# Is the weak interaction Lagrangian invariant under parity transformations?

The weak interaction term in the Lagrangian reads

$$\bar \Psi \gamma_\mu P_L \Psi W^\mu.$$

Under parity transformations, because of $\Psi \rightarrow \gamma_0 \Psi$ and $\gamma_5 \rightarrow - \gamma_5$, which yields $P_L = \frac{1-\gamma_5}{2} \rightarrow \frac{1+\gamma_5}{2}=P_R$ the weak interaction term tranforms into

$$\bar \Psi \gamma_\mu P_L \Psi W^\mu \rightarrow \bar \Psi \gamma_0 \gamma_\mu P_R \gamma_0 \Psi W^\mu = \bar \Psi \gamma_0 \gamma_\mu \gamma_0 P_L \Psi W^\mu = \bar \Psi \gamma_\mu P_L \Psi W^\mu$$

Am I making a computational error here or is the weak interaction Lagrangian invariant under parity transformations?

The error is that $\gamma_5$ doesn't intrinsically change sign under parity. Also, don't forget that under parity the spatial components of $W_\mu$ change sign. And also $\gamma^0 \gamma^\mu \gamma^0 \neq \gamma^\mu$.

• Wow that would be a lot of errors. Firstly, do you have any reference on $\gamma_5 \neq - \gamma_5$? Secondly, I don't think that we have under parity transformations $W_i \rightarrow - W_i$, because following the same logic $A_i \rightarrow - A_i$ and the Lagrangian for electromagnetic interactions, wouldn't be invariant under parity transformations.
– jak
Dec 15 '14 at 10:16
• Okay, we have $W_i \rightarrow - W_i$, which yields an invariant Lagrangian for electromagnetism, too because in the Weyl Basis: $\gamma^0 \gamma^\mu \gamma^0 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & \sigma_\mu \\ \bar{\sigma}_\mu & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} \sigma_\mu & 0 \\ 0 & \bar{\sigma}_\mu \end{pmatrix} = \begin{pmatrix} 0 & \bar{\sigma}_\mu \\ \sigma_\mu & 0 \end{pmatrix}$. This means $\gamma_0 \rightarrow \gamma_0$ and $\gamma_i \rightarrow - \gamma_i$.
– jak
Dec 15 '14 at 10:24
• The minus sign from $\gamma_i \rightarrow - \gamma_i$ cancels the minus sign from $W_i \rightarrow - W_i$.
– jak
Dec 15 '14 at 10:25
• For example in this book: books.google.de/… on page 156 eq. 6.26 it is written that under parity transformations $\gamma_5 \rightarrow - \gamma_5$ . Therefore, my question remains unsanswered...
– jak
Dec 15 '14 at 12:28
• Read the book (and other resources) carefully: none of the gamma matrices on their own change sign under parity. It is the whole bilinear $\bar\psi F \psi$ that transforms, and how it transforms is to be derived from the transformation law $\Psi\rightarrow\gamma^0\Psi$ alone. Dec 15 '14 at 13:58