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Time-energy uncertainty (question after reading FAQ!)

  1. Mar 29, 2013 #1
    Hi,
    The FAQ is a good answer, but it leaves me wondering about something.

    From what I read, the jist is that a short enough measurement sees no difference between an energy eigenstate and a quantum superposition which mostly comprises of that energy eigenstate but also has a few other components from other eigenstates in there.

    But when we make a measurement and get, say an off-centre energy E + dE, what does that mean? Why is the molecule not choosing between all the possible discrete energies based on the probability factors c1, c2..., and returning one of these? The fact that we actually measure an energy very close to but not equal to the energy eigenvalue implies (to me) that somehow we're averaging the energies of all the states with the weighting factors. I don't understand why an average is being taken, if we make an energy measurement then the molecule is forced into an energy eigenstate, and we get an exact energy.

    The uncertainty about the initial state is then removed, and our measurement is an energy distribution, but the distribution is discrete amongst the energy eigenstates, not just a smearing around the state with the largest weighting.

    Thanks for any help
    Mike
     
    Last edited: Mar 29, 2013
  2. jcsd
  3. Mar 29, 2013 #2

    mfb

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    Well, real measurements are never "pure" energy measurements, or need infinite time for that.

    With a finite measurement time, the result can be "anything". You can express this as superposition of energy eigenstates, but that is just a mathematical calculation - you can express every state as such a superposition.
     
  4. Mar 29, 2013 #3
    But we are extracting a real amount of energy out. If that energy does not correspond to any of the energy eigenstates, then what does it correspond to? Does a superposition of these states still have an energy?
     
  5. Mar 29, 2013 #4

    mfb

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    Particles don't have to be in energy eigenstates, where is the problem?
    Sure
     
  6. Mar 29, 2013 #5
    So when we make a measurement of energy, the result doesn't have to be an energy eigenstate because the final state isn't necessarily in that state? Can you see why I'm a bit confused by this idea, it seems to contradict the idea that an energy measurement fixes the state to one of definite energy.

    Would a more accurate expression of that statement ^^ be that a finite-duration energy measurement fixes the state to one with roughly the same energy as an energy eigenstate? And the longer you measure it, the more of the other energy eigenstates go out of phase and can be discounted from the measurement?
     
  7. Mar 29, 2013 #6

    mfb

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    As I said, there is no perfect energy measurement. There are very good approximations, however.
    In the limit of a perfect energy measurement.
    Right
     
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