0
$\begingroup$

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?

$\endgroup$
5
  • 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 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 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. $\endgroup$ – Cosmas Zachos Apr 21 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 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 $\gamma ^5$... $\endgroup$ – Cosmas Zachos Apr 21 at 23:17
2
$\begingroup$

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.

$\endgroup$

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.