Antiunitary Symmetries, First Look

An antiunitary map preserves transition probabilities while conjugating complex scalars. Complex conjugation is the basic example: it sends $i\psi$ to $-iK\psi$, so it cannot be represented by an ordinary complex matrix acting linearly.

Helpful background. Review complex vector spaces for scalar multiplication and inner products for complex conjugation in the first slot. After the algebra below, Antiunitary Symmetries explains Wigner’s theorem and the physical role of these maps.

A map $A$ on a complex Hilbert space is antilinear when

$$A(a\psi+b\phi)=a^*A\psi+b^*A\phi.$$

It is antiunitary when it is bijective and, in the physics inner-product convention,

$$\langle A\psi\mid A\phi\rangle = \langle\psi\mid\phi\rangle^*.$$

Thus it preserves norms and transition probabilities while conjugating complex scalars. In particular $AiA^{-1}=-i$.

Complex conjugation and factorization

After choosing an orthonormal basis, complex conjugation

$$K\sum_n c_n\lvert n\rangle = \sum_n c_n^*\lvert n\rangle$$

is antiunitary and obeys $K^2=I$. The operator $K$ depends on the chosen basis; antiunitarity itself does not.

Every antiunitary operator can be written

$$A=UK,$$

where $U$ is unitary relative to that basis. Conversely, a unitary operator times $K$ is antiunitary. The factorization changes when the reference basis changes, while the map $A$ remains the same.

Composing unitary and antiunitary maps

First map	Second map	Composition
linear unitary	linear unitary	linear unitary
linear unitary	antiunitary	antiunitary
antiunitary	linear unitary	antiunitary
antiunitary	antiunitary	linear unitary

The square $A^2$ is therefore unitary. The square is a linear map and need not equal $I$; the calculation below gives an example with $A^2=-I$.

A two-component calculation

In the standard basis let $\psi=(1,i)^T/\sqrt2$. Then $K\psi=(1,-i)^T/\sqrt2$: both vectors have norm one, although they need not represent the same ray. Moreover,

$$K(i\psi)=-iK\psi.$$

This is the difference from a linear unitary map $U$, for which $U(i\psi)=iU\psi$. Norm preservation alone does not tell you which of these two scalar rules holds.

The factorization $A=UK$ also explains why two antiunitary maps compose to a linear unitary map. In a fixed basis,

$$(UK)(VK)=UV^*,\qquad A^2=UU^*,$$

where $V^*$ means entrywise conjugation, not the adjoint $V^\dagger$. There is no general rule that an antiunitary must square to the identity.

Exercise

For the real unitary matrix $U=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$, show that $A=UK$ is antiunitary and $A^2=-I$. Explain why multiplying $A$ by a constant phase does not change its square.

Solution

The product of a unitary map and conjugation is antiunitary. Since $U$ is real, $A^2=UU^*=U^2=-I$. Antilinearity gives

$$(e^{i\alpha}A)^2=e^{i\alpha}e^{-i\alpha}A^2=A^2.$$

The phase cancels because the second scalar passes through an antilinear map. This algebra prepares the distinction between the two time-reversal squares discussed in Time Reversal.

References

• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013 — unitary and antiunitary implementation of quantum symmetries.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020 — antiunitarity and time reversal.