sector contributes only one new particle, a scalar $h^{0}$ called the Higgs boson. The mass $m_{h}$ of this particle is unknown. However, the couplings of the $h^{0}$ to known fermions and bosons are completely determined by the masses of those particles and the weak interaction coupling constants. Thus, it is possible to compute the amplitudes for production and decay of the $h^{0}$ in some detail. More complicated models of $S U(2) \times U(1)$ symmetry breaking typically contain one or more particles that share some properties with the $h^{0}$. Thus, this study is a useful starting point for the more general analysis of experimental tests of these models.
In this Final Project you will compute the amplitudes for the Higgs boson $h^{0}$ to decay to pairs of quarks, leptons, and gauge bosons. The computations beautifully illustrate the working of perturbation theory for non-Abelian gauge fields. Those decays of the Higgs boson that involve quarks and gluons bring in aspects of QCD. Thus, this exercise reviews all of the important technical methods of Part III. Except for a question raised at the end of part (a), the problem relies only on material from unstarred sections of Part III. The material in Chapter 20 plays an essential role. Material from Chapter 21 enters the analysis only in parts (b) and (f), and the other parts of the problem (except for the final summary in part (h)) do not rely on these. If you have studied Section 19.5, you will have some additional insight into the results of parts (c) and (f), but this insight is not necessary to work the problem.
Consider, then, the minimal form of the Glashow-Weinberg-Salam electroweak theory with one Higgs scalar field $\phi$. The physical Higgs boson $h^{0}$ of
this theory was discussed in Section 20.2, and we listed there the couplings of this particle to quarks, leptons, and gauge bosons. You can now use that information to compute the amplitudes for the various possible decays of the $h^{0}$ as a function of its mass $m_{h}$. You will discover that the decay pattern has a complicated dependence on the mass of the Higgs boson, with different decay modes dominating in different mass ranges. The dependences of the various decay rates on $m_{h}$ illustrate many aspects of the physics of gauge theories that we have discussed in Part III.
In working this exercise, you should consider $m_{h}$ as a free parameter. For the other parameters of weak-interaction theory, you might use the following values: $m_{b}=5 \mathrm{GeV}, m_{t}=175 \mathrm{GeV}, m_{W}=80 \mathrm{GeV}, m_{Z}=91 \mathrm{GeV}, \sin ^{2} \theta_{w}$ $=0.23, \alpha_{s}(m_{Z})=0.12$. If the Higgs boson mass $m_h$ is sufficiently large (specifically, if $m_h > 2m_Z$), it can also decay to $Z^{0} Z^{0}$. Compute the decay width $\Gamma(h^0 \rightarrow Z^0Z^0)$. As context from the original problem, if $m_h \gg m_Z$, this decay width can be approximated by $\Gamma(h^0 \rightarrow Z^0Z^0) \approx \Gamma(h^0 \rightarrow \phi^3\phi^3)$, where $\phi^3$ is a Goldstone boson, and an explanation and verification of this approximation was requested.
|
[
"\\Gamma(h^{0} \\rightarrow Z^{0} Z^{0})=\\frac{\\alpha m_{h}^{3}}{32 \\pi m_{Z}^{2} \\sin ^{2} \\theta_{w}}(1-4 \\tau_{Z}^{-1}+12 \\tau_{Z}^{-2})(1-4 \\tau_{Z}^{-1})^{1 / 2}"
] |
Expression
|
For $h^{0} \rightarrow Z^{0} Z^{0}$ process, it can be easily checked that nothing gets changed in the calculation for $h^0 \rightarrow W^+W^-$ decay (described elsewhere) except that all $m_{W}$ should be replaced with $m_{Z}$, while an additional factor $1 / 2$ is needed to account for the identical particles in final state. Therefore we have
\begin{equation*}
\Gamma(h^{0} \rightarrow Z^{0} Z^{0})=\frac{\alpha m_{h}^{3}}{32 \pi m_{Z}^{2} \sin ^{2} \theta_{w}}(1-4 \tau_{Z}^{-1}+12 \tau_{Z}^{-2})(1-4 \tau_{Z}^{-1})^{1 / 2} \tag{21.60}
\end{equation*}
where $\tau_{Z} \equiv(m_{h} / m_{Z})^{2}$.
|
{"$m_h$": "Higgs boson mass", "$m_Z$": "Z boson mass", "$\\Gamma$": "decay width", "$h^0$": "Higgs boson in the decay process", "$Z^0$": "Z boson", "$\\phi^3$": "Goldstone boson", "$\\alpha$": "fine-structure constant", "$\\theta_{w}$": "Weinberg angle", "$\\tau_{Z}$": "dimensionless parameter related to Z boson"}
|
Final Project III
|
Decays of the Higgs Boson
|
67 |
Others
|
Derive the commutator $[M^{\mu \nu}, M^{\rho \sigma}]$. Hints :
- Denote $\Lambda^{\prime} \mu_{\nu} \approx \delta^{\mu}{ }_{v}+\chi^{\mu}{ }_{v}$. Check $(\Lambda^{-1} \Lambda^{\prime} \Lambda)_{\rho \sigma} \approx \delta_{\rho \sigma}+\chi_{\mu \nu} \Lambda^{\mu}{ }_{\rho} \Lambda^{v}{ }_{\sigma}$.
- Denote $\Lambda^{\mu}{ }_{v} \approx \delta^{\mu}{ }_{v}+\omega^{\mu}{ }_{v}$. Check $U^{-1}(\Lambda) M^{\mu \nu} U(\Lambda) \approx M^{\mu v}+\frac{i}{2} \omega_{\rho \sigma}[M^{\mu \nu}, M^{\rho \sigma}]$.
- Identify this expression with $\wedge^{\mu}{ }_{\rho} \Lambda^{\nu}{ }_{\sigma} M^{\rho \sigma}$ (when simplifying by $\omega_{\rho \sigma}$, one needs to enforce the antisymmetry of the residual factors).
|
[
"[M^{\\mu \\nu}, M^{\\rho \\sigma}]=\\mathfrak{i}(g^{\\nu \\sigma} M^{\\mu \\rho}+g^{\\mu \\rho} M^{v \\sigma}-g^{\\nu \\rho} M^{\\mu \\sigma}-g^{\\mu \\sigma} M^{v \\rho})"
] |
Expression
|
We can write
\begin{aligned}
(\Lambda^{-1} \Lambda^{\prime} \Lambda)_{\rho \sigma} & =\Lambda_{\rho}^{-1 \mu} \Lambda_{\mu}^{\prime}{ }^{\nu} \Lambda_{v \sigma}=\Lambda_{\rho}^{\mu}(\delta_{\mu}{ }^{\nu}+\chi_{\mu}{ }^{\nu}+\mathcal{O}(\chi^{2})) \Lambda_{v \sigma} \\
& =\delta_{\rho \sigma}+\Lambda_{\rho}^{\mu} \Lambda_{\sigma}^{v} \chi_{\mu \nu}+\mathcal{O}(\chi^{2})
\end{aligned}
Therefore, we have
\begin{equation*}
\mathrm{U}(\Lambda^{-1} \Lambda^{\prime} \Lambda)=1+\frac{i}{2} \Lambda_{\rho}^{\mu} \Lambda_{\sigma}^{\nu} \chi_{\mu \nu} M^{\rho \sigma}+\mathcal{O}(\chi^{2}) \tag{*}
\end{equation*}
On the other hand, we also have
\begin{aligned}
\mathrm{U}(\Lambda)^{-1} \mathcal{M}^{\mu \nu} \mathrm{U}(\Lambda) & =(1-\frac{i}{2} \omega_{\rho \sigma} M^{\rho \sigma}+\mathcal{O}(\omega^{2})) \mathcal{M}^{\mu \nu}(1+\frac{i}{2} \omega_{\rho \sigma} M^{\rho \sigma}+\mathcal{O}(\omega^{2})) \\
& =M^{\mu v}+\frac{i}{2} \omega_{\rho \sigma}[M^{\mu \nu}, M^{\rho \sigma}]+\mathcal{O}(\omega^{2})
\end{aligned}
The second term of the right hand side must coincide at order $\omega^{1}$ with the coefficient of $\frac{i}{2} \chi_{\mu \nu}$ in $(*)$ :
\begin{aligned}
\frac{i}{2} \omega_{\rho \sigma}[M^{\mu v}, M^{\rho \sigma}] & =\Lambda_{\rho}^{\mu} \Lambda_{\sigma}^{v} M^{\rho \sigma}|_{\text {order } \omega^{1}} \\
& =(\delta^{\mu}{ }_{\rho} \omega^{v}{ }_{\sigma}+\omega^{\mu}{ }_{\rho} \delta^{v}{ }_{\sigma}) M^{\rho \sigma}=\omega^{v}{ }_{\sigma} M^{\mu \sigma}+\omega_{\rho}^{\mu} M^{\rho v} \\
& =\omega_{\rho \sigma}(g^{v \rho} M^{\mu \sigma}+g^{\mu \sigma} M^{v \rho})
\end{aligned}
At this stage, we cannot simply "divide" by $\omega_{\rho \sigma}$, because $\omega_{\rho \sigma}$ is not linearly independent of $\omega_{\sigma \rho}$. Before we can perform this operation, we should explicitly antisymmetrize the coefficient of $\omega_{\rho \sigma}$ in the right hand side by writing
$$\frac{i}{2} \omega_{\rho \sigma}[M^{\mu \nu}, M^{\rho \sigma}]=\frac{1}{2} \omega_{\rho \sigma}(g^{\nu \rho} M^{\mu \sigma}+g^{\mu \sigma} M^{v \rho}-g^{\nu \sigma} M^{\mu \rho}-g^{\mu \rho} M^{v \sigma})$$
Then, we obtain the following expression for the commutator of two generators of the Lorentz algebra
$$[M^{\mu \nu}, M^{\rho \sigma}]=\mathfrak{i}(g^{\nu \sigma} M^{\mu \rho}+g^{\mu \rho} M^{v \sigma}-g^{\nu \rho} M^{\mu \sigma}-g^{\mu \sigma} M^{v \rho}) $$
|
{"$M^{\\mu \\nu}$": "generator of the Lorentz transformation", "$M^{\\rho \\sigma}$": "generator of the Lorentz transformation", "$\\chi^{\\mu}{ }_{v}$": "small variation parameter in transformation", "$\\omega^{\\mu}{ }_{v}$": "infinitesimal parameter for transformation", "$g^{\\nu \\sigma}$": "Minkowski metric tensor component", "$g^{\\mu \\rho}$": "Minkowski metric tensor component", "$\\delta^{\\mu}{ }_{v}$": "Kronecker delta function component", "$\\Lambda^{\\mu}{ }_{v}$": "Lorentz transformation matrix component", "$\\wedge^{\\mu}{ }_{\\rho}$": "part of a transformation expression", "$\\Lambda_{\\rho}^{\\mu}$": "inverse Lorentz transformation matrix component", "$\\Lambda_{\\sigma}^{v}$": "Lorentz transformation matrix component"}
|
Basics of Quantum Field Theory
|
Basics of Quantum Field Theory
|
68 |
Others
|
Derive the commutators $[\mathrm{M}^{\mu \nu}, \mathrm{P}^{\rho}]$ and $[\mathrm{P}^{\mu}, \mathrm{P}^{\nu}]$. Hints :
- Note that $\Lambda^{-1} \mathrm{a} \Lambda$ is a translation by $(\Lambda^{-1} \mathrm{a})_{\rho}=\mathrm{a}_{\mu} \wedge^{\mu}{ }_{\rho}$.
- Denote $\Lambda^{\mu}{ }_{v} \approx \delta^{\mu}{ }_{v}+\omega^{\mu}{ }_{v}$. Check that $U^{-1}(\Lambda) P^{\mu} U(\Lambda) \approx P^{\mu}+\frac{i}{2} \omega_{\rho \sigma}[P^{\mu}, M^{\rho \sigma}]$.
- Identify this with $\wedge^{\mu}{ }_{\rho} \mathrm{P}^{\rho}$.
