Consider an ensemble of systems where each system is in one of a set of states $|\alpha_i\rangle$, with proportions $w_i$, such that the density operator is

$$ \hat{\rho} = \sum_i w_i |\alpha_i\rangle \langle \alpha_i|. $$

I'd like to derive the von Neumann equation to describe the time evolution of the density matrix. Now, I imagine that the entire point of such an exercise is that the weights as well as the states change in time; the states evolve via the Schrodinger equation, so I would think to stick in the above definition. Doing so, I get

$$ i\hbar\frac{\partial \hat{\rho}}{\partial t} = -[\hat{\rho},\hat{H}] + i\hbar \sum_i \frac{\partial w_i}{\partial t} |\alpha_i \rangle \langle \alpha_i|, $$

which is obviously not the von Neumann equation due to the last term on the right. What happens to that term? Am I missing an important reason that the $w_i$ do not evolve in time?


1 Answer 1


I determined the answer to my question, so I decided to post it.

The weights $w_i$ are constant in time. The reason for this is that all of the time evolution is contained in the kets; all of the $Nw_i$ systems in the $|\alpha_i (t_0)\rangle$ state at time $t_0$ will be in the state $|\alpha_i(t)\rangle$ at a later time $t$, so it is accurate to treat only the time evolution of the kets and keep the $w_i$ constant.

  • 1
    $\begingroup$ As a comment, there are situations where it is useful to consider a time-dependence of the probabilities. E.g. when a system interacts with a reservoir that is not too large. Also, in the case where one considers a sub-system and derive the evolution equation for that, the resulting equation is usually not the von Neumann equations and (I imagine) that sometimes the resulting equation can sometimes be in the form you have written in your question. $\endgroup$ Commented Oct 5, 2021 at 10:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.