\begin{document}

\title{Curvature and torsion of quantum evolution}

\author{V. M. Tkachuk\footnote{tkachuk@ktf.franko.lviv.ua}\\
Department of Theoretical Physics,\\ Ivan Franko National
University of Lviv,\\ 12 Drahomanov St., Lviv, UA-79005, Ukraine}


\maketitle

\begin{abstract}
We study characteristics of quantum evolution which can be called
curvature and torsion. The curvature shows a deviation of the
state vector of the quantum evolution from the geodesic line and
the torsion shows a deviation of state vector from the plane of
evolution (a two-dimensional subspace) at a given time.
\end{abstract}

\vspace{0.5cm}

\noindent {PACS numbers}: 03.65.-w

\noindent {Keywords}: Curvature; Torsion; Quantum evolution;
Geometry of quantum state space

\section{Introduction}
Geometric ideas play an important role in quantum mechanics, in
particular in the studies of the quantum evolution
\cite{Anandan90,Abe93,Grigorenko91}, quantum brachistochrone
problem \cite{Mosta07,Bender08}, entanglement of quantum states
\cite{Brody01,Brody07}, Berry phase has geometric origin \cite{Shapere89}.

In the classical case the curvature and
torsion are important geometric characteristics of the trajectory.
The aim of the present paper is to answer the question: What is
the quantum analogue of these classical geometrical notions?
Partly the answer on this question was given in \cite{Brody96} where the authors from a different perspective than in this paper, namely, considering geometry of quantum statistical interference, derived explicit expression for the curvature of quantum evolution.

First let us consider some facts of the geometry of the space of
quantum states. A distance between two quantum states
$|\psi_1\rangle$ and $|\psi_2\rangle$ can be defined in different
ways. In this paper we will refer to the Fubiny-Study distance and
the Wootters distance defined, respectively, as follows (for a
short review see, for instance, \cite{Dodonov 99})
\begin{eqnarray}
d^{({\rm
FS})}(|\psi_1\rangle,|\psi_2\rangle)=\gamma\sqrt{1-|\langle\psi_1|\psi_2\rangle|^2},
\\
d^{({\rm W
})}(|\psi_1\rangle,|\psi_2\rangle)=\gamma\arccos|\langle\psi_1|\psi_2\rangle|,
\end{eqnarray}
where $\gamma$ is an arbitrary constant.

These distances are equivalent for the neighboring states when
$|\langle\psi_1|\psi_2\rangle|^2=1-\delta^2$, where $\delta$ is a
small value, namely, $d^{({\rm FS})}=d^{({\rm W})}=\gamma\delta$.
Likewise we have for other definitions of distance. As a result of
this the element of length for the family (set) of quantum state vectors
$|\psi(\xi^1,\xi^2,...,\xi^k)\rangle$ parametrized by $k$
parameters $\xi^1,\xi^2,...,\xi^k$ is the same for different
definitions of distance
\begin{eqnarray}
ds^2=g_{ij}d\xi^id\xi^j
\end{eqnarray}
with metric tensor
\begin{eqnarray} \label{metric}
g_{ij}=\gamma^2 {\rm
Re}(\langle\psi_i|\psi_j\rangle-\langle\psi_i|\psi\rangle\langle\psi|\psi_j\rangle),
\end{eqnarray}
where
\begin{eqnarray}
|\psi_i\rangle={\partial\over\partial\xi^i}|\psi(\xi^1,\xi^2,...,\xi^k)\rangle.
\end{eqnarray}
This form of metrics for quantum states was discussed by many
authors \cite{Abe93,Provost80,Revicule97,Aalok06,Aalok07}.

It is convenient to put $\gamma=2$. Then in a two-dimensional case
$g_{ij}$ is metric tensor of a sphere with the radius equal one
(the Bloch sphere).

Considering evolution of a quantum state according to the
Schr\"odinger equation one can introduce the velocity of quantum
evolution \cite{Anandan90}
\begin{eqnarray} \label{velocity}
v={ds\over dt}=\gamma{\sqrt{\langle(\Delta H)^2\rangle}/ \hbar},
\end{eqnarray}
where $\Delta H=H-\langle H\rangle$.

