From Wigner's theorem, we get that a physical symmetry can be described either by a unitary or antiunitary operator, eventually with a phase factor, as in here.

However we have to respect the group structure, so we need to have:

$$O(f)\circ O(g)=e^{i \cdot\phi(f,g)}\cdot O(f*g)\tag{1}$$

where $*$ is the group operation.

For unitary operators this is fine, but for antiunitary operators there is a problem: the left side of the equation is linear while the right side is antilinear.

Here my doubt comes. The argument of linearity is used to show that only unitary operators can describe continuous symmetries (e.g. spatial translations), but why does it not work for symmetries associated with a finite group (e.g. time reversal)?

I'm not asking why time reversal needs to be described by an antiunitary operator, I'm asking how come an antiunitary operator can satisfy $(1)$ if it's the representation of a finite group.

  • $\begingroup$ But it barely is. The group of T is just $Z_2$, abelian, with T and the identity. $\endgroup$ Jun 26, 2021 at 20:17
  • $\begingroup$ The identity is not antiunitary though, so what you are describing is not a group of antiunitary operators. Instead we actually get a group of unitary operators for continuous symmetries. Is this the way to solve the issue? $\endgroup$ Jun 26, 2021 at 20:31
  • $\begingroup$ Possibly useful. The square of an anti unitary operator is unitary. $\endgroup$ Jun 26, 2021 at 21:01

2 Answers 2


I am not sure to understand the problem. What it is clear is that a (unitary or projective unitary) representation is not possible if it is made of antiunitary operators only. That is because, just in view of the representation rules, some operators should be antiunitary and unitary simultaneously as you point out. (In your example, if $f$ and $g$ are represented by anti unitary operators, their composition $f*g$ must be unitarily represented.) In fact, in the concrete cases where the time reversal operation is part of a larger symmetry group, not all operators representing the elements of the group are antiunitary according to the composition rule, even if the time reversal is represented anti unitarily.

Another issue is the possibility of existence of antiunitary elements in the image of the representation. Evidently an element $a$ cannot be represented by an antiunitary operator if, e.g., $a=b^2$, for some other element $b$. This is the case, in particular, when representing a Lie group and when we restrict ourselves to deal with the connected component of the group including the identity element: each element here can be constructed as a finite product of elements $b^2$ as above. That is the reason why, a posteriori, the time reversal symmetry does not belong to that component, as we know that it admits antiunitary representations.


Time reversal is a implemented at the level of the Hilbert space $\mathcal H$ by a (projective) representation $\rho$ of $Z_2$, with $\rho(0) = \mathbb I$ and $\rho(1) = T$. Note that

$$\rho(0+0)=\rho(0)=\mathbb I = \mathbb I\circ \mathbb I = \rho(0)\circ \rho(0)$$ $$\rho(0+1)=\rho(1)=T = \mathbb I \circ T = \rho(0)\circ \rho(1)$$ $$\rho(1+1)=\rho(0)=\mathbb I = e^{i\theta} T\circ T = e^{i\theta}\rho(1)\circ\rho(1)$$

where $\theta=0$ for bosonic systems and $\theta=\pi$ for fermionic systems. The second line of the above is an antiunitary operator, while the first and third are unitary operators.


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