Time Reversal with Operator Domains

Time-reversal invariance of a self-adjoint Hamiltonian requires an antiunitary map that preserves its domain and conjugates the Hamiltonian to itself. It then reverses the unitary evolution parameter through spectral calculus. This page sharpens the operator statement behind Time Reversal, especially where boundary conditions or external parameters affect whether the symmetry belongs to one fixed model.

Required background. Unitary and Antiunitary Operator Transport supplies spectral covariance. Self-Adjoint Operators explains why domains are part of the symmetry test.

State the dynamical symmetry with its domain

Let $H$ be self-adjoint and $\Theta$ antiunitary. The fixed-model condition is

$$\Theta D(H)=D(H), \qquad H\Theta\psi=\Theta H\psi \quad(\psi\in D(H)).$$

Equivalently, $\Theta H\Theta^{-1}=H$ as self-adjoint operators. Spectral calculus then gives

$$\Theta e^{-itH/\hbar}\Theta^{-1} =e^{+itH/\hbar}=U(-t).$$

No power series in the unbounded operator is required. The reversed solution is $\psi_{\rm rev}(t)=\Theta\psi(-t)$. For domain initial data it satisfies the Schrödinger equation strongly; for arbitrary Hilbert vectors the unitary-orbit formulation remains meaningful.

This is different from a commuting unitary symmetry, for which $VU(t)V^{-1}=U(t)$. The antiunitary map conjugates $i$, and that scalar change is responsible for reversing the time parameter.

Spectral subspaces give the invariant statement

For every Borel set $\Delta\subseteq\mathbb R$, invariance implies

$$\Theta E_H(\Delta)\Theta^{-1}=E_H(\Delta).$$

Thus every energy spectral subspace is preserved. If $\Theta^2=-I$, a nonzero finite-dimensional invariant spectral subspace has even dimension. In particular, every finite-multiplicity energy eigenspace has even dimension.

The orthogonal-pair argument and its physical applications are developed in Kramers Degeneracy. In a continuous spectrum, the projection identity remains meaningful without treating generalized eigenvectors as normalizable states. One should not interpret “even dimension” as a finite multiplicity statement about every continuum energy.

Neither antiunitarity alone nor $H$ having real coefficients establishes $\Theta^2=-I$. Spinless conjugation has square $+I$ and need not force any doublet.

Boundary phases can break conjugation symmetry

On $L^2(0,L)$, define the self-adjoint momentum

$$P_\theta=-i\hbar\,\frac{d}{dx}, \qquad D(P_\theta)= \{\psi\in H^1(0,L):\psi(L)=e^{i\theta}\psi(0)\}.$$

For $K\psi=\overline\psi$, conjugating the endpoint condition gives

$$K D(P_\theta)=D(P_{-\theta}), \qquad K P_\theta K=-P_{-\theta}.$$

The local derivative has the expected reversed sign, but the boundary parameter also changes. Momentum’s construction and extension theory are separate from this symmetry calculation.

For the kinetic Hamiltonian $H_\theta=P_\theta^2/(2m)$, the domain is

$$\begin{aligned} D(H_\theta)=\{\psi\in H^2(0,L):\;& \psi(L)=e^{i\theta}\psi(0),\\ &\psi'(L)=e^{i\theta}\psi'(0)\}. \end{aligned}$$

Hence $K H_\theta K=H_{-\theta}$. Complex conjugation is a symmetry of the same operator when $\theta=-\theta$ modulo $2\pi$, namely $\theta=0$ or $\pi$. For other values it relates two models with different boundary data.

This is a precise way a real differential expression $-\hbar^2d^2/(2m\,dx^2)$ can fail the fixed-model time-reversal test.

Check the conclusion against the spectrum

The momentum eigenvalues are $\hbar(2\pi n+\theta)/L$, so

$$E_n(\theta)= \frac{\hbar^2(2\pi n+\theta)^2}{2mL^2}, \qquad n\in\mathbb Z.$$

Conjugation exchanges $n$ at $\theta$ with $-n$ at $-\theta$. At $\theta=0$ the zero-momentum ground state is nondegenerate, consistent with $K^2=I$. At $\theta=\pi$, the label pairing within the same model is $n\leftrightarrow-n-1$.

The latter spectrum has pairs, but their existence is not the $\Theta^2=-I$ Kramers theorem. Additional spatial and spectral structure can create degeneracy even when time reversal squares to $+I$. Conversely, a generic boundary phase shifts the pairing into a comparison between two different operators.

Transform backgrounds before declaring invariance

A parameter-dependent model can satisfy $\Theta H(\lambda)\Theta^{-1}=H(\lambda^\Theta)$ without satisfying invariance at fixed $\lambda$. Magnetic fields and time-odd background parameters provide familiar examples. The relevant question is whether the comparison transforms the background or holds it fixed.

The interval phase above is a clean mathematical diagnostic of this distinction. It prevents a covariance law between models from being advertised as a degeneracy guarantee inside one model.

Exercise: test the boundary conditions

Take $\theta=\pi/3$. Does complex conjugation preserve $D(H_\theta)$? Which spectral relation nevertheless survives?

Solution

It sends both endpoint phases to $e^{-i\pi/3}$, so the domain becomes $D(H_{-\pi/3})$, not $D(H_{\pi/3})$. Nevertheless, $K H_{\pi/3}K=H_{-\pi/3}$ and $E_n(\pi/3)=E_{-n}(-\pi/3)$. The two operators have matching spectra under the antiunitary correspondence; this alone does not give paired eigenstates of one fixed operator.

References

• B. C. Hall, Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013 — self-adjoint domains and spectral calculus.
• K. Landsman, Foundations of Quantum Theory: From Classical Concepts to Operator Algebras, Fundamental Theories of Physics 188, Springer, 2017, Chapter 5 — antiunitary quantum symmetries.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014 — self-adjoint boundary conditions and spectral problems.