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Electromagnetic waves and the correspondence principle

  1. Sep 12, 2014 #1

    In classical electromagnetic theory the energy of an electromagnetic wave is proportional to its amplitude squared.

    In contrast, the quantum mechanical equation ##E = \hbar \nu## states that the energy is proportional to the frequency of the wave (photon).

    Now, according to the correspondence principle, quantum physics must reduce to classical physics as the quantum numbers get sufficiently large: which in this case would amount to having a large enough number of photons.

    So my question is, how does one go from the quantum mechanical description ##E = \hbar \nu## to the classical description of the energy being proportional to the square of the amplitude?
  2. jcsd
  3. Sep 12, 2014 #2


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    A classical electromagnetic wave has not much to do with single photons (Fock states). Quantum-field theoretically a classical em. wave is described by a coherent state with a large average number of photons. It's a superposition of all photon-number (Fock) states. So in this state the photon number is not determined. There's an uncertainty relation between the phase and the photon number, i.e., the more accurately you determine the phase of the state the less determined is the photon number. For a coherent state the photon number is Poisson distributed.

    I'm not aware of any other concept in physics which is represented in as misleading a way as photons! Read a good book on quantum optics to get an understanding what photons really are and for which problems you really need to bother with them. A lot can be achieved in the semiclassical limit, i.e., where the em. field is treated as classical ("c-number") field and only the matter particles (in atomic/condensed-matter physics the electrons) are quantized. A good book on quantum optics is

    Scully, M. O., Zubairy, M. S.: Quantum Optics, Cambridge University Press, 1997
  4. Sep 12, 2014 #3
    Thanks for your answer!

    Often people seem to be equating a photon with a "mode": i.e. a monocromatic sinusoidal electromagnetic wave. Why would you say that this is wrong?
  5. Sep 12, 2014 #4


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    A plane wave is never a state, because it's not square integrable. Of course, the free photon field operator (!!!) can be expanded in terms of creation and annihilation operators wrt. any single-particle basis you like, among them the momentum-helicity eigenstates of the photon. That's of course all fine.

    What's very misleading is, e.g., the idea you find in many textbooks about quantum mechanics (even otherwise very good ones) that the photoelectric effect proves the existence of photons. That's simply not true, because you can understand the photoelectric effect in the semi-classical picture, where the em. field is not quantized. The discrete absorption (emission) energies are due to the quantization of the bound electrons perturbed by the classical oscillating em. field. It immediately comes out of time-dependent perturbation theory that you absorb/emit portions of energy of the size [itex]\hbar \omega [/itex], where [itex]\omega[/itex] is the frequency of the absorbed/emitted radiation. So this does not prove the necessity for quantizing the electromagnetic field and introduce photons.

    The damage done in the minds of many people is that they think they could understand photons as little classical bullets ("particles"). This is justified for massive particles to a certain extent, because it's the classical limit of (non-relativistic) quantum theory, but it's totally misleading concerning photons. Its classical limit are coherent states with high occupancy, i.e., classical electromagnetic (wave) fields and in no case particles. A photon not even formally admits a position operator to begin with (there's a mathematical no-go theorem telling you that for massless particles with spin [itex]\geq 1[/itex] there's no position operator in the usual sense). Also note that photons can only be "seen" by measuring them somehow, i.e., you define their "position" in the sense of detection eventts in some detector consisting of massive particles, for which "position" is a well defined observable.

    A very nice review article about what photons really are is

    M. O. Scully, M. Sargent, The concept of the photon, Physics Today 25, 38 (1972)
  6. Dec 10, 2015 #5
    The photo-electric effect at least requires that the energy of the EM-field is quantized, because the conduction electrons can absorb a continuous spectrum of engery and could continuously climb up the energy ladder until they got enough energy to leave the metal. But this is not the case, so Einstein concluded that Planck's quantization of the EM-field's energy is also necessary to explain the photo-electric effect.
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