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Just want to see if my understanding of the photon is correct

  1. Jul 23, 2010 #1
    The photon is a discrete particle, a point, generating an electromagnetic field which fluctuates with a wavelength depending on the energy of the photon. The wavelength also determines the amount of potential paths the photon can take. The wavelike behaviour in the double-slit experiment is accounted for by the fact that the photon can "detect" possible paths before going through them.


    Is this a fair explanation?
     
  2. jcsd
  3. Jul 23, 2010 #2

    K^2

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    No. There are many, many things wrong with it.

    A photon in electrodynamics is very different from a photon in quantum electrodynamics. You have the two jumbled together. Furthermore, you seem to be confused on basic quantum mechanics as well, so I wouldn't feel comfortable explaining QED in fine detail.

    In electrodynamics, a photon is simply a single frequency component of an electromagnetic wave. It has oscillating electric and magnetic fields. It is spread out through space. It does have a specific wavelength. It can have an arbitrary amplitude/energy.

    In QED, a photon is a point-particle. It's location is undetermined, but unlike the above, it's not simply spread out. It's still a point object. It can only have discrete energies, but because it can be in superposition, that only matters as far as absorption and emission go. It also still behaves like a wave, as any quantum object does.

    Neither of these "detect" all possible path. They literally take all possible paths at once.

    A single photon behaves identically in both descriptions, but when you have a bunch of photons, there are some important distinctions. For example, QED photons may be entangled, while electrodynamics photons are always in a mixed state. This isn't something you need to worry about, though.
     
  4. Jul 23, 2010 #3
    Ah, I think I see where you're coming from. Are you saying that the photon in electrodynamics is "spread out" in the same sense that an electron is "spread out" around the nucleus of an atom?
     
  5. Jul 23, 2010 #4
    You also can't say that a photon 'generates' an electromagnetic field. It IS the electromagnetic field (or at least vibrations in the field). Waves don't generate water.
     
  6. Jul 23, 2010 #5

    K^2

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    No, I'm saying exactly the opposite. Photon in QED is a point-particle that has a probability distribution, like an electron near a nucleus.

    A photon in classical electrodynamics really is a wave that is physically spread out.
     
  7. Jul 23, 2010 #6
    Okay, so a photon in classical electrodynamics has a "volume" of sorts, and the wavelength is the... Length of the wave? Or is it just a movement in a medium, like a sound wave?
     
  8. Jul 23, 2010 #7
    Is it proper to speak of a photon in classical electrodynamics? I think if it as solely a quantum description.
     
  9. Jul 23, 2010 #8

    K^2

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    Yeah, just do the Fourier transform of whatever electromagnetic field you happen to have. The spectrum you get is effectively a photon spectral density distribution.

    Edit: Just to clarify, say you have E(x,t). Do a Fourier Transform to get E(k,ω). That's your amplitude of photon with wave-vector k and angular frequency ω.
     
  10. Jul 23, 2010 #9
    Well I tend to disagree with K^2. IMHO The photon in QFT is the same as in Electrodynamics. The Feynman diagrams are usually in momentum space. The fact that we say "a photon is coming from the left and an electron is coming from the right an they interact" is due to the fact, that the envelopes of the wave packets are so short that it does look like this, but for the cross section calculations it's all plane waves.
     
  11. Jul 23, 2010 #10

    jtbell

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    I've never seen that called a "photon" in classical electrodynamics. Can you provide a reference to a textbook or journal article or academic web site that does this?
     
  12. Jul 23, 2010 #11
    OK, "Corpuscle." Is that better? Are you happy now? Lol, don't make me cite my reference, either, lmao...
     
  13. Jul 23, 2010 #12

    K^2

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    No. I'd have to dig through tons of material to find someone pointing out this trivial fact. But I'm sure I can find plenty of classical treatments of a vibrating lattice that call the same result for lattice displacement a "phonon" without any quantization. You can probably think of these yourself. It's the same thing.
     
  14. Jul 23, 2010 #13
    Isn't this just a transform of the classical field?
    After the transform you still have a continuous (I mean classical) field.
     
    Last edited: Jul 23, 2010
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