- What elementary property of translations implies $[\mathrm{P}^{\mu}, \mathrm{P}^{\nu}]=0$ ?
|
[
"[\\mathrm{M}^{\\mu \\nu}, \\mathrm{P}^{\\rho}] = \\mathfrak{i}(g^{\\rho \\mu} \\mathrm{P}^{\\nu} - g^{\\rho \\nu} \\mathrm{P}^{\\mu})"
] |
Expression
|
The action of $\Lambda^{-1} \mathrm{a} \Lambda$ on a point $x$ can be written explicitly as follows,
\begin{aligned}
{[(\Lambda^{-1} \mathrm{a} \Lambda) x]_{\rho} } & =\Lambda_{\rho}^{-1 \mu}(a_{\mu}+\Lambda_{\mu}{ }^{\nu} \chi_{v})=\Lambda_{\rho}^{-1 \mu} a_{\mu}+\delta_{\rho}{ }^{\nu} \chi_{v} \\
& =x_{\rho}+a_{\mu} \Lambda_{\rho}^{\mu}
\end{aligned}
Therefore, $\Lambda^{-1} \mathrm{a} \Lambda$ acts as a translation of $x_{\rho}$ by an amount $a_{\mu} \Lambda^{\mu}{ }_{\rho}$. Consider now the representation $\mathrm{U}(\Lambda^{-1} \mathrm{a} \Lambda)$. On the one hand, the previous results tells us that
\begin{equation*}
\mathrm{U}(\Lambda^{-1} \mathrm{a} \Lambda)=1+i \mathrm{a}_{\mu} \Lambda_{\rho}^{\mu}{ }_{\rho}^{\rho} \mathrm{P}^{\rho}+\mathcal{O}(\mathrm{a}^{2}) \tag{*}
\end{equation*}
On the other hand, we have
\begin{aligned}
\mathrm{U}(\Lambda^{-1}) \mathrm{P}^{\mu} \mathrm{U}(\Lambda) & =(1-\frac{i}{2} \omega_{\rho \sigma} M^{\rho \sigma}+\mathcal{O}(\omega^{2})) \mathrm{P}^{\mu}(1+\frac{i}{2} \omega_{\rho \sigma} M^{\rho \sigma}+\mathcal{O}(\omega^{2})) \\
& =\mathrm{P}^{\mu}+\frac{i}{2} \omega_{\rho \sigma}[\mathrm{P}^{\mu}, M^{\rho \sigma}]+\mathcal{O}(\omega^{2})
\end{aligned}
The second term of the right hand side must coincide at order $\omega^{1}$ with the coefficient of $i a_{\mu}$ in $(*)$ :
\begin{aligned}
\frac{i}{2} \omega_{\rho \sigma}[\mathrm{P}^{\mu}, M^{\rho \sigma}] & =\Lambda_{\rho}^{\mu} \mathrm{P}^{\rho}|_{\text {order } \omega^{1}} \\
& =\omega_{\rho}^{\mu}{ }_{\rho} \mathrm{P}^{\rho}=-\omega_{\rho \sigma} g^{\mu \sigma} \mathrm{P}^{\rho}=\frac{1}{2} \omega_{\rho \sigma}(g^{\mu \rho} \mathrm{P}^{\sigma}-g^{\mu \sigma} \mathrm{P}^{\rho})
\end{aligned}
where in the last equality we have performed the antisymmetrization on the indices $\rho, \sigma$. Therefore, we conclude that
[P^{\mu}, M^{\rho \sigma}]=\mathfrak{i}(g^{\mu \sigma} P^{\rho}-g^{\mu \rho} P^{\sigma})
The fact that $P^{\mu}$ and $P^{v}$ commute simply follows from the fact that the order in which two successive translations are performed does not matter.
|
{"$\\mathrm{M}^{\\mu \\nu}$": "Lorentz transformation generator with indices (mu, nu)", "$\\mathrm{P}^{\\rho}$": "momentum operator with index rho", "$\\Lambda$": "Lorentz transformation matrix", "$a_{\\mu}$": "translation vector component with index mu", "$\\omega^{\\mu}{ }_{v}$": "infinitesimal rotation parameter with indices (mu, v)", "$x_{\\rho}$": "spatial component with index rho", "$g^{\\mu \\sigma}$": "metric tensor component with indices (mu, sigma)"}
|
Basics of Quantum Field Theory
|
Basics of Quantum Field Theory
|
69 |
Others
|
Calculate the expression in coordinate space of the retarded propagator given in eq. \begin{align}
\tilde{G}_R^0(\kappa) = \frac{i}{(\kappa_0 + i0^+)^2 - (\kappa^2 + m^2)}.
\end{align}
Hint : perform the $\mathrm{k}_{0}$ integral in the complex plane with the theorem of residues. The remaining integrals are elementary.
|
[
"G_{R}^{0}(x, y) = -\\frac{i}{2 \\pi} \\theta(r^{0}) \\delta(r_{0}^{2}-r^{2})"
] |
Expression
|
The free retarded propagator in coordinate space is given by the following Fourier integral:
$$ G_{R}^{0}(x, y)=i \int \frac{d^{4} k}{(2 \pi)^{4}} \frac{e^{-i k \cdot(x-y)}}{(k^{0}+i 0^{+})^{2}-k^{2}}. $$
The integrand has two poles in the complex plane of the variable $k^{0}$, located at $k^{0}= \pm|\mathbf{k}|-\mathfrak{i} 0^{+}$. The integration over $k^{0}$ can be performed with the theorem of residues by completing the real axis with a semi-circle at infinity. Whether this semi-circle should be in the upper or lower half plane is determined by the request that the exponential factor in the numerator does not diverge when going to infinity in the imaginary direction. To see this, write $k^{0}=k_{r}^{0}+i k_{i}^{0}$. Then, we have
e^{-i k^{0}(x^{0}-y^{0})}=e^{-i k_{r}^{0}(x^{0}-y^{0})} e^{k_{i}^{0}(x^{0}-y^{0})} .
The dangerous factor is the second one, since the argument of the exponential is real. If $x^{0}-y^{0}<0$, this exponential remains bounded if we close the contour in the upper half plane. Since the integrand has no poles on this side, the integral is zero. If on the contrary $x^{0}-y^{0}>0$, we must close the contour in the lower half plane, and the theorem of residues gives non-zero contributions from the two poles. By writing
$$\frac{\mathfrak{i}}{(k^{0}+i 0^{+})^{2}-\mathbf{k}^{2}}=\frac{\mathfrak{i}}{2|\mathbf{k}|}[\frac{1}{k^{0}+\mathfrak{i} 0^{+}-|\mathbf{k}|}-\frac{1}{k^{0}+\mathfrak{i} 0^{+}+|\mathbf{k}|}],$$
we obtain easily the corresponding residues, and the propagator now reads
$$ G_{R}^{0}(x, y)=\theta(x^{0}-y^{0}) \int \frac{d^{3} \mathbf{k}}{(2 \pi)^{3}} \frac{e^{i \mathbf{k} \cdot(x-y)}}{2|\mathbf{k}|}[e^{-i|\mathbf{k}|(x^{0}-y^{0})}-e^{i|\mathbf{k}|(x^{0}-y^{0})}] .$$
In order to make the notations more compact, let us denote $r^{0} \equiv x^{0}-y^{0}$ and $\mathbf{r} \equiv \boldsymbol{x}-\mathbf{y}$. It is convenient to perform the integration over $k$ in spherical coordinates with a polar axis in the direction of $r$. This leads to
\begin{aligned}
G_{R}^{0}(x, y) & =\frac{i}{8 \pi^{2} r} \theta(r^{0}) \int_{0}^{\infty} d k[e^{i k r}-e^{-i k r}][e^{i k r^{0}}-e^{-i k r^{0}}] \\
& =\frac{i}{8 \pi^{2} r} \theta(r^{0}) \int_{0}^{\infty} d k[e^{i k(r+r^{0})}+e^{-i k(r+r^{0})}-e^{i k(r-r^{0})}-e^{-i k(r-r^{0})}] \\
& =\frac{i}{8 \pi^{2} r} \theta(r^{0}) \int_{-\infty}^{+\infty} d k[e^{i k(r+r^{0})}-e^{i k(r-r^{0})}] \\
& =\frac{i}{4 \pi r} \theta(r^{0})[\delta(r+r^{0})-\delta(r-r^{0})]=-\frac{i}{2 \pi} \theta(r^{0}) \delta(r_{0}^{2}-r^{2})
\end{aligned}
The proportionality to $\theta(r^{0})$ makes this propagator retarded, while the delta function with support on the light-cone makes it causal. With a mass, the integral over $k$ would be much more complicated (the result is expressible in terms of Bessel functions), with a support restricted to $r_{0}^{2}-r^{2} \geq 0$ (therefore, it is still causal).
|
{"$\\tilde{G}_R^0$": "retarded propagator in momentum space", "$\\kappa$": "four-momentum", "$\\kappa_0$": "energy component of four-momentum", "$m$": "mass", "$G_{R}^{0}$": "free retarded propagator in coordinate space", "$x$": "four-position coordinate", "$y$": "four-position coordinate", "$k$": "momentum", "$k^{0}$": "energy component of momentum", "$\\mathbf{k}$": "spatial components of momentum", "$x^{0}$": "time component of four-position $x$", "$y^{0}$": "time component of four-position $y$", "$r^{0}$": "time component difference between four-positions $x$ and $y$", "$\\mathbf{r}$": "spatial component difference between four-positions $x$ and $y$", "$r$": "spatial distance between four-positions $x$ and $y$"}
|
Basics of Quantum Field Theory
|
Basics of Quantum Field Theory
|
70 |
Others
|
Consider a hypothetical quantum field theory with a kinetic term $$\mathcal{L}_{0} \equiv-\frac{1}{2 \mu^{2}} \phi(\square+m^{2})^{2} \phi,$$ where $\mu$ is a constant with the dimension of mass. What is the expression for Källen-Lehman spectral function for this theory in this theory?
|
[
"2 \\pi \\mu^{2} \\frac{\\partial}{\\partial M^{2}} \\delta(M^{2}-m^{2})"
] |
Expression
|
The free propagator is obtained from the inverse of the operator between the two fields. In the case of the first example, its momentum space expression reads
$$\mathcal{G}_{\mathrm{R}}^{0}(p)=-\frac{\mathfrak{i} \mu^{2}}{((p^{0}+\mathfrak{i} 0^{+})^{2}-\mathbf{p}^{2}-m^{2})^{2}} $$
where the $\mathrm{i}^{+}$prescription we have chosen ensures that all the poles are located below the real axis in the complex plane of the variable $p^{0}$. Therefore, this is indeed the retarded propagator. Recall now that the
free retarded propagator for the usual scalar kinetic term is
$$\mathrm{G}_{\mathrm{R}}^{0}(\mathrm{p})=\frac{\mathrm{i}}{(\mathrm{p}^{0}+i 0^{+})^{2}-\mathrm{p}^{2}-\mathrm{m}^{2}}$$
The two propagators are thus related to one another by a derivative with respect to the squared mass. More precisely, one has
$$\mathcal{G}_{R}^{0}(p)=-\mu^{2} \frac{\partial}{\partial m^{2}} G_{R}^{0}(p)$$
The Källen-Lehman representation of the standard retarded propagator, $G_{R}^{0}(p)$, reads
$$\mathrm{G}_{\mathrm{R}}^{0}(\mathrm{p})=\int_{0}^{\infty} \frac{\mathrm{d} M^{2}}{2 \pi} \rho(M^{2}) \frac{\mathfrak{i}}{(\mathrm{p}^{0}+i 0^{+})^{2}-\mathrm{p}^{2}-\mathrm{M}^{2}}$$
(Of course, in the case of the free propagator, the spectral function is trivial, namely $\rho(M^{2})=2 \pi \delta(M^{2}-$ $\mathrm{m}^{2}$ ). In the interacting case, it would be more complicated, but still positive definite). Note that in this representation, the dependence on the mass $m^{2}$ is carried only by the spectral function $\rho(M^{2})$. Taking a derivative with respect to $\mathrm{m}^{2}$ of this equation, we obtain
$$\mathcal{G}_{\mathrm{R}}^{0}(p)=\int_{0}^{\infty} \frac{\mathrm{d} M^{2}}{2 \pi}[-\mu^{2} \frac{\partial}{\partial m^{2}} \rho(M^{2})] \frac{\mathfrak{i}}{(p^{0}+\mathfrak{i} 0^{+})^{2}-\mathbf{p}^{2}-M^{2}}$$
and the (free) spectral function in the theory with the higher order kinetic term is the factor between the square brackets, i.e.
$$-\mu^{2} \frac{\partial}{\partial m^{2}} \rho(M^{2})=2 \pi \mu^{2} \frac{\partial}{\partial M^{2}} \delta(M^{2}-m^{2})$$
This distribution is not positive definite. Indeed, when applied to a positive definite test function $f(M^{2})$, we obtain by integration by parts
$$ \int_{0}^{\infty} \frac{d M^{2}}{2 \pi} f(M^{2}) 2 \pi \mu^{2} \frac{\partial}{\partial M^{2}} \delta(M^{2}-m^{2})=-\mu^{2} f^{\prime}(m^{2}) $$
that can have either sign. The non-positivity of this spectral function means that this theory must contain states that contribute negatively to the sum in $$\begin{align}
\rho(M^2) \equiv 2\pi \sum_{\text{classes } \alpha} \delta(M^2 - m_\alpha^2)
\left\langle 0_{\text{out}} \left| \phi(0) \right| \alpha_0 \right\rangle
\left\langle \alpha_0 \left| \phi(0) \right| 0_{\text{in}} \right\rangle.
\end{align}$$, and therefore is not unitary in the usual sense.
|
{"$\\mu$": "constant with the dimension of mass", "$\\phi$": "field", "$m$": "mass", "$\\mathcal{G}_{\\mathrm{R}}^{0}$": "retarded propagator in the modified theory", "$G_{\\mathrm{R}}^{0}$": "retarded propagator in the standard theory", "$\\rho(M^{2})$": "spectral function", "$M$": "variable related to mass squared", "$p$": "momentum", "$p^{0}$": "energy component of momentum", "$\\mathbf{p}$": "spatial component of momentum"}
|
Basics of Quantum Field Theory
|
Basics of Quantum Field Theory
|
71 |
Others
|
For the theory with the kinetic term $\mathcal{L}_{0} \equiv-\frac{1}{2 \mu^{2}} \phi(\square+m^{2})^{2} \phi$, what is the relationship between its retarded propagator $\mathcal{G}_{R}^{0}(p)$ and the free retarded propagator $G_{R}^{0}(p)$ of a standard scalar field? (Express $\mathcal{G}_{R}^{0}(p)$ in terms of $G_{R}^{0}(p)$)
|
[
"\\mathcal{G}_{R}^{0}(p)=-\\mu^{2} \\frac{\\partial}{\\partial m^{2}} G_{R}^{0}(p)"
] |
Expression
|
The free propagator is obtained from the inverse of the operator between the two fields. In the case of the first example, its momentum space expression reads
$$ \mathcal{G}_{\mathrm{R}}^{0}(p)=-\frac{\mathfrak{i} \mu^{2}}{((p^{0}+\mathfrak{i} 0^{+})^{2}-\mathbf{p}^{2}-m^{2})^{2}} $$
where the $\mathrm{i}^{+}$prescription we have chosen ensures that all the poles are located below the real axis in the complex plane of the variable $p^{0}$. Therefore, this is indeed the retarded propagator. Recall now that the
free retarded propagator for the usual scalar kinetic term is
$$\mathrm{G}_{\mathrm{R}}^{0}(\mathrm{p})=\frac{\mathrm{i}}{(\mathrm{p}^{0}+i 0^{+})^{2}-\mathrm{p}^{2}-\mathrm{m}^{2}}$$
The two propagators are thus related to one another by a derivative with respect to the squared mass. More precisely, one has
$$ \mathcal{G}_{R}^{0}(p)=-\mu^{2} \frac{\partial}{\partial m^{2}} G_{R}^{0}(p) $$
|
{"$\\mathcal{L}_{0}$": "kinetic term", "$\\mu$": "mass scale factor", "$\\phi$": "scalar field", "$\\square$": "d'Alembertian operator", "$m$": "mass of the field", "$\\mathcal{G}_{R}^{0}(p)$": "retarded propagator in modified theory", "$G_{R}^{0}(p)$": "free retarded propagator of a standard scalar field", "$p$": "momentum", "$p^{0}$": "energy component of momentum", "$\\mathbf{p}$": "spatial momentum vector"}
|
Basics of Quantum Field Theory
|
Basics of Quantum Field Theory
|
72 |
Others
|
For the theory with the kinetic term $\mathcal{L}_{0} \equiv-\frac{1}{2 \mu^{2}} \phi(\square+m^{2})^{2} \phi$, what is its the free retarded propagator?
|
[
"\\frac{\\mu^{2}}{m_{1}^{2}-m_{2}^{2}}[\\frac{i}{(p^{0}+i 0^{+})^{2}-p^{2}-m_{2}^{2}}-\\frac{i}{(p^{0}+i 0^{+})^{2}-p^{2}-m_{1}^{2}}]"
] |
Expression
|
The free propagator is obtained from the inverse of the operator between the two fields.
