Can diffraction be explained by quantisation?

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Main Question or Discussion Point

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. I know that momentum gets quantised on confinement so does the confinement in a slit quantise the momentum so that only certain final directions of the particle are likely?
 

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  • #2
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Not directly, as you don't establish a standing wave in your slit. The concepts are related, of course, and the momentum distribution afterwards is the Fourier transformation of the slit pattern (for small angles, and up to constants).
 
  • #3
Vanadium 50
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You can also get diffraction with purely classical water waves.
 
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  • #4
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Not directly, as you don't establish a standing wave in your slit.
Why is it different to a potential well? Is there just not enough time for a standing wave to establish?
 
  • #5
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  • #6
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Wow that is a great QM education article for learners.

Printed.

Thanks for posting.
 
  • #7
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It is a consequence of the uncertainty relations:
http://cds.cern.ch/record/1024152/files/0703126.pdf

Thanks
Bill
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?
 
  • #8
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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.

Thanks
Bill
 
  • #9
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Momentum is not conserved - it is changed by the observation. It does not bounce around.

Thanks
Bill
I thought momentum was always conserved. Doesnt the confinement introduce the uncertainty into the momentum rather than the observation?
 
  • #10
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Confinement times momentum = constant = Heisenberg UP.
 
  • #11
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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.

Thanks
Bill
 
  • #12
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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.

Thanks
Bill
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.
 
  • #13
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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.
Hmmm. Good point.

Thanks
Bill
 
  • #14
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Are you saying that it's the measuring device rather than the slit that provides the means of compensating for the momentum change?
 
  • #15
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No, there's an interaction of the particles with the material making up the slits and thus the particles that hit this material transfer momentum to it. Of course, for any closed system momentum is conserved, but particles interacting with the material are not a closed system. The usual treatment of course hides the microscopic picture by just imposing semiclassical boundary conditions, which is an effective description of the very complicated microscopic interactions, and it's accurate enough to understand the measured pattern on the screen. In Fraunhofer observation you get (as in the Kirchhoff approximation of diffraction in classical optics) just the Fourier transform of the slits (e.g., a sinc function for the single slit).
 
  • #16
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So there is something akin to bouncing going on and it would seem that some transfers of momentum are more likely than others.
 
  • #17
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So there is something akin to bouncing going on and it would seem that some transfers of momentum are more likely than others.
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.

Thanks
Bill
 
  • #18
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My original question pertained to the diffraction pattern. Equation 8 appears only applicable to the infinitesimally narrow slit. Please see equation 18.
 
  • #19
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My original question pertained to the diffraction pattern. Equation 8 appears only applicable to the infinitesimally narrow slit. Please see equation 18.
Your point being?

It uses the infinitesimal slit to derive the finite one.

Thanks
Bill
 
  • #20
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Your point being?
The point being that some changes in momentum are more likely than others.
 
  • #21
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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.

Thanks
Bill
 
  • #22
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It's the periodicity that's interesting. The HUP doesn't give that. I'm more interested in how the wavefunction becomes modified within the slit.
 
  • #23
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It's the periodicity that's interesting. The HUP doesn't give that. I'm more interested in how the wavefunction becomes modified within the slit.
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.

Thanks
Bill
 
  • #24
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We might discuss how the wavefunction becomes modified in the slit.
 
  • #25
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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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