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Density matrix

  1. May 28, 2014 #1
    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. jcsd
  3. May 28, 2014 #2
    The solution is correct if βis real. Is it real?
     
  4. May 28, 2014 #3
    Yes it is!
     
  5. May 28, 2014 #4

    Fredrik

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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.
     
  6. May 29, 2014 #5
    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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