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Wavefunction collapse and measurement

  1. May 5, 2010 #1
    So, rookie question, I know, but I'm having a little trouble with the idea of wavefunction collapse as it pertains to stationary states:

    If a measurement of energy collapses a wavefunction into an energy eigenstate, it stays there forever (unless perturbed). But my impression is that although position measurements collapse a wavefunction, the wavefunction will begin to evolve rather quickly after the measurement.

    Does this have something to do with the fact that x doesn't commute with H? What are some other measurements that will yield an evolving wavefunction?

    Say I have some operator associated with an observable that has two values: 'happy' and 'sad.' What else do I need to know about this situation to predict whether a measurement of 'happy' will leave the particle in that state for all time?

  2. jcsd
  3. May 8, 2010 #2


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    Yes, it has to do with whether or not the operator A corresponding to the measurement you are doing commutes with the Hamiltonian H. H governs the time evolution of the wave function. A measurement of A will collapse the system into an eigenstate of A. If A commutes with H, the system will remain in that eigenstate of A as it evolves in time as determined by H. If A and H don't commute, then it will be in an eigenstate of A immediately after the measurement, but will then begin evolving into some linear combination of all the eigenstates of A as time passes.
  4. May 9, 2010 #3
    Okay, makes sense. Thanks for your time phyzguy, glad to hear I wasn't too far from home there!
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