The free retarded propagator reads
\begin{aligned}
\mathcal{G}_{\mathrm{R}}^{0}(\mathrm{p})&=-\frac{\mathfrak{i} \mu^{2}}{((\mathrm{p}^{0}+\mathfrak{i 0 ^ { + }})^{2}-\mathrm{p}^{2}-\mathrm{m}_{1}^{2})((\mathrm{p}^{0}+\mathfrak{i 0 ^ { + }})^{2}-\mathrm{p}^{2}-\mathrm{m}_{2}^{2})} \\
& =\frac{\mu^{2}}{m_{1}^{2}-m_{2}^{2}}[\frac{i}{(p^{0}+i 0^{+})^{2}-p^{2}-m_{2}^{2}}-\frac{i}{(p^{0}+i 0^{+})^{2}-p^{2}-m_{1}^{2}}] .
\end{aligned}
|
{"$\\mathcal{L}_{0}$": "kinetic term of the theory", "$\\mu$": "parameter associated with the kinetic term", "$\\phi$": "field in the theory", "$m$": "mass parameter in the kinetic term", "$\\mathcal{G}_{\\mathrm{R}}^{0}$": "free retarded propagator", "$\\mathrm{p}^{0}$": "energy component of momentum", "$\\mathrm{p}$": "momentum", "$\\mathrm{m}_{1}$": "mass parameter (1)", "$\\mathrm{m}_{2}$": "mass parameter (2)"}
|
Basics of Quantum Field Theory
|
Basics of Quantum Field Theory
|
73 |
Others
|
For the theory with the kinetic term $-\frac{1}{2 \mu^{2}} \phi(\square+m_{1}^{2})(\square+m_{2}^{2}) \phi$, what is its Källen-Lehman spectral function?
|
[
"\\frac{2 \\pi \\mu^{2}}{m_{1}^{2}-m_{2}^{2}}[\\delta(M^{2}-m_{2}^{2})-\\delta(M^{2}-m_{1}^{2})]"
] |
Expression
|
The free propagator is obtained from the inverse of the operator between the two fields. In the case of the first example, its momentum space expression reads
\mathcal{G}_{\mathrm{R}}^{0}(p)=-\frac{\mathfrak{i} \mu^{2}}{((p^{0}+\mathfrak{i} 0^{+})^{2}-\mathbf{p}^{2}-m^{2})^{2}}
where the $\mathrm{i}^{+}$prescription we have chosen ensures that all the poles are located below the real axis in the complex plane of the variable $p^{0}$. Therefore, this is indeed the retarded propagator. Recall now that the
free retarded propagator for the usual scalar kinetic term is
\mathrm{G}_{\mathrm{R}}^{0}(\mathrm{p})=\frac{\mathrm{i}}{(\mathrm{p}^{0}+i 0^{+})^{2}-\mathrm{p}^{2}-\mathrm{m}^{2}}
The two propagators are thus related to one another by a derivative with respect to the squared mass. More precisely, one has
\mathcal{G}_{R}^{0}(p)=-\mu^{2} \frac{\partial}{\partial m^{2}} G_{R}^{0}(p)
The Källen-Lehman representation of the standard retarded propagator, $G_{R}^{0}(p)$, reads
\mathrm{G}_{\mathrm{R}}^{0}(\mathrm{p})=\int_{0}^{\infty} \frac{\mathrm{d} M^{2}}{2 \pi} \rho(M^{2}) \frac{\mathfrak{i}}{(\mathrm{p}^{0}+i 0^{+})^{2}-\mathrm{p}^{2}-\mathrm{M}^{2}}
(Of course, in the case of the free propagator, the spectral function is trivial, namely $\rho(M^{2})=2 \pi \delta(M^{2}-$ $\mathrm{m}^{2}$ ). In the interacting case, it would be more complicated, but still positive definite). Note that in this representation, the dependence on the mass $m^{2}$ is carried only by the spectral function $\rho(M^{2})$. Taking a derivative with respect to $\mathrm{m}^{2}$ of this equation, we obtain
\mathcal{G}_{\mathrm{R}}^{0}(p)=\int_{0}^{\infty} \frac{\mathrm{d} M^{2}}{2 \pi}[-\mu^{2} \frac{\partial}{\partial m^{2}} \rho(M^{2})] \frac{\mathfrak{i}}{(p^{0}+\mathfrak{i} 0^{+})^{2}-\mathbf{p}^{2}-M^{2}}
and the (free) spectral function in the theory with the higher order kinetic term is the factor between the square brackets, i.e.
-\mu^{2} \frac{\partial}{\partial m^{2}} \rho(M^{2})=2 \pi \mu^{2} \frac{\partial}{\partial M^{2}} \delta(M^{2}-m^{2})
This distribution is not positive definite. Indeed, when applied to a positive definite test function $f(M^{2})$, we obtain by integration by parts
\int_{0}^{\infty} \frac{d M^{2}}{2 \pi} f(M^{2}) 2 \pi \mu^{2} \frac{\partial}{\partial M^{2}} \delta(M^{2}-m^{2})=-\mu^{2} f^{\prime}(m^{2})
that can have either sign. The non-positivity of this spectral function means that this theory must contain states that contribute negatively to the sum in eq. (1.113), and therefore is not unitary in the usual sense.
Consider now the second example, more general, of theory with higher derivatives in the kinetic term. This time, the free retarded propagator reads
\begin{aligned}
& \mathcal{G}_{\mathrm{R}}^{0}(\mathrm{p})=-\frac{\mathfrak{i} \mu^{2}}{((\mathrm{p}^{0}+\mathfrak{i 0 ^ { + }})^{2}-\mathrm{p}^{2}-\mathrm{m}_{1}^{2})((\mathrm{p}^{0}+\mathfrak{i 0 ^ { + }})^{2}-\mathrm{p}^{2}-\mathrm{m}_{2}^{2})} \\
& =\frac{\mu^{2}}{m_{1}^{2}-m_{2}^{2}}[\frac{i}{(p^{0}+i 0^{+})^{2}-p^{2}-m_{2}^{2}}-\frac{i}{(p^{0}+i 0^{+})^{2}-p^{2}-m_{1}^{2}}] .
\end{aligned}
By the same reasoning as before, we see that this corresponds to the following spectral function,
$$\frac{2 \pi \mu^{2}}{m_{1}^{2}-m_{2}^{2}}[\delta(M^{2}-m_{2}^{2})-\delta(M^{2}-m_{1}^{2})]$$
which is also not positive definite (one delta function contributes positively and the other one negatively). The bottom line of this exercise is that kinetic terms with higher derivatives in general lead to serious problems with unitarity. In the case of the second example, the theory could nevertheless have a practical use if $m_{1} \gg m_{2}$, provided one stays in an energy range much lower than $m_{1}$ in order not to excite the mode that has a negative norm (such a kinetic term is in fact the basis of the Pauli-Villars regularization).
|
{"$\\mu$": "parameter in the kinetic term", "$\\phi$": "scalar field", "$\\square$": "d'Alembertian operator", "$m_1$": "first mass parameter in the kinetic term", "$m_2$": "second mass parameter in the kinetic term", "$p$": "momentum", "$p^0$": "energy component of the momentum", "$\\mathbf{p}$": "spatial momentum", "$m$": "mass parameter", "$M$": "mass variable in the Källen-Lehman spectral function", "$\\rho$": "spectral function"}
|
Basics of Quantum Field Theory
|
Basics of Quantum Field Theory
|
74 |
Others
|
Consider all the 1-loop and 2-loop graphs for the six-point function in a scalar theory with a $\phi^{4}$ interaction. Write the corresponding Feynman integrals.
If the answer exists in an integral, then find the integrand
|
[
"\\frac{(-i \\lambda)^{3} i^{3}}{(\\ell^{2}-m^{2}+i 0^{+})((\\ell+p)^{2}-m^{2}+i 0^{+})((\\ell-q)^{2}-m^{2}+i 0^{+})} \\frac{(-i \\lambda)^{4} i^{2} \\mathfrak{i}^{3}}{(\\ell^{\\prime 2}-m^{2}+i 0^{+})((\\ell+\\ell^{\\prime}+p+r)^{2}-m^{2}+i 0^{+}) (\\ell^{2}-m^{2}+i 0^{+})((\\ell+p)^{2}-m^{2}+i 0^{+})((\\ell-q)^{2}-m^{2}+i 0^{+})} \\frac{(-i \\lambda)^{4} i^{2} \\mathfrak{i}^{3}}{[(\\ell^{2}-m^{2}+i 0^{+})((\\ell+p)^{2}-m^{2}+i 0^{+})((\\ell-q)^{2}-m^{2}+i 0^{+})]^{2}}"
] |
Expression
|
$$\int \frac{d^{D} \ell}{(2 \pi)^{D}} \frac{(-i \lambda)^{3} i^{3}}{(\ell^{2}-m^{2}+i 0^{+})((\ell+p)^{2}-m^{2}+i 0^{+})((\ell-q)^{2}-m^{2}+i 0^{+})}$$
\begin{aligned}
&\frac{1}{2} \int \frac{d^{D} \ell d^{D} \ell^{\prime}}{(2 \pi)^{2 D}} \frac{(-i \lambda)^{4} i^{2}}{(\ell^{\prime 2}-m^{2}+i 0^{+})((\ell+\ell^{\prime}+p+r)^{2}-m^{2}+i 0^{+})} \\
& \times \frac{\mathfrak{i}^{3}}{(\ell^{2}-m^{2}+i 0^{+})((\ell+p)^{2}-m^{2}+i 0^{+})((\ell-q)^{2}-m^{2}+i 0^{+})} .
\end{aligned}
\begin{equation}
\begin{split}
&\frac{1}{2} \int \frac{d^{D} \ell d^{D} \ell^{\prime}}{(2 \pi)^{2 D}}\frac{(-i \lambda)^{4} i^{2}}{(\ell^{2}-m^{2}+i 0^{+})((\ell+p)^{2}-m^{2}+i 0^{+})((\ell-q)^{2}-m^{2}+i 0^{+})} . \\
&\times \frac{\mathfrak{i}^{3}}{(\ell^{2}-m^{2}+i 0^{+})((\ell+p)^{2}-m^{2}+i 0^{+})((\ell-q)^{2}-m^{2}+i 0^{+})}
\end{split}
\end{equation}
Note that there are many different ways of attaching six lines labeled 1 to 6 to these basic topologies.
|
{"$\\phi$": "scalar field", "$\\lambda$": "coupling constant", "$\\ell$": "loop momentum", "$m$": "mass", "$p$": "momentum of the particle", "$q$": "another particle momentum", "$r$": "yet another particle momentum", "$D$": "dimension of spacetime", "$\\ell^{\\prime}$": "additional loop momentum"}
|
Perturbation Theory
|
Perturbation Theory
|
75 |
Others
|
Calculate $\operatorname{tr}(\gamma^{\mu} \gamma^{v} \gamma^{\rho} \gamma^{\sigma})$.
|
[
"4(g^{\\rho \\sigma} g^{\\mu \\nu}-g^{\\nu \\sigma} g^{\\mu \\rho}+g^{\\mu \\sigma} g^{\\nu \\rho})"
] |
Expression
|
With four Dirac matrices, we can write
\begin{align}
\operatorname{tr}(\gamma^{\mu} \gamma^{v} \gamma^{\rho} \gamma^{\sigma})= & -\operatorname{tr}(\gamma^{\mu} \gamma^{v} \gamma^{\sigma} \gamma^{\rho})+2 g^{\rho \sigma} \operatorname{tr}(\gamma^{\mu} \gamma^{\nu}) \\
= & +\operatorname{tr}(\gamma^{\mu} \gamma^{\sigma} \gamma^{v} \gamma^{\rho})+2 g^{\rho \sigma} \operatorname{tr}(\gamma^{\mu} \gamma^{\nu})-2 g^{v \sigma} \operatorname{tr}(\gamma^{\mu} \gamma^{\rho}) \\
= & -\operatorname{tr}(\gamma^{\sigma} \gamma^{\mu} \gamma^{\nu} \gamma^{\rho})+2 g^{\rho \sigma} \operatorname{tr}(\gamma^{\mu} \gamma^{\nu})-2 g^{v \sigma} \operatorname{tr}(\gamma^{\mu} \gamma^{\rho}) \\
& +2 g^{\mu \sigma} \operatorname{tr}(\gamma^{\nu} \gamma^{\rho}) .
