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Shape of the wave of a photon

  1. Oct 11, 2013 #1
    With a double slit measurement a single photon is a wave which goes throught both slits. To extinguish each other at certain places the wave must have the same amplitude at both slits, also at large distance between the slit. The positions also extinguish with 0.5, 1,5, ... periode difference, so also in time the amplitude is the same. If taken to the extreme it would have the same amplitude everywhere in the universe.
    What is this wave? Outside the double explanations I don't see it elsewhere in publications. Is this the wave of a photon, freely travelling in space?
     
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  3. Oct 11, 2013 #2

    mfb

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    "Yes", but don't extend that answer too far.

    It will not. The amplitude is (nearly) the same in the region where you get the interference pattern if the slits have the same width and get the same intensity from the light source, but that won't happen everywhere.
     
  4. Oct 11, 2013 #3
    How does the amplitude degrade in space-time?

    Or is there is formulea for this wave?
     
  5. Oct 11, 2013 #4

    jtbell

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    The amplitude of the wave decreases as 1/r where r is the distance from the source, if r is large enough that the source "looks" like a point.
     
  6. Oct 11, 2013 #5
    Don't you mean $$\frac{1}{r^{d-1}}$$ where d is the dimension of space (3 in our case)?
     
  7. Oct 11, 2013 #6

    jtbell

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    I assumed we were discussing the propagation of light/photons in 3-dimensional space.
     
  8. Oct 11, 2013 #7
    Right. So it should 1/r2 so it expands outward as a sphere, shouldn't it?
     
  9. Oct 11, 2013 #8
    But: suppose the source is not far above the slits. For detector positions right or left from the slits the waves through both slits always have a different r. If that would mean a different amplitude they can never extinguish each other.
     
  10. Oct 11, 2013 #9

    Bill_K

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    The wave amplitude goes as 1/r. The energy ∝ |amplitude|2 ∝ 1/r2.
     
  11. Oct 11, 2013 #10
    Last edited: Oct 11, 2013
  12. Oct 11, 2013 #11

    mfb

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    That is true.
    Single-slit effects (you always have them) are another complication that we did not consider yet here.
     
  13. Oct 11, 2013 #12
    You don't need complete destructive interference to observe an interference pattern, so it is quite forgiving of your experimental setup, in that respect. The human eye is pretty good at picking out contrast. Even if the amplitude through one slit is significantly lower than through the other, the interference pattern is still visible. The peaks just aren't as high and the troughs aren't as deep, on the interference pattern.
     
    Last edited: Oct 11, 2013
  14. Oct 11, 2013 #13
    I did, and if I had found (or understood) the solution for a photon I would not be on this forum. I am here because I hope someone can explain without all formulae of wikipedia (which therefore is only readable for you guys). If the forum is not intended for that, I wil leave.
    Although I have the expression there are more here who don't know yet the answer
     
  15. Oct 11, 2013 #14
    According Feynman (QED) there is a complete destructive interference. I don't see 1/r2 in his arrows.
     
  16. Oct 11, 2013 #15

    Cthugha

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    Please reread the response you have been given earlier, especially the "if"-part:
    Going from the standard electrical field picture at high intensities to the single photon level does not change much. You just move on from discussing fields interpreted as real entities to discussing probability amplitudes for the detection of a single photon. When averaging over many of those events, the interference patterns seen will be the same for single photons and bright light fields in the same geometry. The addition you get when discussing probability amplitudes is simply that you cannot detect a single photon twice.

    Having said that, the shape of the "wave" part about the photon (the probability amplitude) just depends on geometry. The realistic thing you can get in the lab is a point like source having a single atom or a single quantum dot as the emitter which will give you the 1/r decay. In principle and especially in theory any geometry is possible. You can have a plane wave, something looking like a standard beam or shaped beams. There is no intrinsic shape of the "wave of a photon". So if Feynman does not have any 1/r terms, it is very likely that he considered a plane wave geometry for single photons. To allow us to give a more well defined answer, you might need to quote the exact text in Feynman's book.
     
  17. Oct 11, 2013 #16
    Don't think for a second that your questions aren't welcome. If they weren't, we wouldn't respond.

    You should also expect that many people reading them are familiar with what it's like to try to get to grips with these things. Asking questions on here will give you some good answers, but it's not necessairily the fastest route to understanding the problems. I think many of us know what it's like to be given unsatisfying answers, even if they are well thought out answers given by people who are experts in their field.

    Well written books can take you through things in a matter of hours that took others many decades of questioning, because they are written by people who know the right questions to ask to arrive at the level of understanding that you want to get to. That said, I wholeheartedly understand anyone who doesn't find book learning particularly appealing and never let anything discourage your inquisition.
     
