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How do I calculate this, $\beta$ is a parameter, $H$ a matrix:

$(1-\beta \partial_\beta)\ln(Tr(e^{-\beta H})) = \ln(Tr(e^{-\beta H})) - \beta \frac{\partial_\beta Tr(e^{-\beta H})}{Tr(e^{-\beta H})}=?$

Are there identities to simplify $\ln(Tr)$ and my derivative?

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  • $\begingroup$ If you know the eigenvalues $\lambda_1,...\lambda_n$ of the matrix $H$, $$Tr \big(e^{-\beta \hat H}\big)=Tr \big(I-\beta \,diag(\lambda_1,...,\lambda_n)+\frac{\beta^2}{2!} \,diag(\lambda_1^2,...,\lambda_n^2)+...\big)=diag\big(e^{-\beta\lambda_1}, ...,e^{-\beta\lambda_n}\big)$$ $\endgroup$
    – Svyatoslav
    Dec 19, 2022 at 14:11

1 Answer 1

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$ \def\a{\alpha}\def\b{\beta}\def\l{\lambda} \def\qiq{\quad\implies\quad} \def\tr{\operatorname{Tr}} $For typing convenience, let a dot denote derivatives wrt $\b$ and define the variables $$\eqalign{ E &= \exp(-\b H) &\qiq \dot E = -HE \\ \a &= \tr(E) &\qiq \dot\a = \tr(\dot E)\\ \l &= \log(\a) &\qiq\dot\l = \frac{\dot\a}{\a} = \left(\frac{-\tr(HE)}{\tr(E)}\right) \\ }$$ Substituting this into your equation yields $$\eqalign{ \big(\l - \b\dot\l\big) = \l \;+\; \frac{\tr(\b HE)}{\tr(E)} \\ }$$

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