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?

  • 3
    $\begingroup$ 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. $\endgroup$
    – Hannes
    Apr 21, 2021 at 19:41
  • $\begingroup$ @Hannes, It's maybe better if you promote your comment to an answer. Of course, if your time permits. $\endgroup$
    – SG8
    Apr 21, 2021 at 19:50
  • $\begingroup$ 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. $\endgroup$ Apr 21, 2021 at 20:11
  • $\begingroup$ @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 $\gamma_0$. $\endgroup$
    – Hannes
    Apr 21, 2021 at 23:12
  • $\begingroup$ Fair enough, any function of just $\gamma ^5$, without further noncommuting matrices, as in the question. Works for braiding past just $\gamma ^5$... $\endgroup$ Apr 21, 2021 at 23:17

1 Answer 1


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.

Edit: I see this has also already been pointed out in the comments by Hannes too.


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