    Last edited: Oct 11, 2013
  18. Oct 12, 2013 #17
    Feynman is talking about amplitudes of single photons to follow a certain path (in his book QED). In fig 20 (a mirror) he say (translated back from Dutch):"According quantum theory light has an amplitude for reflection which is equal for every position on the mirror". Different paths give the same amplitude but different phase (direction of the arrow). Fig 5 shows reflection in glas depending on thicknes, which is a cosine between 0-16%, even with glass of more then 50 meters thick.
    Then in fig 49 he shows the double slit and mentions (agains translated) "sometimes we get for a certain distance between the holes more ticks then expected, with a somewhat different distance the dector does not tick at all" (the detector is a photon counter).
    According Wikipedia: "and when they are in anti-phase, i.e. the path difference is equal to half a wavelength, one and a half wavelengths, etc., then the two waves cancel and the summed intensity is zero"
     
  19. Oct 12, 2013 #18

    vanhees71

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    I don't know what reference of Feynman's you are referring to. His book on QED is pretty conventional using the old Fermi formulation in the operator formalism. This surprised me a bit, because the path-integral quantization method is most convenient particularly for gauge theories as compared to the operator formalism.

    Anyway, one cannot stress often enough that a naive particle picture for massless quanta is almost always wrong. The first thing is that for massless quanta with spin [itex]\geq 1[/itex] there is not even a position operator in the strict sense.

    By definition a single-photon state is an asymptotically free one-particle photon state with definite norm.

    The double-slit experiment with single photons is described as any double-slit experiment with single quanta as a scattering process with asymptotically free single-particle states coming in and asymptotically free single-particle states coming out. This leads to the probabilities to register an asymptotically free particle beyond the double slit (e.g., using a photo plate). A single quantum never makes an intereference pattern but only a single dot on the screen. The interference pattern, reflecting the detection probabilities as a function of position, can be found by repeating this scattering experiment several times.

    Note that nowhere I have used the idea of position of a single photon but only the possition of a registration of a photon at the screen. This position of a registration of a photon is a well defined physical quantity that can be measured, the position of a photon in the strict sense of an observable cannot even be defined in principle! For details, see

    http://www.mat.univie.ac.at/~neum/physfaq/topics/position.html
     
  20. Oct 12, 2013 #19
    Do you remember in the other thread we talked about needing a surface for specular reflection?

    Feynman is describing, using path integrals, how the different paths across this surface, superpose to cancel each other out, with the exception of the apparent reflection path from Fermat's law of reflection. He does mention that he is making an approximation when he says that the the amplitude is the same for all points on the section of surface. The approximation holds when the section of mirror that he is treating is very small, hence the distance that the light travels to each point is approximately the same. This is very similar to the approximation used in calculus, where you take 2 points on a curve which are close together to calculate its gradient.

    The cool thing about this is that he's demonstrating that the reflection from a mirror can be considered an interference patten itself.
     
    Last edited: Oct 12, 2013
  21. Oct 12, 2013 #20
    Hi DParlevliet...getting a 'picture' of this stuff takes some time, so be patient....Expect your head to spin a few more times!!!
    I'll see if I can put some pieces together to clarify and summarize some subtle points...no math because I remember relatively little of it!!

    From the Wikipedia link already provided:

    Interpretation of the wave function

    Here is another very insightful tidbit from another Wikipedia article:

    So this IS crazy: a deterministic expression for quantum behavior! Further, after almost 100 years, arguments still ensue about what the wave function 'really' represents.

    Post #2:
    [This 'wave' can be thought of as representing some distributed photon behaviors, a probability distribution relating to likely detection location, but says nothing about observables of this mode. When detected, photons are always pointlike as are all particles in the Standard Model of particle physics. As wiki says, oddly enough, it is a deterministic expression, yet measurements/observables based on it are NOT deterministic.]
    edit: Particle 'wave' characteristics are always detected as pointlike objects.

    post #9:
    Turns out the probability of locating a particle is also proportional the the amplitude squared.

    post #15:


    Haven't seen that before...I like it...Bravo!!

    In Wikipedia terminology, the 'standard electrical field picture' is deterministic, the single photon is a quantum particle, and measurements [averaging over many events ] turns out to be NON deterministic. Nobody knows why.

    Analogously, here is what Roger Penrose says:

    [and THAT changes things from deterministic to probabilistic!!]


    [It depends on geometry.]

    Post#16:
    a great point.....so when you mumble to yourself [as many of us have at times] "That seems crazy." it probably is. It was unlikely not scientists first choice of interpretation...Feynman says something like this about that:

    A good physicist is one who has the stubbornness to make all the mistakes possible before finally arriving at the correct conclusion.
     
    Last edited: Oct 12, 2013
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