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 $U=\eta\sigma_2$?

  • 1
    $\begingroup$ Excuse me there are two = symbols :-). $\endgroup$
    – Sebastiano
    Commented Aug 10, 2019 at 11:25

1 Answer 1


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}

  • $\begingroup$ Why is $T\sigma_a T^{-1} = - \sigma_a$? As far as I know, $T$ is antiunitary. That only implies $T(-i)T^{-1} = i$. $\endgroup$
    – Kimari
    Commented Aug 11, 2019 at 11:25
  • $\begingroup$ 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. $\endgroup$ Commented Aug 11, 2019 at 15:26
  • $\begingroup$ But, does the spin flip really have something to do with the sign? I believe that we can take into account the spin flip by just switching the matrix (1,0) to (0,1) completely without worrying about the sign. $\endgroup$
    – Kimari
    Commented Aug 11, 2019 at 21:14

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