# Question about the parity violation of weak interaction Lagrangian

In the textbook of A. Zee, Quantum Field Theory in a Nutshell, the author states that the following Lagrangian:

$$\mathcal{L} = G (\overline{\psi}_{1L} \gamma^\mu \psi_{2L})(\overline{\psi}_{3L} \gamma^\mu \psi_{4L})$$

Which describes the weak interaction, violates the parity invariance. I've tried to prove this but with no success. Before I start showing what I've done, it's good to give you some context of the notation I'm familiar with. First, the parity operator is defined as $$P\psi := \gamma^0 \psi$$, where $$\gamma^0 := \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} .$$

And that

$$\psi_{L,R} := \frac{1}{2}(1 \mp \gamma^5) \psi = P_{L,R} \psi \quad \text{and} \quad \overline{\psi}_{L,R} : = \overline{\psi} \frac{1}{2} (1 \pm \gamma^5) = \overline{\psi}P_{R,L},$$

where

$$\gamma^5 := \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} .$$

And $$\overline{\psi} := \psi^\dagger \gamma^0$$. So, what I've tried to do is, basically:

$$\begin{split} P(\overline{\psi}_L \gamma^\mu \psi_L) = \gamma^0 \psi^\dagger \gamma^0 P_R \gamma^\mu \gamma^0 P_L \psi &= \gamma^0 \psi^\dagger P_L \gamma^0 \gamma^\mu \gamma^0 P_L \psi \\ &= \gamma^0 \psi^\dagger P_L (\gamma^\mu)^\dagger P_L \psi \end{split}$$

But I really don't know to proceed from here or if there are some mistakes in my calculations. Any help would be really welcomed.

• Did you reverse the order when getting the parity image of $\psi^\dagger$? Show what you have! May 16, 2022 at 19:28
• I'd start with "what is $P \psi_{L}$?" and go from there. May 16, 2022 at 19:44
• @CosmasZachos I don't think I truly understand the question, but to compute the parity of $\psi^\dagger$ I've used that $\mathcal{P} \overline{\psi}_L = \psi^\dagger \gamma^0 \gamma^0 P_R = \psi^\dagger P_R$, however I don't know how to use this result May 16, 2022 at 21:42
• @JerrySchirmer For $P \psi_L$ I got that $P \psi_L = P_L \gamma^0 \psi = P_L \begin{pmatrix} \psi_R \\ \psi_L \end{pmatrix} = \psi_R$. Is this enough o prove that the lagrangian violates the parity invariance? I mean, if a plug that into the lagrangian I think I'll get something like $\overline{\psi}_{iL} \gamma^0 \gamma^\mu \psi_{iR}$, but I don't know if that's enough for the proof May 16, 2022 at 21:48
• @GustavoB.Silva: all you need for the proof is that the parity transformed Lagrangian is not equal to the original Lagrangian, which will definitely be the case if you're not left with any $\phi_L$s when you're done, right? May 16, 2022 at 22:08

$$\gamma^0 P_L \gamma^0 = P_R, \\ \gamma^0 P_R \gamma^0 = P_L,$$
$$P: \qquad \psi(x) \longrightarrow \gamma^0 \psi(-x) ~~~~\leadsto \\ P: \qquad \psi(x)^\dagger \longrightarrow \psi(-x)^\dagger \gamma^0 ~~~~\leadsto \\ P: \qquad \overline \psi(x)\gamma^0\psi'(x) \longrightarrow \overline \psi(-x)\gamma^0 \gamma^0 \gamma^0 \psi'(-x) =\overline \psi(-x) \gamma^0 \psi'(-x), \\ P: \qquad \overline \psi(x)\vec \gamma\psi'(x) \longrightarrow \overline \psi(-x)\gamma^0 \vec \gamma \gamma^0 \psi'(-x) =-\bar\psi(-x) \vec \gamma \psi'(-x).$$
One projector suffices for each spinor bilinear, given the intercalated γ s, so that $$P: \qquad \overline \psi(x)\gamma^0P_L\psi'(x) ~~ \overline \psi''(x)\gamma^0P_L\psi'''(x)\longrightarrow \\ \overline \psi(-x)\gamma^0 \gamma^0 P_L\gamma^0 \psi'(-x)~~ \overline \psi(-x)\gamma^0 \gamma^0 P_L\gamma^0 \psi'(-x)=\overline \psi(-x) \gamma^0 P_R \psi'(-x)~~\overline \psi''(-x) \gamma^0 P_R \psi'''(-x),$$ and similarly for the spacelike γ s... You do it.