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The Mach-Zehnder interferometer experiment

  1. Jun 28, 2008 #1
    hi all, i'm new here :shy:

    i'm italian, so i apologize for my english

    i never studied quantum physics (i only read scientific books) but i have a question that concerns the mach-zehnder interferometer experiment

    i hope you all know how this interferometer works, otherwise http://en.wikipedia.org/wiki/Mach-Zehnder

    let's consider the case of a single foton instead of a light beam: if we don't know the photon's path we know wich detector will detect the photon, viceversa if we know the photon's path the two detectors have a 50% chance to detect the photon.

    a little digression:
    correct me if i'm wrong, but, at a quantum level, reflection and refraction are viewed as scattering events, and a scattering event can cause decoherence, right?

    well, why doesn't decoherence occurs after the photon passes through the first beam splitter (that of course refracts the photon)?
    or, in other words, why is this experiment supposed to work?
     
  2. jcsd
  3. Jun 30, 2008 #2
    help :(
     
  4. Jun 30, 2008 #3

    Mentz114

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    I don't think decoherence is relevant here. Why do you think it should be ?
     
  5. Jun 30, 2008 #4
    thank you for your answer :)

    if we don't know the photon's path, after the final beam splitter occurs an interference effect (because we only have a wave function, not a particle): in particular, a destructive interference on the way of one detector and a constructive interference on the way of the other detector (that's why always the same detector detects the photon).

    when i say "if we know the photon's path the two detectors have a 50% chance to detect the photon" i mean that we measure the position of the photon (causing decoherence) somewhere between the first beam splitter and the final beam splitter: so after the measure the photon acts like a particle and we have no longer interference effects (and that's why in this case both detectors can detect the photon).

    but since rifraction and reflection represent scattering events (and scattering events cause decoherence) i don't understand how it is possible the first scenario to occur
     
  6. Jul 1, 2008 #5

    Cthugha

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    If decoherence occurs, you just made an experiment, which enables you to tell, which path the photon took. Now the question is, whether the recoil of the event of reflection at the beam splitter is enough to give you that information. Usually, this is not the case, as you can easily see in almost any published experiment in optics, which uses mirrors.

    However, there have been experiments using movable mirrors or mirrors connected to springs, which show, that the interference pattern disappears as you changed the state of the mirror. Of course, there is also momentum transfer using rigid mirrors, but in this case the momentum goes to the electrons of the mirror, which transfer it to the lattice ions, which transfer it to the holder of the mirror, which transfer it to the table it is mounted on, which transfers it to other components on this table and so on and so on. The state of the mirror will be unchanged and you won't be able to trace any changes due to the reflection at the beam splitter.
     
  7. Jul 1, 2008 #6
    if momentum transfer occurs, maybe i'm not able to trace this event, but shurely the electrons of the mirror "know" they've been hit by something that changed their momentum. the result is that the information about photon's path becomes available for those electrons, and i don't think that physics cares about wich informations men can and can't reach.

    anyway, stop thinking about the experiment for a moment: is refraction (and reflection) a series of scattering events? i'm still not shure about that.
    but if it is so:
    wave function meets the mirror-->wave function passes through the mirror-->this passing causes refraction-->refraction means scattering-->scattering causes decoherence-->wave function collapses-->we no longer have a wave, we have a particle-->no destructive or constructive interference-->each detector has a 50% chance to detect the photon
     
  8. Jul 1, 2008 #7

    Cthugha

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    Of course it does - well at least it cares about whether the information is available in principle. Only irreversible interactions and state changes cause decoherence. Also you have to take the proper system into account. The photon does not simply interact with a single electron. At the beam splitter the photon gets entangled with the system. Now it makes a huge difference, whether this system is just a single electron or a huge mirror connected to a massive table. In one case the resulting state will be orthogonal to the initial state, in the other case it will be almost the same.

    By the way, this is a standard example of complementarity. You can either have which path information or an interference pattern. The more which path information you have, the less visible the interference pattern will be. There is even an own kind of uncertainty relation called Englert-Greenberger duality relation for these quantities. I suppose, you might be able to watch the interference pattern gradually disappear by changing the mirrors (using smaller ones, not fasten them in a rigid manner, attaching them to a spring and so on).
     
    Last edited: Jul 1, 2008
  9. Jul 1, 2008 #8
    you mean that the more the information is "universe-accessible", the more the interference pattern fades away?
     
  10. Jul 1, 2008 #9

    Cthugha

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    Universe-accessible? Well, I suppose, you could call it that way.

