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Momentum eigenfunction question

  1. Feb 1, 2010 #1
    If one measures a system for momentum, it will collapse into a delta in momentum space. (right?) How would that look in position space, and how would the function evolve in time, if no subsequent measurements are made, in both position and momentum space? A mathematical answer would be appreciated.
     
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  3. Feb 1, 2010 #2

    Matterwave

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    This question is actually a bit subtle. In position space, the momentum eigenfunctions, if you just do the math, are just sines and cosines.

    The problem is that sines and cosines are non-normalizable over all space, so they can't actually represent real particles.

    If you measure the momentum of a particle, you actually don't collapse the momentum down to one single value, there will always be uncertainties. In that sense, the real description of a particle should always be wave-packets (which are sharply peaked around some momentum value p, but not delta functions).
     
  4. Feb 1, 2010 #3

    SpectraCat

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    Isn't this phrased a bit ambiguously? It is true from an experimental point of view that there will always be measurement uncertainty, but this is independent from the HUP uncertainty that you seem to be implying. So, it is in principle possible to make a perfectly precise *measurement* of the momentum of the particle, however that will be a value from inside a distribution, the width of which is controlled by the HUP, right?
     
  5. Feb 2, 2010 #4

    Matterwave

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    I don't believe HUP would allow you to measure a perfectly precise momentum, as that would imply, somehow, an infinite uncertainty in position. If there is an infinite uncertainty in the position...what are you measuring?

    Dropping the HUP for a moment, a sine or cosine wave cannot represent a real particle because they are not normalizable. In more fancy terms, the eigenstate of momentum (and position, actually), are not within the Hilbert space.

    We usually expand our definition of Hilbert space to include these eigenstates...but that doesn't mean that they can actually represent a real particle.

    At least, this is my understanding of it. I may be wrong. I'm not a QM professor after all. :P
     
  6. Feb 2, 2010 #5
    Normalization/measurement considerations aside, and ignoring subtle points, the short answer is

    psi(t,x) = exp(i*(p*x-E*t))

    where E = p^2/2m. This is non-relativistic of course. Also assuming no varying potential energy, so it is a free state.

    Torquil
     
  7. Feb 2, 2010 #6

    SpectraCat

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    This has been discussed here recently, but I can't find the thread just now, so I will summarize the content.

    This is one of those subtle issues, but it is real and not just semantic. Say you have a free particle represented by some wavepacket with a finite width [tex]\Delta p[/tex]. That momentum distribution represents all of the possible momentum eigenstates that could be measured. However, it is a fundamental postulate of QM that when the measurement is conducted, a single eigenstate is projected out of the wavefunction, [tex]\left\langle k | \psi\right\rangle[/tex], where [tex]k=2\pi/\lambda[/tex] indexes a free particle momentum eigenstate. This measurement can be conducted with arbitrary precision, independent of the width of the starting wavepacket in momentum space. Where the HUP comes in is in determining the probability of measuring that state, which is given by [tex]\left|\left\langle k | \psi\right\rangle\right|^{2}[/tex]. So, if you conducted the same measurement again from another identical wavepacket, there is an overwhelming probability that you would measure a different eigenstate.

    Thus the HUP tells us only about the probability of a single measurement, or what to expect from a series of measurements with identical starting conditions. It says nothing about the value, or the measurement uncertainty, of a particular measurement.

    That is a fair question, but note that it refers to the situation *after* the measurement has occurred. So, it would have no relevance in the case where the particle was detected by splatting it onto a detector screen, since that event "destroyed" the wavefunction of the particle, or at least perturbed it severely into some completely different space. On the other hand, if the particle momentum was measured in a scattering experiment, then it does make sense to talk about the wavefunction after the measurement. However in that case, you defer the actual measurement to the probe particle, which again can be measured with arbitrary precision. The momentum of the target particle after the scattering event will not be known precisely in this case, but will be represented by a distribution that can be determined from a set of such measurements to reveal the momentum distribution of the probe particle, and we are in the same situation as above for the direct measurement.

    I hope this is clear ...
     
  8. Feb 2, 2010 #7
    Given that the measuring apparatus is localised, the uncertainty principle guarantees there is some uncertainty in the momentum of the measuring apparatus. There is then a minimum uncertainty in the result of any measurement of momentum performed by the localised piece of apparatus.
    Only measuring apparatus of infinite extent could give a perfect measurement of momentum.
     
  9. Feb 2, 2010 #8

    SpectraCat

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    A fair point with regard to the gedanken experiment with "perfect" instrumentation, and it clearly places a lower bound on the meaning of "arbitrary" with respect to the measurement precision. In essence, I guess it means that even detections where particles are "splatted" on a detector screen have an aspect of the scattering I described above to them, with a finite width of the detector momentum. However, since a detector screen is a macroscopic object, its momentum, and any associated uncertainty, can typically be ignored in such considerations.:wink:
     
  10. Feb 2, 2010 #9
    Thanks guys, those answers help alot! So what I understand is that in PRACTICE, the wave function will have some uncertainty built into it, so the resulting function in position space will still be a superpostion of the momentum eigenfunctions. What I'm really struggling with is how multiple measurements on observables with continuous spectra are represented. Could someone give an example of a momentum measurement, then a position measurement, on something really simple, like a one-d infinite potential well?
     
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