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We are asked to show that

$$\not A\not B = A\cdot B - i\sigma_{\mu\nu}A^\mu B^\nu $$

I know that

$$\not A = A_\mu \gamma^\nu $$by definition, and:

$$\sigma_{\mu \nu}=\frac{i}{2} [\gamma_\mu, \gamma_\nu]$$

Thus,

$$A_\mu \gamma^\mu B_\nu \gamma^\nu = A_\mu B^\mu - i\sigma_{\mu\nu}A^\mu B^\nu$$ $$A_\mu \gamma^\mu B_\nu \gamma^\nu = A_\mu B^\mu + \frac{1}{2} [\gamma_\mu, \gamma_\nu]A^\mu B^\nu$$

$$A_\mu \gamma^\mu B_\nu \gamma^\nu = A_\mu B^\mu + \frac{1}{2} (\gamma_o \gamma_1 - \gamma_1\gamma_o) A^\mu B^\nu$$

I am stuck here, I don't have an idea of how to show that both sides are indeed the same.

Edit: See Daniel's and user1504's post below. Here is the thought process: I need

$$\text{I need } \gamma^\mu \gamma^\nu \text{to be something that lowers and changes the index of B and adds a} -\sigma_{\mu\nu} \text{ term.}$$

So we essentially want to show that $$\gamma^\mu \gamma^\nu = g^{\mu\nu} - i\sigma^{\mu\nu}$$ Verify this gives you the correct expression on both sides. Then use the commutator and anticommutator expression for the gamma matrices. How do you need to add or subtract them to get to this expression?

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  • $\begingroup$ What does $\gamma^\mu\gamma^\nu$ need to equal for your last equation to be satisfied? $\endgroup$
    – Daniel
    Sep 19 '20 at 1:28
  • $\begingroup$ Hint: Forget about $A$ and $B$. You want to show that $\gamma^\mu\gamma^\nu = g^{\mu\nu} -i \sigma^{\mu\nu}$. You've got a relationship between $\sigma^{\mu\nu}$ and the $\gamma$'s. Can you think of one between $g^{\mu\nu}$ and the $\gamma$s? $\endgroup$
    – user1504
    Sep 19 '20 at 1:32
  • $\begingroup$ @user1504 Thank you. I just derived that by adding the commutator of the gamme matrices and the anticommutator relationship that I had in my original post. $\endgroup$ Sep 19 '20 at 2:55
  • $\begingroup$ @Daniel Thank you. That gave me good intuition for tackling problems like this in the future. Problem is solved thanks to both of your help :). $\endgroup$ Sep 19 '20 at 2:58
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    $\begingroup$ Can I suggest that if you have solved the problem that you post an answer. This is well accepted practice and stops the system having an apparently unanswered question listed. It might even get you some up votes. $\endgroup$
    – StephenG
    Sep 19 '20 at 12:49

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