Proving Statistical Nature of $\hat{\rho}$ with Hermitian $\hat{H}$

[LagrangeEuler]

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

 

[dauto]

The solution is correct if βis real. Is it real?

 

[LagrangeEuler]

Yes it is!

 

[Fredrik]

If H is bounded, then
$$\left(e^{-\beta\hat H}\right)^\dagger =\left(\lim_{n\to\infty}\sum_{k=0}^n \frac{1}{k!}(-\beta\hat H)^k\right)^\dagger =\lim_{n\to\infty}\left(\sum_{k=0}^n \frac{1}{k!}(-\beta\hat H)^k\right)^\dagger =\dots =e^{-\beta\hat H^\dagger} =e^{-\beta\hat H}$$ The first equality is just the definition of the exponential of a bounded operator. The second follows from the fact that the $\dagger$ operation is a continuous function. The rest is easy.

 

[Oxvillian]

Perhaps it's worth pointing out in passing that this is the world-famous density operator used for dealing with thermal ensembles