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I Modified double-slit experiment - two photons

  1. Nov 28, 2017 #1
    On the following link https://physics.stackexchange.com/q...ent-two-electron-sources-instead-of-two-slits
    there is a discussion of the modification of double-slit experiments where two electrons sources are put in place of the slits. The conclusion is - there will be no interference in that case.

    I understand, the same argument should generally stay for photons, i.e. 2 (coherent) photons should not interfere in such an experiment -- so that if we conduct thousands of experiments with 2 (coherent) photons emitted simultaneously, the result screen points on which photons are absorbed will not show interference pattern.

    If that is right, then how two coherent laser beams (instead of 2 singular photons) may interfere?
     
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  3. Nov 28, 2017 #2

    PeterDonis

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    Yes, because it's stipulated that only one electron source fires at a time, so it's the same as just firing a single electron source and observing where it hits on a screen, with nothing else present.

    2 coherent photons is not the correct analogy. The correct analogy is one coherent photon fired at a detector, with nothing else present. Yes, there would be two photon sources, but it would be stipulated that only one fires at a time, just as it is for the electron case (see above).

    Yes, it will. Correspondingly, if you have two "coherent" electron sources that both fire at the same time, you will see interference. Practically speaking, AFAIK this experiment would not be possible to run with our current technology, because we don't know how to set up two coherent electron sources the way we know how to set up two coherent photon sources. But the prediction of QM is unequivocal.
     
  4. Nov 28, 2017 #3
    Alas, my fault, my bad. I brought the wrong reference to illustrate my question.

    The setup I actually meant to consider is two photon (or electron) guns firing at the same time.
    Unfortunately, I cannot fix the original post.

    So, in the case of photons/electrons, do you mean that if we fire two of them in the same time from the near standing guns, that will (after accumulating the data from thousands of experiments) give us the interference pattern? I personally fail to see how this works. I.e. in the normal double slit experiment we add two complex exponents with phase difference which is dependent on the shift on the screen, so when we add them we get interference. In my case with double guns, we IMO have a wave function of two electrons $$\psi(x,y)=\psi_1(x)\psi_2(y)-\psi_1(y)\psi_2(x)$$ where x and y are the coordinates of the first and second electron and I do not see how this may give us an interference .
     
  5. Nov 28, 2017 #4

    PeterDonis

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    It will if the two sources are coherent with each other. We know how to set that up in the case of photons (for example, use two lasers tuned to the same frequency). AFAIK we don't know how in the case of electrons.

    No, we have a wave function of one quantum system consisting of two electrons. So for each run of the experiment you will have two detections on the screen instead of one. But you still have one quantum wave function, and if you set up the electron sources appropriately, that wave function will have interference terms in it. "Interference" here would mean that the pattern of detections on the screen, after a large number of trials, would not be the same as for two independent electron sources.
     
  6. Nov 28, 2017 #5
    I agree with all you wrote. It is one quantum system, there will be two detections per experiment. And... I did not mean or write anything different. I meant to consider the $$\psi(x,y)$$ which depends on coordinates of each electron.

    I just do not understand how to see the interference terms in the wave function - may be because of generally little training in math and qm, may be because of other reasons.
     
  7. Nov 28, 2017 #6

    PeterDonis

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    Think about two coherent laser sources tuned to the same frequency. Don't even think about the quantum aspects if you're not comfortable with that: just think of two classical waves with the same wavelength and frequency, coming from two different sources. Will they interfere with each other?
     
  8. Nov 28, 2017 #7
    Yes they will.
     
  9. Nov 28, 2017 #8

    PeterDonis

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    Then you would expect two coherent lasers to interfere, correct? Putting the quantum aspect back in just means thinking about a quantum wave function that acts just like the classical wave--but it tells you the probability of getting detections at various points on the screen. So that probability should show an interference pattern, just like the classical wave does.

    And if you can set up the wave function that way for photons, you can in principle set it up that way for electrons too--it's just that practically speaking we don't know how to actually do it for electrons the way we do for photons.
     
  10. Nov 28, 2017 #9
    I would say that classical interference of two lasers is analogous to quantum double slit experiment. We just add two wave functions (from the two slits) the same way as we add classical waves.

    My concern is related to the case of two quantum paritcles, say electrons. It is only for their wave function I fail to see how the interference appears.

    I.e. I can (if I am right) write a wave function of two electrons as $$\psi(x,y)=\psi_1(x)\psi_2(y)-\psi_2(x)\psi_1(y)$$, but I do not see how interference may be possible in such a function.

