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Momentum transferred even though no collapse?

  1. May 9, 2009 #1

    jaketodd

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    For example, in the double slit experiment, is momentum transferred to say, the wall in between the slits even when the particle manifests on the wall behind the slits? In other words: Is momentum transferred to places where the particle does not manifest?

    Thanks!

    Jake
     
  2. jcsd
  3. May 11, 2009 #2
    Sure. The momentum conservation requires that.
     
  4. May 11, 2009 #3
    That's a good question. Here's my educated guess of the answer:

    Let me make sure I understand your question. You're talking about a single particle approaching a wall with two slits, where some of its wavefunction passes through both slits and reaches the back wall and produces an interference pattern. Here's a diagram:

    ~~~particle~~~-------A----------B

    (I think one dimension's enough to think things through.) "A" is the wall where the slits are, and "B" is the back wall. The particle approaches from the left. It has a certain probability of passing through wall A, but none of going beyond wall B. You're asking if the particle can transfer momentum to both "A" and "B" at the same time.

    I don't think so.

    There are two things which we could measure here: position and momentum. If we know the particle's either at A or B, that would be a position measurement.

    If we measure the momentum transferred to some wall, that would be a momentum measurement of the particle. It would, in practice, also be a position measurement. For example, if the particle is an electron, and the wall is negatively charged, then the existence of a significantly large change in momentum of the wall implies a force between the electron and the wall, which would imply that they were close together. Therefore, if we assume the interaction between the wall and the particle is short-range, measuring a change in momentum of one wall implies that the particle's position is measured, too. For example, if we know wall A was impacted by the particle, and we know the wall's force is short-range, then we know the particle was near wall A.

    To summarize, the particle approaches from the left as a broad wavefunction. If it transfers momentum to wall A, it must be localized near A, and then it can't be close enough to wall B to transfer momentum to that wall as well. And vice versa.
     
  5. May 11, 2009 #4

    jaketodd

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    Thanks for your replies, guys. But the replies are contradictory to each other. Does anyone (including you guys) want to add their opinion, perhaps with a solid reference to back it up? Thanks
     
  6. May 11, 2009 #5
    The momentum conservation is not optional. It is required. That fact alone all by itself tells you that the first answer you got from 'ueit' is correct. But I will elaborate a little more. The collapse requires a measurement that tells us what state the photon is (or it was more precisely.) In this case there are two relevant states for this experiment: it either went through one slit, or the other. Can you tell which slit the photon went through by measuring the transverse momentum it deposited on the wall where the slits are located? The answer is no. Even if you could measure such a tiny momentum transfer on a macroscopic object, it still would not tell tell you which slit the photon went through. That's why you get the interference pattern. If the answer had been yes instead, then the wave function would have collapsed, and the interference pattern you had expected to observe would have been been destroyed.

    In short, momentum can be transferred without collapsing the wave function. Double slit experiment is a good example, but by no means the only one.
     
  7. May 11, 2009 #6
    I'm confused. I thought we were talking about momentum applied perpendicular to the walls, not transversely. Jaketodd, could you clarify your example?
     
  8. May 12, 2009 #7
    I don't think it's helpful to focus on the double slit aspect of this problem. All the essential features of this question are present in the single slit case. You have light passing through the first wall and making a mark on the second wall. If the slit is verry narrow, the impact on the target can be almost anywhere. From the location of the mark we can see how much the momentum of the light was changed when it passed through the slit. So exactly when was the momentum transferred?

    The second wall may be light years away from the first wall. Was the momentum transferred when the light passed through the slit, or not until the spot appeared on the target?

    There are two ways of describing what happens. In Scenario A, we can say that until the moment of detection, both the phton momentum and the recoil momentum of the wall are indeterminate; but once either one of them is measured, the second one instantaneously takes on the complementary value in accordance with Conservation of Momentum. Hence, the wave collapse. The other picutre is Scenario B: that the wall recoils at the instant the photon passes through, and the photon's path is thereafter determined until its moment of impact. But thanks to the Uncertainty Principle, we can never measure the wall's recoil accurately enough to tell us where the photon is going to land. Therefore Scenario A and Scenario B are experimentally undistinguishable.
     
  9. May 13, 2009 #8
    I think that the situation is analogous with the case of an asteroid approaching Jupiter from outside our galaxy. When does the momentum get transferred? There is no definite time point for momentum transfer because gravity acts at an infinite range. The correct way to approach the problem is to study the asteroid's trajectory in the common field of all objects (asteroid itself, all objects in our galaxy and even some other galaxies). Of course one can simplify the problem to a certain degree by accepting an increased error in calculations.

    The same is true for EM interaction. In order to truly understand it one must consider the resultant field produced by all particles that take part into the experiment. The difficulty here is the fact that, unlike gravity, QM does not give us a well-defined law of motion for an individual particle therefore one can only guess what is significant and what is not for the particle's motion. Unfortunately, the most common "guess" is that particles behave like Newtonian billiard balls, moving in straight lines between direct collisions. Such a view was already known to be false even before QM was discovered.
     
  10. May 13, 2009 #9
    I have to totally disagree. It's true that I used the word "instantaneous" referring to Scenario B but I didn't mean it in the sense of billiard ball collisions but in the sense of your Jupiter analogy, where at every instant in the trajectory momentum is being transferred from one to another. In contrast with Scenario A, where both the wall and the particle have indeterminate momentum until one of them is measured, at which point the other "instantaneously" takes on the opposite momentum.
     
