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QED and thin films.

  1. Jun 15, 2013 #1
    I've been reading a bit around Quantum Electrodynamics (QED) which says that when you shine monochromatic light onto a thin film, the fact of whether it produces a dark or bright band is purely a probability function. It seems to say you use Feynman arrows, one for the first surface and one for the second. You set the arrows spinning in a circle and then stop them as soon as the light is detected from each surface. The vector difference in the two arrows represents the radius of a circle and the bigger the area of the circle the higher the probability of getting a bright band. My textbook says that thin film patterns are caused by wave interference of the light and so it depends on the phase difference of the light waves. I was just wondering if these two concepts agree with each other. It seems to imply that, in terms of thin films, QED has nothing to do with wave interference when explaining thin films? Does this mean wave interference is a dated classical theory?
     
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  3. Jun 15, 2013 #2

    Jano L.

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    Two different but related explanations of the interference phenomena. The Feynman explanation is probabilistic, it gives you probability that "photon" will be detected at some place. The wave theory is deterministic, it gives you exactly the value of the electromagnetic field at that point.
     
  4. Jun 15, 2013 #3
    Thanks for your reply. But this surely means that the wave theory is an incorrect model for dealing with thin film interference? If you have a photomultiplier to detect reflection from the two surfaces (or single surface) you get a lumpy signal which means that it is impossible for it to be a wave, it must be particle. So is the wave interference model scientifically incorrect? If so, why is it in textbooks?
     
  5. Jun 15, 2013 #4

    Jano L.

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    No, that would be a rather reckless conclusion. Lumpy signals do not disprove the electromagnetic ("wave") theory of radiation. The microscopic motions inside the detector play a great role in detection of weak radiation. The detection process involves interaction of light with matter consisting of many atoms. The lumpiness of the detector behaviour can be due to complicated detection process involving these atoms and their fields, triggered by the primary wave and the inner state of the detector. Basically you can imagine the detector as one big trigger, which can be triggered if proper conditions are satisfied - wave is present, and the microscopic state of the detector and chaotic electromagnetic fields are right so that it can trigger.
     
  6. Jun 15, 2013 #5

    DevilsAvocado

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    What happen when you throw a stone in the water? :wink:

    Classical wave interference works perfect, both before and after Feynman:

    https://www.youtube.com/watch?v=egRFqSKFmWQ


    Both the Schrödinger equation and the Feynman path integral give you the probabilities of finding a particle in a given place at a given time. Feynman's "probability amplitude" obeys the Schrödinger equation, which is a wavefunction corresponding to the probability [density], and the Schrödinger wavefunction behaves like water waves or waves on a string.

    waves/path = [microscopic] probabilities
    particle = [macroscopic] measurement

    The Schrödinger wavefunction works perfect for describing electrons in the atom, but it can’t describe relativistic phenomena and systems where the particle number change over time, including particle creation/annihilation.

    This is the reason for QED – and it works with exceptional precision.

    Feynmans_QED_probability_amplitudes.gif
     
    Last edited by a moderator: Sep 25, 2014
  7. Jun 15, 2013 #6
    Thanks for your responses, they are greatly appreciated. Still a bit confused. I have just managed to find a Feynman lecture on youtube where he implicity says that you cannot explain the reflection of light from differing surfaces treating the light as a wave. He goes on to say that if light was a wave it would not respond the way it does when detected by the photomultiplier. Doesn't the photomultiplier work by the progressive release of electrons by the photoelectric effect which is purely the result of photon (particle) interaction? I totally understand that you can treat light as a wave and it works fine but I just want to know which one is correct or are they both right or does no-one know?
     
  8. Jun 16, 2013 #7

    Jano L.

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    Can you post a link or quote his exact words?

    There are various approaches to understanding the interaction of light and matter. In one of them, the radiation is described by an electromagnetic field as in classical theory, and the matter is described by Schroedinger's equation. There are no photons in this picture. The explanation of the basic photoelectric effect does not need the concept of "quantum of light".

    There are other approaches, for example one in which the electromagnetic field is quantized and people use the word photon there from time to time. The meaning of the word varies depending on the context, but usually it is used as a name of the object which "gets absorbed" by the atom when the latter gains energy and changes its state from one Hamiltonian eigenstate to another. However, this is just a simplified and inaccurate picture of what happens according to more complicated mathematics. The relevant equations are differential equations and there are no jumps between states or sudden absorptions; everything evolves continuously.
     
  9. Dec 8, 2015 #8
    Is necro-threading considered bad practice in this forum? I've just had the same questions as the OP as I was watching a lecture by Feynman, which I believe could be the one the OP mentioned. See here: . He is talking about how Newton was puzzled with the results of thin film reflection, and raises and quickly discards the idea that light might be wave instead.
     
  10. Dec 9, 2015 #9

    DrClaude

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    Yes. Although, in this case, the OP is still around. Anyway, next time please open a new thread if you have a question.
     
  11. Dec 9, 2015 #10
    oh, okay, sorry about that. I'm still in early stages of learning QED. I have a lot of questions, but I'd need to study more before I can feel comfortable asking.
     
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