# Density matrix

1. May 28, 2014

### LagrangeEuler

1. The problem statement, all variables and given/known data
If Hamiltonian $\hat{H}$ is hermitian show that operator $\hat{\rho}=\frac{e^{-\beta\hat{H}}}{Tr(e^{-\beta\hat{H}})}$ is statistical.

2. Relevant equations
In order to be statistical operator $\hat{\rho}$ must be hermitian and must have trace equal $1$.
$Tr(\hat{\rho})=1$

3. 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?
$Tr(\hat{\rho})=\frac{Tr(e^{-\beta\hat{H}})}{Tr(e^{-\beta\hat{H}})}=1$

2. May 28, 2014

### dauto

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

3. May 28, 2014

### LagrangeEuler

Yes it is!

4. May 28, 2014

### Fredrik

Staff Emeritus
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.

5. May 29, 2014

### Oxvillian

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