Antiunitary Symmetries

An antiunitary symmetry is represented by a surjective antilinear norm-preserving operator. Antiunitary transformations are less familiar than unitary ones, but they are unavoidable in quantum mechanics. Time reversal is the central example. For the operator-theoretic first pass, see Antiunitary Symmetries, First Look; for the dynamical derivation, see Antiunitary Time Reversal.

Antilinearity

An antiunitary operator $T$ is antilinear:

$$T(a\lvert\psi\rangle+b\lvert\phi\rangle) = a^*T\lvert\psi\rangle+b^*T\lvert\phi\rangle.$$

This is the first point to remember. Complex scalars are conjugated when they pass through $T$.

Inner Products and Probabilities

Antiunitary transformations conjugate inner products:

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

Therefore they preserve transition probabilities:

$$\lvert\langle T\phi|T\psi\rangle\rvert^2 = \lvert\langle\phi|\psi\rangle\rvert^2.$$

This is why antiunitary transformations are allowed by Wigner’s theorem.

Wigner’s theorem is a statement about ray symmetries: under its standard hypotheses, a bijection of rays that preserves transition probabilities can be implemented by either a unitary or an antiunitary operator. The implementing operator is defined only up to an overall phase. A continuous one-parameter symmetry connected to the identity stays in the unitary branch; antiunitary implementations therefore characteristically represent disconnected or discrete operations such as time reversal.

Complex Conjugation Is Basis Dependent

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

$$T=UK,$$

where $U$ is unitary and $K$ complex-conjugates coefficients in that basis. This is useful, but it can be misleading if stated without context. The operation $K$ depends on the basis chosen. The antiunitary operator $T$ is the invariant object.

The factorization is general. Choose an orthonormal basis and its conjugation $K$. Since the product of two antiunitary maps is linear, $U:=TK$ is linear. Moreover,

$$\langle U\phi\mid U\psi\rangle = \langle TK\phi\mid TK\psi\rangle = \langle K\phi\mid K\psi\rangle^* = \langle\phi\mid\psi\rangle,$$

so $U$ is unitary and $T=UK$. A change of reference basis changes $U$ and $K$ separately but not their product.

To perform that change explicitly, let the columns of a unitary matrix $V$ be the new basis vectors in the old coordinates. The same state has new coordinates $c'=V^\dagger c$, as in Active and Passive Transformations. Since $T$ sends $c$ to $Uc^*$, its action in the new coordinates is

$$V^\dagger U(Vc')^* =V^\dagger U V^*c'^*.$$

Thus $T$ is represented by $U'K$ with $U'=V^\dagger U V^*$, where $K$ now conjugates the new coordinate column and $V^*$ denotes entrywise conjugation. The right-hand factor is $V^*$, whereas a linear operator’s matrix transforms as $H'=V^\dagger H V$.

In these coordinates, time-reversal invariance is the matrix condition $U'H'^*U'^\dagger=H'$. A real Hamiltonian matrix is therefore the correct test only in a basis where this particular time-reversal operation is $K$. An imaginary matrix entry alone does not demonstrate broken time reversal; the exercise below gives a two-level example.

Composition and Operator Transformations

The product of two antiunitary maps is unitary; multiplying a unitary and an antiunitary map in either order is antiunitary. Consequently $T^{-1}$ is antiunitary and $T^2$ is unitary.

Conjugation by an antiunitary map preserves operator products but conjugates scalar coefficients:

$$T(cA+dB)T^{-1} = c^*TAT^{-1}+d^*TBT^{-1},$$ $$T(AB)T^{-1} = (TAT^{-1})(TBT^{-1}).$$

It follows that

$$T[A,B]T^{-1} = [TAT^{-1},TBT^{-1}].$$

The scalar conjugation is essential in algebra checks. For example, if $[x,p]=i\hbar I$, a time-reversal action with $x\mapsto x$ and $p\mapsto-p$ gives $[x,-p]=-i\hbar I$, exactly matching the transformed right-hand side $T(i\hbar I)T^{-1}=-i\hbar I$.

Time Reversal

Time reversal must conjugate $i$. The Schrödinger equation

$$i\hbar\frac{\partial}{\partial t}\lvert\psi(t)\rangle = H\lvert\psi(t)\rangle$$

contains $i$ and a time derivative. Reversing the time direction requires an operation that sends $i$ to $-i$ so that the transformed equation has the correct sign structure.

For a spinless particle in a position basis, a simple time-reversal action is complex conjugation:

$$T\psi(x)=\psi(x)^*.$$

This leaves $x$ unchanged and reverses momentum:

$$TxT^{-1}=x, \qquad TpT^{-1}=-p.$$

The scalar-particle representation and the $\Theta^2=+I$ consequence are developed in Time Reversal for Spinless Particles.

