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I am attempting to prove a set of results for the products of gamma matrices and traces of products of gamma matrices, but got stuck on this particular one.

$$Tr(\gamma^{\mu_1}...\gamma^{\mu_n})=g^{\mu_1\mu_2}Tr(\gamma^{\mu_3}...\gamma^{\mu_n})-g^{\mu_1\mu_3}Tr(\gamma^{\mu_2}\gamma^{\mu_4}...\gamma^{\mu_n})+...+g^{\mu_1\mu_n}Tr(\gamma^{\mu_2}...\gamma^{\mu_{n-1}}).$$

My previous strategy to get the metric in expressions has been to exploit the anti-commutation relation, writing $\gamma^{\mu\nu}$ as $$\gamma^{\mu\nu}+\gamma^{\nu\mu}-\gamma^{\nu\mu}=\{\gamma^{\mu},\gamma^{\nu}\}-\gamma^{\nu}\gamma^{\mu}=2g^{\mu\nu}-\gamma^{\nu}\gamma^{\mu}.$$ I then feel that I will have to use the cyclical property of the trace to get the desired expression: I have tried with the case for three gamma matrices to see if it can be extended, but not sure how to do this. For example, if I have three gamma matrices in the trace we have

$Tr(\gamma^{\mu_1}\gamma^{\mu_2}\gamma^{\mu_3})=Tr(\{\gamma^{\mu_1},\gamma^{\mu_2}\}-\gamma^{\mu_2}\gamma^{\mu_1})\gamma^{\mu_3})=Tr(2g^{\mu_{1}\mu_{2}}\gamma^{\mu_3}-\gamma^{\mu_2}\gamma^{\mu_1}\gamma^{\mu_3})$

From linearity of the trace, I can write this as two traces. The second one is 0 because it's the trace of a product of an odd number of gamma matrices.

$$Tr(2g^{\mu_{1}\mu_{2}}\gamma^{\mu_3})$$

The metric is symmetric so we can re-write:

$$Tr((g^{\mu_{1}\mu_{2}}+g^{\mu_{2}\mu_{1}})\gamma^{\mu_3})=Tr(g^{\mu_{1}\mu_{2}}\gamma^{\mu_3})+Tr(g^{\mu_{2}\mu_{1}}\gamma^{\mu_3})$$

This looks partly right but not sure how to get the metrics out of the trace and some of it is not in the right order anyway (also for even $n$ we can't use the trick where part of the trace went to 0).

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    $\begingroup$ Trace of an odd number of gamma's is zero, just use $\gamma^1 \gamma^i = 2g^{1i} - \gamma^i \gamma^1$ continually for the even case. $\endgroup$
    – bolbteppa
    Apr 8 '18 at 11:58
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    $\begingroup$ Yes of course, what am I talking about, I was analysing the case for 3 matrices when I said myself that the trace for an odd number is 0. $\endgroup$
    – Tom
    Apr 8 '18 at 15:08
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You're on the right track. What you have here are components of the metric. If you're using the Minkowski metric in rectangular coordinates, these components are constant (±1 or 0), so you can take them out of the trace by exploiting the linearity. You will indeed require the cyclic property of the trace to prove the general result, but the case for three or any odd number of gamma matrices isn't particularly illuminating for the same. Try it with four to see if your result agrees with the well known result. Once you do that, you'll be able to see how to extend the result to the general case a bit more clearly.

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