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What is the physical meaning of [H,ρ ]=0 for a stationary state?

  1. Nov 9, 2011 #1
    If ρ is the desity operator of a ensemble.
    We get (ih/2π) ∂ρ/∂ t = [H,ρ ].
    for a stationary state [H,ρ ]=0 . So H and ρ can share the same eigenvectors. Can someone explain what does this mean?
    :smile:
     
  2. jcsd
  3. Nov 9, 2011 #2
    let suppose you have of H:

    [itex]\hat{H}|E_i\rangle=E_{H,i}|E_i\rangle[/itex]

    If you can diagonalize H and rho together
    there is no reason to think that the set of [itex]|E_i\rangle[/itex] that can be diagonalized at the same time with H are ALL the [itex]|E_i\rangle[/itex] that are diagonal for [itex]\hat{H}[/itex]. in general the eigenstates in common is a subset of the eigenstates of H.
    But let consider only the sub set of eigenstates of H that are also eigenstates of rho, you have

    [itex]\hat{\rho}|E_i\rangle=E_{p,i}|E_i\rangle[/itex]

    and so you can write rho as

    [itex]\hat{\rho}=\Sigma_i \lambda_i |E_i\rangle\langle E_i|[/itex]

    that is the representation of your rho in the eigenstates of hamiltonian.
     
  4. Nov 9, 2011 #3

    kith

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    The density matrix being diagonal corresponds to a completely incoherent mixture. So the eigenvectors of the density matrix are the constituents of this mixture.
     
  5. Nov 14, 2011 #4
    But the expression
    [itex]\hat{\rho}=\Sigma_i \lambda_i |E_i\rangle\langle E_i|[/itex]
    is of the same form to the defination of desity operator in a mixture state. I still don't know the specail characteristics of thie kind of rho. I am sorry, I am so stupid to understand the physical mean of [H, ρ ]=0.
    Thank you for your reply.
     
  6. Nov 14, 2011 #5
    Thank you, Kith! You mean that if [H,ρ]=0,thenthe eigenvectors of the ρ are the constituents of a incoherent mixture.
     
  7. Nov 14, 2011 #6

    vanhees71

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    The complete equation of motion for the Statistical Operator is the von Neumann equation,

    [tex]\left (\frac{\partial \hat{\rho}}{\partial t} \right )_{\text{exp}} + \frac{1}{\mathrm{i} \hbar} [\hat{\rho},\hat{H}]=0.[/tex]

    Since a stationary state is defined as one for which the Statistical Operator is not explicitly dependent on time, for this case you arrive at the simplified equation

    [tex][\hat{\rho},\hat{H}]=0.[/tex]

    This means that for a stationary (or equilibrium) state, the Statistical Operator of the system must be a function of the conserved quantities and not explicitly dependent on time.
     
  8. Feb 14, 2013 #7
    Now I see, the equation
    [tex]\left (\frac{\partial \hat{\rho}}{\partial t} \right )_{\text{exp}} + \frac{1}{\mathrm{i} \hbar} [\hat{\rho},\hat{H}]=0.[/tex]
    plays similar role in quantum statistical mechanics as the Liuville's Equation does in Statistical mechanics.
     
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