# Time reversal symmetry and T^2 = -1

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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?

This post imported from StackExchange Physics at 2014-04-05 17:29 (UCT), posted by SE-user Kevin Walker
asked Jan 13, 2012
Now cross-listed on TP.SE: theoreticalphysics.stackexchange.com/q/843/189

This post imported from StackExchange Physics at 2014-04-05 17:29 (UCT), posted by SE-user Qmechanic

## 1 Answer

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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 $T^2_{quasi}=-1$.

This post imported from StackExchange Physics at 2014-04-05 17:29 (UCT), posted by SE-user Xiao-Gang Wen
answered May 27, 2012 by (3,485 points)

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