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I want to show that tr($\gamma^\mu \gamma^\nu$) = 4$\eta^{\mu \nu}$. I know that {$\gamma^\mu , \gamma^\nu$} = 2$\eta^{\mu\nu}I_4$ and tr($\gamma^\mu \gamma^\nu$) = tr($\gamma^\nu \gamma^\mu$),

so tr($\gamma^\mu \gamma^\nu$) = tr(2$\eta^{\mu\nu} I_4$ - $\gamma^\nu \gamma^\mu$) $\implies$ 2tr($\gamma^\mu \gamma^\nu$) = 2tr($\eta^{\mu\nu} I_4$), and ony way to get the answer is by pulling out $\eta^{\mu\nu}$ out of the trace.

I want to ask how we can do that? since $\eta^{\mu\nu}$ is itself a matrix, then why it is taken out of the trace and its diagonal elements are not summed?

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    $\begingroup$ This is understandable confusion coming from the fact that $\eta^{\mu\nu}$ is not a matrix, it is the components of a matrix. On the other hand $\gamma^\mu$ is a matrix ($\eta^{01}$ is a number, $\gamma^0$ is a matrix). In the definition of the Clifford algebra $\{\gamma^\mu,\gamma^\nu\}=\eta^{\mu\nu}1$ the $\eta^{\mu\nu}$ is multiplied by the identity matrix with spinor components. So when you see $\mathrm{tr}(\eta^{\mu\nu})$ what this actually means is $\eta^{\mu\nu}{1^\alpha}_\alpha$ where $\alpha$ are the spinor components of the identity. $\endgroup$
    – Charlie
    May 18 at 17:45
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    $\begingroup$ As a slight amendment to what I've written above, $\mathrm{tr}(\eta^{\mu\nu})$ can mean the trace of the matrix whose components are $\eta^{\mu\nu}$, but not in this context. Unfortunately this is common notation. $\endgroup$
    – Charlie
    May 18 at 17:51
  • $\begingroup$ Your comment is really helpfull. I need to get some knowledge of Clifford algebra too. $\endgroup$
    – sawan kt
    May 18 at 18:24
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    $\begingroup$ It does not help that in physics we generally: (1) Straight up don't write spinor indices on things, (2) Call $\eta^{\mu\nu}$ "the metric" when it is really the ($\mu$,$\nu$)'th component of the metric when written as a matrix and (3) Don't specify the domain over which the trace function is being defined (from which the question arises, are we taking the trace over the spinor indices or over the 4-vector indices?). $\endgroup$
    – Charlie
    May 18 at 19:00
  • $\begingroup$ Thankyou, now it is completely clear. @Charlie $\endgroup$
    – sawan kt
    May 19 at 11:08
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The trace is over the matrix indices which represent the internal degrees of freedom of the spinor, not over the space-time indices.

Each $\gamma^\mu$ is a matrix, so if we want to write out all the indices explicitely, it would be $(\gamma^\mu)^{i}_{~~j}$. Then $\{\gamma^\nu,\gamma^\mu\} = 2\eta^{\mu\nu}$ becomes : $$(\gamma^\mu)^i_{~~j}(\gamma^{\nu})^j_{~~k} + (\gamma^\nu)^i_{~~j}(\gamma^{\mu})^j_{~~k} = 2 \eta^{\mu\nu}\delta^i_k$$

and what you want to show is : $$(\gamma^\mu)^i_{~~j}(\gamma^\nu)^j_{~~i} = 2 \eta^{\mu\nu}$$

So the short answer is : the $\eta^{\mu\nu}$ goes out of the trace freely.

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