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Non quantized light

  1. Aug 28, 2009 #1
    An electronic circuit that produces electromagnetic waves in the visible light range, would such an experiment be able to produce electromagnetic waves with energy below the photon's energy ?

    I know it is technically hard to conduct such an experiment, but ... what would happen then?
     
  2. jcsd
  3. Aug 28, 2009 #2
    No that would not be possible. At a low enough intensity, a detector would be counting individual photons.

    This does bring up an interesting point. As you move toward longer wavelength, the energy of each photon goes down.

    E=hf E=hc/wavelength

    So at some long wavelength (out of the visual), it becomes technically hard to empirically see individual photons. We think they're still there though.
     
  4. Aug 29, 2009 #3
    If you want to describe what would happen, you have to describe your electronic circuit using quantum mechanics. So, you can't treat the currents and voltages in your circuit classically, they become operators that don't commute.

    In case of an LC circuit with angular resonance frequency omega, what you find is that this circuit has energy levels of:

    E_n = (n+1/2)hbar omega

    If this circuit is coupled to the electromagnetic field, then you see that the emission of a photon of angular frequency is accompanied by a transition of the circuit to a lower energy level.
     
  5. Aug 30, 2009 #4
    You say we must use the quantum harmonic scillator...
    so, aren't there any deviations from that?
    Are those circuits of radio waves unable to create divisions of photons, experimentally ?
    An answer to that question would tell us either we are forced to quantize all oscillators or not.
     
  6. Aug 30, 2009 #5
    Everything, including your circuit is described by quantum mechanics. If it is not a harmonic oscilator, it is still described by some Hamiltonian. It is also coupled to the electromagnetic radiation field. Then if you sart with some initial conditions, in which the radiation field in the ground state (no photns present), and you apply the time evolution operator to compute the future state, you'll see that the radiation field is no longer in the ground state.
     
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