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Schrodinger equation - stationary states

  1. Aug 17, 2015 #1
    two questions:

    1. besides using Ehrenfests theorem, is there another way of showing that the expectation value of momentum is zero in a stationary state ? (I don't see it when simply applying the definition on the stationary solution)

    2. If we have a state that is a superposition of stationary states and we measure the energy, the wavefunction collapses to a single state, does it mean it will be stationary forever ?
  2. jcsd
  3. Aug 17, 2015 #2


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    1. Honestly, I don't know any more general way of answering this question other than using Ehrenfest theorem, but if you restrict the problem to the one electron system and centrosymmteric potential, ##V(\mathbf{r}) = V(-\mathbf{r})##, you can easily show that the stationary states must have definite parity. Consequently the expectation value of any odd operator such as momentum should vanish.

    2. Yes it will be.
  4. Aug 17, 2015 #3
    Thanks !

    I was using the integral definition of the observation value of momentum, the two exponents with the time dependence cancel out, but I don't see why the remaining integral is zero. I was hoping there was a general way of showing it without using a specific example for the potential.

    [QUOTE="2. Yes it will be.[/QUOTE]

    This is so counter intuitive, everything around us is not stationary, it's hard to imagine something becomes stationary forever simply since we measured its energy...
  5. Aug 17, 2015 #4


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    In idealized sense where the effect of measurement process can be completely omitted (not even perturbatively), the state will really collapse to the one corresponding to the energy level being the outcome of the measurement. Further evolution will only affect the temporal phase of the state but the state itself remains itself.
  6. Aug 17, 2015 #5


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    You may find it less counterintuitive if you think of it in terms of conservation of energy - it would be more surprising if the energy, once measured, did change over time.
  7. Aug 17, 2015 #6

    Thanks ! that does help a lot !
  8. Aug 17, 2015 #7


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    In order to calculate these things, you neglect all other parts of the world which don't influence your system at the moment. This may change in the future. So if the interaction between the system and some part which you didn't include in its approximate description can't be neglected anymore, the states need not remain stationary.

    For example, if you calculate the stationary states of the hydrogen atom, you make the assumption that the interactions with all other particles can be neglected. But if a hydrogen atom in such a stationary state collides with another particle, that assumption isn't really true and the interaction between them may change the state of the atom.

    Another example along a related but slightly different line is spontaneous emission. You probably know that except for the ground state, the excited states of the hydrogen atom decay by emitting photons. This seems to contradict the standard calculation which tells you that the excited states should be stationary. But since we don't include the electromagnetic field in this calculation, we shouldn't be too surprised if it doesn't capture all of the behavior of hydrogen.
    Last edited: Aug 17, 2015
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