Antiunitary Time Reversal

Time reversal in quantum mechanics is represented by an antiunitary operator, not by an ordinary unitary operator. This is not a matter of convention. It is forced by the sign of $i$ in the Schrödinger equation, by the canonical commutation relations, and by the requirement that time reversal relate forward evolution to backward evolution.

Write the time-reversal operator as $\Theta$. Antiunitarity means two things:

$$\Theta(a\lvert\psi\rangle+b\lvert\phi\rangle) = a^*\Theta\lvert\psi\rangle+b^*\Theta\lvert\phi\rangle,$$

and

$$\langle \Theta\phi|\Theta\psi\rangle = \langle \phi|\psi\rangle^*.$$

The first line is antilinearity; the second says transition probabilities are preserved. In particular,

$$\Theta i\Theta^{-1}=-i.$$

That one sign is the algebraic heart of time reversal.

Two Questions to Separate

There are two closely related questions that are often blended together:

• What operator represents time reversal on Hilbert space?
• When is a Hamiltonian invariant under that operator?

The first question is answered by antiunitarity and the transformation of observables such as position, momentum, orbital angular momentum, and spin. The second question is dynamical:

$$\Theta H\Theta^{-1}=H$$

for a time-independent Hamiltonian with no external parameters held fixed in a time-reversal-odd way. A Hamiltonian can fail the second test even though the time-reversal operator itself is perfectly well defined.

The conceptual overview is Time Reversal. This page gives the working derivation of antiunitarity.

The Schrödinger Equation Argument

Start with a state satisfying

$$i\hbar\frac{d}{dt}\lvert\psi(t)\rangle = H\lvert\psi(t)\rangle.$$

The time-reversed candidate is the state

$$\lvert\psi_T(t)\rangle = \Theta\lvert\psi(-t)\rangle.$$

Let $\tau=-t$. Since

$$\frac{d}{d\tau}\lvert\psi(\tau)\rangle = -\frac{i}{\hbar}H\lvert\psi(\tau)\rangle,$$

we get

$$\frac{d}{dt}\lvert\psi(-t)\rangle = \frac{i}{\hbar}H\lvert\psi(-t)\rangle.$$

Now apply $\Theta$. Because $\Theta$ is antilinear, the scalar $i$ becomes $-i$:

$$\begin{aligned} \frac{d}{dt}\lvert\psi_T(t)\rangle &= \Theta\left( \frac{i}{\hbar}H\lvert\psi(-t)\rangle \right) \\ &= -\frac{i}{\hbar} \Theta H\lvert\psi(-t)\rangle \\ &= -\frac{i}{\hbar} (\Theta H\Theta^{-1}) \lvert\psi_T(t)\rangle. \end{aligned}$$

Therefore

$$i\hbar\frac{d}{dt}\lvert\psi_T(t)\rangle = (\Theta H\Theta^{-1}) \lvert\psi_T(t)\rangle.$$

If $\Theta H\Theta^{-1}=H$, the transformed state obeys the same Schrödinger equation. The antilinear conjugation of $i$ is exactly what makes the sign come out correctly.

What Would Go Wrong for a Unitary Operator

Suppose one tried to represent time reversal by a unitary operator $U_T$ that leaves position fixed and reverses momentum:

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

Apply this to the canonical commutator. The transformed left-hand side is

$$[X,-P] = -i\hbar I.$$

But if $U_T$ were unitary, it would leave the scalar $i$ untouched:

$$U_T[X,P]U_T^{-1} = U_T(i\hbar I)U_T^{-1} = i\hbar I.$$

These two answers contradict each other. For an antiunitary operator $\Theta$, the same calculation is consistent because

$$\Theta(i\hbar I)\Theta^{-1} = -i\hbar I.$$

Thus the antiunitary sign is already forced by the basic position-momentum algebra.

Propagators and Reversed Evolution

The same point appears in the time-evolution operator

$$U(t)=e^{-iHt/\hbar}.$$

For an antiunitary $\Theta$, the exponential transforms as

$$\Theta e^{-iHt/\hbar}\Theta^{-1} = e^{+i(\Theta H\Theta^{-1})t/\hbar}.$$

This follows term by term from the power series: each scalar coefficient is complex-conjugated. If $H$ is time-reversal invariant, then

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

This is the clean operator statement that time reversal maps forward time evolution to backward time evolution.

Antiunitary Does Not Mean Arbitrary

In a chosen orthonormal basis, any antiunitary operator can be written as

$$\Theta=UK,$$

where $U$ is unitary and $K$ complex-conjugates components in that basis. This formula is useful, but it must be read with care:

• $K$ depends on the chosen basis.
• $U$ carries the physical part not supplied by plain conjugation.
• The product $\Theta=UK$ is the meaningful symmetry operator.

For a spinless particle in a standard position representation, one often has $\Theta=K$; the concrete scalar-particle consequences are collected in Time Reversal for Spinless Particles. For a spin-$1/2$ degree of freedom in the usual $S_z$ basis, a common convention is

$$\Theta=-i\sigma_yK.$$

The spinor convention and the proof of $\Theta^2=-I$ are developed in Time Reversal for Spin-1/2 Particles.

