Time Reversal for Spinless Particles

For a spinless particle with scalar wavefunction $\psi(\mathbf x)$, the simplest time-reversal operator is complex conjugation in the position representation:

$$\Theta=K, \qquad (\Theta\psi)(\mathbf x)=\psi(\mathbf x)^*.$$

This is the prototype of time reversal with

$$\Theta^2=+I.$$

It is simpler than the spin-$1/2$ case because there is no intrinsic spinor index that must rotate when time is reversed. The physical content is still nontrivial: $\Theta$ is antiunitary, it reverses momentum and orbital angular momentum, and it constrains Hamiltonians to be real in an appropriate basis.

Scalar Wavefunctions Under Time Reversal

“Spinless” here means a single scalar particle in nonrelativistic quantum mechanics, with no internal spin degree of freedom and no extra internal matrix on which time reversal acts. The Hilbert space is typically

$$L^2(\mathbb R^d)$$

with a self-adjoint Hamiltonian domain, or a corresponding space on a finite region. The boundary conditions are part of the symmetry test. In the usual position basis, position eigenkets can be chosen so that time reversal leaves the spatial point unchanged:

$$\Theta\lvert \mathbf x\rangle = \lvert \mathbf x\rangle.$$

Then the wavefunction action is just complex conjugation.

This does not mean every effective “spinless” model is time-reversal invariant. Magnetic fluxes, complex hopping phases, absorbing optical potentials, and rotating frames can break the symmetry even without spin.

Position-Space Action

Position acts by multiplication:

$$(X_i\psi)(\mathbf x)=x_i\psi(\mathbf x).$$

Since $x_i$ is real,

$$KX_iK^{-1}=X_i.$$

Momentum is

$$P_i=-i\hbar\frac{\partial}{\partial x_i}.$$

Complex conjugation changes $i$ to $-i$, so

$$KP_iK^{-1}=-P_i.$$

Thus spinless time reversal has the expected even and odd variables:

$$\Theta\mathbf X\Theta^{-1} = \mathbf X, \qquad \Theta\mathbf P\Theta^{-1} = -\mathbf P.$$

Orbital angular momentum is therefore time-reversal odd:

$$\Theta\mathbf L\Theta^{-1} = \Theta(\mathbf X\times\mathbf P)\Theta^{-1} = -\mathbf L.$$

This is a useful contrast with parity. Parity reverses both $\mathbf X$ and $\mathbf P$, so $\mathbf L$ is parity even; time reversal leaves $\mathbf X$ fixed and reverses $\mathbf P$, so $\mathbf L$ is time-reversal odd.

Momentum Representation

The compact formula $\Theta=K$ belongs to the position representation. In momentum space, time reversal must also flip the momentum label. If

$$\widetilde\psi(\mathbf p) = \langle \mathbf p|\psi\rangle,$$

then the transformed wavefunction is

$$(\Theta\widetilde\psi)(\mathbf p) = \widetilde\psi(-\mathbf p)^*,$$

up to the phase convention chosen for momentum eigenkets. This is the same physical operator written in a different representation.

The basis dependence of $K$ is discussed more generally in Antiunitary Time Reversal.

The Square Is Plus One

For scalar wavefunctions,

$$\Theta^2\psi = K(K\psi) = \psi.$$

Hence

$$\Theta^2=+I.$$

This is why spinless time reversal does not force Kramers degeneracy. A time-reversal-invariant spinless Hamiltonian can have nondegenerate bound states. Degeneracies may still occur, but they must come from other symmetries, boundary conditions, or accidental structure rather than from the $\Theta^2=-I$ Kramers mechanism.

Real Hamiltonians

Consider

$$H = \frac{\mathbf P^2}{2m}+V(\mathbf X),$$

with real scalar potential $V$. Then

$$KHK^{-1}=H.$$

The kinetic term is invariant because $\mathbf P^2$ is unchanged, and the potential is invariant because $V(\mathbf x)$ is real.

In a finite-dimensional basis where spinless time reversal is represented by $K$, the condition

$$KHK^{-1}=H$$

is simply

$$H^*=H.$$

If $H$ is also Hermitian, this means $H$ is a real symmetric matrix in that basis. This is the elementary finite-dimensional version of the spinless time-reversal constraint.

Real Eigenfunctions

If $H$ is invariant under $K$ and

$$H\psi=E\psi,$$

then complex conjugating gives

$$H\psi^*=E\psi^*$$

for real $E$. Thus $\psi^*$ is another eigenfunction with the same energy.

If the eigenvalue is nondegenerate, then

$$\psi^*=c\psi, \qquad \lvert c\rvert=1.$$

By multiplying $\psi$ by an overall phase, one can choose a representative with

$$\psi^*=\psi.$$

If $c=e^{i\chi}$, the rephased state $e^{i\chi/2}\psi$ is real. This is why bound-state wavefunctions for real one-dimensional potentials and conjugation-invariant boundary conditions are often chosen real. In degenerate subspaces, the statement is not that every vector is real, but that one can choose a basis adapted to the antiunitary symmetry.

