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Matterwave

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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).

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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?

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).

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Matterwave

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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

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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

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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.I don't believe HUP would allow you to measure a perfectly precise momentum, as that would imply, somehow, an infinite uncertainty in position.

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

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 theIf there is an infinite uncertainty in the position...what are you measuring?

I hope this is clear ...

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Only measuring apparatus of infinite extent could give a perfect measurement of momentum.

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SpectraCat

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A fair point with regard to the

Only measuring apparatus of infinite extent could give a perfect measurement of momentum.

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