Observables and Signs

Time reversal leaves position even and makes momenta odd:

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

Orbital angular momentum is odd because $\mathbf L=\mathbf X\times\mathbf P$:

$$\Theta\mathbf L\Theta^{-1} = -\mathbf L.$$

Spin is also time-reversal odd:

$$\Theta\mathbf S\Theta^{-1} = -\mathbf S.$$

The antiunitary nature of $\Theta$ is not a separate decorative feature added after these sign rules. It is what makes the sign rules compatible with the algebra of quantum observables.

Hamiltonian Tests

For a spinless particle in a real scalar potential,

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

time reversal is a symmetry because $P^2$ and $V(X)$ are unchanged. In a basis where $\Theta=K$, this Hamiltonian is real in the position representation.

With magnetic fields, the test is more delicate. A Zeeman term

$$H_Z = -\gamma\mathbf S\cdot\mathbf B$$

transforms as

$$\Theta H_Z(\mathbf B)\Theta^{-1} = H_Z(-\mathbf B).$$

The family of Hamiltonians is covariant if the magnetic field is also reversed. A single Hamiltonian with a fixed nonzero external $\mathbf B$ is not time-reversal invariant. This distinction is essential in applications to spin dynamics, Kramers degeneracy, and symmetry classification.

The Square of Time Reversal

Antiunitarity alone does not determine $\Theta^2$. Depending on the Hilbert-space sector,

$$\Theta^2=+I \qquad\text{or}\qquad \Theta^2=-I$$

can occur. Spinless systems often have $\Theta^2=+I$. A single spin-$1/2$ degree of freedom has $\Theta^2=-I$. More generally, a pure angular-momentum multiplet with quantum number $j$ has the familiar pattern

$$\Theta^2=(-1)^{2j}I.$$

The sign cannot be changed by multiplying $\Theta$ by a phase. If $\Theta'=e^{i\chi}\Theta$, then antilinearity gives

$$(\Theta')^2 = e^{i\chi}e^{-i\chi}\Theta^2 = \Theta^2.$$

This invariant square is what makes Kramers Degeneracy possible when $\Theta^2=-I$.

Common Mistakes

• Treating $\Theta$ as a unitary operator that happens to reverse momenta.
• Forgetting that $\Theta(a\lvert\psi\rangle)=a^*\Theta\lvert\psi\rangle$.
• Saying “time reversal means $t\mapsto -t$” without specifying the Hilbert-space operator.
• Treating $K$ as basis-independent.
• Testing a magnetic-field Hamiltonian while keeping a time-reversal-odd external field fixed.
• Assuming that antiunitary time reversal always has the same square in every physical sector.

Cross-Links

• Time Reversal
• Antiunitary Symmetries
• Antiunitary Symmetries, First Look
• Time Reversal for Spinless Particles
• Time Reversal for Spin-1/2 Particles
• Kramers Degeneracy
• Symmetry Constraints on Hamiltonians
• Projective Representations
• Symmetry Classification Preview
• Time-Reversal Operator

References

• E. P. Wigner, Group Theory and Its Application to the Quantum Mechanics of Atomic Spectra, Academic Press, 1959.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• A. Messiah, Quantum Mechanics, Dover, 1999.
• M. Tinkham, Group Theory and Quantum Mechanics, Dover, 2003.

Exercises

1. Use the canonical commutator to show that a unitary operator cannot leave $X$ fixed while reversing $P$.

Solution

If $U_T$ were unitary with $U_TXU_T^{-1}=X$ and $U_TPU_T^{-1}=-P$, then

$$U_T[X,P]U_T^{-1} = [X,-P] = -i\hbar I.$$

But unitary conjugation leaves the scalar $i$ unchanged, so the same expression is

$$U_T(i\hbar I)U_T^{-1} = i\hbar I.$$

This contradiction is removed when the operator is antiunitary, because antiunitarity sends $i$ to $-i$.

2. Prove the propagator identity

$$\Theta e^{-iHt/\hbar}\Theta^{-1} = e^{+i(\Theta H\Theta^{-1})t/\hbar}$$

for antiunitary $\Theta$.

Solution

Expand the exponential:

$$e^{-iHt/\hbar} = \sum_{n=0}^\infty \frac{1}{n!} \left(-\frac{it}{\hbar}\right)^n H^n.$$

Antiunitarity conjugates the scalar coefficient and conjugates the operator product:

$$\Theta \left[ \left(-\frac{it}{\hbar}\right)^nH^n \right] \Theta^{-1} = \left(+\frac{it}{\hbar}\right)^n (\Theta H\Theta^{-1})^n.$$

Summing the series gives the stated identity.

3. A single spin-$1/2$ in a fixed magnetic field has Hamiltonian $H=-\gamma\mathbf S\cdot\mathbf B$. Is this Hamiltonian time-reversal invariant?

Solution

Spin is time-reversal odd:

$$\Theta\mathbf S\Theta^{-1} = -\mathbf S.$$

If the external field is kept fixed, then

$$\Theta H(\mathbf B)\Theta^{-1} = \gamma\mathbf S\cdot\mathbf B \neq H(\mathbf B)$$

unless $\mathbf B=0$. The family is covariant under $\mathbf B\mapsto-\mathbf B$, since

$$\Theta H(\mathbf B)\Theta^{-1} = H(-\mathbf B).$$

So a fixed nonzero magnetic field breaks time-reversal symmetry, while the field-reversed comparison is time-reversal covariant.