\section{Geodesic in the space of quantum state vectors}
The geodesic line (one-parametric set of the quantum state vectors) that connects two state vectors $|\psi_0\rangle$
and $|\psi_1\rangle$ can be defined as their linear combination
\begin{eqnarray}\label{geodesic}
|\psi(\xi)\rangle=C[(1-\xi)|\psi_0\rangle+\xi|\psi_1\rangle
e^{i\phi}],
\end{eqnarray}
where $\xi$ is a real parameter changing from $0$ to $1$. We
choose the phase multiplier $e^{i\phi}$ from the following
condition. Note that $|\psi_0\rangle$ and
$e^{i\phi_0}|\psi_0\rangle$ describe the same quantum state,
similarly, $|\psi_1\rangle$ and $e^{i\phi_1}|\psi_1\rangle$
describe the same quantum state. Therefore, we require that
geodesic lines defined between the states $|\psi_0\rangle$,
$|\psi_1\rangle$ and between the states
$e^{i\phi_0}|\psi_0\rangle$, $e^{i\phi_1}|\psi_1\rangle$ coincide.
This requirement is satisfied if we choose
\begin{eqnarray} \label{phi}
e^{i\phi}={\langle\psi_1|\psi_0\rangle\over
|\langle\psi_1|\psi_0\rangle|}.
\end{eqnarray}
The normalization condition $\langle\psi(\xi)|\psi(\xi)\rangle=1$
gives
\begin{eqnarray}
C={1\over\sqrt{1-2\xi(1-\xi)(1-|\langle\psi_1|\psi_0\rangle|)}}.
\end{eqnarray}

The geodesic line (\ref{geodesic}) is a one-parametric set of
states and there exist many possibilities to parametrize it. One can
show that the length of the curve in quantum space does not depend
on the way of its parametrization. For the calculation of the length
of the geodesic line it is convenient to write its equation as
follows
\begin{eqnarray}\label{geodesic2}
|\psi(\theta)\rangle=C[\sin({\theta/2})|\psi_0\rangle+\cos({\theta/2})|\psi_1\rangle
e^{i\phi}],
\end{eqnarray}
here new parameter $0 \le\theta\le \pi$ and normalization constant $ C=1/ \sqrt {1+
|\langle\psi_1|\psi_0\rangle|\sin\theta}$.
Stress once more that (\ref{geodesic}) and (\ref{geodesic2})
describe the same one-parametric family of quantum state vectors, namely, the geodesic line.

Using (\ref{metric}) for the one-parameter set of states
(\ref{geodesic2}) we obtain
\begin{eqnarray}
ds={\gamma\over 2}{\sqrt{1-|\langle\psi_1|\psi_0\rangle|^2} \over
1+|\langle\psi_1|\psi_0\rangle|\sin\theta}d\theta.
\end{eqnarray}
Then the length of the geodesic line connecting the states
$|\psi_0\rangle$ and $|\psi_1\rangle$ is
\begin{eqnarray}
s=\int ds=\gamma\arccos|\langle\psi_1|\psi_0\rangle|.
\end{eqnarray}
As it should be, the length of the geodesic line is equal to the
Wootters distance between the corresponding state vectors.

In conclusion of this section let us note that we can calculate
the length of curve (\ref{geodesic}) connecting the states
$|\psi_0\rangle$ and $|\psi_1\rangle$ for an arbitrary phase
$\phi$. Then the geodesic line can be defined as the one having
minimal length. One can find that minimal length is achieved for
$\phi$ given in (\ref{phi}) and is equal to the Wootters distance.

\section{Curvature}
The state vector of the quantum evolution depends on one parameter, namely, on time $t$ and
is one-parametric set of state vectors $|\psi(t)\rangle=\exp(-iHt)|\psi_0\rangle$ generated
by the Hamiltonian of the system. The deviation of evolution state vector $|\psi(t)\rangle$
from the geodesic connecting
the same two state vectors is related with the curvature of quantum evolution.

In order to introduce the curvature as well as
the torsion we consider the evolution in two stages. At first, we
consider the evolution during the time $\Delta t$ from an initial
state $|\psi_0\rangle$ to
\begin{eqnarray} \label{psip}
|\psi'\rangle=e^{-iH\Delta t/\hbar}|\psi_0\rangle
\end{eqnarray}
and then -- during $\Delta t'$ from $|\psi'\rangle$ to
\begin{eqnarray}\label{psi1}
|\psi_1\rangle=e^{-i H\Delta t'/\hbar}|\psi'\rangle=e^{-i H
(\Delta t+\Delta t')/\hbar}|\psi_0\rangle.
\end{eqnarray}
where $H$ is a time independent Hamiltonian. In this section
without the loss of generality we put $\Delta t=\Delta t'$.

