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In Zee's book on QFT, p.103, he showed that the spin-1/2 state has $T^2=-1$ by finding that the $T = UK$ has a matrix $U\propto \sigma_2$ and a complex conjugation $K$.

However, how do we know that we need $T = UK$ has a matrix $U\propto \sigma_2$ instead of just $U\propto \sigma_1$?

In his argument, he just needs to switch spin-up with spin-down states. But $T = UK$ with $U\propto \sigma_1$ works equally well. But $T^2=+1$ in that case instead.


From Zee's book on QFT, p.103:

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Zee's discussion here is sloppy, in that it only considers spin flips of $z$-eigenstates. More generally, we want our time reversal operator $T$ to satisfy $$ T^{\dagger} \vec{S} T = -\vec{S} $$ In other words, $T$ flips the spin of not just $S^z$ eigenstates, but $S^x$ and $S^y$ eigenstates as well. You can see immediately with this observation that you need to choose $S^y$ instead of $S^x$: $S^y$ is the only spin operator with imaginary components. The operator $K S^x$ would simply commute with $S^x$, and thus does not satisfy the above condition; on the other hand, $K S^y$ anticommutes with $S^x$ and $S^z$ via the anticommutivity of the Pauli matrices, and anticommutes with $S^y$ via complex conjugation.

For a fantastic self-contained discussion of time reversal in quantum mechanics, see this set of notes.

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