I wanted to check the electroweak standard model Lagrangian is invariant under CPT transformation by first checking how the bilinears of $\bar{\psi} \psi$, $\bar{\psi}\gamma_5 \psi$ $\bar{\psi}\gamma^{\mu} \psi$, $\bar{\psi}\gamma^{\mu}\gamma_5 \psi$, $\bar{\psi}\sigma^{\mu \nu} \psi$ transform under discrete symmetries (C, P, T) separately.

I seem to have a problem with the time-reversal symmetry only. I define the T operator as acting

$$\psi \to i\gamma_1 \gamma_3 \psi^*$$ $$\psi^* \to -i\gamma_1^* \gamma_3^* \psi = -i\gamma_1 \gamma_3 \psi$$ $$\bar{\psi} = {\psi^*}^T \gamma^0 = (-i\gamma_1 \gamma_3 \psi)^T \gamma^0$$

Hence, I get

$$\bar{\psi} \psi = (-i\gamma_1 \gamma_3 \psi)^T \gamma^0 (i\gamma_1 \gamma_3 \psi^*)= (\psi^T \gamma_1^T \gamma_3^T \gamma^0 \gamma_1 \gamma_3 \psi^*)$$

Then I use $\gamma_3^T \gamma_1^T = \gamma_3 \gamma_1$ by the definition of gamma matrices and $\gamma^0 \gamma_{1,3} = -\gamma_{1,3} \gamma^0$ from their anticommutation relations and $\gamma_{1,3}^2 = -1$. Hence I got

$$\bar{\psi} \psi = (\psi^T \gamma_1 \gamma_3 \gamma^0 \gamma_1 \gamma_3 \psi^*) = -(\psi^T \gamma^0 \psi^*) = -(\psi^{\alpha} \gamma^0_{\alpha \beta} \psi^{\beta*}) = +\psi^{\beta*}\gamma^0_{\alpha \beta} \psi^{\alpha}$$

now use the fact that $\gamma_0 = \gamma_0^T$ so

$$\bar{\psi} \psi = -\psi^{\beta*}\gamma^0_{\alpha \beta} \psi^{\alpha} = + \bar{\psi}\psi$$

As expected this scalar term under T transform as $\bar{\psi}\psi = \bar{\psi}\psi$. The axial part

$$\bar{\psi} \gamma_5 \psi = (-i\gamma_1 \gamma_3 \psi)^T \gamma^0 \gamma_5 (i\gamma_1 \gamma_3 \psi^*)= (\psi^T \gamma_1^T \gamma_3^T \gamma^0 \gamma_5\gamma_1 \gamma_3 \psi^*) = (\psi^T \gamma_1 \gamma_3 \gamma^0 \gamma_5\gamma_1 \gamma_3 \psi^*) = (\psi^T \gamma_3 \gamma^0 \gamma_5 \gamma_3 \psi^*)= - (\psi^T \gamma^0 \gamma_5 \psi^*)$$

By using the anticommutation of the fields as done for the scalar term $\{{\psi, \psi^*}\}=0$, from the axial term:

$$\bar{\psi} \gamma_5 \psi = - (\psi^T \gamma^0 \gamma_5 \psi^*) = (\bar{\psi} \gamma_5 \psi)$$

However this should be with an opposite sign and I can't figure out where I did wrong.

  • $\begingroup$ As long as you are aware that your definition of T is neither $\hat T$, nor T of Chapter 11 of M Schwartz's QFT text. He goes painstakingly illustrating the point, and your very first formula has an extra i which turns around to bite you. $\endgroup$ Jan 15, 2023 at 23:02
  • $\begingroup$ is $\gamma^5$ purely imaginary? what basis are you using? $\endgroup$ Jan 16, 2023 at 21:47
  • 1
    $\begingroup$ It is in the Dirac basis so it's real. $\endgroup$
    – Monopole
    Jan 16, 2023 at 23:28

1 Answer 1


I think your second = goes wrong. Looks like you need to reverse $\gamma_1^T$ and $\gamma_3^T$. I found the rest of your working helpful.


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