# Question about choosing the time reversal operator

According to the "QFT Nutshell" page 102-104 about Time reversal, the author said about applying the time reversal to the spin 1/2 system.

Let $$T=UK$$, where $$K$$ complex conjugates everything to its right and $$U$$ is a unitary operator. Acting with $$T$$ on the spin-up state, we obtain the spin-down state. Thus, we need a nontricial matrix $$U=\eta\sigma_2$$ to flip the spin. Note that $$T^2=\eta\sigma_2 K\eta \sigma_2 K = \eta\sigma_2\eta^*\sigma_2^*KK=-1$$. This is the origin of Kramer's degeneracy

My question is that, it seems we can choose a different choice of the time reversal to be $$U=\eta\sigma_1$$ equally well. This also clearly flips the spin-up state to the spin-down state and vice versa. With this choice, we have

$$T^2=\eta\sigma_1 K\eta \sigma_1 K = \eta\sigma_1\eta^*\sigma_1^*KK=1$$

because $$\sigma_1$$ is real. We won't see the Kramer degeneracy anymore. But physics should not depend on the choice. Is there any other constrains we have to impose in order to get $$U=\eta\sigma_2$$?

• Excuse me there are two = symbols :-). – Sebastiano Aug 10 '19 at 11:25

We know that the action of time-reversal on $$\sigma_a$$ is $$\begin{equation} T\sigma_aT^{-1}=-\sigma_a. \end{equation}$$ Now, we have that $$\begin{equation} T\sigma_aT^{-1}=(UK)\sigma_a (K U^{-1})=U \sigma_a^{*}U^{-1}=(-1)^{a+1}U\sigma_aU^{-1}, \quad (a=1,2,3). \end{equation}$$ which thus means that $$U$$ must satisfy $$\begin{equation} U\sigma_a U^{-1}=(-1)^a\sigma_a. \end{equation}$$ If we write $$U=e^{i d \sum_a n_a\sigma_a},$$ where $$d>0$$ and $$\sum_a n^2_a=1,$$ then we have that $$\begin{equation} U\sigma_a U^{-1}=\sigma_a\cos 2d+\sum_{bc}\epsilon_{abc}n_b \sigma_c \sin 2d+n_a\sum_{b}n_b\sigma_b(1-\cos 2d). \end{equation}$$ This form of the transformation allows us to obtain a set of equations for $$d,n_a$$ by evaluating the trace $$\begin{equation} \text{Tr}\left[\sigma_b U\sigma_a U^{-1}\right]=2(-1)^{a}\delta_{ab}. \end{equation}$$ Inspection of these equations lead to the only solution $$n_a=\delta_{a2}$$ and $$d=\pi/2.$$ We thus conclude that $$\begin{equation} T=e^{i\frac{\pi}{2}\sigma_2}K=i\sigma_2 K. \end{equation}$$
• Why is $T\sigma_a T^{-1} = - \sigma_a$? As far as I know, $T$ is antiunitary. That only implies $T(-i)T^{-1} = i$. – TangBear Aug 11 '19 at 11:25
• Time-reversal should also reverse the sign of angular momenta vectors, such as spin $S_a=\frac{\hbar}{2}\sigma_a.$ This is equivalent to the condition you cite that $T$ should flip spin states. From this perspective of flipping states, what is missing to be able to exclude the $T=\sigma_1K$ possibility is that $T$ must flip spin states with respect to \textit{any} quantization axis. This leads to $T\propto \sigma_2 K$ as the only possibility. – Ian Mondragon Shem Aug 11 '19 at 15:26