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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 ?

Community
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user35952
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1 Answers1

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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: $$ {\gamma^{\mu}}^{+} \to (\hat {U}^{+} \gamma^{\mu} \hat {U})^{+} = \hat {U}^{+} {\gamma^{\mu}}^{+}\hat {U} = \hat {U}^{+} \gamma_{0}\gamma^{\mu}\gamma_{0}\hat {U} = \hat {U}^{+}\gamma_{0} \hat {U} \hat {U}^{+} \gamma^{\mu}\hat {U} \hat {U}^{+}\gamma^{0}\hat {U}= $$ $$ =\tilde {\gamma}^{0}\tilde {\gamma}^{\mu}\tilde {\gamma}^{0}. $$

Andrew McAddams
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