# Zee's book on QFT on the spin-1/2 state has $T^2=-1$, why not $T^2=+1$

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:

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.