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Is a photon an EM wave packet?

  1. Nov 24, 2004 #1
    The similarities between the classical wave packet and the QM photon are striking. They both move at the group velocity. They both are subjected to an uncertainty principle. A classical wave packet is like a photon a region of concentrated energy, ...

    So is a photon something like the quantised version of a wavepacket?
    Last edited: Nov 24, 2004
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  3. Nov 24, 2004 #2


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    Quantum mechanics deals with classical electromagnetic field and therefore "photon" is a strange notion to it.The photon has nothing to do with the uncertainty principle,as his velocity,momentum,energy,wavelength are given.Trying to figure out where is a photon is a nonsense...It travels with the speed of light.
    No,a photon is a particle just like the electron...Wave behavior of light is similar to the one of particles...So,their particle behaviors are the same as well.
    Last edited: Nov 24, 2004
  4. Nov 25, 2004 #3
    I have an interesting observation here.Suppose you have two photons(represented by wavepackets) close by.In the region of superposition how should the fields add? Suppose you have E1(x) and E2(x) as the two wavepackets at a given instant.One may naively expect the field to be E1(x) + E2(x) in the region of superposition--but this is not correct.Energy goes as E^2, so the linear superposition does not conserve energy.So one should have E in the region of superposition given by E^2 =E1^2 + E2^2.Comments please.
  5. Nov 25, 2004 #4

    If the two wave E1 and E2 are totally uncorrelated (in phase), then the power term is also the sum of their respective power, the cross-term E1*E2*cos(phase) is statistically zero.

    Seen in QM, two statistically uncorrelated waves represent in fact two different photons. Think to the two-holes-in-a-screen experiment: two different photons emitted from each hole have uncorrelated phases, and their energy will also add-up like their amplitudes.

    On the contrary, if their phase are correlated, then the two waves are actually only one photon. Think to the two-holes-in-a-screen experiment: two correlated waves add-up after the screen and correspond to only photon to be detected.
  6. Nov 26, 2004 #5
    More simply (at least for me :yuck: ),
    In QM you add states |psi>=|psi1>+|psi2>, but you always compute the results (energy, momentum, etc...) with the square of these states: there is no surprise (think on the probability conservation law).
    For example, when you have a doubt, just think with the density matrix rho=|psi><psi| and <a>=tr(rho.A) (where A is any observable, ie for example the projector of an energy value measurement).
    Therefore you have the simple result: rho=|psi><psi|<>|psi1><psi1|+|psi2><psi2|=rho1+rho2.

    This is the mathematical consistency of QM and it explains somewhat why we find a lot of squares in the results!

  7. Nov 27, 2004 #6
    In a laser you have photons in phase(this does not amount to saying there is just one photon as you say) and the cross-term is not zero.If E1 ~ E2 = E then you get 4E^2 as the answer for energy if you superpose the amplitudes of the 'wavepackets' in a linear way(whereas the correct answer is 2E^2).
  8. Nov 27, 2004 #7

    You are right.
    We should go to the limit of very low intensity, single photon counting.
    Then, of course, two photons have uncorrelated waves.
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