Aspects of time reversal

Time reversal in quantum mechanics

Here we follow a paper by Eugene Paul Wigner from 1932, Über die Operation der Zeitumkehr in der Quantenmechanik .

Schrödinger’s equation is

\[H\left|\Psi(t)\right\rangle = {\rm i}\hbar\partial_t\left|\Psi(t)\right\rangle,\]

connecting an operator in Hilbert space with a time derivative acting onto a parameter of a wavefunction, definitely not an operator in Hilbert space. If we just replace $t\to-t$, the equation reads $H\left|\Psi(-t)\right\rangle=-{\rm i}\hbar\partial_t\left|\Psi(-t)\right\rangle$. For real $H$, the original form is restored by complex conjugation:

\[H\left|\Psi(-t)\right\rangle^* = {\rm i}\hbar\partial_t\left|\Psi(-t)\right\rangle^*,\]

so we are searching for an antilinear operator $K$ transforming $\left|\Psi(-t)\right\rangle\to\left|\Psi_\text{umk}(t)\right\rangle$.

Definition (by Wigner): An antilinear operator $K$ obeys the following rules:

\[\begin{aligned} K^2\left|\phi\right\rangle &= \left|\phi\right\rangle, & K^2 &=1,\\ K\left(a\left|\phi\right\rangle+b\left|\psi\right\rangle\right) &= a^*K\left|\phi\right\rangle+b^*K\left|\psi\right\rangle, \\ \left(\left\langle\phi\right|K\right)\left|\psi\right\rangle &= \left[\left\langle\phi\right| \left(K\left|\psi\right\rangle\right)\right]^* = \left(\left\langle\psi\right|K\right)\left|\phi\right\rangle.\end{aligned}\]

We note that in contrast to linear operators, the «direction» in which $K$ is applied matters, indicated by the parentheses in the last line of the above equations.

But time reversal in quantum mechanics is not just complex conjugation: We can only require that probabilites (generalized scalar products) don’t change upon time reversal, or

\[\begin{aligned} \left\langle\phi_\text{umk}(t)\right| \left.\psi_\text{umk}(t)\right\rangle &= \left(\left\langle\phi(-t)\right| \left.\psi(-t)\right\rangle\right)^* = \left(\left\langle\phi(-t)\right|UU^\dagger \left|\psi(-t)\right\rangle\right)^* \\ \Rightarrow \left|\psi_\text{umk}(t)\right\rangle &= U\left|\psi(-t)\right\rangle^* = \underbrace{ UK }_{T} \left|\psi(-t)\right\rangle\end{aligned}\]

with $U$ being some unitary transformation and $T$ being the (antiuntary) time-reversion transformation. We always have

\[K^2=1 \Rightarrow K=K^{-1} \Rightarrow T^{-1}=(UK)^{-1}=K^{-1}U^{-1}=KU^\dagger.\]

If such a unitary operator $U$ exists with

\[TH(-t)T^{-1} = UH^*(-t)U^\dagger = H(t),\]

we can write Schrödinger’s equation like

\[\begin{aligned} UH^*(-t)U^\dagger U\left|\psi(-t)\right\rangle^* &= {\rm i}\hbar\partial_tU\left|\psi(-t)\right\rangle^* \\ \Rightarrow H(t)\left|\psi_\text{umk}(t)\right\rangle &= {\rm i}\hbar\partial_t\left|\psi_\text{umk}(t)\right\rangle.\end{aligned}\]

It follows that if $\left|\psi(t)\right\rangle$ is a solution, $\left|\psi_\text{umk}(t)\right\rangle=T\left|\psi(-t)\right\rangle$ is a solution, too.

Corollary: A time-reversal invariant Hamiltonian is real. Because $\left|\phi\right\rangle$ and $T\left|\phi\right\rangle$ are eigenfunctions, we can replace these by the real functions $\left|\phi\right\rangle+T\left|\phi\right\rangle$ and ${\rm i}\left(\left|\phi\right\rangle-T\left|\phi\right\rangle\right)$. Therefore all eigenfunctions can be assumed to be real. Therefore also the Hamiltonian $H$ (called Energieoperator by Wigner) must be real. [A function $\left|\phi\right\rangle$ is real if $T\left|\phi\right\rangle=\left|\phi\right\rangle$, imaginary if $T\left|\phi\right\rangle=-\left|\phi\right\rangle$].

The same holds for all spatial symmetries: It is irrelevant whether we first transform spatially, then reverse time, or vice versa. That is, $T$ commutes with the spatial transformation $Q$ up to a possible multiplicative phase (constant $c$ of unit modulus). Linearity for $Q$ requires that this constant must be identical for all wavefunctions $Q$ is applied to, so setting $O=\sqrt cQ$ indeed makes $O$ a real operator (which commutes with $T$ and $H$).

Comparison with classical mechanics

We require $TxT^{-1}=T$ and $TpT^{-1}=-p$ for the position and momentum operators, respectively. This requrement has its origin in the requirement that Lagrange’s equation of motion,

\[\frac{\partial L}{\partial x} -\frac{\text d}{\text dt}\frac{\partial L}{\partial\dot x} =0\]

is invariant under $t\to-t$. Thus angular momentum $\ell=x\times p$ and spin $s$ should transform as $T\ell T^{-1}=-\ell$ and $TsT^{-1}=-s$.

Parametrizing the spin using Pauli’s matrices as $s_i=(1/2)\sigma_i$ in their standard form, the following relations hold:

\[\begin{aligned} T\sigma T^{-1} &= UK\sigma K^{-1}U^{-1} \Rightarrow \sigma \stackrel!= -T\sigma T^{-1} = -UK\sigma K^{-1}U^{-1} = -U\sigma^*U^{-1} \nonumber\\ \Rightarrow\sigma_x &= -U\sigma_xU^{-1}, \quad \sigma_y=U\sigma_yU^{-1}, \quad \sigma_z=-U\sigma_zU^{-1}. \label{eqn:time:spin}\end{aligned}\]

This corresponds to a rotation of the spin $s$ by an angle $\pi$ around the $y$ axis. A common choice (up to an irrelevant phase factor) is the real matrix

\[U = \exp\left({\rm i}\pi s_y\right) = \begin{pmatrix} 0&1\\-1&0 \end{pmatrix} = {\rm i}\sigma_y, \quad U^2 = -1. \label{eqn:time:u}\]

For $n$ spin-$1/2$ particles, we obtain

\[\begin{aligned} T\left|x_1,\dots,x_n;\sigma^{(1)},\dots,\sigma^{(n)}\right\rangle &= {\rm i}^n \sigma_y^{(1)}\cdots\sigma_y^{(n)} \left|x_1,\dots,x_n;\sigma^{(1)},\dots,\sigma^{(n)}\right\rangle \\ &\stackrel!= {\rm i}^n \sigma_y^{(1)}\cdots\sigma_y^{(n)} \left|x_1,\dots,x_n;-\sigma^{(1)},\dots,-\sigma^{(n)}\right\rangle^*\end{aligned}\]