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Compatibility of qunatum measurements

  1. Dec 16, 2005 #1
    The question is "two operators A,B do not commute". Is it true that
    ([A,B]psi) does not equal zero for any state psi?"

    I'm not sure if there are any cases in which a state psi could produce this result (except for the obvious one where psi = 0 and there is no particle).

    Thanks
     
  2. jcsd
  3. Dec 16, 2005 #2

    Physics Monkey

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    Here is a hint: think about the commutator [tex] [x^2 , p] [/tex]. Can you find a state that gives zero when you hit it with the commutator even though commutator isn't identically zero?
     
  4. Dec 17, 2005 #3
    I've tried applying the commutator to an arbitrary psi and i get i*h bar * 2x* psi. The only case where I can think of this equalling zero is when psi = 0, which is the obvious case of having no particle.
     
  5. Dec 17, 2005 #4

    Physics Monkey

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    You can't think of any state that gives zero when you hit it with the position operator? Hint: think about position eigenstates.
     
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