A deviation of the quantum evolution from the geodesic line
connecting $|\psi_0\rangle$ and $|\psi_1\rangle$ can be
characterized by the maximum value of
$|\langle\psi'|\psi(\xi)\rangle|^2$ with respect to $\xi$. When
${\rm max}|\langle\psi'|\psi(\xi)\rangle|^2=1$ then the state
$|\psi'\rangle$ belongs to the geodesic. A deviation of
$|\psi'\rangle$ from the geodesic is lager when ${\rm
max}|\langle\psi'|\psi(\xi)\rangle|^2$ is smaller.  It is
convenient to introduce the following expression $1-{\rm
max}|\langle\psi'|\psi(\xi)\rangle|^2 = {\rm
min}(1-|\langle\psi'|\psi(\xi)\rangle|^2)$ which equals zero when
the deviation is zero and is positive otherwise becoming larger as
the deviation is larger. This expression multiplied by some
constant $\gamma^2$ is just a squared Fubiny-Study distance. So, a
deviation of the quantum evolution from the geodesic line
connecting $|\psi_0\rangle$ and $|\psi_1\rangle$ can be
characterized by the distance between the state $|\psi'\rangle$
and the geodesic line $|\psi(\xi)\rangle$
\begin{eqnarray}
d^2=\min d^2(\xi)=
\min\gamma^2(1-|\langle\psi'|\psi(\xi)\rangle|^2).
\end{eqnarray}
The minimal value of this expression is achieved at $\xi=1/2$.
Taking into account the terms of order $(\Delta t)^4$ we find
\begin{eqnarray} \label{dmax}
d^2={\gamma^2\over4}\left(\langle(\Delta
H)^4\rangle-\langle(\Delta H)^2\rangle^2\right){(\Delta
t)^4\over\hbar^4}={\gamma^2\over4}\kappa{(\Delta
t)^4\over\hbar^4}.
\end{eqnarray}
Here, the multiplier
\begin{eqnarray}
\kappa=\langle(\Delta H)^4\rangle-\langle(\Delta H)^2\rangle^2
\end{eqnarray}
can be called the curvature coefficient or, simply, curvature. It
is convenient to introduce a dimensionless curvature coefficient
\begin{eqnarray} \label{barcurv}
\bar \kappa={\langle(\Delta H)^4\rangle-\langle(\Delta
H)^2\rangle^2\over\langle(\Delta H)^2\rangle^2}.
\end{eqnarray}
For the first time this result was obtained in \cite{Brody96} in the frame of study of the geometry
of quantum statistical interference.

Now we show that the curvature of the quantum evolution can also
be obtained using the geometric treatment. For small time the
classical motion along a given curve can be treated as a motion
along a circle with radius $R$ for which we can write
\begin{eqnarray} \label{1R}
{1\over R}={2d\over(s/2)^2},
\end{eqnarray}
where $s$ is the length of the curve between two neighboring
points on it, which can be considered as an arc of the circle, and
$d$ is the distance between the middle point of an arc and the
chord connecting these two points. Similarly to (\ref{1R}) we now
define the radius of curvature for the quantum evolution. In our
case $d$ is given by (\ref{dmax}) and
\begin{eqnarray}
s=v2\Delta t=\gamma{\sqrt{\langle(\Delta H)^2\rangle}\over
\hbar}2\Delta t
\end{eqnarray}
is the length that a quantum system passes during the time
$2\Delta t$ of the evolution. Here $v$ is the velocity of quantum
evolution given in (\ref{velocity}). As a result we have
\begin{eqnarray}
{1\over R^2}={1\over\gamma^2}{\langle(\Delta
H)^4\rangle-\langle(\Delta H)^2\rangle^2\over\langle(\Delta
H)^2\rangle^2}={\bar \kappa\over\gamma^2}.
\end{eqnarray}

\section{Torsion}
Torsion is related with the deviation of evolution state vector from the plane of evolution
(a two-dimensional subspace) at a given time.

In order to find torsion we consider
the evolution in the two stages given by (\ref{psip}) and
(\ref{psi1}). Two vectors $|\psi_0\rangle$ and $|\psi'\rangle$
form the first stage
define the plane of evolution. Using these vectors we can
construct the orthogonal ones
\begin{eqnarray}
|\phi_1\rangle={1\over
\sqrt{2(1+a)}}\langle(|\psi_0\rangle+e^{-i\alpha}|\psi'\rangle),\\
|\phi_2\rangle={1\over
\sqrt{2(1+a)}}\langle(|\psi_0\rangle-e^{-i\alpha}|\psi'\rangle),
\end{eqnarray}
where $a$ and $\phi$ are defined by
$\langle\psi_0|\psi'\rangle=ae^{i\alpha}$. Then the unit operator
in a two-dimensional subspace spanned by $|\phi_1\rangle$ and
$|\phi_2\rangle$ is
\begin{eqnarray}
\hat
I_2=|\phi_1\rangle\langle\phi_1|+|\phi_2\rangle\langle\phi_2|.
\end{eqnarray}
Note that it is the projection operator of an arbitrary state
vector on a two-dimensional subspace.

