# Trace -logarithm - matrix

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?

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

$$\def\a{\alpha}\def\b{\beta}\def\l{\lambda} \def\qiq{\quad\implies\quad} \def\tr{\operatorname{Tr}}$$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)} \\ }