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I read some books about high energy physics and string theory and I saw the following formula is used very frequently

I am not good at mathematics.

So can any one help me derive this formula?

See Brown, Quantum Field Theory.

Rewrite this as $\DeclareMathOperator{\tr}{tr} \ln \det \gamma = \tr \ln \gamma$. Then, we can switch to an eigenbasis of the matrix $\gamma$ and

$\ln \det \gamma = \sum_i \ln \lambda_i$

where $\lambda_i$ are the eigenvalues of $\gamma$. Since the eigenvalues of $\ln \gamma$ are $\ln \lambda_i$ we have

$\sum_i \ln \lambda_i = \tr \ln \gamma$

and thus

$\ln \det \gamma = \tr \ln \gamma$.

Thanks a lot @trof

$\gamma$ may not have an eigenbasis. However, the result can still be seen as true by writing $\gamma = PJP^{-1}$, where $J$ is the Jordan normal form of $\gamma$.

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