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