# A question involving chiral transformations and gamma matrices

I'm looking at a calculation that involves an infinitesimal transformation of a Dirac fermion field:

$$\Psi \rightarrow e^{i \beta \gamma^5} \Psi.$$

Then the conjugate field $$\bar{\Psi} = \Psi^{\dagger} \gamma^0$$ transforms as $$\bar{\Psi} \rightarrow (e^{i \beta \gamma^5} \Psi)^\dagger \gamma^0$$. Then from here we get:

$$\Psi^\dagger e^{-i \beta \gamma^5} \gamma^0.$$

So far I understand the steps, but I don't how from here one jumps to $$\Psi^\dagger \gamma^0 e^{i \beta \gamma^5}.$$

Why does the sign in the exponential changes and the gamma matrix is suddenly on the right?

• Try expanding the exponential to its series and apply the anti-commutation relation to pull the $\gamma_0$ to the left of the exponential series. Apr 21 at 19:41
• @Hannes, It's maybe better if you promote your comment to an answer. Of course, if your time permits.
– SG8
Apr 21 at 19:50
• When you commute $\gamma^0$ past any function of $\gamma^5$, you obtain the very same function of $-\gamma^5$, instead. Can you prove that? No expansions. Apr 21 at 20:11
• @CosmasZachos That is not true for any function. Try pulling $\gamma_0$ through $\gamma_1\gamma_5$ or most other multiplications of $\gamma_5$ with a matrix that is non-commuting with $\gamma_0$. Apr 21 at 23:12
• Fair enough, any function of just $\gamma ^5$, without further noncommuting matrices, as in the question. Works for braiding past just $\gamma ^5$... Apr 21 at 23:17

As one usually does, write the exponential term with a power series expansion, $$\Psi^{\dagger} \big(1 - i \beta \gamma^{5} + \mathcal{O}(\beta^2) \big) \gamma^0$$ then using the anticommutative properties $$\{\gamma^5,\gamma^{\mu} \} = 0$$ you can move $$\gamma^{0}$$ through the $$\gamma^{5}$$ terms, picking up a minus sign in the process. You can check the higher order terms too.