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mfb

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Vanadium 50

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You can also get diffraction with purely classical water waves.

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bhobba

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I was wondering if the diffraction pattern through a single slit could be explained as a consequence of quantisation of momentum transverse to the main direction of travel.

It is a consequence of the uncertainty relations:

http://cds.cern.ch/record/1024152/files/0703126.pdf

Thanks

Bill

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Wow that is a great QM education article for learners.

Printed.

Thanks for posting.

Printed.

Thanks for posting.

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The author's approach isn't one generally encountered, so thanks for posting it. In the single slit there is obviously a change in the transerve momentum of the particle. How is the momentum conserved? Does the the particle start bouncing around between the sides of the well?It is a consequence of the uncertainty relations:

http://cds.cern.ch/record/1024152/files/0703126.pdf

Thanks

Bill

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bhobba

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The author's approach isn't one generally encountered, so thanks for posting it. In the single slit there is obviously a change in the transerve momentum of the particle. How is the momentum conserved? Does the the particle start bouncing around between the sides of the well?

Momentum is not conserved - it is changed by the observation. It does not bounce around.

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Bill

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I thought momentum was always conserved. Doesnt the confinement introduce the uncertainty into the momentum rather than the observation?Momentum is not conserved - it is changed by the observation. It does not bounce around.

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Bill

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Confinement times momentum = constant = Heisenberg UP.

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bhobba

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I thought momentum was always conserved.

In QM it isn't because of the HUP. By Ehrenfests theorem its conserved on the average. The uncertainty is introduced by the fact going through a slit means just behind the slit it's position is very certain so by the HUP its momentum is unknown. But since KE is conserved this means its velocity is unchanged so it's direction is uncertain. That's why you get a diffraction pattern.

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Bill

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mfb

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Momentum is conserved exactly in every interaction. Your measurement system will provide any apparent change of momentum, and you rarely care about the momentum of the system, so it looks like your momentum changed. But total momentum was always constant.In QM it isn't because of the HUP. By Ehrenfests theorem its conserved on the average. The uncertainty is introduced by the fact going through a slit means just behind the slit it's position is very certain so by the HUP its momentum is unknown. But since KE is conserved this means its velocity is unchanged so it's direction is uncertain. That's why you get a diffraction pattern.

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Bill

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bhobba

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Your measurement system will provide anyapparentchange of momentum, and you rarely care about the momentum of the system, so it looks like your momentum changed. But total momentum was always constant.

Hmmm. Good point.

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Bill

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bhobba

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How you draw that conclusion I can't follow.

Its as I said, and the link I gave said, due to the uncertainty relations the momentum is uncertain - in fact any value is equally likely - see equation 8. The particle is not a closed system so its momentum can change, although as MFB correctly pointed out in the interaction overall momentum must be conserved.

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Bill

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bhobba

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Your point being?

It uses the infinitesimal slit to derive the finite one.

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Bill

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The point being that some changes in momentum are more likely than others.Your point being?

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bhobba

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The point being that some changes in momentum are more likely than others.

For a finite slit sure. It pretty much follows from the HUP. The wider the slit the less likely the momentum will change ie a large change is less likely than a small one. For an infinitesimal one any momentum is equally likely - and that can be used to derive the finite case.

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Bill

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bhobba

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It is the same as passing through an infinitesimal slit at each point and you take the superposition. Seems rather obvious to me. I don't know what more can be said.

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Bill

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We might discuss how the wavefunction becomes modified in the slit.

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bhobba

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We might discuss how the wavefunction becomes modified in the slit.

Because each point acts like an infinitesimal slit - I don't see there is anything more to it.

You are trying to read more into it than there is.

Thanks

Bill

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