The following screenshot comes from this book: Nonequilibrium many-body theory of quantum systems: A modern introduction.

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In the discussion, the authors have performed the spectrum decomposition for the density matrix $\widehat{\rho}$ and also the Hamiltonian $\hat{H}^M$ with the same basis $|\Psi_k\rangle$, which are the eigenkets of the density matrix. But we all know the density matrix satisfies the Von-Neumann equation: $$i \dfrac{d}{dt} \hat{\rho} = [\hat{H},\hat{\rho}]$$ So my question is that: Why can they choose the same basis to perform the spectrum decomposition? In general, the density matrix does not commute with the Hamiltonian.


Not completely sure, you may want to check further: but I think $H^M$ is not directly the real Hamiltonian --- in your reference "construct the operator" means it's a newly-defined operator with $|\Psi_k\rangle$ and $\rho_k$; and $\beta$ is also newly-defined there, not necessary meaning the temperature.

I guess the reason you take it as the Hamiltonian is that the final "partition function": $Z=\text{Tr}\big[e^{-\beta H^M}\big]$ looks really like the partition function for canonical ensemble. Then the question now is why these two things are "the same"? To answer it, let us forget about definition $H^M$ for a second, which is directly related to $\rho$'s eigensystems $|\psi_k\rangle$ and $E_k^M$, and go back to the definition in quantum statstics of canonical ensemble.

Recall that, by definition (actually equal-distribution-theorem), when equilibrium is reached, the probability of the system being in an energy window around $E_{k}$ is proportional to $e^{-\beta E_k}$, but be careful here $E_k$ is the Hamiltonian $H$'s eigenvalue. Now we ask the question: what is the density matrix of such an equilibrium ensemble? The result turns out to be: \begin{align} \rho_{eq} \propto \sum_k e^{-\beta E_k}| \phi_k\rangle\langle \phi_k| \end{align} "$_{eq}$" means this is for equilibrium . Remember, again, here we only use real Hamiltonian $H$, and $|\phi_k\rangle$ and $E_k$ are corresponding eigensystem of it. Therefore, our logic is that: from original Hamiltonian, we construct the equilibrium density matrix $\rho_{eq}$ which satisfies the probability distribution desired. In this way, for a canonical ensemble, the equilibrium density matrix is directly diagonalized in Hamiltonian's eigen-basis, by definition. Then, like you said, that means: \begin{align} i\frac{d}{dt}\rho_{eq} = \big[H, \rho_{eq}\big] = 0 \end{align} and $\rho_{eq}$ doesn't evolve any more -- which makes sense since it's equilibrium.

Now, what about the logic in your reference? It turns out it's constructed in an inverse way: from an arbitrary density matrix $\rho$, construct an operator $H^M$, which is not necessarily the real Hamiltonian of the system, and also define a $\beta$ from $\rho_k$, which is not necessarily the temperature. So what's their meaning? Well, if you want, you could say if you design a system with Hamiltonian $H^M$ and temperature $\beta$, then the given density matrix $\rho$ happens to be the equilibrium density matrix $\rho_{eq}$: \begin{align} \rho_{eq}\big(H^M,\beta\big) = \rho \end{align}

In a short word, this is not surprising: they just design a system with equilibrium being the given $\rho$. And it's not contradict to the von Neumann equation, since in von Neumann equation $\rho$ and $H$ are real density matrix and real Hamiltonian of the system, other than designed from each other --- which is the logic of canonical ensembles or the logic in your reference.

  • $\begingroup$ So if the system stays in the thermal equilibrium state, then the derivative of density matrix will be zero? $\endgroup$ – Jack Dec 18 '17 at 4:07
  • $\begingroup$ @Jack Or, I think it could also be stated inversely: when time evolution (derivative) is zero, then we call it as equilibrium -- that's also a reasonable definition (though this is not from thermal probability law). And in canonical ensemble logic, we construct an equilibrium $\rho_{eq}$ according to probability distribution, and then we could check the time derivative is zero from von Neumann equation -- in this sense, we could say the definition of thermal equilibrium and time-evolution principle from von Neumann equation are consistent. $\endgroup$ – Kite.Y Dec 18 '17 at 4:44

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