# Why one can swap the product of a Lorentz transformation and a Dirac $\gamma^\mu$ matrix?

Ashok tries to prove Lorentz invariance of the Dirac equation. If the spinor follows the transformation rule $$\Psi' = S\Psi$$, then

$$(i\gamma^\mu\partial_\mu-m)\Psi = 0\to (i\gamma^\mu\Lambda^\nu_{\;\mu}\partial'_\nu-m)S^{-1}\Psi = 0.$$

Afterwards he writes

$$(i\Lambda^\mu_{\;\nu}\gamma^\nu\partial'_\mu-m)S^{-1}\Psi = 0.$$

It may appear at first glance that he just commute the Lorentz transformation and the Dirac gamma matrix and swap indexes $$\mu \leftrightarrow\nu$$. Is this correct or is it an errata or is there something here more involved?

First, $$\gamma^\mu$$ and $$\Lambda^\mu_{\phantom{\mu}\nu}$$, at fixed $$\mu$$ and $$\nu$$, can be commuted because they are just numbers from the point of view of the Lorentz indices (the matrix nature of $$\gamma^\mu$$ is only a spectator).
Second, $$\mu$$ and $$\nu$$ are summed over, so they can be renamed
$$\gamma^\mu \Lambda^\nu_{\phantom{\nu}\mu} \,\partial'_\nu \underset{\substack{\mu\to\rho\\\nu\to\lambda}}{=} \gamma^\rho \Lambda^\lambda_{\phantom{\lambda}\rho} \,\partial'_\lambda\underset{\substack{\rho\to\nu\\\lambda\to\mu}}{=}\gamma^\nu \Lambda^\mu_{\phantom{\mu}\nu}\,\partial'_\mu\,.$$ It's not conceptually necessary to do it in two steps, I just did it for clarity.
• I think that the "spectator" thing is what is really tricky. I agree with you, but Ashok's book treat $\gamma^\mu$ really as matrices, hence the source of confusion. – user2820579 Jun 20 at 2:08
• You can put explicit Dirac indices $(\gamma^\mu)_{\alpha\beta}\ldots \Psi^\beta$ and see that they don't play any role whatsoever in this identity. – MannyC Jun 20 at 5:09