Gaussian97 said:
Still in 4. It is better to resum all the terms so, what is the value of the sum
$$\sum_{n=0}^{\infty}(\mp)^n\frac{(i\alpha)^n}{n!}$$
Ahhh I assume that what you meant is this: you want to get back to exponents
$$\sum_{n=0}^{\infty}(\mp)^n\frac{(i\alpha)^n}{n!}=e^{\mp i\alpha}$$
So we end up with 4. as follows
$$\psi_{L,R}\rightarrow e^{i\alpha\gamma_5}\psi_{L,R}=\sum_{n=0}^\infty \frac{(i \alpha \gamma_5)^n}{n!}\psi_{L,R} = \sum_{n=0}^\infty (\mp)^n \frac{(i \alpha)^n}{n!}\psi_{L,R}= e^{\mp i\alpha}\psi_{L,R}$$
Gaussian97 said:
Also, I think I will never insist sufficiently in that ##e^{i\alpha \gamma^5}## is a matrix and ##\psi_{L,R}## is a vector, therefore an expression like $$e^{-i\alpha\gamma^5}\psi^\dagger_{L,R}$$ makes no sense.
My bad, I always forget it. Actually the transformation I provided for ##\psi_{L,R}^{\dagger}## was wrong because of that. Now I think it should be OK.
Regarding 5.
5. Substitute in 1. and should be straightforward to see that is not invariant unless ##m=0##.
Let's first show that the Lagrangian is invariant when ##m=0##
The Lagrangian when ##m=0## is
$$\mathscr{L} = i\bar{\psi_L} \not{\!\partial}\psi_L + i\bar{\psi_R} \not{\!\partial}\psi_R=i\psi_L^{\dagger} \gamma^0 \gamma^{\mu}\partial_{\mu}\psi_L + i\psi_R^{\dagger} \gamma^0 \gamma^{\mu}\partial_{\mu}\psi_R$$
We analogously get the transformation for ##\psi_{L,R}^{\dagger}##
$$\psi_{L,R}^{\dagger}\rightarrow e^{-i\alpha\gamma_5}\psi_{L,R}=\sum_{n=0}^\infty \frac{(-i \alpha \gamma_5)^n}{n!}\psi_{L,R} = \sum_{n=0}^\infty (\pm)^n \frac{(i \alpha)^n}{n!}\psi_{L,R} = e^{\pm i\alpha} \psi_{L,R}$$
We now apply the transformations we got to the Lagrangian (note that we can commute ##e^{+ i\alpha}## as we wish, because it is not a matrix anymore (EDIT: this is probably a wrong assumption and thus the reason why I am not getting the desired result; a hint would be appreciated if that is the case :) ).
$$\mathscr{L'} = i e^{+ i\alpha} \psi_{L} \gamma^0 \gamma^{\mu}\partial_{\mu}e^{- i\alpha}\psi_{L} + ie^{- i\alpha}\psi_{R} \gamma^0 \gamma^{\mu}\partial_{\mu}e^{+ i\alpha}\psi_{R}=i \psi_{L} \gamma^0 \gamma^{\mu}\partial_{\mu} \psi_{L} + i \psi_{R} \gamma^0 \gamma^{\mu}\partial_{\mu} \psi_{R}$$
Mmm...
almost there!
I have to be missing something here; of course
$$i \psi_{L} \gamma^0 \gamma^{\mu}\partial_{\mu} \psi_{L} + i \psi_{R} \gamma^0 \gamma^{\mu}\partial_{\mu} \psi_{R} \neq i\psi_L^{\dagger} \gamma^0 \gamma^{\mu}\partial_{\mu}\psi_L + i\psi_R^{\dagger} \gamma^0 \gamma^{\mu}\partial_{\mu}\psi_R$$
Due to the fact
$$\psi_{L,R} \neq \psi_{L,R}^{\dagger}$$
...