In order to find the deviation of the state vector
$|\psi_1\rangle$ obtained on the second stage from the plane of evolution we calculate the mean
value of $\hat I_2$
\begin{eqnarray}
I_2=\langle\psi_1|\hat I_2|\psi_1\rangle
=|\langle\phi_1|\psi_1\rangle|^2+|\langle\phi_2|\psi_1\rangle|^2.
\end{eqnarray}
When $I_2=1$ then $|\psi_1\rangle$ belongs to the subspace spanned
by $|\phi_1\rangle$ and $|\phi_2\rangle$. It means that three state vectors
$|\psi_0\rangle$, $|\psi'\rangle$, $|\psi_1\rangle$ belong to the
same plane (the two-dimensional subspace) of evolution and thus
the torsion is zero. The expression $1-I_2$ gives the magnitude of
the torsion. Considering small $\Delta t$ and $\Delta t'$ and
taking into account the terms up to fourth order we find
\begin{eqnarray}
1-I_2=\left(\langle(\Delta H)^4\rangle-\langle(\Delta
H)^2\rangle^2-{\langle(\Delta H)^3\rangle^2\over\langle(\Delta
H)^2\rangle}\right){\Delta t^2(\Delta t+\Delta t')^2\over
4\hbar^4}.
\end{eqnarray}
The coefficient
\begin{eqnarray}
\tau=\langle(\Delta H)^4\rangle-\langle(\Delta
H)^2\rangle^2-{\langle(\Delta H)^3\rangle^2\over\langle(\Delta
H)^2\rangle}
\end{eqnarray}
does not depend on $\Delta t$ and $\Delta t'$ and can be called
the torsion coefficient. For simplicity we put $\Delta t=\Delta
t'$ and then
\begin{eqnarray}
1-I_2=\tau{\Delta t^4\over \hbar^4}.
\end{eqnarray}
Similarly to the dimensionless curvature coefficient we introduce
a dimensionless torsion coefficient
\begin{eqnarray}\label{bartor}
\bar\tau={\tau\over\langle(\Delta H)^2 \rangle^2} =
{\langle(\Delta H)^4\rangle-\langle(\Delta
H)^2\rangle^2\over\langle(\Delta H)^2\rangle^2}-{\langle(\Delta
H)^3\rangle^2\over\langle(\Delta H)^2\rangle^3}=\bar
\kappa-{\langle(\Delta H)^3\rangle^2\over\langle(\Delta
H)^2\rangle^3}.
\end{eqnarray}

Now let us show that $1-I_2$ has a geometrical meaning, namely, it
is proportional to the squared distance of the state
$|\psi_1\rangle$ to the plane of quantum evolution spanned by
$|\phi_1\rangle$ and $|\phi_2\rangle$. The distance between a
given state $|\psi_1\rangle$ and the plane is that between
$|\psi_1\rangle$ and the normalized projection of this vector onto
the plane. The normalized projection of $|\psi_1\rangle$ on the
plane is
\begin{eqnarray}
|\tilde\psi_1\rangle=c \hat I_2|\psi_1\rangle,
\end{eqnarray}
where from the condition
$\langle\tilde\psi_1|\tilde\psi_1\rangle=1$ we find
$c=1/\sqrt{\langle\psi_1|\hat I_2\hat I_2|\psi_1\rangle}=
1/\sqrt{\langle\psi_1|\hat I_2|\psi_1\rangle}$. Here we use that
$(\hat I_2)^2=\hat I_2$. Then the squared distance between the
state $|\psi_1\rangle$ and the plane is
\begin{eqnarray}
d^2=\gamma^2(1-|\langle\psi_1|\tilde\psi_1\rangle|^2)=
\gamma^2(1-|\langle\psi_1|\hat
I_2|\psi_1\rangle|)=\gamma^2(1-I_2),
\end{eqnarray}
where we use that $|I_2|=I_2$.

