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Time-varying expectation values

  1. Mar 26, 2009 #1
    Hi all.

    I have a question which arose from the answer of a homework problem. A particle is in the state given by

    [tex]
    \left| \psi \right\rangle = \frac{1}{{\sqrt 3 }}\left[ {\left| \psi \right\rangle _1 + \left| \psi \right\rangle _2 + \left| \psi \right\rangle _3 } \right],
    [/tex]

    where [itex]{\left| \psi \right\rangle }_i[/itex] is a stationary state of the Hamiltonian. Finding the expectation value of an operator Q in this state, it turns out that <Q> is time-dependent (in fact it oscillates). I was wondering how it is possible to have a time-dependent expectation value? If there was a time-dependent perturbation, then I would be convinced, since the state [itex]{\left| \psi \right\rangle }[/itex] would change over time, but in this case [itex]{\left| \psi \right\rangle }[/itex] doesn't change.
     
  2. jcsd
  3. Mar 26, 2009 #2
    the three states has different energies (eigenvalue of the hamiltonian) and thus they evolve in a different way.
     
  4. Mar 26, 2009 #3
    When you have just a single state, calculating the expectation value of any operator requires you to multiply the wavefunction by it's complex conjugate. For that reason, the time dependency (the complex exponent) cancels out.
    [tex]e^{ix} (e^{ix})^* = e^{ix}e^{-ix} = 1[/tex]

    If however you have multiple states like now, the complex exponents do not cancel out. Actually, a few terms cancel out like usual, but you are left with cross terms that do not, since the states have different energies.
    [tex]e^{ix_1} (e^{ix_2})^* = e^{ix_1}e^{-ix_2} \neq 1[/tex] (in general)
     
  5. Mar 26, 2009 #4
    But what does this mean physically? I'm having a hard time interpreting how this can physically be possible without any external (time-varying) perturbation.
     
  6. Mar 26, 2009 #5
    think of every eigenstate as a wave with different frequency, in this way the modulo has different value at different times.
     
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