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Double slit intensity - QM description

  1. Feb 3, 2005 #1

    I wondered how one treats the double slit experiment quantum mechanically,
    that is, using probability amplitudes,wave functions,etc....to calculate the
    intensity on the screen.

    If anyone knows a link or a paper or a book where I can find it, that'd be nice.

  2. jcsd
  3. Feb 3, 2005 #2


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    I'd cite the Marcelo paper once again, but even I am getting bored at reading my own posting on this.

  4. Feb 4, 2005 #3

    can you give me the link to that paper? Or a link to where you
    have discussed it before? Thanks

  5. Feb 4, 2005 #4


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  6. Feb 4, 2005 #5
    Thanks alot. Unfortunately I have to wait till Monday because I can (hopefully)
    access the paper by Marcella
    from my university's computers.

    The reason why I asked is because
    (i) in the Feynman lectures Vol. 3 the double slit experiment is really explained
    in a wonderful way. But I wanted to know how to really calculate the pattern
    if the width and the distance of the slits is given.

    (ii) I read about a problem: You have the following experimental
    setup: single slit (with given width) and then behind it a double-slit (with given width and distance of the slits). What does the pattern look like? There was a solution but I wasn't comfortable with it (in the solution, the single slit didnt play a role).

    So I went to the library and read in Hecht's optics book that you obtain
    the intensity-forumula by using the Fraunhofer approximation. (A really wonderful explanation).
    I thought for QM, I just interpret the E-Field in the Fraunhofer calculation as the probability amplitude and sum up the different paths up by integration (like Feynman described in Vol. 3).

    I went home and put the integral into Mathematica and yeah,
    it yielded an interference pattern. I didn't really expect to obtain it,
    because I wasn't even sure that I interpreted Feynman's rule (when to add and when to multiply probabilty amplitudes) and Fraunhofer's
    approximation correctly.

    I played a little bit with the width of the single slit and it changed the pattern
    which I found quite interesting.

    If I understood correctly, Feynman path integrals work in that way
    (summing up probabitly amplitudes).
  7. Feb 8, 2005 #6
    Ok, I read the article by Marcella (http://www.iop.org/EJ/abstract/0143-0807/23/6/303),
    and I found it very interesting. Especially the part where he talks about the state preperation (I wasn't aware that the slit causes a certain state preparation).

    But I've still got some questions:
    1) In equation (18) there's a factor [itex] 1/ \sqrt{2} [/itex]. Where
    has it gone in equation (21) ?

    2) In equation (7) and (13), the probability has no unit, that's correct.
    But in equation (17) and (24) the probability has the unit of a length.

    3) How do I get [itex] 1/ \sqrt{a} [/itex] in equation (19) and (20).
    Isn't that supposed to be

    [itex] \frac{1}{\sqrt{a}} e^{i \beta} [/itex]

    At least that's what I get from normalization condition. I know,
    the phase factor can usually be neglected. But we add
    probability amplitudes in eq (21) first line. A phase factor would play
    a role.

    However, with respect to 3) I am only asking for mathematical
    reasons, because mathematically, there would be a phase factor.
    Since we do physics here, one can choose the [itex] \beta [/itex]
    to be zero (at least I hope).

  8. Feb 20, 2005 #7
    I got more questions:

    4) Before measurement, there's no y-component for the momentum.
    But after the double-slit, there is a certain probability for
    [itex] p_{y} = p \cdot \mbox{sin}\theta[/itex].
    Where does this "momentum-kick" come from? (momentum conservation)

    5) Why do I not detect any photons in front of the slit?
    That is, why is [itex] \theta [/itex] restricted to [itex][-\pi/2, \pi/2][/itex] ?
  9. Feb 20, 2005 #8


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    The most popular explanation for the emergence of an extra momentum here is due to the "interaction" of the photon with the slit. I put the world interaction in quotes because this should not be confused with photons scattering, reflecting, etc. off the slit, because it isn't. This is because we know how they should behave when they interact with the slits like that. The interaction that's relevant here has more to do a "decoherence reservoir", in which a quantum state is coupled to a macroscopic, measuring device. This is what always happens when we make a measurement.

    So without the slit to take up the recoil momentum, you would not have such effect.

  10. Feb 22, 2005 #9
    Another question:

    We see interference in the double-slit experiment.
    It also works, if there's air between the double-slit and the screen.
    Why doesn't the air destroy the superposition state?
    Isn't the photon absorbed and then reemitted by the air-molecules?
    I am asking because if one molecule does absorb and reemit the photon,
    then this would be a position measurement, right? This measurement would
    destroy the superposition state.
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