\end{align}
This gives:
\begin{align}
\operatorname{tr}(\gamma^{\mu} \gamma^{\nu} \gamma^{\rho} \gamma^{\sigma}) & =g^{\rho \sigma} \operatorname{tr}(\gamma^{\mu} \gamma^{\nu})-g^{\nu \sigma} \operatorname{tr}(\gamma^{\mu} \gamma^{\rho})+g^{\mu \sigma} \operatorname{tr}(\gamma^{\nu} \gamma^{\rho}) \\
& =4(g^{\rho \sigma} g^{\mu \nu}-g^{\nu \sigma} g^{\mu \rho}+g^{\mu \sigma} g^{\nu \rho}) .
\end{align}
|
{"$\\gamma^{\\mu}$": "Dirac gamma matrix with index mu", "$\\gamma^{v}$": "Dirac gamma matrix with index v", "$\\gamma^{\\rho}$": "Dirac gamma matrix with index rho", "$\\gamma^{\\sigma}$": "Dirac gamma matrix with index sigma", "$g^{\\rho \\sigma}$": "metric tensor component with indices rho and sigma", "$g^{v \\sigma}$": "metric tensor component with indices v and sigma", "$g^{\\mu \\nu}$": "metric tensor component with indices mu and nu", "$g^{\\nu \\sigma}$": "metric tensor component with indices nu and sigma", "$g^{\\mu \\rho}$": "metric tensor component with indices mu and rho", "$g^{\\mu \\sigma}$": "metric tensor component with indices mu and sigma", "$g^{\\nu \\rho}$": "metric tensor component with indices nu and rho"}
|
Quantum Electrodynamics
|
Quantum Electrodynamics
|
76 |
Others
|
Consider two coherent states $|\chi_{\text {in }}\rangle$ and $|\vartheta_{\text {in }}\rangle$. The state $|\chi_{\text {in }}\rangle$ is defined by the function $\chi(\mathbf{k}) \equiv(2 \pi)^{3} \chi_{0} \delta(\mathbf{k})$, and the state $|\vartheta_{\text {in }}\rangle$ is defined by the function $\vartheta(\mathbf{k}) \equiv(2 \pi)^{3} \vartheta_{0} \delta(\mathbf{k})$, where $\chi_0$ and $\vartheta_0$ are constants, and $\delta(\mathbf{k})$ is the 3-dimensional Dirac delta function. Calculate the overlap $\langle\vartheta_{\text {in }} \mid \chi_{\text {in }}\rangle$.
|
[
"\\exp (-\\frac{V|\\chi_{0}-\\vartheta_{0}|^{2}}{4 m})"
] |
Expression
|
Consider now two such coherent states, $|\chi_{\text {in }}\rangle,|\vartheta_{\text {in }}\rangle$, in the special case where their defining functions have only a zero mode: $\chi(\mathbf{k}) \equiv(2 \pi)^{3} \chi_{0} \delta(\mathbf{k}), \vartheta(\mathbf{k}) \equiv(2 \pi)^{3} \vartheta_{0} \delta(\mathbf{k})$. The overlap of these two states is given by
\begin{align}
\langle\vartheta_{\text {in }} \mid \chi_{\text {in }}\rangle= & \exp (-\frac{1}{2} \int \frac{d^{3} \mathbf{k}}{(2 \pi)^{3} 2 \mathrm{E}_{\mathrm{k}}}[|\chi(\mathbf{k})|^{2}+|\vartheta(\mathbf{k})|^{2}]) \\
& \times\langle 0_{\text {in }}| \exp (\int \frac{d^{3} \mathrm{k}}{(2 \pi)^{3} 2 \mathrm{E}_{\mathrm{k}}} \vartheta^{*}(\mathbf{k}) \mathrm{a}_{\mathbf{k}, \text { in }}) \exp (\int \frac{\mathrm{d}^{3} \mathbf{k}}{(2 \pi)^{3} 2 \mathrm{E}_{\mathrm{k}}} \chi(\mathbf{k}) \mathrm{a}_{\mathrm{k}, \text { in }}^{\dagger})|0_{\text {in }}\rangle \\
= & \exp (-\frac{1}{2} \int \frac{d^{3} \mathbf{k}}{(2 \pi)^{3} 2 \mathrm{E}_{\mathrm{k}}}|\chi(\mathbf{k})-\vartheta(\mathbf{k})|^{2}) \\
= & \exp (-\frac{V|\chi_{0}-\vartheta_{0}|^{2}}{4 m})
\end{align}
Therefore, spatially homogeneous coherent states (i.e. ground states of quadratic theories shifted by a uniform field), have an exponentially suppressed overlap, and the argument of the exponential is proportional to the volume. Thus, distinct coherent states of this type are orthogonal if the volume is infinite.
|
{"$|\\chi_{\\text {in }}\\rangle$": "coherent state defined by chi", "$|\\vartheta_{\\text {in }}\\rangle$": "coherent state defined by vartheta", "$\\chi_{0}$": "constant characterizing chi", "$\\vartheta_{0}$": "constant characterizing vartheta", "$V$": "volume", "$m$": "mass"}
|
Spontaneous Symmetry Breaking
|
Spontaneous Symmetry Breaking
|
77 |
Others
|
Using Weyl's prescription for quantization, where a classical quantity $f(q, p)$ is mapped to an operator $F(Q,P)$ by
$$ F(Q, P) \equiv \int \frac{d p d q d \mu d v}{(2 \pi)^{2}} f(q, p) e^{i(\mu(Q-q)+v(P-p))}, $$
calculate the quantum operator corresponding to $f(q,p) = qp$.
(You may use the results from previous parts that for $g(q,p) = (\alpha q + \beta p)^k$, its Weyl map is $(\alpha Q + \beta P)^k$. Specifically, $q^2$ maps to $Q^2$, $p^2$ maps to $P^2$, and $(q+p)^2$ maps to $(Q+P)^2$.)
|
[
"\\frac{1}{2}(QP+PQ)"
] |
Expression
|
To obtain the Weyl mapping of $qp$, we use the polarization identity: $q p=\frac{1}{2}((q+p)^{2}-q^{2}-p^{2})$.
From the previous result $(\alpha q+\beta p)^{n} \rightarrow (\alpha Q+\beta P)^{n}$, we have the following specific mappings:
For $q^2$: set $\alpha=1, \beta=0, n=2$. So, $q^{2} \rightarrow Q^{2}$.
For $p^2$: set $\alpha=0, \beta=1, n=2$. So, $p^{2} \rightarrow P^{2}$.
For $(q+p)^2$: set $\alpha=1, \beta=1, n=2$. So, $(q+p)^{2} \rightarrow (Q+P)^{2}$.
We know that $(Q+P)^2 = (Q+P)(Q+P) = Q^2 + QP + PQ + P^2$, because $Q$ and $P$ do not generally commute.
Now, we apply the Weyl mapping to $qp$ using its expression in terms of squares:
\text{Weyl}[qp] = \text{Weyl}[\frac{1}{2}((q+p)^{2}-q^{2}-p^{2})]
Due to the linearity of the Weyl mapping:
\text{Weyl}[qp] = \frac{1}{2}(\text{Weyl}[(q+p)^{2}] - \text{Weyl}[q^{2}] - \text{Weyl}[p^{2}])
Substituting the mapped operators:
\text{Weyl}[qp] = \frac{1}{2}((Q+P)^{2} - Q^{2} - P^{2})
\text{Weyl}[qp] = \frac{1}{2}((Q^{2}+QP+PQ+P^{2}) - Q^{2} - P^{2})
\text{Weyl}[qp] = \frac{1}{2}(QP+PQ)
Therefore, the Weyl mapping of $qp$ is $\frac{1}{2}(QP+PQ)$.
|
{"$f$": "classical quantity", "$q$": "position variable", "$p$": "momentum variable", "$Q$": "quantum position operator", "$P$": "quantum momentum operator", "$\\alpha$": "parameter in linear combination", "$\\beta$": "parameter in linear combination", "$k$": "exponent in Weyl map", "$\\mu$": "integration variable for position", "$v$": "integration variable for momentum"}
|
Functional Quantization
|
Functional Quantization
|
78 |
Others
|
Consider the fermionic integral,
$$\langle\chi_{j_{1}} \cdots \chi_{j_{p}} \bar{\chi}_{i_{1}} \cdots \bar{\chi}_{i_{q}}\rangle \equiv \operatorname{det}^{-1}(\boldsymbol{M}) \int \prod_{k=1}^{n}[d \chi_{k} d \bar{\chi}_{k}] \chi_{j_{1}} \cdots \chi_{j_{p}} \bar{\chi}_{i_{1}} \cdots \bar{\chi}_{i_{q}} \exp (\bar{\chi}^{\top} \boldsymbol{M} \boldsymbol{\chi}) .$$
Assuming $p=q$, compute the expectation value for $p=q=1$, i.e., find an expression for $\langle \chi_{j_1} \bar{\chi}_{i_1} \rangle$.
|
[
"-M_{j_1 i_1}^{-1}"
] |
Expression
|
To calculate these integrals, we introduce the generating function:
$$Z[\overline{\boldsymbol{\eta}}, \boldsymbol{\eta}] \equiv \operatorname{det}^{-1}(\boldsymbol{M}) \int \prod_{k=1}^{n}[d \chi_{k} d \bar{\chi}_{k}] \exp (\overline{\boldsymbol{\eta}}^{\top} \boldsymbol{\chi}-\bar{\chi}^{\top} \boldsymbol{\eta}) \exp (\bar{\chi}^{\top} \mathbf{M} \boldsymbol{\chi})$$
This is a Gaussian Grassmann integral, which evaluates to:
$$Z[\bar{\boldsymbol{\eta}}, \boldsymbol{\eta}] = \exp (\overline{\boldsymbol{\eta}}^{\top} \boldsymbol{M}^{-1} \boldsymbol{\eta})$$
Expanding this exponential:
$$Z[\bar{\boldsymbol{\eta}}, \boldsymbol{\eta}] = 1 + \sum_{a,b} \bar{\eta}_{a} M_{ab}^{-1} \eta_{b} + \mathcal{O}(\eta^2 \bar{\eta}^2)$$
Alternatively, by expanding the factor $\exp (\overline{\boldsymbol{\eta}}^{\top} \boldsymbol{\chi}-\bar{\chi}^{\top} \boldsymbol{\eta})$ in the definition of $Z[\bar{\boldsymbol{\eta}}, \boldsymbol{\eta}]$:
$$\exp (\overline{\boldsymbol{\eta}}^{\top} \boldsymbol{\chi}-\bar{\chi}^{\top} \boldsymbol{\eta}) = (1 + \sum_a \bar{\eta}_a \chi_a + \dots ) (1 - \sum_b \bar{\chi}_b \eta_b + \dots ) = 1 + \sum_a \bar{\eta}_a \chi_a - \sum_b \bar{\chi}_b \eta_b - \sum_{a,b} \bar{\eta}_a \chi_a \bar{\chi}_b \eta_b + \dots$$
So,
$$Z[\bar{\boldsymbol{\eta}}, \boldsymbol{\eta}] = \langle 1 \rangle + \sum_a \bar{\eta}_a \langle \chi_a \rangle - \sum_b \langle \bar{\chi}_b \rangle \eta_b - \sum_{a,b} \bar{\eta}_a \langle \chi_a \bar{\chi}_b \rangle \eta_b + \dots$$
Since $\langle \chi_a \rangle = 0$ and $\langle \bar{\chi}_b \rangle = 0$ (as $p
eq q$ for these terms), we have:
$$Z[\bar{\boldsymbol{\eta}}, \boldsymbol{\eta}] = \langle 1 \rangle - \sum_{a,b} \bar{\eta}_a \langle \chi_a \bar{\chi}_b \rangle \eta_b + \dots$$
(Note: $\langle 1 \rangle = 1$ by normalization). Comparing the coefficients of $\bar{\eta}_{j_1} \eta_{i_1}$ in both expansions of $Z[\bar{\boldsymbol{\eta}}, \boldsymbol{\eta}]$:
From $\exp (\overline{\boldsymbol{\eta}}^{\top} \boldsymbol{M}^{-1} \boldsymbol{\eta})$, the coefficient of $\bar{\eta}_{j_1} \eta_{i_1}$ is $M_{j_1 i_1}^{-1}$.
From the series expansion in terms of expectation values, the coefficient of $\bar{\eta}_{j_1} \eta_{i_1}$ is $-\langle \chi_{j_1} \bar{\chi}_{i_1} \rangle$.
