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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.

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$T^2=-1$ is only true for states with half integer spin. –  Thomas Jan 20 '12 at 18:25
    
Thanks, that's helpful. With that hint I was able to do a more fruitful web search and came across Kramer's theorem. I'll edit the question accordingly. –  Kevin Walker Jan 20 '12 at 18:54
    
It is not unheard of that if the OP figures out the answer to their question, they post it and can even accept it if they think it is correct. –  user566 Jan 20 '12 at 19:42
    
Note that signature is important. Space reflections correspond to unitary operators whereas time reflections correspond to anti unitary ones –  Squark Jan 21 '12 at 12:31
    
I could better understand a question $Pin(3,1)$ vs $Pin(1,3)$. How we could answer what we are using in Euclidean case? Even in Lorentzian case we sometimes have to talk about experimental data (e.g. discovery of antiparticles, search for Majorana neutrino, etc.) to clarify such questions. –  Alex 'qubeat' Jan 21 '12 at 13:21
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1 Answer

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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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