For spin-$1/2$, time reversal also acts nontrivially on spin. A common convention is

$$T=-i\sigma_y K,$$

which gives

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

Sign conventions for this formula vary by phase choice, but physical predictions do not.

In the $\sigma_z$ basis, $K$ is entrywise conjugation and

$$T=-i\sigma_yK.$$

Because $-i\sigma_y$ is real,

$$T^2 = (-i\sigma_y)K(-i\sigma_y)K = (-i\sigma_y)^2 = -I.$$

This two-dimensional example also shows why multiplying an antiunitary operator by a phase does not change $T^2$: the second occurrence of the phase is conjugated.

The detailed spinor convention and the proof of $T^2=-I$ are treated in Time Reversal for Spin-1/2 Particles.

Kramers Degeneracy Preview

When an antiunitary time-reversal symmetry satisfies

$$T^2=-I,$$

every normalizable energy eigenstate of a self-adjoint time-reversal-invariant Hamiltonian has an orthogonal partner at the same energy. Here $T$ must preserve the Hamiltonian domain and commute with $H$. This is Kramers Degeneracy; it applies to bound-state eigenspaces as well as finite matrices, and does not by itself make a claim about nonnormalizable scattering states.

A symmetry is not an implementable quantum gate

Antiunitarity describes a relation between states and experiments. It does not supply a deterministic physical gate that conjugates every unknown input state. On density matrices, conjugation in a fixed basis gives transposition, which fails complete positivity when extended to an entangled system. Time reversal therefore has a different operational meaning from a unitary pulse applied by an experimentalist.

Exercises

1. Let $T$ be antiunitary. Show that $T(i\lvert\psi\rangle)=-iT\lvert\psi\rangle$.

Solution

By antilinearity,

$$T(i\lvert\psi\rangle) = i^*T\lvert\psi\rangle = -iT\lvert\psi\rangle.$$

2. Show that the inverse of an antiunitary operator is antiunitary.

Solution

If $T$ is antiunitary, it is bijective. For $\chi=T\psi$ and $\eta=T\phi$,

$$\langle T^{-1}\chi\mid T^{-1}\eta\rangle = \langle\psi\mid\phi\rangle = \langle\chi\mid\eta\rangle^*.$$

Antilinearity of $T^{-1}$ follows by applying $T$ to a linear combination and using bijectivity.

3. In an orthonormal basis let $H=E_0\sigma_x$, where $E_0$ is a nonzero real energy, and let the time-reversal operation be $T=K$. Rephase the second basis vector by $i$, so $V=\operatorname{diag}(1,i)$. Find $H'$ and $U'$, then verify time-reversal invariance and $T^2=I$ in the new coordinates.

Solution

The original coefficient matrix is $U=I$. The passive transformation gives

$$H'=V^\dagger H V =E_0\begin{pmatrix}0&i\\-i&0\end{pmatrix} =-E_0\sigma_y, \qquad U'=V^\dagger V^*=\sigma_z.$$

Although $H'^*=-H'$, the complete time-reversal operation is now $\sigma_zK$, and

$$U'H'^*U'^\dagger =\sigma_z(E_0\sigma_y)\sigma_z =-E_0\sigma_y=H'.$$

Its square is $U'U'^*=\sigma_z^2=I$. The physical symmetry and its square have survived the coordinate change. Applying $K$ alone to $H'$ would test a different operation. This example models a two-level system with $T^2=+I$, rather than the physical spin-$1/2$ time reversal discussed above.

4. Let $T$ be antiunitary. Show that

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

for real $t$ and a self-adjoint Hamiltonian. Use a power series for a bounded $H$, and explain how the spectral theorem extends the result.

Solution

For bounded $H$, expand the norm-convergent power series. Conjugation by $T$ preserves products and conjugates every scalar, so $(-i)^n$ becomes $(+i)^n$ and $H^n$ becomes $(THT^{-1})^n$. For an unbounded self-adjoint $H$, use its spectral resolution instead: $T$ transforms each spectral projection and complex-conjugates the bounded function $e^{-iEt/\hbar}$. A power series on arbitrary vectors is not justified in that case.

5. For spinless time reversal $T=K$ in the position representation, show that $TpT^{-1}=-p$ for $p=-i\hbar\,d/dx$.

Solution

Apply the transformed operator to a test wavefunction $\psi$:

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

Thus $KpK^{-1}=-p$ since $K^{-1}=K$.

References

• V. Bargmann, “Note on Wigner’s theorem on symmetry operations,” Journal of Mathematical Physics 5, 862–868 (1964). doi:10.1063/1.1704188.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, chapter 4.
• E. P. Wigner, Group Theory and Its Application to the Quantum Mechanics of Atomic Spectra, Academic Press, 1959.