In the lecture notes accompanying a course I'm following, it is stated that $$\DeclareMathOperator{\Tr}{Tr} \Tr\left[\gamma^{\mu}\gamma^{\nu}\right] = 4 \eta^{\mu\nu} $$

Yet when I try to prove this, I find something different as follows: $$ \begin{eqnarray} \Tr\left[\gamma^{\mu}\gamma^{\nu}\right] & = & \gamma^\mu\gamma_\mu \\ & = &\eta_{\nu\mu}\gamma^\nu\gamma^\mu \\ & = &\frac12\left(\eta_{\nu\mu}+\eta_{\mu\nu}\right)\gamma^\nu\gamma^\mu \\ &=&\frac12\left(\eta_{\nu\mu}\gamma^\nu\gamma^\mu+\eta_{\mu\nu}\gamma^\nu\gamma^\mu\right) \\ &=&\frac12\eta_{\mu\nu}\left(\gamma^\nu\gamma^\mu+\gamma^\mu\gamma^\nu\right) \\ &=&\eta_{\mu\nu}\eta^{\mu\nu}I_4 = 4I_4 \end{eqnarray} $$ Where I used the anticommutator relation $\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}I_4$. With $I_4$ the identity matrix.

I assume that I made a mistake somewhere, or are these statements equivalent?

  • 1
    $\begingroup$ The mistake is in the first step: $\Tr\left[\gamma^{\mu}\gamma^{\nu}\right] = \gamma^\mu\gamma_\mu$ $\endgroup$ – Thomas Fritsch Jan 26 '19 at 11:30
  • $\begingroup$ Oh dear, I see.. I'll give it another go! Thank you $\endgroup$ – Simon Jan 26 '19 at 11:37

As Thomas Fritsch already mentioned in his comment, your first step is erroneous. The trace is over the spinor indices, not over the spacetime indices.

The key ingredients are the cyclicity of the trace and the anti-commutation relations of the $\gamma$ matrices. It is \begin{eqnarray} \Tr\left[\gamma^{\mu}\gamma^{\nu}\right] & = & \Tr\left[2\eta^{\mu\nu}I_4-\gamma^{\nu}\gamma^{\mu}\right] \\ & = &8\eta^{\mu\nu}-\Tr\left[\gamma^{\mu}\gamma^{\nu}\right], \end{eqnarray} where in the first line we used the anti-commutation relation $\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}I_4$ and in the second line the cyclicity of the trace $\Tr\left[\gamma^{\mu}\gamma^{\nu}\right]=\Tr\left[\gamma^{\nu}\gamma^{\mu}\right]$. The sought for expression then follows by rearranging the terms.

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