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I want to know if gamma matrices commute with the Dirac spinor field, i.e., are the following equalities correct?

$$ \psi\gamma^{\mu}\overset{?}{=}\gamma^{\mu}\psi $$

$$ \psi^{\dagger}\gamma^{\mu}\overset{?}{=}\gamma^{\mu}\psi^{\dagger} $$

$$ \overline{\psi}\gamma^{\mu}=\psi^{\dagger}\gamma^{0}\gamma^{\mu}=\psi^{\dagger}\left(2\eta^{0\mu}-\gamma^{\mu}\gamma^{0}\right)\overset{?}{=}2\eta^{0\mu}\psi^{\dagger}-\gamma^{\mu}\overline{\psi} $$

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    $\begingroup$ It is not clear what e.g. the l.h.s. of your first equation is even supposed to be - the r.h.s. is $\gamma$ applied to a spinor, producing another spinor, but the l.h.s. is not, on the face of it, a spinor. Don't forget that there are operations (mostly applications of a matrix to a vector) here that are not written explicitly. You're effectively asking "Does a matrix commute with a column vector", which does not seem like a well-defined question without more information. $\endgroup$ – ACuriousMind♦ Jun 30 '18 at 12:30
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No. The Dirac matrices are just that — matrices. The quantity $\gamma^{\mu}\psi$ is defined as a matrix acting on a vector, as $\psi$ is a vector. However, $\psi\gamma^{\mu}$ is a kind of nonsensical expression. The same goes for your other examples.

To rephrase your own question: Do matrices commute with column vectors? Clearly, they don’t, since one order of multiplication isn’t even well defined.

I hope this helps!

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  • $\begingroup$ Take the conjugate Dirac equation,$\gamma^{\mu}\partial_{\mu}\overline{\psi}=im\overline{\psi}$. If $\gamma^{\mu}$ is a matrix and $\overline{\psi}=\psi^{\dagger}\gamma^{0}$ is a line vector, how does the l.h.s make sense if it represents a matrix multiplied by a line vector? $\endgroup$ – johani Jun 30 '18 at 13:04
  • $\begingroup$ Often it helps to make explicit all indices: $\psi'=\gamma^\mu\psi$ means $(\psi')^a=(\gamma^\mu)^a_{\ b}\psi^b$ so you can write $\psi^b(\gamma^\mu)^a_{\ b}$ but not $\psi\gamma^\mu$. In the same way, $\bar\psi'=\bar\psi\gamma^\mu$ is a 1-form (or covector) ans means $\bar\psi'_b=\bar\psi_a(\gamma^\mu)^a_{\ b}$. $\endgroup$ – Christophe Jun 30 '18 at 13:40
  • $\begingroup$ @johani That's not the conjugate Dirac equation. The conjugate Dirac equation reads $\partial_{\mu}\overline{\psi}\gamma^{\mu}=im\overline{\psi}$. $\endgroup$ – Bob Knighton Jun 30 '18 at 13:56

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