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

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?

#### Attachments

• image.jpg
18.6 KB · Views: 381

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

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

dextercioby
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.

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.

, 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?

dextercioby
Homework Helper
Yes, if H=H(Q) such as e.g. H = Q^2, then [Q^2,Q]=0 on all vectors on which the commutator makes sense.

Yes, if H=H(Q) such as e.g. H = Q^2, then [Q^2,Q]=0 on all vectors on which the commutator makes sense.

So whenever I want to use Ehrenfest theorem, the operator H= H(Q), eg Q = momentum. If Q = position then I can't use Ehrenfest because H is not f(x). Correct?

One more to ask, I read from somewhere says: when one operator commute another, that means they have no eigenfunction(wavefunction together) . So we can't measure both at the same time.
Why they didn't commute mean we can't measure them at the same time?