How does the von Neumann equation relate to Schrödinger's equation?

[TeethWhitener]

TL;DR

Formalism for going from the von Neumann equation to the Schrödinger equation

I was trying to show how to get Schrödinger’s equation from the von Neumann equation and I’m not quite confident enough in my grasp of the functional analysis formalism to believe my own explanation. Starting from
$$i\hbar\frac{\partial}{\partial t}\rho=[H,\rho]$$
We have
$$i\hbar\left(\frac{\partial |\psi\rangle}{\partial t}\langle \psi| +|\psi\rangle\frac{\partial \langle\psi|}{\partial t}\right) =H|\psi\rangle\langle\psi|-|\psi\rangle\langle\psi|H$$
Collecting bra and ket terms,
$$\left(i\hbar\frac{\partial |\psi\rangle}{\partial t}-H|\psi\rangle\right)\langle\psi|=|\psi\rangle\left(-i\hbar\frac{\partial\langle\psi|}{\partial t}-\langle\psi|H\right)$$
My hand-wavy explanation is to put everything on one side and zero on the other and claim that the bra and ket coefficients have to go to zero separately. If I just had any two independent vector spaces, this explanation would suffice, but is this still true when the bra space and ket space are dual to one another? Apologies if I’m missing something obvious.

 

[A. Neumaier]

One cannot derive it without aligning the phases. One just shows that given a wave function satisfying the SE, the corresponding density operator satisfies the von Neumann equation. Assuming the initial-value problem to be uniquely solvable (which amounts to adding smoothness requirements) it is the only solution with a density operator of rank 1. But of course, the wave funcgtion's phase is not yet determined. Again one needs to make some smoothness assumption to get that.

 

[A. Neumaier]

$$\left(i\hbar\frac{\partial |\psi\rangle}{\partial t}-H|\psi\rangle\right)\langle\psi|=|\psi\rangle\left(-i\hbar\frac{\partial\langle\psi|}{\partial t}-\langle\psi|H\right)$$
Apologies if I’m missing something obvious.

What you can conclude from your equation is that $i\hbar\frac{\partial |\psi\rangle}{\partial t}-H|\psi\rangle$ is parallel to $|\psi\rangle$. Call the (complex) factor $\alpha(t)$. Write down the resulting equation and take the adjoint. Insert the two equations obtained into your equation to get a necessary and sufficient condition on this factor. But it will not be uniquely determined.

 

[TeethWhitener]

Thanks for your response! I’ll work through your suggestion when I get a moment.

Also, it’s clear from my original post (and @A. Neumaier ’s response), but I wanted to point out that I edited the summary to correctly reflect my question.

 

[vanhees71]

The upshot is that the evolution of the state ket of a pure state is not unique since always a maybe timedependent phase is indetermined. That doesn't affect the uniqueness of the state, because the state is not represented uniquely by the state-ket, but only the state-ket up to this indetermined phase factor.

You can formalize this by either saying that a pure state is represented by a (unit) ray in Hilbert space, i.e., a state ket modulo a phase factor (which can depend on time) or, and that's imho a bit simpler in practice, you define states generally to be presented by statistical operators, and the special case of a pure state is that it is a projector, i.e., fulfilling the equation $\hat{\rho}^2=\hat{\rho}$ in addition to the general constraints ($\hat{\rho}$ self-adjoint, $\mathrm{Tr} \hat{\rho}=1$, and positive semidefiniteness).