# Time reversal symmetry on bilinears

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.

• 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. Jan 15, 2023 at 23:02
• is $\gamma^5$ purely imaginary? what basis are you using? Jan 16, 2023 at 21:47
• It is in the Dirac basis so it's real. Jan 16, 2023 at 23:28

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.