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Is the moon sorry, the mirror there when no one looks?

  1. Feb 19, 2006 #1

    nrqed

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    Considering (again!) which way type of experiments (the interference type of set ups when a photon may go through two different routes and then one recombines the paths and checks for the presence of absence of interference (as used in quantum erasers, etc)).

    Usually, people say that if there is no way to detect the path taken by the photon, there will be interference.
    But I am bothered by the mirrors in the experiment. Don't they act as some kind of measurement device? (even if we do not use them to detect the path of the photons)??
    I mean, doesn't the mirror (one of them at least) ''knows'' whether the photon was reflected off from it?? I know this sounds silly, but it seems that for the photon to be reflected, it must interact with the atoms in the mirror, and that would seem to represent a type of microscopic measurement. Even if the mirror is perfect.

    Pushing the argument further, what distinguished what we usually consider a measurement from what the mirror does by reflecting the photon (or not reflecting it...I am not sure what to think!!)?
    I mean, at waht point do we consider the interaction of the photon with a system a ''measurement'' which will lead to the lost of the interference pattern? There is something maybe about the irreversibility of the interaction with a measuring instrument that makes this qualititatively different from the interaction with the mirror...It is puzzling to me (but may be obvious to other people here).

    Also, usually the distinction between interference vs non interference seems to be a yes/no issue... a ''digital'' question.
    But can there be situations where the lost of interference is only partial? maybe 80% of the photons show no interference effects whereas the other 20% do..... This would be possible if the intreaction with a measuring device would lead to a partial lost of information...

    I know that this is very vague...But I hope someone(s) will help clarify or focus the discussion.
    Thanks


    Pat
     
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  3. Feb 19, 2006 #2

    vanesch

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    For the mirror to "know" whether the photon reflected or not, it must be in a different final state after reflection than when no reflection occured, and this is not the case. If there is such a thing as sufficient momentum transfer (for instance) that the mirror is in a macroscopically different state (hence in a quantum - mechanically orthogonal state), then the interference will disappear (classically explained by the total uncertainty of the position of the mirror over one wavelength). But if there is so little transfer as to have the quantum state of the mirror before and after the reflection to be essentially in total overlap, then this will mostly not influence the interference pattern.

    It's one of the problems (I find) with copenhagen kind of views. If you give a quantum state to the mirror, you could see it as follows:

    (|photon> + |nophoton>)|mirrorbefore>

    during interaction

    |photon>|bendmirror>+ |nophoton>|mirrorbefore>

    after reflection

    |reflectedphoton>|mirrorbefore> + |nophoton>|mirrorbefore>
    = (|reflectedphoton> + |nophoton>) |mirrorbefore>

    (as a naive presentation of course: the "time during interaction" is to be taken symbolically).

    The point is that if the quantum state of the mirror after reflection is the same (or almost the same) as the quantum state before (or without photon), then the mirror doesn't KNOW about whether it reflected the photon: it forgot about it perfectly.

    Well, the concept of measurement IS confusing of course in Copenhagen (that's why I don't like it). But the essential difference is that after a measurement, "things" are left in a different state according to the outcome of measurement (the apparatus, the environment, the guy looking ...).

    Of course! That, you have when the (quantum) state of the mirror after interaction is not orthogonal, but is not identical either with the state without interaction. If it is orthogonal, then you have perfect entanglement and hence no interference. If it is identical, you have "full" interference. If it is in between (some angle, but not orthogonal), you have diminished the IP.

    cheers,
    patrick.
     
  4. Feb 20, 2006 #3

    nrqed

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    Thanks for replying...I know it's very basic but is the kind of thing that I never took the time to really think about.

    Part of my problem is that I am not sure about the boundary between "macroscopic" and "microscopic" here.
    Presumably, if the photon was reflected by the mirror, it has (presumably) transferred momentum to at least one atom. Even if we do not observe the atom. Now, maybe there is some overlap between the new wavefunction of the atom and its old wavefunction, and that may be the whole point (that, if there is some overlap, there there is some "chance" that the mirror does not "know" if the photon was there). But then it would seem that interference could never be completely "saved". There would always be partial destruction of the interference pattern.

    I am probably completely in the left field...


    But the amound of momentum transfer is dictated by conservation of momentum, no?

    I know I will be slapping my head once I finally understand :rolleyes:
    Meanwhile, thansk for your patience.

    Pat
     
  5. Feb 20, 2006 #4

    vanesch

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    That's why I personally prefer not to make such a distinction, and consider everything on the same level (MWI view). But it shouldn't matter too much.

    No, I don't think so (at least if it was to be reflected coherently). It should be reflected by the "entire sea of electrons" of the mirror who are to act as one entity.

    That's indeed what would happen if there were an interaction with a single atom, and the reason why that cannot be the case!
     
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