    Maybe this tricky experiment using a Mach-Zehnder interferometer with single photons might be interesting for you:


    Delayed-Choice Test of Quantum Complementarity with Interfering Single Photons
    (Phys. Rev. Lett. 100, 220402 (2008))

    (Vincent Jacques, E Wu, Frédéric Grosshans, François Treussart, Philippe Grangier, Alain Aspect, and Jean-François Roch)

    This paper is also available on ArXiv:

    http://aps.arxiv.org/pdf/0801.0979
     
  11. Jul 1, 2008 #10

    Mentz114

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    Cthugha,
    thanks for the excellent explanation.

    M
     
  12. Jul 1, 2008 #11
    ya i didn't know how to say :) i hope the meaning is clear

    anyway thank you very much...i'll shurely read the article :)
     
  13. Jul 2, 2008 #12
    another question:
    the wave function of the photon takes, in the same time, both paths, right?
    let's consider path number 1: the photon (or its wave function) transfers, for example, 3/4 of its momentum to the mirror.
    let's now consider path number 2: the photon transfers 3/4 of its momentum to the mirror.
    something here is wrong, because the photon can't transfer 6/4 of its momentum.
     
  14. Jul 2, 2008 #13

    Cthugha

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    This "the photon takes both paths"-notion is a bit misleading. Remember that the square of the wave function gives you the probability density. So, if you have two paths with equal probability, the photon taking both paths means, that the photon is in a superposition. It does not trans fer 3/4 of its momentum to both mirrors, but it transfers 3/4 of its momentum and you do not know exactly to which mirror the momentum went. As soon as you are able to find that out in principle, the photon is not in a superposition of paths anymore and it takes just one path.
     
  15. Jul 3, 2008 #14
    yes but again, maybe i can't figure out to which mirror the mometum went, but the electrons of the mirror know if their momentum is changed or not.
    are the electrons of the mirror in a superposition too?
     
  16. Jul 3, 2008 #15

    Cthugha

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    QM works a bit different. Even the momentum of the mirror (although it is a macroscopic object) has no precisely defined state, but only an average value with a certain amount of fluctuations. As long as the amount of momentum transferred is smaller than the amount of the fluctuations, you will find, that the momentum of the electrons is changed, but you can't distinguish, whether this happened due to photon scattering or due to the usual fluctuations.

    Another interesting overview paper on Arxiv might be this one:
    http://arxiv.org/pdf/cond-mat/0405023

    Chapter II about the Debye-Waller factor for macroscopic objects deals with elastic photon scattering and mirrors.
     
  17. Jul 3, 2008 #16
    even using an highly energetic photon?
    (with "fluctuations" you mean the uncertainty due to heisenberg priciple?)
     
  18. Jul 3, 2008 #17

    Cthugha

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    Well, it is really complicated to find mirrors for em radiation with extremely high energy anyway. Remember, that we were talking about elastic scattering, where the energy transfer is assumed to be close to zero anyway.

    And yes, I mean the fluctuations due to uncertainty. In QM you can calculate the variance of any quantity of interest and usually it will be nonzero.

    For example even the photon number in a coherent beam of light has some uncertainty as well. As an even better example thermal light is just a consequence of such fluctuations as the average value of the em field in a thermal state is 0, but the expectation value of the variance is not, which means, that thermal light is some kind of "noise effect" due to the fluctuations being correlated. This result impressed me pretty much, when I first read about it.
     
  19. Jul 3, 2008 #18

    Mentz114

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    I'd like to read about this too - have you got a reference ?

    M
     
  20. Jul 3, 2008 #19

    Cthugha

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    Oh, I am pretty sure I first read it in the standard reference of quantum optics:
    "Optical Coherence and Quantum Optics" by Mandel and Wolf


    Another way to illustrate this phenomenon is by using Wigner functions, which are quasiprobability distributions of the phase space of a state. So you have the amplitude on one axis, phase on the other axis and the probability density in z-direction.

    A gallery can be found here:
    http://www.iqis.org/quantech/wiggalery.html

    For example you can see here that a thermal state looks pretty much like a vacuum state. It is centered at zero amplitude and distributed evenly in all "directions", so the average of the field is zero. The intensity of the light is however proportional to the square of the amplitude and the expectation value of the square is of course not zero. So the variance - the difference between the average of the square and the square of the average - is the cause for thermal light. You can even see this in the underlying photon number statistics: No matter how large the average photon number of thermal light is, the underlying photon number distribution, which is the Bose-Einstein distribution will always have a maximum at a photon number of 0.

    I know this sounds a bit strange, but a look at the Wigner function gallery might clear things up.
     
  21. Jul 3, 2008 #20

    Mentz114

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    Cthugha,
    thanks a lot. The link is very good. I have more on this in Gerry and Knight, 'Quantum Optics'.

    M
     
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