1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $\gamma^0 \gamma^\mu \gamma^0 \neq \gamma^\mu$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – jak Dec 15 '14 at 10:16
  • $\begingroup$ 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 $\gamma_i \rightarrow - \gamma_i$. $\endgroup$ – jak Dec 15 '14 at 10:24
  • $\begingroup$ The minus sign from $\gamma_i \rightarrow - \gamma_i$ cancels the minus sign from $W_i \rightarrow - W_i$. $\endgroup$ – jak Dec 15 '14 at 10:25
  • $\begingroup$ 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... $\endgroup$ – jak Dec 15 '14 at 12:28
  • $\begingroup$ 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. $\endgroup$ – QuantumDot Dec 15 '14 at 13:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.