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

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  • #1
Outrageous
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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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  • #2
dextercioby
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You question is not quite inteligible. You need to specify who the symbols stand for. Who's H, Q ?
 
  • #3
Outrageous
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You question is not quite inteligible. You need to specify who the symbols stand for. Who's H, Q ?

Thanks for replying~
H is Hamiltonian , Q is an operator
 
  • #4
dextercioby
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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.
 
  • #5
Outrageous
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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.
 
  • #6
Outrageous
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, 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?
 
  • #7
dextercioby
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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.
 
  • #8
Outrageous
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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?
 

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