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B Thought experiment violates Heisenberg? (of course not)

  1. Apr 11, 2016 #1
    Suppose we have an elementary double-slit experiment: A laser fires individual photons through a double slit at a detectionscreen made of atoms.

    As we fire photons, an interference pattern emerges, exposing the momentum of the photons (the frequency of the laser).

    So, we have registration of all the individual positions of the photons that impacted the screen, along with their momenta. Doesn't this violate the Heisenberg uncertainty relation? How can there emerge an interferencepattern anyway in that case?

    One could argue that a large number of measurements approaches classical mechanics. But the atoms that make up the detectionscreen register the impact (momenta) of individual photons also. So this seems to violate the HUR too!

    I am not seeing the solution to this clearly.
     
  2. jcsd
  3. Apr 11, 2016 #2

    Nugatory

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    Have you learned the position and the momentum of the particle before it interacted with the slit, after it interacted with the slit, or do you know the value of one before that interaction and the value of the other after that interaction?

    And are you paying attention to the exact time that each spot appeared on your detector? How precisely are you nailing down the exact position of the particle when it interacts with the detector?
     
  4. Apr 11, 2016 #3
    Hello Nugatory. I take your questions not for retorical. In my example, there is nothing known about the future positions of the photons on the detectorscreen (a screen made of for instance: gold foil?) until they hit the detector. The momenta of the photons however can be deduced right at the beginning from the frequency of the laser. The exact time the photons hit the screen I don't take in consideration. (and I am curious why you ask the latter, for I have always wondered if the positions of interfering photons depend on time :smile: )
     
    Last edited: Apr 11, 2016
  5. Apr 11, 2016 #4

    Nugatory

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    That gives you the momentum before the interaction with the slit. The interaction with the slit helps narrow down the position, but it also changes the momentum.

    If you're not paying attention to when the spot appears at the detector you don't have much of a position measurement - all you have is "dunno where it is, but a while ago it was somewhere between the slit and the spot".

    You might want to get hold of Ghirardi's "Sneaking a look at God's cards", which has a thorough but layman-friendly discussion of what's going on as the particle interacts with the slits.

    (You also want to be a bit cautious about doing this thought experiment with photons instead of massive particles - it hasn't mattered yet, but the difficulties of assigning a position to a photon in flight will start to get in the way as you dig deeper into the problem).
     
  6. Apr 11, 2016 #5
    The uncertainty in the momentum is almost at right angles to the initial momentum as is the uncertainty as to which slit it went through.
     
  7. Apr 11, 2016 #6
    Well, you could just take a laser and fire it directly at the detector (no slits). While the photon is "in flight" you know its momentum quite precisely (from the label on the side of the laser) but its position is not even defined (no position operator for photon). When it hits the detector, you know its position quite precisely; but the photon no longer exists, so its momentum is no longer defined.

    Ok, you say, but just before it hit (a picosecond, let's say) I know its momentum; plus I can "retrodict" its position, on the line between the laser and the detection point, a light-picosecond in front of the detector. No: there's no position operator. You're not allowed to suppose this apparently logical conclusion.

    Ok, let's do it with an electron instead. Now I'm allowed to retrodict the position, because an electron in flight does have a position. But I have to know the exact time it hit. The extra experimental apparatus to determine that, however it works, will inevitably introduce extra uncertainty into the momentum and/or position. Plus, the electron source can't be as accurate as a laser; there will be a considerably greater uncertainty in the electron's momentum (and speed) than with the photon.

    Without doing detailed calculations we can be quite certain Heisenberg will not be violated by our experiment.

    Of course I could be misunderstanding something.
     
  8. Apr 12, 2016 #7

    atyy

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    Part of the resolution is that the momentum you measure from the interference pattern is the momentum of the particle just after the slit.

    The position you measure is the position at the screen. You have not measured the position of the particle just after the slit. In other words, you have not measured canonically conjugate position and momentum, which are the position and momentum in the uncertainty principle.

    In very special cases (when one has some knowledge of the state), one can make a simultaneous "measurement" of position and momentum. However, what is forbidden is to do so for an arbitrary unknown state.
     
    Last edited: Apr 12, 2016
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