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Spectrum broadening and conservation of energy?

  1. Mar 21, 2015 #1
    So, this is something I've never understood in detail.

    If an excited system decays and emits a photon, the lifetime of the decay will broaden the spectrum of the photon right?
    Basically just a Fourier transform of the "shape" of the emission in time to get the frequency components of the emitted light.

    As this emitted photon is a superposition of a continuous range of frequencies, if it passes trough a spectrometer, I can detect it as having one of those frequencies with respective probability right?

    But if I detect a photon with 'more' energy than I thought was present in the excited system, where does that energy come from? Was this uncertainty in energy already present in the system?


    What's wrong with the following thought experiment?

    1) I generate a photon in a very good optical cavity. With this single photon, I can associate a very well defined frequency and energy.
    2) I now pass an atom (or 2 level system with same energy difference) in ground state trough this optical cavity. It stays inside the optical cavity for half a Rabi cycle and absorbs the photon with 100% certainty.
    3) The atom is now out of the cavity and is excited with 100% certainty.
    4) The excited atom now starts decaying and emits a photon. But due to the decay the spectrum is "broad".
    5) I can now detect the photon with a slightly different energy? At what step did extra energy get in or get out?


    I'm missing something. I've asked it to people before, but I never got a good answer. (or I didn't get it).

    Thanks!
     
  2. jcsd
  3. Mar 21, 2015 #2

    HallsofIvy

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    No, there will be no change in energy. If the spectrum is "broad" then the energy is spread over many different wave lengths but the total energy is the same. Your error is that with a "broad" spectrum, you can no longer detect a single photon.
     
  4. Mar 21, 2015 #3
    I probably have a wrong way of looking at this, but I'm thinking of a wave packet in terms of a superposition of photons of different wavelengths and phase. So a superposition of a continuous range of states with different energy.
    And it is in the collapse that I don't understand what the energy does. It is probably true that I don't detect a single photon but what do i detect then?

    Consider for example two 2-level atoms A and B.
    A has an energy difference between states α and B a difference β.
    If the "wave packet" emitted from A is sent to atom B. If I observe B, there is a probability of finding it in the excited state.


    I just figured I can reduce my problem to the situation of a "detuned optical cavity".
    If I have a 2-level photon with energy α inside an optical cavity with a mode frequency/energy β.
    As the system evolves you have some probability of finding a photon in the cavity. If I then take the atom A out and observe that it is in it's ground state, I know there has to be a photon with energy β in the cavity.
    And spontaneous decay is just the same but with a continuous range of frequencies (Weisskopf Wigner theory)


    I also starting to realize that i never understood quantum mechanics :(
     
    Last edited: Mar 21, 2015
  5. Mar 21, 2015 #4

    bhobba

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    Welcome to the club.

    Thanks
    Bill
     
  6. Mar 21, 2015 #5
    I don't think you reach 100% transition probability unless the photon frequency is exactly equal to the transition frequency? If there is some detuning I think the transition probability oscillates with a maximum less than 1.

    Anyway the atom will sometimes be excited and you can ask about where the extra energy comes from in that case.

    I think the energy comes from the motion of the atom. Suppose the atom moves through the cavity very very slowly. Then you can argue by the adiabatic theorem that the atom cannot absorb the photon. For example, suppose the photon has less energy than the atom's transition energy. Then the ground state of the overall system is to have the photon in the cavity and the atom in its ground state. By the adiabatic theorem, if the system starts in the ground state and we only change things very slowly it will end up in the ground state. I think another way of understanding this result is that since the atom has very little kinetic energy we can't "steal" enough kinetic energy to excite the atom. [Actually I'm not sure this argument is very good because it's ignoring states like "empty cavity + ground state atom + free photon" which could have the same energy as the initial "ground" state.]

    If the atom moves through the cavity quickly enough it can be excited; then I expect any energy excess or deficit at the end to be made up by a change in the kinetic energy of the atom.

    You can probably understand this semiclassically by saying that the energies of the energy eigenstates of the overall system depend on the strength of the coupling between the atom and the cavity. This coupling strength depends on how far the atom has moved into the cavity. So the energies depend on the position of the atom, which means that the atom will experience forces as it enters and leaves the cavity. If the atom enters the cavity in the ground state but leaves in the excited state, it will experience different forces on entering and leaving. So it will end up with a changed kinetic energy.
     
    Last edited: Mar 21, 2015
  7. Mar 21, 2015 #6
    A detuned cavity has indeed the same sort of energy problem that I don't get. I regret not putting it literally in my first post, it's too late to edit now.

    The dynamics of an atom in a detuned cavity happens exactly how you describe, it oscillates, but the atom (or cavity depending on your initial state) doesn't reach a pure excited state. But if you observe it, you can collapse your system in that state.
    But I don't see how the energy would come from the movement of the atom as equations (The Jaynes cummings model) don't need movement of the atom.
     
  8. Mar 22, 2015 #7

    Vanadium 50

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    Yes.

    The answers have gone quite far afield, but the important thing is that the answer is "yes". The energy uncertainty is always there.
     
  9. Mar 22, 2015 #8
    I can accept that.
    But it is not true that this uncertainty is already present in for example a photon in high quality optical cavity.
    If I make one mirror transparent, the photon starts leaking out => it's spectrum broadens.

    So, I assume the energy uncertainty seeps in my system simply by coupling it to the vacuum.

    So, to actually remove a mirror from the cavity (or reduce it's reflectivity) I would require a random amount of energy distributed by frequency distribution as the photon.

    I can accept that, but I have to think about it.
     
  10. Mar 25, 2015 #9
    To describe your system the coupling between the cavity and the atom in the Jaynes-Cummings model has to change with time. Since the Hamiltonian is not time-invariant, energy will not be conserved if you try to model the situation with the Jaynes-Cummings model.

    Why is energy not conserved? Because we are not considering the entire system. We need to consider the position degree of freedom of the atom as well. In reality the overall Hamiltonian does not change with time. Rather the coupling depends on the position of the atom, and the atom's position changes with time. So the simple Jaynes-Cumming system is coupled to the position degree of freedom of the atom. Energy conservation is not violated; rather there is an exchange of energy between the Jaynes-Cumming system and the kinetic energy of the atom
     
    Last edited: Mar 25, 2015
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