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

  1. Oct 16, 2013 #1
    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. jcsd
  3. Oct 16, 2013 #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 ?
     
  4. Oct 16, 2013 #3
    Thanks for replying~
    H is Hamiltonian , Q is an operator
     
  5. Oct 16, 2013 #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.
     
  6. Oct 17, 2013 #5
    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.
     
  7. Oct 17, 2013 #6
    What do you mean? H is a function of Q ? Then they will commute?
     
  8. Oct 17, 2013 #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.
     
  9. Oct 19, 2013 #8
    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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