Currents Reverse

For a scalar particle without vector potential, the probability current is

$$\mathbf j = \frac{\hbar}{m}\operatorname{Im} \left(\psi^*\nabla\psi\right).$$

Under $\Theta=K$,

$$\psi\mapsto\psi^*, \qquad \mathbf j\mapsto-\mathbf j.$$

Plane waves show this directly:

$$e^{i\mathbf k\cdot\mathbf x} \mapsto e^{-i\mathbf k\cdot\mathbf x}.$$

The momentum and current reverse, while the probability density is unchanged.

Magnetic Fields and Vector Potentials

A spinless particle can still couple to a magnetic field through minimal coupling:

$$H(\mathbf A) = \frac{1}{2m} \left(\mathbf P-q\mathbf A(\mathbf X)\right)^2 +q\Phi(\mathbf X).$$

For real $\mathbf A$ and $\Phi$,

$$\Theta H(\mathbf A)\Theta^{-1} = H(-\mathbf A),$$

up to the usual gauge convention. Equivalently, the magnetic field is time-reversal odd:

$$\mathbf B\mapsto-\mathbf B.$$

Therefore a fixed nonzero magnetic field generally breaks ordinary time reversal even without spin. A nonzero vector potential by itself is not enough: a pure gauge can be removed, and some fluxes are equivalent to their negatives by a single-valued gauge transformation. The test is physical field and holonomy together with the operator domain, not merely whether $\mathbf A$ appears in a formula.

Real Differential Expression, Nonreal Boundary Condition

Consider the free Hamiltonian on a ring of length $L$, represented on $[0,L]$ with a twist:

$$H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2},\qquad \psi(L)=e^{i\alpha}\psi(0),\quad \psi'(L)=e^{i\alpha}\psi'(0).$$

Its differential expression is real, but $K$ changes the twist to $-\alpha$. The domain is invariant only when $e^{i\alpha}=e^{-i\alpha}$, meaning $\alpha=0$ or $\pi$ modulo $2\pi$. This is a simple example of why checking the coefficients of $H$ does not finish the symmetry test. In a flux description these cases correspond to an integer or half-integer flux quantum, where reversing the flux gives a gauge-equivalent ring problem.

Scattering and Traveling Waves

Spinless time reversal maps an outgoing wave to an incoming wave with reversed momentum. For a real potential, this relates time-reversed scattering processes. For example, in one dimension,

$$e^{ikx} \mapsto e^{-ikx}.$$

A scattering state includes its incoming and outgoing boundary conditions. Complex conjugation reverses their roles as well as the current, so it generally produces a different scattering state rather than a real representative of the original one.

Common Mistakes

• Saying $\Theta=K$ without specifying the representation.
• Forgetting that $\Theta=K$ is still antiunitary, not linear.
• Confusing spinless time reversal with parity. Time reversal leaves position fixed; parity reverses position.
• Assuming $\Theta^2=+I$ implies no degeneracies. It only means there is no Kramers-enforced degeneracy.
• Inferring time-reversal breaking solely from a nonzero vector potential, or ignoring a flux-dependent boundary condition when the local differential expression is real.
• Concluding every eigenfunction must be real even inside a degenerate subspace or with scattering boundary conditions.

Exercises

1. Verify $KPK^{-1}=-P$ for $P=-i\hbar\,d/dx$.

Solution

Since $K^{-1}=K$,

$$(KPK^{-1}\psi)(x) = K\left[ -i\hbar\frac{d}{dx}\psi(x)^* \right].$$

Complex conjugating the result gives

$$(KPK^{-1}\psi)(x) = i\hbar\frac{d}{dx}\psi(x) = -P\psi(x).$$

Thus $KPK^{-1}=-P$.

2. Show that a plane wave reverses momentum and current under spinless time reversal.

Solution

For

$$\psi_k(x)=e^{ikx},$$

complex conjugation gives

$$K\psi_k(x)=e^{-ikx}=\psi_{-k}(x).$$

The momentum eigenvalue changes from $\hbar k$ to $-\hbar k$. The current

$$j = \frac{\hbar}{m}\operatorname{Im}(\psi^*\partial_x\psi)$$

changes from $\hbar k/m$ to $-\hbar k/m$.

3. Let

$$H=aI+b_x\sigma_x+b_y\sigma_y+b_z\sigma_z$$

be a Hermitian two-state Hamiltonian in a basis where spinless time reversal is $K$. Which terms are allowed by $KHK^{-1}=H$?

Solution

The matrices $I$, $\sigma_x$, and $\sigma_z$ are real, while $\sigma_y$ is imaginary. Therefore

$$K\sigma_xK^{-1}=\sigma_x, \qquad K\sigma_zK^{-1}=\sigma_z, \qquad K\sigma_yK^{-1}=-\sigma_y.$$

The time-reversal-invariant Hamiltonian has

$$b_y=0,$$

while $a$, $b_x$, and $b_z$ are allowed. Equivalently, $H$ must be real symmetric in this basis.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon Press, 1977.
• A. Messiah, Quantum Mechanics, Dover, 1999.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.