Therefore, $M_{j_1 i_1}^{-1} = -\langle \chi_{j_1} \bar{\chi}_{i_1} \rangle$, which gives:
$$\langle \chi_{j_1} \bar{\chi}_{i_1} \rangle = -M_{j_1 i_1}^{-1}$$
|
{"$\\chi_{j_1}$": "fermionic variable with index j_1", "$\\bar{\\chi}_{i_1}$": "conjugate fermionic variable with index i_1", "$\\boldsymbol{M}$": "matrix M", "$M_{j_1 i_1}^{-1}$": "inverse component of matrix M with indices j_1 and i_1", "$n$": "number of fermionic variables"}
|
Path Integrals for Fermions and Photons
|
Path Integrals for Fermions and Photons
|
79 |
Others
|
Consider the fermionic integral,
$$
\langle\chi_{j_{1}} \cdots \chi_{j_{p}} \bar{\chi}_{i_{1}} \cdots \bar{\chi}_{i_{q}}\rangle \equiv \operatorname{det}^{-1}(\boldsymbol{M}) \int \prod_{k=1}^{n}[d \chi_{k} d \bar{\chi}_{k}] \chi_{j_{1}} \cdots \chi_{j_{p}} \bar{\chi}_{i_{1}} \cdots \bar{\chi}_{i_{q}} \exp (\bar{\chi}^{\top} \boldsymbol{M} \boldsymbol{\chi}) .
$$
Assuming $p=q$, compute the expectation value for $p=q=2$, i.e., find an expression for $\langle \chi_{j_1} \chi_{j_2} \bar{\chi}_{i_1} \bar{\chi}_{i_2} \rangle$.
|
[
"M_{j_1 i_2}^{-1} M_{j_2 i_1}^{-1} - M_{j_1 i_1}^{-1} M_{j_2 i_2}^{-1}"
] |
Expression
|
We use the generating function $Z[\overline{\boldsymbol{\eta}}, \boldsymbol{\eta}]$ as defined in the previous problem. The expansion of $Z[\bar{\boldsymbol{\eta}}, \boldsymbol{\eta}] = \exp (\overline{\boldsymbol{\eta}}^{\top} \boldsymbol{M}^{-1} \boldsymbol{\eta})$ to second order in $\bar{\eta}$ and $\eta$ is: $Z[\bar{\boldsymbol{\eta}}, \boldsymbol{\eta}] = 1 + \sum_{a,b} \bar{\eta}_{a} M_{ab}^{-1} \eta_{b} + \frac{1}{2!} ( \sum_{a,b} \bar{\eta}_{a} M_{ab}^{-1} \eta_{b} )^2 + \dots = 1 + \sum_{a,b} \bar{\eta}_{a} M_{ab}^{-1} \eta_{b} + \frac{1}{2} \sum_{a,b,c,d} \bar{\eta}_{a} M_{ab}^{-1} \eta_{b} \bar{\eta}_{c} M_{cd}^{-1} \eta_{d} + \dots$ Using anticommutation $\eta_b \bar{\eta}_c = - \bar{\eta}_c \eta_b$: $= 1 + \sum_{a,b} \bar{\eta}_{a} M_{ab}^{-1} \eta_{b} - \frac{1}{2} \sum_{a,b,c,d} \bar{\eta}_{a} \bar{\eta}_{c} M_{ab}^{-1} M_{cd}^{-1} \eta_{b} \eta_{d} + \dots$ As noted, this can be written by antisymmetrizing the $M^{-1}$ terms: $Z[\bar{\boldsymbol{\eta}}, \boldsymbol{\eta}] = \dots - \frac{1}{4} \sum_{a,c,b,d} \bar{\eta}_{a} \bar{\eta}_{c} (M_{ab}^{-1} M_{cd}^{-1} - M_{ad}^{-1} M_{cb}^{-1}) \eta_{b} \eta_{d} + \dots$ Expanding $\exp (\overline{\boldsymbol{\eta}}^{\top} \boldsymbol{\chi}-\bar{\chi}^{\top} \boldsymbol{\eta})$ gives terms with two $\bar{\eta}$ and two $\eta$ variables. Comparing coefficients, the result matches $\langle \chi_{j_1} \chi_{j_2} \bar{\chi}_{i_1} \bar{\chi}_{i_2} \rangle = M_{j_1 i_2}^{-1} M_{j_2 i_1}^{-1} - M_{j_1 i_1}^{-1} M_{j_2 i_2}^{-1}$.
|
{"$\\chi_{j_{1}}$": "fermionic field component with index j1", "$\\chi_{j_{2}}$": "fermionic field component with index j2", "$\\bar{\\chi}_{i_{1}}$": "conjugate fermionic field component with index i1", "$\\bar{\\chi}_{i_{2}}$": "conjugate fermionic field component with index i2", "$\\bar{\\eta}_{a}$": "component of the conjugate auxiliary field with index a", "$\\eta_{b}$": "component of the auxiliary field with index b", "$M_{ab}^{-1}$": "element of the inverse matrix of M at a,b"}
|
Path Integrals for Fermions and Photons
|
Path Integrals for Fermions and Photons
|
80 |
Others
|
Consider a general function of a single Grassmann variable $\chi$, which can be written as $f(\chi) = a + \chi b$, where $a$ and $b$ are c-numbers (or objects that commute with Grassmann variables). Introduce a second Grassmann variable $\eta$ that anticommutes with $\chi$. Calculate explicitly the integral $\widetilde{f}(\eta) \equiv \int d \chi e^{\chi \eta} f(\chi)$. (Recall Grassmann properties: $e^{\chi \eta} = 1 + \chi \eta$, $\chi^2 = 0$, and Berezin integration rules $\int d\chi = 0$, $\int d\chi \chi = 1$).
|
[
"\\eta a+b"
] |
Expression
|
Given $f(\chi) = a + \chi b$ and $e^{\chi \eta} = 1 + \chi \eta$ (since higher powers of $\chi\eta$ would involve $\chi^2$ or $\eta^2$ which are zero if $\chi, \eta$ are single components, or this is the truncated expansion for Grassmann variables).
The integral is:
\begin{align}
\widetilde{f}(\eta) & =\int d \chi e^{\chi \eta}(a+\chi b) \\& =\int d \chi(1+\chi \eta)(a+\chi b) \\
& =\int d \chi(a + \chi b + \chi \eta a + \chi \eta \chi b)
\end{align}
Since $\chi$ and $\eta$ are Grassmann variables, they anticommute ($\chi \eta = -\eta \chi$), and $\chi^2 = 0$. Thus, the term $\chi \eta \chi b = -\eta \chi^2 b = -\eta (0) b = 0$.
The integral simplifies to:$$\widetilde{f}(\eta) = \int d \chi(a+\chi(\eta a+b))$$
Using the Berezin integration rules $\int d\chi = 0$ and $\int d\chi \chi = 1$:
$$\widetilde{f}(\eta) = a \int d\chi (1) + (\eta a+b) \int d\chi \chi = a \cdot 0 + (\eta a+b) \cdot 1 = \eta a+b$$
|
{"$\\chi$": "Grassmann variable", "$f$": "general function of a Grassmann variable", "$a$": "c-number that commutes with Grassmann variables", "$b$": "c-number that commutes with Grassmann variables", "$\\eta$": "second Grassmann variable that anticommutes with $\\chi$"}
|
Path Integrals for Fermions and Photons
|
Path Integrals for Fermions and Photons
|
81 |
Others
|
Consider two Grassmann variables $\theta_\pm$.
For the operator $\tau_3 \equiv \frac{1}{2}(\theta_{+} \frac{\partial}{\partial \theta_{+}}-\theta_{-} \frac{\partial}{\partial \theta_{-}})$, find an eigenfunction corresponding to the eigenvalue $1/2$. Express it using $1, \theta_+, \theta_-, \theta_+\theta_-$ and normalize.
|
[
"$\\theta_{+}"
] |
Expression
|
From the eigenvalue equation $\tau_3 f = \lambda f$, setting $\lambda=1/2$:
$0 = \frac{1}{2} a \implies a=0$
$(\frac{1}{2}-\frac{1}{2})b=0 \implies 0 \cdot b = 0$ (no constraint on $b$)
$(-\frac{1}{2}-\frac{1}{2})c=0 \implies -c=0 \implies c=0$
$0 = \frac{1}{2} d \implies d=0$
So, eigenfunctions for $\lambda=1/2$ are of the form $b\theta_+$. Choosing $b=1$ for normalization, an eigenfunction is $\theta_+$.
|
{"$\\theta_{+}$": "Grassmann variable", "$\\theta_{-}$": "Grassmann variable", "$\\tau_3$": "operator associated with the problem"}
|
Path Integrals for Fermions and Photons
|
Path Integrals for Fermions and Photons
|
82 |
Others
|
Consider two Grassmann variables $\theta_\pm$.
For the operator $\tau_1 \equiv \frac{1}{2}(\theta_{+} \frac{\partial}{\partial \theta_{-}}+\theta_{-} \frac{\partial}{\partial \theta_{+}})$, find an eigenfunction corresponding to the eigenvalue $1/2$. Express it using $1, \theta_+, \theta_-, \theta_+\theta_-$ and normalize.
|
[
"$\\theta_{+} + \\theta_{-}"
] |
Expression
|
For $\lambda=1/2$: $a=0, d=0$. $\frac{1}{2}c = \frac{1}{2}b \implies c=b$. Eigenfunctions are $b\theta_+ + b\theta_- = b(\theta_+ + \theta_-)$. Normalized: $\theta_+ + \theta_-$.
|
{"$\\theta_{+}$": "Grassmann variable theta plus", "$\\theta_{-}$": "Grassmann variable theta minus", "$\\tau_1$": "operator tau one"}
|
Path Integrals for Fermions and Photons
|
Path Integrals for Fermions and Photons
|
83 |
Others
|
Consider two Grassmann variables $\theta_\pm$.
For the operator $\tau_1 \equiv \frac{1}{2}(\theta_{+} \frac{\partial}{\partial \theta_{-}}+\theta_{-} \frac{\partial}{\partial \theta_{+}})$, find an eigenfunction corresponding to the eigenvalue $-1/2$. Express it using $1, \theta_+, \theta_-, \theta_+\theta_-$ and normalize.
|
[
"$\\theta_{+} - \\theta_{-}"
] |
Expression
|
For $\lambda=-1/2$: $a=0, d=0$. $\frac{1}{2}c = -\frac{1}{2}b \implies c=-b$. Eigenfunctions are $b\theta_+ - b\theta_- = b(\theta_+ - \theta_-)$. Normalized: $\theta_+ - \theta_-$.
|
{"$\\theta_{+}$": "Grassmann variable (positive component)", "$\\theta_{-}$": "Grassmann variable (negative component)", "$\\tau_1$": "operator involving Grassmann variables"}
|
Path Integrals for Fermions and Photons
|
Path Integrals for Fermions and Photons
|
84 |
Others
|
Consider two Grassmann variables $\theta_\pm$. For the operator $\tau_2 \equiv \frac{i}{2}(\theta_{-} \frac{\partial}{\partial \theta_{+}}-\theta_{+} \frac{\partial}{\partial \theta_{-}})$, find an eigenfunction corresponding to the eigenvalue $-1/2$. Express it using $1, \theta_+, \theta_-, \theta_+\theta_-$ and normalize.
|
[
"$\\theta_{+} - i\\theta_{-}"
] |
Expression
|
For $\lambda=-1/2$: $a=0, d=0$. $-\frac{i}{2}c = -\frac{1}{2}b \implies ic=b \implies c=-ib$. Eigenfunctions are $b\theta_+ - ib\theta_- = b(\theta_+ - i\theta_-)$. Normalized: $\theta_+ - i\theta_-$.
|
{"$\\tau_2$": "operator", "$\\theta_{+}$": "variable theta plus", "$\\theta_{-}$": "variable theta minus", "$i$": "imaginary unit"}
|
Path Integrals for Fermions and Photons
|
Path Integrals for Fermions and Photons
|
85 |
Others
|
Consider two Grassmann variables $\theta_\pm$. Given the operators $\tau_{1 \equiv \frac{1}{2}(\theta_{+} \frac{\partial}{\partial \theta_{-}}+\theta_{-} \frac{\partial}{\partial \theta_{+}})$ and $\tau_{2} \equiv \frac{i}{2}(\theta_{-} \frac{\partial}{\partial \theta_{+}}-\theta_{+} \frac{\partial}{\partial \theta_{-}})$, calculate the action of the operator $(\tau_1 - i\tau_2)$ on the Grassmann variable $\theta_+$.}
|
[
"\\theta_{-}"
] |
Expression
|
$\tau_1 \theta_+ = \frac{1{2}\theta_-$. $\tau_2 \theta_+ = \frac{i}{2}\theta_-$. $(\tau_1 - i\tau_2)\theta_+ = \frac{1}{2}\theta_- - i(\frac{i}{2}\\theta_-) = \frac{1}{2}\theta_- + \frac{1}{2}\theta_- = \theta_-$.}
|
{"$\\tau_{1}$": "operator involving Grassmann variables", "$\\tau_{2}$": "operator involving Grassmann variables", "$\\theta_{+}$": "Grassmann variable", "$\\theta_{-}$": "Grassmann variable"}
|
Path Integrals for Fermions and Photons
|
Path Integrals for Fermions and Photons
|
86 |
Others
|
Assume that $[X, Y]=i c Y$ where c is a numerical constant. Use the all-orders Baker-CampbellHausdorff formula to calculate $\ln (e^{i X} e^{i Y})$.
|
[
"i X+\\frac{i c}{e^{c}-1} Y"
] |
Expression
|
In this case, we have
$$
\operatorname{ad}_{x}(Y)=-i[X, Y]=c Y, \quad e^{\operatorname{ad}_{X}} Y=e^{c} Y
$$
Since $Y$ commutes with itself, this implies that
$$
e^{\operatorname{tad}_{Y}} e^{\mathrm{ad}_{x}} \mathrm{Y}=e^{\mathrm{c}} \mathrm{Y}
$$
Using this in the general Baker-Campbell-Hausdorff formula gives
$$\ln (e^{i X} e^{i Y})=i X+i \int_{0}^{1} d t \frac{\ln (e^{c})}{e^{c}-1} Y=i X+\frac{i c}{e^{c}-1} Y$$
(Obviously, this result goes to $i(X+Y)$ when $\mathrm{c} \rightarrow 0$.)
|
{"$X$": "operator X", "$Y$": "operator Y", "$i$": "imaginary unit", "$c$": "numerical constant"}
|
Non-Abelian Gauge Symmetry
|
Non-Abelian Gauge Symmetry
|
87 |
Others
|
Calculate the one-loop $\beta$-function of a scalar field theory with cubic interactions in six spacetime dimensions.
|
[
"\\beta=-\\frac{3 \\lambda^{3}}{4(4 \\pi)^{3}}"
] |
Expression
|
Scalar field theory with cubic interactions is renormalizable in six dimensions (the coupling constant of this cubic interaction is dimensionless in six dimensions), and the power counting indicates that the divergences are in the 2-point and 3-point functions. In order to calculate the one-loop $\\beta$-function, we need fist to calculate at this order the self-energy and vertex counterterms. Let us start with the self-energy, given by
$$
\Sigma(p) \equiv \frac{\mathfrak{i} \lambda^{2} \mu^{2 \epsilon}}{2} \int \frac{d^{\mathrm{D}} \ell}{(2 \pi)^{\mathrm{D}}} \frac{1}{\ell^{2}(\ell+p)^{2}}=\frac{\mathfrak{i} \lambda^{2} \mu^{2 \epsilon}}{2} \int_{0}^{1} d x \int \frac{d^{\mathrm{D}} \ell}{(2 \pi)^{\mathrm{D}}} \frac{1}{[\ell^{2}-\Delta]^{2}}
$$
where $\Delta \equiv-x(1-x) p^{2}$ (note that the loop momentum $\ell$ has been shifted in the last expression). Performing a Wick's rotation and using standard results for D-dimensional integral gives
$$
\Sigma(p)=-\frac{\lambda^{2} \mu^{2 \epsilon}}{2(4 \pi)^{\mathrm{D} / 2}} \Gamma(2-\frac{\mathrm{D}}{2}) \int_{0}^{1} \mathrm{~d} x \Delta^{\mathrm{D} / 2-2} .
$$
Then, define $D \equiv 6-2 \epsilon$, and set $\epsilon$ to zero in all factors, except the $\Gamma$ function that has a simple pole (with residue -1 ) and the factor $(\mu^{2} / p^{2})^{\epsilon}$ that carries the scale dependence. We arrive at
$$
\Sigma(p)=-\frac{\lambda^{2} p^{2}}{12(4 \pi)^{3} \epsilon}[-\frac{\mu^{2}}{p^{2}}]^{\epsilon}
$$
At this order, the field renormalization constant is
\begin{equation*}
\mathrm{Z}=1+\frac{\partial \Sigma}{\partial \mathrm{p}^{2}}=1-\frac{\lambda^{2}}{12(4 \pi)^{3} \epsilon}[\frac{\mu^{2}}{M^{2}}]^{\epsilon} \tag{*}
\end{equation*}
where we have chosen the renormalization point $p^{2}=-M^{2}$.