    It is no problem to consider a boson case instead of fermions

    $$\psi(x,y)=\psi_1(x)\psi_2(y)+\psi_2(x)\psi_1(y)$$

    I still do not see how the interference appears for such a case
     
    Last edited: Nov 28, 2017
  11. Nov 28, 2017 #10

    PeterDonis

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    Yes. So quantum interference of two lasers should also be analogous to the quantum double slit experiment.

    Photons are also quantum particles. Just set up the lasers so that, on average, each one only fires one photon into the experiment at a time. They will still interfere. If we knew how to make "lasers" that fired electrons instead of photons, we could do the same thing with electrons.

    No, it isn't, because you're claiming you can write the same sort of wave function for bosons, and you don't see how interference shows up there either. Photons are bosons.

    You're not right. Your wave function assumes that each electron has a definite position; it basically says "there is an electron at position x and an electron at position y". That is only a small subset of the possible two-electron states, and this subset does not contain any interference. But a more general two-electron state can. (Similar remarks apply to two-photon states, although the sign change--symmetry vs. antisymmetry--does change the details of the interference pattern.)
     
  12. Nov 28, 2017 #11
    What I meant to say is I do not see how interference is possible for boson and fermion two particle wave function. I understand how interference is possible for a single particle wave function of the electron passing the double slit, yet if I substitute this double slit with two electron or photon guns firing simultaneously, then I do not see how the interference may appear in the case of two-particle-wave-function.

    I agree that my two-electron state described by my wave function is not general. I assumed that two electrons in such an experiment are not entangled (except for trivial fermion entanglement), so it may be factorized into the product of two single-particle wave functions.
     
  13. Nov 28, 2017 #12

    PeterDonis

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    That's because you're writing down a wave function that rules out interference a priori, without stopping to consider whether that wave function is general enough to capture all possibilities.

    First of all, the wave function you wrote down does not factorize into the product of two single-particle wave functions. It is the sum of two terms that factorize that way, but you can't factorize the whole wave function that way.

    Second, the key point is not whether the wave function is entangled, but the fact that the wave function you wrote down describes two particles each of which has a definite position. That is what rules out interference a priori. A more general wave function, in which the individual particles do not have definite positions, would allow interference.
     
  14. Nov 28, 2017 #13
    I did not define my wave functions of individual particles as delta-functions, so they are not functions with a definite position.

    Anyway, you may want to share the wave function in the form you find correct for this particular setup.
    It looks that discussing of what is wrong with my wave function without bringing your version is not much productive.
     
    Last edited: Nov 28, 2017
  15. Nov 29, 2017 #14
    I think I came (to some extent) to the resolution of the problem.

    For bosons $$\psi(x,y)=\psi_1(x)\psi_2(y)+\psi_2(x)\psi_1(y)$$

    Then, the probability density to detect the first boson at a particular coordinate x is:
    $$p_1(x) = \int\limits_{-\infty}^{\infty}\psi^*(x,y)\psi(x,y)dy=\\
    =\int\limits_{-\infty}^{\infty}(\psi_1^*(x)\psi_2^*(y)\psi_1(x)\psi_2(y)
    +\psi_2^*(x)\psi_1^*(y)\psi_2(x)\psi_1(y)
    +\psi_1^*(x)\psi_2^*(y)\psi_2(x)\psi_1(y)
    +\psi_1(x)\psi_2(y)\psi_2^*(x)\psi_1^*(y))dy=\\
    =\psi_1^*(x)\psi_1(x)+\psi_2^*(x)\psi_2(x)+2 Re(\int\limits_{-\infty}^{\infty}\psi_1^*(x)\psi_2^*(y)\psi_2(x)\psi_1(y)dy)=\\
    =\psi_1^*(x)\psi_1(x)+\psi_2^*(x)\psi_2(x)+2Re(\psi_1^*(x)\psi_2(x)\int\limits_{-\infty}^{\infty}\psi_2^*(y)\psi_1(y)dy)$$

    What we got looks generally like a probability density for the double slit experiment, if not the integral ## \int\limits_{-\infty}^{\infty}\psi_2^*(y)\psi_1(y)dy##

    for instance, if this integral is zero, there will be no interference.
    It looks to me that this integral never allows the full interference as its absolute value reaches one only when ##\psi_1## and ##\psi_2## are linearly dependent, but then we will not have interference anyway...
     
    Last edited: Nov 29, 2017
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