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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?

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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.

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  • $\begingroup$ 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. $\endgroup$ Jun 20, 2019 at 2:08
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    $\begingroup$ But they are matrices acting on different vector spaces, which is how I would explain that they commute. $\endgroup$
    – G. Smith
    Jun 20, 2019 at 2:45
  • $\begingroup$ 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. $\endgroup$
    – MannyC
    Jun 20, 2019 at 5:09
  • $\begingroup$ Yes, that's what I meant with the tricky part. This was provided by the link of Accidentaletc. The point is that there is a tensor product hidden there that nobody writes and usually nobody says. $\endgroup$ Jun 20, 2019 at 15:47

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