Let us now turn our attention to the 3-point function. Denoting $p, q$ two of the momenta entering in the graph, the one-loop vertex correction reads
\begin{align}
\mu^{\epsilon} \Gamma(p, q) & =i \lambda^{3} \mu^{3 \epsilon} \int \frac{d^{\mathrm{D}} \ell}{(2 \pi)^{\mathrm{D}}} \frac{1}{\ell^{2}(\ell+p)^{2}(\ell-q)^{2}} \\
& =2 i \lambda^{3} \mu^{3 \epsilon} \int_{\substack{0 \leq x, y \leq 1 \\
x+y \leq 1}} d x d y \int \frac{d^{\mathrm{D}} \ell}{(2 \pi)^{\mathrm{D}}} \frac{1}{[\ell^{2}-\Delta]^{3}},
\end{align}
with $-\Delta \equiv x(1-x) p^{2}+y(1-y) q^{2}+2 x y p \cdot q$. The reason why we denote the left hand side $\mu^{\epsilon} \Gamma$ is that this loop integral is a correction to the 3-point coupling, that has dimension (mass) ${ }^{\epsilon}$ in $\mathrm{D}=6-2 \epsilon$ dimensions. By writing it in this way, $\Gamma$ is a correction to the dimensionless coupling $\lambda$. After a Wick's rotation, the pole in this integral reads
\begin{equation*}
\Gamma(p, q)=\frac{\lambda^{3}}{2(4 \pi)^{3} \epsilon}[\frac{\mu^{2}}{M^{2}}]^{\epsilon} \tag{**}
\end{equation*}
where we have chosen the special kinematical configuration $p^{2}=q^{2}=(p+q)^{2}=-M^{2}$ as renormalization point.
From the results $(*)$ and $(* *)$, we get the following wavefunction and vertex counterterms,
$$
\delta_{z}=-\frac{\lambda^{2}}{12(4 \pi)^{3} \epsilon}[\frac{\mu^{2}}{M^{2}}]^{\epsilon}, \quad \delta_{\lambda}=-\frac{\lambda^{3}}{2(4 \pi)^{3} \epsilon}[\frac{\mu^{2}}{M^{2}}]^{\epsilon}
$$
and the $\beta$-function in six dimensions is then given by
$$
\beta=\lim _{\epsilon \rightarrow 0} M \frac{\partial}{\partial M}[\frac{3 \lambda^{3}}{8(4 \pi)^{3} \epsilon}(1-2 \epsilon \ln (\frac{M}{\mu})+\mathcal{O}(\epsilon^{2}))]=-\frac{3 \lambda^{3}}{4(4 \pi)^{3}} .
$$
|
{"$\\beta$": "beta function", "$\\lambda$": "coupling constant", "$\\mu$": "renormalization scale", "$\\epsilon$": "dimensional regularization parameter", "$D$": "spacetime dimensions", "$p$": "momentum", "$\\Delta$": "shifted square of momentum", "$\\Sigma$": "self-energy", "$\\Gamma$": "vertex correction", "$Z$": "field renormalization constant", "$M$": "renormalization point"}
|
Renormalization of Gauge Theories
|
Renormalization of Gauge Theories
|
88 |
Others
|
Calculate the one-loop $\beta$ function in quantum electrodynamics. How does the electromagnetic coupling strength vary with distance? What is the physical interpretation of this behaviour?
|
[
"\\beta(e)=\\frac{e^{3}}{12 \\pi^{2}}"
] |
Expression
|
In order to follow the same procedure as in Exercise 10.1 for calculating the $\beta$ function in QED, we would need to calculate the one-loop electron self-energy, the one-loop photon self-energy, and the one-loop correction to the electron-photon vertex. An alternative is to use the relation $Z_{1} Z_{2}^{-1} Z_{3}^{-1 / 2} e_{r}=e_{b}$ and the fact that $Z_{1}=Z_{2}$ due to Ward identities. Therefore, the relationship between the bare and renormalized couplings can also be obtained from the photon wavefunction renormalization, i.e. from a unique loop diagram. At one-loop, the photon polarization tensor is given by
$$\Pi^{\mu u}(q)=-i e^{2} \mu^{2 \epsilon} \int \frac{d^{D} \ell}{(2 \pi)^{D}} \frac{\operatorname{tr}(\gamma^{\mu} \ell \gamma^{v}(\ell+ot q))}{\ell^{2}(\ell+q)^{2}}$$
From gauge invariance, we expect the tensor structure to be $\Pi^{\mu v}(q) \equiv(g^{\mu v} q^{2}-q^{\mu} q^{v}) \Pi(q^{2})$, and the function $\Pi(q^{2})$ can be obtained from the trace, since $(D-1) q^{2} \Pi(q^{2})=\Pi_{\mu}^{\mu}(q)$, that reads
\begin{align}
\Pi_{\mu}^{\mu}(\mathrm{q}) & =4(\mathrm{D}-2) i e^{2} \mu^{2 \epsilon} \int \frac{\mathrm{~d}^{\mathrm{D}} \ell}{(2 \pi)^{\mathrm{D}}} \frac{\ell \cdot(\ell+\mathrm{q})}{\ell^{2}(\ell+\mathrm{q})^{2}} \\
& =4(\mathrm{D}-2) i e^{2} \mu^{2 \epsilon} \int_{0}^{1} \mathrm{dx} \int \frac{\mathrm{~d}^{\mathrm{D}} \ell}{(2 \pi)^{\mathrm{D}}} \frac{\ell^{2}+\Delta+\text { terms odd in } \ell}{[\ell^{2}-\Delta]^{2}}
\end{align}
where $\Delta \equiv-x(1-x) q^{2}$, and the $\ell$ in the last expression is a shifted integration variable. Performing a Wick's rotation and integrating over $\ell$ gives
\begin{align}
\Pi(q^{2}) & =e^{2} \mu^{2 \epsilon} \frac{4(\mathrm{D}-2)}{(\mathrm{D}-1) \mathrm{q}^{2}(4 \pi)^{\mathrm{D} / 2}}[\Gamma(1-\frac{\mathrm{D}}{2})-2 \Gamma(2-\frac{\mathrm{D}}{2})] \int_{0}^{1} \mathrm{~d} x \Delta^{\mathrm{D} / 2-1} \\
& =\frac{4 \mathrm{e}^{2}}{3(4 \pi)^{2} \epsilon}[-\frac{\mu^{2}}{\mathrm{q}^{2}}]^{\epsilon}
\end{align}
The photon wavefunction renormalization factor is
$$
Z_{3}=\frac{1}{1+\Pi}=1-\Pi+\mathcal{O}(e^{4})=1-\frac{4 e^{2}}{3(4 \pi)^{2} \epsilon}[\frac{\mu^{2}}{M^{2}}]^{\epsilon}+\mathcal{O}(e^{4})
$$
the last equality being at the point $q^{2}=-M^{2}$. From the relation $Z_{3}^{-1 / 2} e_{r}=e_{b}$, the scale dependence of the renormalized coupling constant can be determined from that of $Z_{3}$ (since the bare coupling is by definition scale independent),
$$
\beta(e)=\frac{e}{2 Z_{3}} M \frac{\partial Z_{3}}{\partial M}=\frac{e^{3}}{12 \pi^{2}},
$$
which is often written as
$$
\beta(\alpha) \equiv M \frac{\partial \alpha}{\partial M}=\frac{2 \alpha^{2}}{3 \pi} .
$$
Since the $\beta$ function of QED is positive, the coupling strength increases with the energy scale, or at shorter distances. Conversely, it decreases at larger distances. This can be understood physically as a screening phenomenon: a test charge polarizes the vacuum around it by arranging virtual electron-antielectron pairs in such a way that they screen the Coulomb potential seen at a distance. This is very similar to Debye screening in plasma physics (but there it is due to the polarization of on-shell charged particles, instead of vacuum fluctuations). As the probe of the electrical field approaches the test charge, the charge it sees is closer and closer to the naked charge of the test particle. For a point-like charge, the latter is infinite since it is not hidden by any screening.
|
{"$\\beta$": "beta function", "$e$": "electron charge", "$Z_{1}$": "renormalization constant for vertex", "$Z_{2}$": "renormalization constant for electron field", "$Z_{3}$": "renormalization constant for photon field", "$e_{r}$": "renormalized electron charge", "$e_{b}$": "bare electron charge", "$\\mu$": "regularization scale", "$q$": "momentum transfer", "$\\ell$": "loop momentum", "$\\Delta$": "shifted integration variable", "$M$": "momentum scale", "$\\alpha$": "fine-structure constant"}
|
Renormalization of Gauge Theories
|
Renormalization of Gauge Theories
|
89 |
Others
|
The special conformal transformation is given by $y^{\mu}=\frac{x^{\mu}+b^{\mu} x^{2}}{1+2 b \cdot x+b^{2} x^{2}}$. For an infinitesimal 4-vector $b^{\mu}$, this transformation can be expanded to first order in $b^{\mu}$ as $y^{\mu} \approx x^{\mu}+\delta x^{\mu}$, where $\delta x^{\mu} = (x^{2} g^{\mu \rho}-2 x^{\rho} x^{\mu}) b_{\rho}$ (using $g^{\mu\rho}$ to raise the index of $b$). The generator $T_{\rho}$ (associated with parameter $b^{\rho}$) is defined such that the transformation on coordinates is $x'^{\mu} = \exp (i b^{\rho} T_{\rho}) x^{\mu} \approx x^{\mu} + i b^{\rho} (T_{\rho} x^{\mu})$, from which $\delta x^{\mu} = i b^{\rho} (T_{\rho} x^{\mu})$. The provided text then gives an expression for an object $T^{\mu}$ (note the upper index) as $T^{\mu}=-\mathfrak{i}(x^{2} g^{\mu \rho}-2 x^{\rho} x^{\mu}) \partial_{\rho}$. What is this expression for $T^{\mu}$?
|
[
"T^{\\mu}=-\\mathfrak{i}(x^{2} g^{\\mu \\rho}-2 x^{\\rho} x^{\\mu}) \\partial_{\\rho}"
] |
Expression
|
Given the infinitesimal transformation for $y^{\mu}$:
$$y^{\mu}=\frac{x^{\mu}+b^{\mu} x^{2}}{1+2 b \cdot x+b^{2} x^{2}}$$
For infinitesimal $b^{\mu}$, we expand $(1+2 b \cdot x+b^{2} x^{2})^{-1} \approx 1 - (2 b \cdot x+b^{2} x^{2}) + \mathcal{O}(b^2) \approx 1 - 2 b \cdot x + \mathcal{O}(b^2)$.
So, $y^{\mu} \approx (x^{\mu}+b^{\mu} x^{2})(1-2b \cdot x) + \mathcal{O}(b^2)$
$y^{\mu} \approx x^{\mu} - 2(b \cdot x)x^{\mu} + b^{\mu}x^2 + \mathcal{O}(b^2)$.
Let $b^{\mu} = g^{\mu\sigma}b_{\sigma}$. Then $b \cdot x = b_{\sigma}x^{\sigma}$.
$y^{\mu} \approx x^{\mu} + (g^{\mu\sigma}x^2 - 2x^{\mu}x^{\sigma})b_{\sigma} + \mathcal{O}(b^2)$.
So, $\delta x^{\mu} = (x^2 g^{\mu \sigma} - 2x^{\sigma}x^{\mu})b_{\sigma}$. (The problem statement uses $\rho$ for the summed index of $b$).
$\delta x^{\mu} = (x^{2} g^{\mu \rho}-2 x^{\rho} x^{\mu}) b_{\rho}$.
From the definition of how generators act on fields $\Phi(x)$, an infinitesimal transformation $\delta \Phi = i \epsilon^a (G_a \Phi)$ involves the generator $G_a$. For coordinate transformations $x_\mu \rightarrow x_\mu + \delta x_\mu$, where $\delta x_\mu = \epsilon^a \xi_{a \mu}(x)$, the generators are often written as $G_a = -i \xi_a^
u (x) \partial_
u$.
The text states that from $\delta x^{\mu}$, we can read off the generator $T_{\rho}$ such that $\exp (i b^{\rho} T_{\rho}) x^{\mu}}= x^{\mu}+\delta x^{\mu}+\cdots$.
It then provides the expression: $T^{\mu}=-\mathfrak{i}(x^{2} g^{\mu \rho}-2 x^{\rho} x^{\mu}) \partial_{\rho}$.
