2
$\begingroup$

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

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

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

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

$\endgroup$
  • 2
    $\begingroup$ 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
4
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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

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.