Density matrix

  • #1
691
17

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##
 

Answers and Replies

  • #2
1,948
201
The solution is correct if βis real. Is it real?
 
  • #3
691
17
Yes it is!
 
  • #4
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,872
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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
196
22
Perhaps it's worth pointing out in passing that this is the world-famous density operator used for dealing with thermal ensembles :smile:
 

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