This expression has a free index $\mu$ on $T$ and summation over $\rho$ on the right hand side. This represents a set of four differential operators, one for each value of $\mu=0,1,2,3$. This $T^{\mu}$ is the specific quantity requested, as derived/stated in the source material.
|
{"$y^{\\mu}$": "transformed coordinate vector", "$x^{\\mu}$": "original coordinate vector", "$b^{\\mu}$": "infinitesimal 4-vector parameter", "$x^{2}$": "square of the coordinate vector", "$b^{2}$": "square of the vector $b$", "$b \\cdot x$": "dot product of $b^{\\mu}$ and $x^{\\mu}$", "$g^{\\mu\\rho}$": "metric tensor (raising the index)", "$T_{\\rho}$": "generator associated with the parameter $b^{\\rho}$", "$T^{\\mu}$": "specific generator operator for infinitesimal transformations", "$\\delta x^{\\mu}$": "change in the coordinate vector $x^{\\mu}$", "$b_{\\rho}$": "component of the vector $b$ with a lowered index", "$\\partial_{\\rho}$": "partial derivative with respect to the coordinate $x^{\\rho}$"}
|
Quantum Anomalies
|
Quantum Anomalies
|
90 |
Others
|
In Yang-Mills theory in the temporal gauge $A^{0}=0$ (with coupling $g=1$ for simplicity), what is the expression for the conjugate momentum $\Pi_a^i$ of the gauge field component $A_a^i$?
|
[
"\\Pi_{a}^{i} = \\partial_{0} A_{\\mathrm{a}}^{i}"
] |
Expression
|
By using the temporal gauge $A^{0}=0$, we circumvent the problem that $A_{a}^{0}$ has a vanishing conjugate momentum, because $A^{0}$ is not a dynamical variable in this gauge. Regarding the other components of the gauge potential, the conjugate momentum $\Pi^{i}$ of $A^{i}$ is given by
$$\Pi_{a}^{i} \equiv \frac{\partial \mathcal{L}_{\mathrm{YM}}}{\partial \partial_{0} \mathcal{A}_{\mathrm{a}}^{i}}=\partial_{0} A_{\mathrm{a}}^{i}$$
|
{"$A^{0}$": "gauge field component (0)", "$g$": "coupling constant", "$\\Pi_a^i$": "conjugate momentum of gauge field component (a, i)", "$A_a^i$": "gauge field component (a, i)", "$\\mathcal{L}_{\\mathrm{YM}}$": "Yang-Mills Lagrangian", "$\\partial_{0}$": "partial derivative with respect to time"}
|
Localized Field Configurations
|
Localized Field Configurations
|
91 |
Others
|
For Yang-Mills theory in the temporal gauge $A^{0}=0$ (with $g=1$), derive the Hamiltonian density $\mathcal{H}$. Express it first in terms of the chromo-electric fields $E_a^i = \Pi_a^i$ and chromo-magnetic fields $B_a^i = \frac{1}{2} \epsilon_{ijk} F_a^{jk}$, and then in terms of $\Pi_a^i$ and $F_a^{ij}$.
|
[
"\\mathcal{H}=\\frac{1}{2}(E_{a}^{i} E_{a}^{i}+B_{a}^{i} B_{a}^{i}) = \\frac{1}{2} \\Pi_{a}^{i} \\Pi_{a}^{i}+\\frac{1}{4} F_{a}^{i j} F_{a}^{i j}"
] |
Expression
|
It is convenient to introduce the chromo-electric and chromo-magnetic fields by
$$
\mathrm{E}_{\mathrm{a}}^{i} \equiv \mathrm{~F}_{\mathrm{a}}^{0 i}=\Pi_{a}^{i} \quad(\text { in } A^{0}=0 \text { gauge }), \quad B_{a}^{i} \equiv \frac{1}{2} \epsilon_{i j k} F_{a}^{j k}
$$
in terms of which the Lagrangian density can be written as follows,
\begin{align}
-\frac{1}{4} F_{a}^{\mu \nu} F_{\mu v}^{a} & =-\frac{1}{4}(-F_{a}^{O i} F_{a}^{O i}-F_{a}^{i 0} F_{a}^{i O}+F_{a}^{i j} F_{a}^{i j})=\frac{1}{4}(2 E_{a}^{i} E_{a}^{i}-\epsilon_{i j k} \epsilon_{i j l} B_{a}^{k} B_{a}^{l}) \
& =\frac{1}{4}(2 E_{a}^{i} E_{a}^{i}-(\delta_{j j} \delta_{k l}-\delta_{j k} \delta_{j l}) B_{a}^{k} B_{a}^{l})=\frac{1}{2}(E_{a}^{i} E_{a}^{i}-B_{a}^{i} B_{a}^{i}) .
\end{align}
Then, the Hamiltonian is given by
$$\mathcal{H}=\underbrace{\Pi_{a}^{i}(\partial_{0} A_{a}^{i})}_{E_{a}^{i} E_{a}^{i}}-\mathcal{L}=\frac{1}{2}(E_{a}^{i} E_{a}^{i}+B_{a}^{i} B_{a}^{i})=\frac{1}{2} \Pi_{a}^{i} \Pi_{a}^{i}+\frac{1}{4} F_{a}^{i j} F_{a}^{i j} .$$
|
{"$A^{0}$": "temporal gauge field component", "$g$": "coupling constant, set to 1", "$\\mathcal{H}$": "Hamiltonian density", "$E_a^i$": "chromo-electric field component", "$\\Pi_a^i$": "canonical momentum, equivalent to chromo-electric field component", "$B_a^i$": "chromo-magnetic field component", "$\\epsilon_{ijk}$": "Levi-Civita symbol", "$F_a^{jk}$": "field strength tensor (spatial components)", "$F_a^{ij}$": "field strength tensor", "$\\partial_{0}$": "time derivative"}
|
Localized Field Configurations
|
Localized Field Configurations
|
92 |
Magnetism
|
A point charge $e$ is located at point $O$ near a system of grounded conductors, inducing charges $e_{a}$ on these conductors. If the charge $e$ is absent and one of the conductors (the $a$-th) has a potential $\varphi_{a}^{\prime}$ (with the remaining conductors still grounded), then the potential at point $O$ is $\varphi_{0}^{\prime}$. Express the charge $e_{a}$ in terms of $\varphi_{a}^{\prime}$ and $\varphi_{0}^{\prime}$.
|
[
"e_a = -\\frac{e \\varphi_0'}{\\varphi_a'}"
] |
Expression
|
If the charge $e_{a}$ on the conductor gives the conductor a potential $\varphi_{a}$, while the charge $e_{a}^{\prime}$ gives the conductor a potential $\varphi_{a}^{\prime}$, then we have:
$$\sum_{a} \varphi_{a} e_{a}^{\prime}=\sum_{a, b} \varphi_{a} C_{a b} \varphi_{b}^{\prime}=\sum_{a} \varphi_{a}^{\prime} e_{a}$$
We apply this relationship to the two states of a system consisting of all conductors and a point charge $e$ (the latter viewed as a limiting case of a small-sized conductor). In one state, there is charge $e$ with charge on the conductors being $e_{a}$ and the potential $\varphi_{a}=0$. In another state, the charge $e=0$, but one conductor has a potential $\varphi_{a}^{\prime} \neq 0$. Thus we get $e \varphi_{0}^{\prime}+e_{a} \varphi_{a}^{\prime}=0$, from which follows
$$e_{a}=-\frac{e \varphi_{0}^{\prime}}{\varphi_{a}^{\prime}}$$
For example, if the charge $e$ is at a distance $r(r>a)$ from the center of a grounded spherical conductor of radius $a$, then $\varphi_{0}^{\prime}=\varphi_{a}^{\prime} \frac{a}{r}$, and the induced charge on the sphere is
$$
e_{a}=-\frac{e a}{r}
$$
As another example, consider a charge $e$ between two grounded concentric spheres of radii $a$ and $b$ respectively (the charge is at a distance $r$ from the center, where $a<r<b$). If the outer sphere is grounded and the inner sphere is charged to a potential $\varphi_{a}$, then the potential at a distance $r$ is given by
$$
\varphi_{0}^{\prime}=\varphi_{a}^{\prime} \frac{1 / r-1 / b}{1 / a-1 /} b.
$$
Thus, the induced charge by $e$ on the inner sphere is
$$
e_{a}=-e \frac{a(b-r)}{r(b-a)}
$$
Similarly, the induced charge on the outer sphere is
$$
e_{b}=-e \frac{b(r-a)}{r(b-a)}
$$
|
{"$e$": "point charge", "$e_{a}$": "induced charge on the a-th conductor", "$\\varphi_{a}^{\\prime}$": "potential of the a-th conductor", "$\\varphi_{0}^{\\prime}$": "potential at point O", "$r$": "distance from the center", "$a$": "radius of the inner sphere", "$b$": "radius of the outer sphere"}
|
Electrostatics of Conductors
|
Electrostatic field energy of conductor
|
93 |
Magnetism
|
Determine the capacitance $C$ of a circular ring made of a thin wire with a circular cross-section (ring radius is $b$, and the radius of the wire cross-section is $a$, where $b \gg a)$.
|
[
"C=\\frac{\\pi b}{\\ln (8 b / a)}"
] |
Expression
|
Since the ring is very thin, the electric field near its surface is equal to that produced by the same charge distribution along the axis of the ring (this is exact for a straight cylinder). Therefore, the potential of the ring is
$$
\varphi_{a}=\frac{e}{2 \pi b} \oint \frac{\mathrm{~d} l}{r},
$$
where $r$ is the distance from a given point on the ring surface to its axis element $\mathrm{d} l$, and the integral is along the axis. We divide the integral into two regions: $r<\Delta$ and $r>\Delta$, where $\Delta$ is a distance such that $a \ll \Delta \ll b$. Hence, when $r<\Delta$, it can be assumed that this section of the ring is straight, yielding
$$
\int_{\Delta>r} \frac{\mathrm{~d} l}{r}=\int_{-\Delta}^{\Delta} \frac{\mathrm{d} l}{\sqrt{l^{2}+a^{2}}} \approx 2 \ln \frac{2 \Delta}{a} .
$$
In the region $r>\Delta$, the thickness of the wire can be neglected, which is to say $r$ is assumed to be the distance between two points along the axis of the ring. Thus
$$
\int_{r>\Delta} \frac{\mathrm{d} l}{r}=2 \int_{\varphi_{0}}^{\pi} \frac{b \mathrm{~d} \varphi}{2 b \sin (\varphi / 2)}=-2 \ln \tan \frac{\varphi_{0}}{4}
$$
In this expression, $\varphi$ is the angle subtended by chord $r$ at the center of the ring, and the lower limit of integration is derived from $2 b \sin (\frac{\varphi_{0}}{2})=\Delta$, yielding $\varphi_{0} \approx \Delta / b$. Adding the two parts of the integral, the quantity $\Delta$ cancels automatically, and finally, the expression for the capacitance $C$ of the ring is given by
$$C=\frac{\pi b}{\ln (8 b / a)}$$
|
{"$C$": "capacitance", "$b$": "ring radius", "$a$": "radius of the wire cross-section", "$\\varphi_{a}$": "potential of the ring", "$r$": "distance from a point on the ring to its axis", "$\\Delta$": "intermediate distance scale", "$l$": "axis element", "$\\varphi$": "angle subtended at the center of the ring"}
|
Electrostatics of Conductors
|
Electrostatic field energy of conductor
|
94 |
Magnetism
|
An infinitely long cylindrical conductor with radius $R$ is immersed in a uniform transverse electric field with strength $\mathfrak{C}$. Find the potential distribution $\varphi(r, \theta)$ outside the cylinder.
|
[
"\\varphi=-\\mathfrak{C} r \\cos \\theta(1-\\frac{R^{2}}{r^{2}})"
] |
Expression
|
Using polar coordinates in the plane perpendicular to the axis of the cylinder. Thus, the solution to the two-dimensional Laplace equation involving a constant vector is
$$
\varphi_{1}=\text { const } \cdot \mathfrak{C} \cdot \nabla \ln r=\text { const } \cdot \frac{\mathfrak{C} \cdot \boldsymbol{r}}{r^{2}} .
$$
Adding the above with $\varphi_{0}=-\boldsymbol{r} \cdot \mathfrak{C}$ and letting const $=R^{2}$, we obtain
$$
\varphi=-\mathfrak{C} r \cos \theta(1-\frac{R^{2}}{r^{2}}) .
$$
|
{"$R$": "radius of the cylindrical conductor", "$\\mathfrak{C}$": "strength of the uniform transverse electric field", "$\\varphi$": "potential distribution", "$r$": "radial distance in polar coordinates", "$\\theta$": "angular coordinate in polar coordinates"}
|
Electrostatics of Conductors
|
Solutions to electrostatic problems
|
95 |
Magnetism
|
An infinitely long conducting cylinder with a radius $R$ is immersed in a uniform transverse electric field with a strength of $\mathfrak{C}$. Find the induced surface charge density $\sigma(\theta)$ on the surface of the cylinder.
|
[
"\\sigma=\\frac{\\mathfrak{C}}{2 \\pi} \\cos \\theta"
] |
Expression
|
The potential is
\begin{equation}
\varphi=-\mathfrak{C} r \cos \theta(1-\frac{R^{2}}{r^{2}}).
\end{equation}
Thus, the surface charge density is
$$\sigma= -\frac{1}{4\pi} \frac{\partial \varphi}{\partial r} = \frac{\mathfrak{C}}{2 \pi} \cos \theta. $$
|
{"$R$": "radius of the conducting cylinder", "$\\mathfrak{C}$": "strength of the uniform transverse electric field", "$\\sigma(\\theta)$": "induced surface charge density", "$\\theta$": "polar angle"}
|
Electrostatics of Conductors
|
Solutions to electrostatic problems
|
96 |
Magnetism
|
An infinitely long conducting cylinder with a radius of $R$ is immersed in a uniform transverse electric field with a strength of $\mathfrak{C}$. Find the induced dipole moment $\mathscr{P}$ per unit length of the cylinder.
|
[
"\\mathscr{P}=\\mathfrak{C} R^{2} / 2"
] |
Expression
|
The potential is
\begin{equation*}
\varphi=-\mathfrak{C} r \cos \theta(1-\frac{R^{2}}{r^{2}}).
