I just started to learn how to quantise Dirac field. Meanwhile, as we can write the Dirac equation in terms of gamma matrices :

$$ (i\hbar\gamma^\mu\partial_\mu - m)\psi = 0 $$ where $\gamma_\mu$ matrices obey Clifford algebra $$ \{\gamma^\mu,\gamma^\nu\} = \eta^{\mu\nu} $$

Now I just came across that the adjoint of the gamma matrices can be written as :

$$ \gamma^{\mu\dagger} = \gamma^0\gamma^\mu\gamma^0 $$ I already checked this question, but it doesn't suggest a way to prove it in a representation independent manner. Also the Wikipedia page suggests that gamma matrices are chosen such that they satisfy the above relations, since they are arbitrary upto a similarity transformation.

So is that the way this identity comes, or is there something else ?

  • $\begingroup$ Thanks ! So this is true because that these relations satisfy the commutation relation in the {+---} notation, right ? $\endgroup$
    – user35952
    Feb 7, 2014 at 12:07
  • 5
    $\begingroup$ It's not clear exactly what you mean by "prove it in a representation independent manner". What are you prepared to take as hypotheses? It's important to note that the anticommutation properties alone are not enough to prove that formula, because they have no relation at all to any inner product (and thus adjointness) structure. You need some additional structure, such as unitarity of a representation, to be able to prove it at all. The proof will then depend on what that additional structure is. $\endgroup$ Feb 7, 2014 at 12:07
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    $\begingroup$ No, that question is ill-defined. See my comments there: #1, #2. $\endgroup$ Feb 7, 2014 at 12:10

1 Answer 1


You may use one of representations, then - prove the relation $\gamma_{\mu}^{+} = \gamma_{0}\gamma_{\mu}\gamma_{0}$ in this representation, and finally prove that it is correctly for an arbitrary representation (as I think that you want to see just that).

Here I will use the spinor representation: $$ \gamma^{\mu} = \begin{pmatrix} 0 & \sigma^{\mu} \\ \tilde {\sigma}^{\mu} & 0 \end{pmatrix}, \quad \sigma^{\mu} = (\hat {\mathbf E }, \sigma ), \quad \tilde {\sigma}^{\mu} = (\hat {\mathbf E }, -\sigma ). $$ It's not hard to show that in this representation $$ {\gamma^{\mu}}^{+} = \gamma^{0}\gamma^{\mu}\gamma^{0}. \qquad (1) $$ Then let's introduce the unitary transformation $\Psi \to \hat {U} \Psi, \bar {\Psi} \to \bar {\Psi}\hat {U}^{+}$, which connects spinor and some other basis. Corresponding transformation of the gamma-matrices is $\gamma^{\mu} \to \hat {U}^{+}\gamma^{\mu}\hat {U}$. Lets see how does $(1)$ change under this transformation: $$ {\gamma^{\mu}}^{+} \to (\hat {U}^{+} \gamma^{\mu} \hat {U})^{+} = \hat {U}^{+} {\gamma^{\mu}}^{+}\hat {U} = \hat {U}^{+} \gamma_{0}\gamma^{\mu}\gamma_{0}\hat {U} = \hat {U}^{+}\gamma_{0} \hat {U} \hat {U}^{+} \gamma^{\mu}\hat {U} \hat {U}^{+}\gamma^{0}\hat {U}= $$ $$ =\tilde {\gamma}^{0}\tilde {\gamma}^{\mu}\tilde {\gamma}^{0}. $$


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