  11. May 13, 2009 #10

    jaketodd

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    I'm talking about the classic double slit experiment. The waves oscillate in a curved (wavefront) plane perpendicular to the propagation direction of the wavefront and the face of the curved plane faces the direction of propagation.
     
  12. May 14, 2009 #11
    You said:

    "The other picture is Scenario B: that the wall recoils at the instant the photon passes through, and the photon's path is thereafter determined until its moment of impact."

    This is clearly a case of billiard-ball like collision, incompatible with the gravitational analogy, but if you didn't mean it that way it's OK.

    Scenario A can be ruled out experimentally by performing a diffraction experiment at a single edge of an object. The object will acquire a momentum in a single direction and I doubt that it will start moving only after the photons are detected (say a year after, on the surface of a distant comet).
     
    Last edited: May 14, 2009
  13. May 14, 2009 #12

    Just to refresh: My "scenario A" was where the momentum of the wall and the photon were both indeterminate until either one was measured. You say it can be ruled out experimentally but I don't think it's so easy. If my history is correct these questions were the topics of debate between Bohr and Einstein. The uncertainty principle frustrates most attempts to easily settle this question.
     
  14. May 14, 2009 #13
    I am not saying that you can beat the HUP. I am saying that you can perform an experiment where the transferred momentum is visible before the particles are detected.
     
  15. May 14, 2009 #14

    jaketodd

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    And you're sure momentum has been detected on an object the particle did not manifest on? Will you provide a reference?
     
  16. May 14, 2009 #15
    Yes you can. But once you measure the recoil momentum of the wall, the particle (photon or electron) momentum is also determined.

    The paradox lies in the fact that until you measured either one of them, they could have been anything. Once either of them is measured, they are both determined. The "realistic" model explains this by saying the exact momentums existed from the moment of interaction and were merely unknown until measured. That was my Scenario B. Scenario A says they were both truly indeterminate until one of them was measured. The fact is it is extremely difficult to distinguish experimentally between Scenario A and Scenario B.
     
  17. May 14, 2009 #16
    In the decoherence approach - why aren't photon wavefunctions collapsing upon hitting the wall behind the slits and thus destroy the interference pattern?
     
  18. May 14, 2009 #17

    jaketodd

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    You mean the wall in between the slits?
     
  19. May 14, 2009 #18
    I'm going to expand on my previous comments just a little. There is no controversy about the fact that momentum is transferred when a particle diffracts through a slit. Since the trajectory is changed, there must be recoil. There is also no controversy about the fact that until one or the other is measured (the particle deflection or the barrier recoil) that both are unknown, and the result of the measurement could be almost anything. Finally, there is no controversy over the fact that once one of them is measured, the other is also known as a consequence of Conservation of Momentum.

    The obvious explanation for this state of affairs is that an interaction takes place at the slit, and the results of that interaction are simply unknown until detection occurs. The whole hulaballo over Quantum Mechanics is the fact that certain things about the theory compel us to reject this obvious interpretation and say that even after the particle passes the slit, there is no definite momentum transfer until later, sometimes very much later.

    As much as the theory seems to demand this non-local type of outcome, it remains very difficult indeed to actually set up an experiment to verify this conclusion. In fact this is what what the Bell's Inequality type of experiments claim to do.
     
  20. May 14, 2009 #19
    Because the wavefunction of the wall is very broad in momentum space. I explained this in detail in some other thread here some time ago. If you did an interference experiment with some mirror at one of the slits, and the width of the mirror's wavefunction in momentum space were smaller than the recoil, then you would have perfect "which way" information. The wavefunction of the rest of the universe corresponding to the photon taking one path would be orthogonal to what it would be if the photon took the other path.

    But note that for the wavefunction of the mirror's total momentum to be sharper than the photon momentum requires the center of mass uncertainty to be larger than the wavelength of the photon. So, it is quite obvious why you don't have an interference pattern in that case.

    The mistake people make is thus to pretend that quantum mechanics does not apply to macroscopic objects like mirrors, walls etc. etc. You can't on the one hand locate the surface of the mirror to a fraction of the wavelength of the photon and make sure the momentum of the mirror has no inherent quantum uncertainty that is larger than the photon momentum.

    In fact, since the mirror is a large object with strong interactions with the envoironment, its center of mass wavefunction will always almost "collapse" in the position basis. The density matrix in the position basis will be of the form rho(x-y) = exp[-(x-y)^2/lambda^2] where lambda is of the order of the thermal de-Broglie wavelength. This then means that in momentum space the wavefunction of the mirror will have a width of the order of sqrt[M k T].


    So, if the photon reflects off the mirror, its wavefunction will shift but the overlap with what the wavefunction would have been had the photon not bounced off the mirror, would be very close to 1. So, the visibility of the fringes will not be affected
     
  21. May 15, 2009 #20
    In the double-slit experiment or diffraction at a single slit, the interference/diffraction patterns are symmetrical, therefore the average momentum transferred to the wall is zero. For the diffraction at an edge the average momentum cannot be zero because all particles go (more or less) in a single direction (say left). Momentum conservation requires the wall to acquire a similar momentum to the right. So, even if the noise is big enough to not be able to infer where each particle goes (so that the wave function does not collapse), the average non-zero signal should be measurable.
     
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