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More basic photon questions

  1. Jan 31, 2006 #1
    Four more photon questions.

    So photons are ‘particles’ in momentum eigenspace with (discrete, perfectly well-defined) momentum p and energy E= hf.
    1. Does that definite energy translate into the complete uncertainty of the photon in the position space?
    2. What happens when the momentum eigenstates are superposed? Doesn’t we have localized waves in position space now that correspond to a ‘size’ of a photon?
    3. But when the well-defined energy is gone, are we still allowed to call it a photon?
    Even if this the quantum world, wasn’t the definition of a photon its discreteness in energy space?
    4. When a single photon interacts with matter (gets absorbed by an atom, hits a plate and
    leaves a spot) , well-defined values of energy and momentum get transmitted. But are not these interactions (and so the photon) also precisely localized in position space?

  2. jcsd
  3. Jan 31, 2006 #2


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    they are actually bookkeeping units of the momentum/energy (P,H) eigenstates of the free quantum field.

    Yes, because they correspond to the "classical modes" of the field. For instance, in EM they correspond to the harmonic plane waves. These fill all of space.

    You can of course make superpositions of the eigenstates of (P,H), and these will be possible states of the quantum field. You can now argue all day whether or not you call such solutions "photons". It's really a semantic issue.

    As I said, it is a semantic issue. I find it ok to call superpositions of 1-photon states also 1-photon states. Some object to that. Take your pick.

    Well, the better they are localized in position space, the more "blurry" they are in "pure photon" space (or better, the more "neighbouring" pure photon states also have finite amplitude for the interaction). That's why I consider it a good idea to call a superposition of "pure" one-photon states also a 1-photon state. People are way less picky when it comes, for instance, to electrons. Are superpositions of pure energy and momentum "1-electron" states also "1-electron" states, or not ?
  4. Feb 28, 2006 #3
    When atoms only emit EM waves in certain well-defined frequencies, how can we speak about blurry pure photon states?

    Do atoms emit normally superpositions of pure one-photon states instead of sharp frequencies/ energies?
  5. Feb 28, 2006 #4


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    First question: recoil, thermal agitation, and radiative line broadening -- simply put, the excited states of an atom are not exact eigenstates of the atom-radiation system, and the are, metaphorically speaking, somewhat fuzzy. (This can be made quite rigorous,as was done by Wigner and Weiskopf.Their work is all over the place, particulalry in nuclear and particle physics, and, probably, Google) (It's another matter whether or not the various forms of line broadening are practically important.)

    Second Question: vanesch gave all the info you need to answer this one by yourself.

    Reilly Atkinson
  6. Feb 28, 2006 #5
    Thanks Reilly. Always learn a lot reading your posts.

    I also found a lovely thread. It is called "Length of wavetrain of a single photon". http://www.lns.cornell.edu/spr/1999-02/threads.html#0014350

    I wasn't able to read everything yet but it looks like a gold mine. Saw also Vanesch posting there.

    Especially this post (http://www.lns.cornell.edu/spr/1999-02/msg0014640.html) is hitting the nail to my posted question. I hope no one minds if I quote it here.

    Last edited: Feb 28, 2006
  7. Mar 1, 2006 #6
    I have one more thing to clear. I apologize.

    Reilly said 'the excited states of an atom are not exact eigenstates of the atom-radiation system'. So the blurrieness of E does not only stem from the fact that we don't know in which energy eigenstate the state will fall after measurement but also we don't know the exact eigenstates. Before measurement we are not only ignorant of if the photon comes from electron jumping 1, 2, 3 or 4, but also don't know exactly what kind of photons are produced by each individual quantum jumping.

    Also, as has been said, photons of definite energy do not exist in the real universe.

    So here my question: Is the inexactness of the eigenstates a (theoretical and empirical) necessity for systems that emit and absorb photons?
    (As opposed to bound systems that don't radiate, like particle in the box.)

    Because otherwise infinte long coherent length, i.e. unphysical photons.

  8. Mar 3, 2006 #7


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    Indeed, and they better don't be, because otherwise they could not go to the ground state. The point is that if two systems, S1 and S2, would have a hamiltonian H (of the combined system), such that its eigenstates take on the form of a product: |s1_e1a> x |s2_e1b> with eigenvalue E = e1a + e1b, then this would be a stationary state, and it could not evolve into something else. This would be the case, for instance, if H = H1 + H2 with H1 acting only on S1 and H2 acting only on S2. This is the case of non-interacting systems. If we take the eigenstates of H1, call them |s1_ena> and take the eigenstates of H2, call them |s2_emb> then the eigenstates of H take on the form |s1_ena> x |s2_enb>. And they are strict eigenstates (stationary states) of the overall hamiltonian. So they don't change over time.

    However, if there is an extra interaction term: H = H1 + H2 + lambda H12 which couples both systems, then one could hope for |s1_ena> x |s2_enb> to be GOOD APPROXIMATIONS but not exact eigenstates of H. The fact that they are good approximations is that they are *almost* stationary states (in that we can "recognize" them as relatively stable states of the subsystems, and not as some wildly evolving random state). But they will "slowly" evolve because they are not strictly eigenstates.
    This is why we can "recognize" the excited atom state as an "almost" stable state (and not some very chaotic random state of an atom) - it would be a stable state if we could switch off the coupling to the EM field - but we can't - - but why it finally IS NOT a stable state because it slowly evolves into other states. This is because of the interaction term H12, it is NOT a strict eigenstate of the overall hamiltonian.
    So the initial state |excited atom> |vacuum field state> will not remain so, and will get a growing component with |ground state> |emitted photon>.
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