TITLE: Dirac field energy operator
QUESTION [1 upvotes]: When we quantise the Dirac field in the Heisenberg picture the resulting field and hamiltonian are:
\begin{equation}
\psi(\vec{x},t) = \int{\frac{d^3\vec{p}}{(2\pi)^3} \sum_{s=1}^{2}{\frac{1}{\sqrt{2E_\vec{p}} } (a_{\vec{p},s} u(\vec{p},s)e^{-ipx}+b_{\vec{p},s}^\dagger v(\vec{p},s)e^{-ipx}} })
\end{equation}
\begin{equation}
H = \int{\frac{d^3\vec{p}}{(2\pi)^3} \sum_{s=1}^{2} E_{\vec{p}}(a_{\vec{p},s}^\dagger a_{\vec{p},s}+b_{\vec{p},s}^\dagger b_{\vec{p},s}})
\end{equation}
with the commutation relations $\{a_{p,r},a^\dagger_{q,s}\} = \{b_{p,r},b^\dagger_{q,s}\}= (2\pi)^3\delta^{(3)}(\vec{p}-\vec{q})\delta_{r,s}$
What is the energy of a one particle state which is created by $a_{\vec{p},s}^\dagger$?
I understand that a one particle state would have the form 
\begin{equation}
\sqrt{2E_{\vec{p}}}a_{\vec{p},s}^\dagger | 0 \rangle
\end{equation}
but I'm not sure how to derive it's energy using these definitions. Sorry if this is super trivial.

REPLY [2 votes]: You need to know one more commutator, $\{b_{\vec{p},r},a^\dagger_{\vec{q},s}\} = 0$ and use the fact that $a_{\vec{p},s}|0\rangle=b_{\vec{p},s}|0\rangle=0$.
Note that $$a^\dagger_{\vec{q},r}a_{\vec{q},r}a^\dagger_{\vec{p},s}|0\rangle = a^\dagger_{\vec{q},r}\big(-a^\dagger_{\vec{p},s}a_{\vec{q},r}+(2\pi)^3\delta^3(\vec{q}-\vec{p})\delta_{rs}\big)|0\rangle = \\ =(2\pi)^3\delta^3(\vec{q}-\vec{p})\delta_{rs} a^\dagger_{\vec{q},r}|0\rangle = \\
= (2\pi)^3\delta^3(\vec{q}-\vec{p})\delta_{rs} a^\dagger_{\vec{p},s}|0\rangle $$
$$b^\dagger_{\vec{q},r}b_{\vec{q},r}a^\dagger_{\vec{p},s}|0\rangle = b^\dagger_{\vec{q},r}\big(-a^\dagger_{\vec{p},s}b_{\vec{q},r}\big)|0\rangle = 0 $$
From this you can get that $$ H a^\dagger_{\vec{p},s}|0\rangle = E_{\vec{p}}a^\dagger_{\vec{p},s}|0\rangle$$
Alternatively, you can try to prove that $$ [H,a^\dagger_{\vec{p},s}] = E_{\vec{p}}a^\dagger_{\vec{p},s}$$
it will give the same result.
{"set_name": "stack_exchange", "score": 1, "question_id": 519891}