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I am reading An Introduction to Quantum Field Theory by Peskin & Schroeder, and I am confused about what is the square $T^2$ of time reversal operator $T$.

My guess is that for $P^2$, $C^2$ and $T^2$ they assume they are the identity, though they don't say it explicitly.

As an example on page 69 they calculate the time reversal on $\bar{\psi}\psi$ using the formula calculated before for the time reversal on $\psi$ and $\bar{\psi}$ (that is, $T\bar{\psi}T$ and $T\psi T$). To me mathematically this is possible only through the following step that they have omitted: $$T\bar{\psi}\psi T = (T\bar{\psi}T)(T\psi T) = \ldots,$$ where I have used $T^2=1$.

I am correct to say that they assume $T^2=1$? If I am wrong, how can they use the formula calculated before for the time reversal on $\psi$ and $\bar{\psi}$ to calculate $T\bar{\psi}\psi T$? I am confused also because based on some other resources it seems that $T^2=-1$.

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    $\begingroup$ That section is a huge mess, so you should probably skip it altogether for your own sanity. But if you insist, what they really mean is $T\bar\psi\psi T^{-1}$, with a $-1$ on the second $T$. You are indeed correct that, in general, $T^2$ need not be $+1$, see e.g. physics.stackexchange.com/q/735842/84967 for more information. $\endgroup$
    – AccidentalFourierTransform
    Commented Mar 18 at 12:40
  • $\begingroup$ Indeed usually symmetry operate as $U\psi U^{-1}$ and you can replace $U^{-1}$ with $U$ if $U$ is unitary and you set $U^2=1$ cause in this case you have $U^{\dagger}=U^{-1}=U$. $\endgroup$
    – Andrea
    Commented Mar 18 at 13:36
  • $\begingroup$ You also have the careful, because $T$, being a nonlinear operator, is not associative. $\endgroup$
    – Buzz ♦
    Commented Mar 18 at 15:08
  • $\begingroup$ You can have $T^2=-1$ acting on states. As others suggest, it's good to be careful about $T$ and its inverse, but if you apply it twice to both sides of an operator, it'll square to one. As @Buzz notes, you also have to be careful because $T$ is antilinear / antiunitary. For quantum mechanics, $T O T^{-1}$ achieves complex conjugation of $O$ and $\langle T a , T b \rangle = \langle b, a \rangle$ and $\langle a, T b \rangle = \langle b, T^\dagger a \rangle$, etc. You can work out most relations from that. I think this all extends to QFT too. $\endgroup$
    – just a phase
    Commented Mar 29 at 3:35

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