- #1

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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?

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- Thread starter jiadong
- Start date

- #1

- 14

- 0

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?

- #2

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[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

Science Advisor

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- #4

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But the expression

[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.

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

- #6

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[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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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]

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.

[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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