# Ehrenfest theorem, is there any condition for the operator Q?

Outrageous
For commutator, HQ-QH = 0 .
But for this case as shown below, complex ψQHψ - HcomplexψQψ= 0?
If the operator Q is in term of (∂/∂t) and (∂/∂x) ,then the HQ-QH may not be zero.
Is there any restriction for Q operator?

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Homework Helper
You question is not quite inteligible. You need to specify who the symbols stand for. Who's H, Q ?

Outrageous
You question is not quite inteligible. You need to specify who the symbols stand for. Who's H, Q ?

H is Hamiltonian , Q is an operator

Homework Helper
Well, if Q strongly commutes with H, this is enough restriction as it stands. Assuming psi(x) is in an invariant dense everywhere domain, H=f(Q) is a trivial solution of [Q,H]=0.

Outrageous
For commutator, HQ-QH = 0 .
But for this case as shown below, complex ψQHψ - HcomplexψQψ= 0?
If the operator Q is in term of (∂/∂t) and (∂/∂x) ,then the HQ-QH may not be zero.
Is there any restriction for Q operator?

I think I should ask ,why the H and Q commute each other, how do we know they will commute in general?
The thumbnail I posted is one part for the derivation of ehrenfest's theorem.

Outrageous
, H=f(Q) is a trivial solution of [Q,H]=0.

What do you mean? H is a function of Q ? Then they will commute?