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

  • Thread starter jiadong
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  • #1
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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:
 

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  • #2
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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.
 
  • #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.
 
  • #4
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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.
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.
 
  • #5
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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.
Thank you, Kith! You mean that if [H,ρ]=0,thenthe eigenvectors of the ρ are the constituents of a incoherent mixture.
 
  • #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.
 
  • #7
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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.
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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