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  Time reversal symmetry and T^2 = -1

+ 6 like - 0 dislike
1266 views

I'm a mathematician interested in abstract QFT. I'm trying to undersand why, under certain (all?) circumstances, we must have T2=1 rather than T2=+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 T2=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)SO(n) cover on SO(n); they are called Pin+(n) and Pin(n). If RO(n) is a reflection and ˜RPin±(n) covers R, then ˜R2=±1. So saying that T2=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 in Theoretical Physics by Kevin Walker (65 points) [ no revision ]
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 T2=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 Tphy which satisfy T2phy=1.

b) If we define the time reversal symmetry to be realized as a regular representation in a many-body systems with T2=1, the symmetry operations that act on fractionalized quasiparticles may be realized projectively, with T2quasi=1.

This post imported from StackExchange Physics at 2014-04-05 17:29 (UCT), posted by SE-user Xiao-Gang Wen
answered May 26, 2012 by Xiao-Gang Wen (3,485 points) [ no revision ]

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