Reading a into QFT I recently came across basically this (Kaku p.94):

If $\Psi (x)$ is a solution to the massless Dirac equation in Weyl representation, also $\Phi (x) = \exp(i \Lambda \gamma^5) \cdot \Psi (x)$ will be a solution.

Can someone elaborate on why this is? Expanding the exponential gives something like $c_1 \cdot 1 + i c_2 \cdot \gamma^5$, but from there on I'm stuck.


1 Answer 1


Probably figured it out myself, so if anyone's interested, here is my solution:

$(\gamma^5)^2 = 1 $

$ \gamma^\mu \gamma^5 = - \gamma^5 \gamma^\mu$

$\Rightarrow \gamma^\mu e^{i \Lambda \gamma^5} = \gamma^\mu \sum \frac{(i \Lambda \gamma^5)^n}{n!} = (\sum_{even} \frac{(i\Lambda)^n}{n!})\gamma^\mu+ \gamma^\mu(\sum_{odd}\frac{(i\Lambda)^n\gamma^5}{n!})$

$= (\sum_{even} \frac{(i\Lambda)^n}{n!})\gamma^\mu - (\sum_{odd}\frac{(i\Lambda)^n\gamma^5}{n!})\gamma^\mu$

$= (\sum_{even} \frac{(-i\Lambda)^n}{n!})\gamma^\mu + (\sum_{odd}\frac{(-i\Lambda)^n\gamma^5}{n!})\gamma^\mu$

$= e^{-i\Lambda \gamma^5}\gamma^\mu $

Massless Dirac Eqn.:

$i \hbar \gamma^\mu \partial_\mu \Psi = 0$

$\Rightarrow i \hbar \gamma^\mu \partial_\mu e^{i\Lambda \gamma^5}\Psi = i \hbar e^{-i\Lambda \gamma^5} \gamma^\mu \partial_\mu \Psi = 0$


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.