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## Homework Statement

If Hamiltonian ##\hat{H}## is hermitian show that operator ##\hat{\rho}=\frac{e^{-\beta\hat{H}}}{Tr(e^{-\beta\hat{H}})}## is statistical.

## Homework Equations

In order to be statistical operator ##\hat{\rho}## must be hermitian and must have trace equal ##1##.

##Tr(\hat{\rho})=1##

## The Attempt at a Solution

##\hat{\rho}^{\dagger}=\frac{e^{-\beta\hat{H}^{\dagger}}}{Tr(e^{-\beta\hat{H}})}=\frac{e^{-\beta\hat{H}}}{Tr(e^{-\beta\hat{H}})}=\hat{\rho}##

Not sure about this. If some operator is hermitian then I suppose also exponential function of this operator is also hermitian. Right?

##Tr(\hat{\rho})=\frac{Tr(e^{-\beta\hat{H}})}{Tr(e^{-\beta\hat{H}})}=1##