\end{equation*}
The dipole moment $\mathscr{P}$ per unit length of the cylinder can be determined by comparing $\varphi$ with the potential of a two-dimensional dipole field, which is given by
$$2 \mathscr{P} \cdot \nabla \ln r=\frac{2 \mathscr{P} \cdot \boldsymbol{r}}{r^{2}},$$
Therefore, $\mathscr{P}=\mathfrak{C} R^{2} / 2$.
|
{"$R$": "radius of the conducting cylinder", "$\\mathfrak{C}$": "strength of the uniform transverse electric field", "$\\mathscr{P}$": "induced dipole moment per unit length of the cylinder", "$r$": "distance from the axis of the cylinder", "$\\theta$": "angle in polar coordinates"}
|
Electrostatics of Conductors
|
Solutions to electrostatic problems
|
97 |
Magnetism
|
Determine the attraction energy between an electric dipole and a planar conductor surface.
|
[
"\\mathscr{U}=-\\frac{1}{8 x^{3}}(2 \\mathscr{P}_{x}^{2}+\\mathscr{P}_{y}^{2})"
] |
Expression
|
Choose the $x$-axis perpendicular to the conductor surface, passing through the point where the dipole is located; let the dipole moment vector $\mathscr{P}$ lie in the $xy$ plane. The 'image' of the dipole is at the point $-x$ and has a dipole moment $\mathscr{P}_{x}^{\prime}=\mathscr{P}_{x}, \mathscr{P}_{y}^{\prime}=-\mathscr{P}_{y}$. Therefore, the required attraction energy is equivalent to the interaction energy between the dipole and its 'mirror image', and it equals
$$\mathscr{U}=-\frac{1}{8 x^{3}}(2 \mathscr{P}_{x}^{2}+\mathscr{P}_{y}^{2}) .$$
|
{"$x$": "position coordinate perpendicular to the conductor surface", "$\\mathscr{P}$": "dipole moment vector", "$\\mathscr{P}_{x}$": "component of the dipole moment vector along the x-axis", "$\\mathscr{P}_{y}$": "component of the dipole moment vector along the y-axis", "$\\mathscr{P}_{x}^{\\prime}$": "component of the mirror dipole moment vector along the x-axis", "$\\mathscr{P}_{y}^{\\prime}$": "component of the mirror dipole moment vector along the y-axis", "$\\mathscr{U}$": "attraction energy between the dipole and its mirror image"}
|
Electrostatics of Conductors
|
Solutions to electrostatic problems
|
98 |
Magnetism
|
Try to find an expression for the electric dipole moment of a conductive thin cylindrical rod (length $2l$, radius $a$, where $a \ll l$) placed in an electric field $\mathfrak{C}$, expressed in terms of the parameter $L=\ln (2 l / a)-1$. The electric field is parallel to the axis of the rod.
|
[
"\\mathscr{P} = \\mathfrak{C} \\frac{l^{3}}{3 L}[1+\\frac{1}{L}(\\frac{4}{3}-\\ln 2)]"
] |
Expression
|
Let $\tau(z)$ be the charge induced per unit length on the rod's surface; $z$ is the coordinate along the axis of the cylindrical rod, with the origin at the midpoint of the rod axis. The condition of constant potential on the conductor surface takes the form
$$-\mathfrak{C} z+\frac{1}{2 \pi} \int_{0}^{2 \pi} \int_{-l}^{l} \frac{\tau(z^{\prime}) \mathrm{d} z^{\prime} \mathrm{d} \varphi}{R}=0, \quad R=[(z-z)^{2}+4 a^{2} \sin ^{2} \frac{\varphi}{2}]^{1 / 2},$$
where $\varphi$ is the angle between planes through the rod axis and points on its surface separated by distance $R$. In the integral, using the identity $\tau(z^{\prime})=\tau(z)+[\tau(z^{\prime})-\tau(z)]$, the integration is divided into two parts. In the first part of the integration, noting $l \gg a$, for points not very close to the ends of the rod, we have
$$\frac{\tau(z)}{2 \pi} \iint \frac{\mathrm{~d} z^{\prime} \mathrm{d} \varphi}{R} \approx \frac{\tau(z)}{2 \pi} \int_{0}^{2 \pi} \ln \frac{l^{2}-z^{2}}{a^{2} \sin ^{2}(\varphi / 2)} \mathrm{d} \varphi=\tau(z) \ln \frac{4(l^{2}-z^{2})}{a^{2}}$$
(using the known integral value $\int_{0}^{\pi} \ln \sin \varphi \mathrm{d} \varphi=-\pi \ln 2$). In the second part of the integral, containing the difference $\tau(z^{\prime})-\tau(z)$, terms containing $a^{2}$ in $R$ can be neglected as they do not cause the integral to diverge. Thus,
$$
\mathfrak{C} z=\tau(z) \ln \frac{4(l^{2}-z^{2})}{a^{2}}+\int_{-l}^{l} \frac{\tau(z^{\prime})-\tau(z)}{|z^{\prime}-z|} \mathrm{d} z^{\prime}.
$$
The dependence of $\tau$ on $z$ is essentially proportional to $z$; in this approximation, the integral in the above equation gives $-2 \tau(z)$, resulting in
$$
\tau(z)=\frac{\mathfrak{C} z}{\ln [4(l^{2}-z^{2}) / a^{2}]-2}
$$
This expression is not applicable near the end points of the rod, but the region of $z$ values for calculating the requested dipole moment is not significant. At the precision adopted here, we have
\begin{align}
\mathscr{P} & =\int_{-l}^{l} \tau(z) z \mathrm{~d} z=\frac{\mathfrak{C}}{L} \int_{0}^{l}[z^{2}-\frac{z^{2}}{2 L} \ln (1-\frac{z^{2}}{l^{2}})] \mathrm{d} z \\
& =\mathfrak{C} \frac{l^{3}}{3 L}[1+\frac{1}{L}(\frac{4}{3}-\ln 2)]
\end{align}
(where $L=\ln (2 l / a)-1$ is a large number).
|
{"$l$": "half the length of the cylindrical rod", "$a$": "radius of the cylindrical rod", "$\\mathfrak{C}$": "electric field strength", "$L$": "parameter defined as $\\ln (2 l / a) - 1$", "$\\tau(z)$": "charge induced per unit length at position $z$", "$z$": "coordinate along the axis of the cylindrical rod", "$\\varphi$": "angular coordinate around the rod", "$\\mathscr{P}$": "electric dipole moment of the rod"}
|
Electrostatics of Conductors
|
Solutions to electrostatic problems
|
99 |
Magnetism
|
Attempt to find another approximate expression for the electric dipole moment of a conductive thin cylindrical rod (length $2l$, radius $a$, and $a \ll l$) in an electric field $\mathfrak{C}$, expressed directly in the logarithmic form of $l$ and $a$ $(\ln (4l/a))$. The field is parallel to the rod's axis.
|
[
"\\mathscr{P}=\\frac{\\mathfrak{C l} l^{3}}{3[\\ln (4 l / a)-7 / 3]}"
] |
Expression
|
Let $\tau(z)$ be the induced charge per unit length on the surface of the rod; $z$ is the coordinate along the axis of the cylinder, with the origin chosen at the midpoint of the rod. The condition of constant potential on the conductor's surface is
$$-\mathfrak{C} z+\frac{1}{2 \pi} \int_{0}^{2 \pi} \int_{-l}^{l} \frac{\tau(z^{\prime}) \mathrm{d} z^{\prime} \mathrm{d} \varphi}{R}=0, \quad R=[(z-z)^{2}+4 a^{2} \sin ^{2} \frac{\varphi}{2}]^{1 / 2},
$$
where $\varphi$ is the angle between the plane through the rod's axis and two points on its surface separated by distance $R$. In the integral, use the identity $\tau(z^{\prime})=\tau(z)+[\tau(z^{\prime})-\tau(z)]$, splitting the integral into two parts. In the first part, noting that $l \gg a$, for points not very close to the ends of the rod, we have
$$
\frac{\tau(z)}{2 \pi} \iint \frac{\mathrm{~d} z^{\prime} \mathrm{d} \varphi}{R} \approx \frac{\tau(z)}{2 \pi} \int_{0}^{2 \pi} \ln \frac{l^{2}-z^{2}}{a^{2} \sin ^{2}(\varphi / 2)} \mathrm{d} \varphi=\tau(z) \ln \frac{4(l^{2}-z^{2})}{a^{2}}
$$
(using the known integral result $\int_{0}^{\pi} \ln \sin \varphi \mathrm{d} \varphi=-\pi \ln 2$). In the second part of the integral, containing the difference $\tau(z^{\prime})-\tau(z)$, terms involving $a^{2}$ in $R$ can be omitted as they do not cause divergence of the integral. Thus,
$$
\mathfrak{C} z=\tau(z) \ln \frac{4(l^{2}-z^{2})}{a^{2}}+\int_{-l}^{l} \frac{\tau(z^{\prime})-\tau(z)}{|z^{\prime}-z|} \mathrm{d} z^{\prime} .
$$
The dependence of $\tau$ on $z$ essentially reduces to being proportional to $z$; in this approximation, the integral in the above expression gives $-2 \tau(z)$, resulting in
$$
\tau(z)=\frac{\mathfrak{C} z}{\ln [4(l^{2}-z^{2}) / a^{2}]-2}
$$
Near the endpoints of the rod, this expression is not applicable, but for the calculation of the desired dipole moment, this region of $z$-values is not important. To the precision adopted here, we have
\begin{aligned}
\mathscr{P} & =\int_{-l}^{l} \tau(z) z \mathrm{~d} z=\frac{\mathfrak{C}}{L} \int_{0}^{l}[z^{2}-\frac{z^{2}}{2 L} \ln (1-\frac{z^{2}}{l^{2}})] \mathrm{d} z \\
& =\mathfrak{C} \frac{l^{3}}{3 L}[1+\frac{1}{L}(\frac{4}{3}-\ln 2)]
\end{aligned}
(where $L=\ln (2 l / a)-1$ is a large number), or to the same precision
$$
\mathscr{P}=\frac{\mathfrak{C l} l^{3}}{3[\ln (4 l / a)-7 / 3]} .
$$
|
{"$l$": "half-length of the cylindrical rod", "$a$": "radius of the cylindrical rod", "$\\mathfrak{C}$": "electric field parallel to the rod's axis", "$z$": "coordinate along the axis of the cylinder", "$\\tau(z)$": "induced charge per unit length at position $z$", "$\\varphi$": "angle on the rod's surface", "$\\mathscr{P}$": "electric dipole moment of the rod", "$L$": "a large number, defined as $L=\\ln (2 l / a)-1"}
|
Electrostatics of Conductors
|
Solutions to electrostatic problems
|
100 |
Magnetism
|
Under the influence of a uniform external electric field, consider an uncharged ellipsoid. When the external electric field is only along the $x$ axis of the ellipsoid, find the charge distribution on its surface $\sigma$.
|
[
"\\sigma=\\mathfrak{C} \\frac{\\nu_{x}}{4 \\pi n^{(x)}}"
] |
Expression
|
We first have
$$
\sigma=-\frac{1}{4 \pi} \frac{\partial \varphi}{\partial n}|_{\xi=0}=-(\frac{1}{4 \pi h_{1}}-\frac{\partial \varphi}{\partial \xi})_{\xi=0}
$$
(According to equation
\begin{align}
d l^2 &= h_1^2 d \xi^2 + h_2^2 d \eta^2 + h_3^2 d \zeta^2, \\
h_1 &= \frac{\sqrt{(\xi - \eta)(\xi - \zeta)}}{2 R_\xi}, \quad
h_2 = \frac{\sqrt{(\eta - \zeta)(\eta - \xi)}}{2 R_\eta}, \quad
h_3 = \frac{\sqrt{(\zeta - \xi)(\xi - \eta)}}{2 R_\zeta},
\end{align}
the length element along the normal direction of the ellipsoid surface is $h_{1} \mathrm{~d} \xi$.) With the help of equation
\begin{align}
\varphi = \varphi_0 \left\{ 1 - \frac{\displaystyle \int_{\xi}^{\infty} \frac{ds}{(s + a^2) R_s}}{\displaystyle \int_{0}^{\infty} \frac{ds}{(s + a^2) R_s}}, \right\}
\end{align}
and considering
$$
\nu_{x}=\frac{1}{h_{1}} \frac{\partial x}{\partial \xi}|_{\xi=0}=\frac{x}{2 a^{2} h_{1}}|_{\xi=0},
$$
when the external electric field is along the x-axis direction of the ellipsoid, we obtain
$$
\sigma=\mathfrak{C} \frac{\nu_{x}}{4 \pi n^{(x)}}
$$
|
{"$x$": "coordinate axis", "$\\sigma$": "charge distribution on the surface", "$\\varphi$": "electric potential", "$h_{1}$": "metric coefficient associated with $\\xi$", "$\\xi$": "ellipsoidal coordinate", "$\\eta$": "ellipsoidal coordinate", "$\\zeta$": "ellipsoidal coordinate", "$R_{\\xi}$": "radius of curvature associated with $\\xi$", "$R_{\\eta}$": "radius of curvature associated with $\\eta$", "$R_{\\zeta}$": "radius of curvature associated with $\\zeta$", "$\\varphi_0$": "reference electric potential", "$a$": "characteristic length parameter", "$s$": "integration variable", "$\\mathfrak{C}$": "normalization constant", "$\\nu_{x}$": "component of normal vector along x-axis", "$n^{(x)}$": "normalization factor for x-direction"}
|
Electrostatics of Conductors
|
conducting ellipsoid
|
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