I'm a mathematician interested in abstract QFT. I'm trying to undersand why, under certain (all?) circumstances, we must have $T^2 = -1$ rather than $T^2 = +1$, where $T$ is the time reversal operator. I understand from the Wikipedia article that requiring that energy stay positive forces $T$ to be represented by an anti-unitary operator. But I don't see how this forces $T^2=-1$. (Or maybe it doesn't force it, it merely allows it?)

Here's another version of my question. There are two distinct double covers of the Lie group $O(n)$ which restrict to the familiar $Spin(n)\to SO(n)$ cover on $SO(n)$; they are called $Pin_+(n)$ and $Pin_-(n)$. If $R\in O(n)$ is a reflection and $\tilde{R}\in Pin_\pm(n)$ covers $R$, then $\tilde{R}^2 = \pm 1$. So saying that $T^2=-1$ means we are in $Pin_-$ rather than $Pin_+$. (I'm assuming Euclidean signature here.) My question (version 2): Under what circumstances are we forced to use $Pin_-$ rather than $Pin_+$ here?

(I posted a similar question on physics.stackexchange.com last week, but there were no replies.)

EDIT: Thanks to the half-integer spin hint in the comments below, I was able to do a more effective web search. If I understand correctly, Kramer's theorem says that for even-dimensional (half integer spin) representations of the Spin group, $T$ must satisfy $T^2=-1$, while for the odd-dimensional representations (integer spin), we have $T^2=1$. I guess at this point it becomes a straightforward question in representation theory: Given an irreducible representation of $Spin(n)$, we can ask whether it is possible to extend it to $Pin_-(n)$ (or $Pin_+(n)$) so that the lifted reflections $\tilde R$ (e.g. $T$) act as an anti-unitary operator.

  • 5
    $\begingroup$ $T^2=-1$ is only true for states with half integer spin. $\endgroup$
    – Thomas
    Jan 20, 2012 at 18:25
  • $\begingroup$ Note that signature is important. Space reflections correspond to unitary operators whereas time reflections correspond to anti unitary ones $\endgroup$
    – Squark
    Jan 21, 2012 at 12:31
  • 1
    $\begingroup$ Kramer's Theorem and the issue of antiunitary time reversal symmetries are both dealt with in some detail by the first chapter of Haake, for those interested in reading more on the subject. $\endgroup$ Jan 26, 2012 at 18:28
  • $\begingroup$ possible duplicate of Time reversal symmetry and T^2 = -1 $\endgroup$
    – Qmechanic
    Jul 11, 2012 at 8:56

2 Answers 2


There are two possible answers to why $T^2=-1$:

a) Why not. The total phase of a quantum state is unphysical. So a symmetry may be realized as a projective representation. Here T may be viewed as a projective representation of time reversal $T_{phy}$ which satisfy $T^2_{phy}=1$.

b) If we define the time reversal symmetry to be realized as a regular representation in a many-body systems with $T^2=1$, the symmetry operations that act on fractionalized quasiparticles may be realized projectively, with $T^2_{quasi}=-1$.


The answer for CPT transformations is obvious. A CPT transformation is a 180 degree rotation in the Euclidean theory, so CPT followed by CPT is a 360 degree rotation, which gives you a minus sign on fermionic states, and a plus sign on bosonic states. This is true for all Lorentz invariant theories, or even theories that spontaneously violate Lorentz invariance and keep a CPT unbroken.

The case where you have a T symmetry can be understood from the above. It is not always true that the T symmetry squares to 1 on bosons and -1 on fermions, it is only true for those fermions and bosons whose CP symmetry doesn't square to -1. This is true for normal realizations of CP symmetry, but not for unusual cases, where some parity symmetry can have crazy phases, because it mixes with another discrete symmetry of the theory.

For example, start with electromagnetism with the usual parity, and consider a real pseudoscalar $\phi_3$, which can be made a pseudoscalar by coupling it to a parity invariant fermion using $\gamma_5$, so there is a term in the action.

$$ \phi_3 \bar{\psi}\gamma_5 \psi $$

Consider a second complex scalar $\phi_1$ coupled to $\phi_3$ as follows:

$$ (\phi_1^2 \phi_3 + \bar\phi^2 \phi_3) $$

Now the square of $\phi_1$ needs to get a minus sign under any hypothetical parity transformation. This can be arranged either by multiplying by i, or by swapping the real and imaginary parts of $\phi$. To exclude the latter, you can add a term to the action which is

$$ A(\phi^4 + \bar\phi^{4}) + iB(\phi^4 - \bar{\phi}^4) $$

This is invariant under a parity transformation, but only by sending $\phi\rightarrow \pm i \phi$. The square of parity is -1, so that the T operator on this field $\phi$ squares with a negative sign.

This type of thing is easy to cook up in nonrenormalizable theories. The theory above shows that even renormalizability doesn't exclude these shenanigans.


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