\section{Discussion}
In this paper we have obtained the curvature
and torsion coefficients (\ref{barcurv}) and (\ref{bartor}) for
the quantum evolution which is governed by a time independent
Hamiltonian. In this case the curvature and torsion coefficients
are constant.

The evolution is going along the geodesic when $\bar
\kappa=0$  or explicitly
\begin{eqnarray}\label{Eqk0}
\langle(\Delta H)^4\rangle-\langle(\Delta H)^2\rangle^2=0.
\end{eqnarray}
Introducing operator $\hat A= (\Delta H)^2$ we rewrite this
condition in the form $\langle(\Delta \hat A)^2\rangle=0$. Then we
find that (\ref{Eqk0}) is equivalent to the equation $\Delta \hat
A|\psi\rangle=0$ which explicitly reads
\begin{eqnarray} \label{Eig}
(\Delta H)^2|\psi\rangle=\langle(\Delta H)^2\rangle|\psi\rangle.
\end{eqnarray}
The solution of this equation is
\begin{eqnarray} \label{psi0}
|\psi\rangle={1\over\sqrt
2}\left(|E_1\rangle+e^{i\alpha}|E_2\rangle\right),
\end{eqnarray}
where $\alpha$ is an arbitrary phase, $|E_1\rangle$ and
$E_1\rangle$ are two eigenstates of the Hamiltonian $H$ with
eigenenergies $E_1$ and $E_2$. Considering (\ref{psi0}) as an initial state for time
dependent state that evolves along the geodesic we find
\begin{eqnarray} \label{psit}
|\psi(t)\rangle={1\over\sqrt
2}\left(e^{-iE_1t/\hbar}|E_1\rangle+e^{i\alpha}e^{-iE_2t/\hbar}|E_2\rangle\right).
\end{eqnarray}
One can find that for an arbitrary fixed time the evolution state vector (\ref{psit}) satisfies
equation  (\ref{Eig}) and
the curvature for this evolution is zero.
It is worth stressing that the state vector of the geodesic
evolution contains only two eigenstates of the Hamiltonian and
lies in a two-dimensional subspace.

Now let us show that the torsion of the geodesic is zero. Using
(\ref{Eig}) we have
\begin{eqnarray}
\langle(\Delta H)^3\rangle=\langle\psi|\Delta H(\Delta
H)^2|\psi\rangle=\langle(\Delta H)^2\rangle\langle\psi|\Delta
H|\psi\rangle=0.
\end{eqnarray}
Then according to (\ref{bartor}) and remembering that for the
geodesic $\bar\kappa=0$ we find that the torsion $\bar\tau=0$.

Let us verify that for two-dimensional space the torsion given by
(\ref{bartor}) is zero as this is abiding by the definition. The
most general Hamiltonian for a two-dimensional case reads
\begin{eqnarray}\label{HtooD}
H= \omega(\mbox{\boldmath $\sigma$}{\bf n}) + \epsilon,
\end{eqnarray}
where $\bf n$ is a unit vector.
Note that the curvature and torsion depend on $\Delta H$, where $\epsilon$ is
canceled. So, with out loss the generality we put $\epsilon=0$. Then
using the properties of Pauli
matrices we find for Hamiltonian (\ref{HtooD}) with $\epsilon=0$  the following results
$\langle(\Delta
H)^2\rangle=\omega^2-\langle H\rangle^2$,
$\langle(\Delta
H)^4\rangle-\langle(\Delta H)^2\rangle^2
=4\langle H\rangle^2\langle(\Delta
H)^2\rangle$ and $\langle(\Delta
H)^3\rangle=-2\langle H\rangle\langle(\Delta
H)^2\rangle$. Then one can find that the torsion (\ref{bartor}) in
the two-dimensional case is always zero.
For curvature in this case we have $\bar\kappa=4\langle H\rangle^2/\langle(\Delta
H)^2\rangle$. Thus in two-dimensional case the quantum evolution is going along
the geodesic line when
$\langle H\rangle=\omega\langle(\mbox{\boldmath $\sigma$}{\bf n})\rangle=0$.

In conclusion let us note an interesting fact which follows from
(\ref{bartor}). Namely, for symmetric states when $\langle(\Delta
H )^3\rangle=0$ we find that $\bar\kappa=\bar\tau$. It means that
the curvature and torsion during the evolution of symmetric states
are strongly related.

\section*{Acknowledgment}
I would like to thank Dr. Yu. Krynytskyi for constructive
discussions and useful comments.
{"config": "arxiv", "file": "1006.4447.tex"}