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Models of Photon

  1. Jan 1, 2008 #1
    I believe, nobody knows exactly what photon is. Instead we have several MODELS of photon. Let us summarize these models.

    1. Classical Physics.
    There is NO photon in classical physics. There reason is that there is NO QUANTIZATION of field in classical physics. However there is an electromagnetic radiation in classical physics that is closely related to photon.

    2. Atomic Physics (or Bohr model of atom)
    There is some preliminary idea of photon. It is known that atoms can radiate during transition from level of energy E2 to E1. Total radiated energy should be (E2 - E1), however the radiated wave has not particle-like behavior.

    3. Schrodinger model of atom.
    Schrodinger equation does not describe photon, but photon does exist in the Schrodinger model.
    Energy of Photon is hw,
    photons are created when atom goes down from E2 to E1,
    photons are annihilated when atom absorbs them and goes up from E1 to E2
    FREE photons exist as electromagnetic radiation

    4. Heisenberg model
    I think that is similar to the Schrodinger model, except there is NO FREE photons. Because one of ideas of Heisenberg was to make a theory that describes ONLY what can be measured. We cannot measure photonic field without destroying it.

    5 QFT
    Inherits basic features of the Heisenberg model and generalize it for relativistic case.

    I suggest add more details to these models. I think in result we will see clearly that what we are talking about is not REAL PHOTON, but our simplified understanding of it, because a specific theory put some limit on what we can think about photon and what we cannot.
     
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  3. Jan 1, 2008 #2
    The electromagnetic wave travels with the speed of light, has energy hw and will excite an electron when it gets the chance to interact with it. Why not call this electromagnetic wave the photon? What exactly is the difference between the wave and the photon? Or, since we do not know what the photon is: why would we not want to call the wave the photon? Apart, maybe, from the fact that the wave may be many photons at the same time (can it?)?

    Harald.
     
  4. Jan 1, 2008 #3
    First, I would like to emphasize, that we are talking about MODELS, not about reality behind those models. That is why we should follow the RULES of those models

    So, in classical physics a wave of frequency w may have energy, for example, hw/10, hw/100, 10*hw, hw! and so on.

    In classical physics a wave has NO CHANCE to excite an electron, because there are NO electrons (inside atoms) in classical physics. Electrons (inside atoms) for the first time appeared only in ATOMIC physics.

    Because the expression 'photon' was already used in another branches of physics (not classical) for a very special object. This object has energy EXACTLY hw. And this object appears in very special processes like atomic radiation.

    Because Occam told us "entia non sunt multiplicanda praeter necessitatem", or "entities should not be multiplied beyond necessity".

    Besides that, using expression 'photon' in classical physics may lead to misunderstanding, because for photon E = hw, but for classical wave E not necessarily = hw.
     
  5. Jan 1, 2008 #4

    jtbell

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    When you "split" a classical wave (e.g. a light wave through a beamsplitter), and send parts of it towards two detectors, you can detect the wave in both detectors. When you send a single photon through a beamsplitter, you always detect the photon in either one detector or the other, never both. See for example, the experiment of Grangier, Roger and Aspect in 1986.
     
  6. Jan 1, 2008 #5

    olgranpappy

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    There are plenty more details available in the numerous textbooks on the topics of "Advanced Quantum Mechanics" and "Quantum Field Theory". A good resource is Messiah's "Quantum Mechanics" volume 2, the final chapter of which deals with field quantization. The meaning of a "photon" in this context is quite clear when quantization is performed in the Columb gauge; the electromagnetic vector potential becomes an operator which is a sum over (momentum and physical polarization dependant) photon creation and annihilation operators (weighted by appropriate factors). It is only after quantizing the electromagnetic field in this way, and in introducing photon creation/annihilation operators, that one can properly understand fully non-classical emmision processes such as spontanious emmision.

    ...Although, to his great credit, Einstein was able to determine the spontanious emmision rate of an atom long ago without recourse to field quantization by a beautiful little argument (sketched in Griffiths quantum mechanics book) that utilized Planck's results about "photons" in equilibrium.
     
  7. Jan 1, 2008 #6

    olgranpappy

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    Occam was just a man. One often does well to question the statements of other men rather than to simply nod along...

    But, anyways, I hope my previous post sheds some light on your question.
     
  8. Jan 1, 2008 #7
    Photon = A Ripple in the electromagnetic field
     
  9. Jan 1, 2008 #8
    I am curious about how the beam splitter KNOWS when it should split light into two parts and sent parts to TWO different detectors, and when it should sent all the light to ONE of the detectors.

    Note: this question is different from OLD question about how the beam splitter KNOWS which of detectors to choose to sent all the light to it.

    :bugeye:
     
    Last edited by a moderator: Jan 1, 2008
  10. Jan 1, 2008 #9

    jtbell

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    When the light is a single-photon state, the beamsplitter sends it one way or the other, choosing at random (as far as we can tell). When the light consists of many photons, about half of them go one way and about half of them go the other way, subject to statistical fluctuations as when tossing a coin.

    Or have I missed the point of your question?
     
  11. Jan 1, 2008 #10
    An example of spontaneous emission is the 2p-1s transition of the hydrogen atom in the absence of an external field. You can calculate the transition time (the reciprocal of what I think you call the probability or the alpha coefficient?) using classical wave theory simply by treating the superposition of the two states as a classical antenna. The dipole moment and frequency come out from the solution of the Schroedinger equation, and the transmitted power of the antenna is a very classical calculation.

    Why do you need photons to explain this?

    Marty
     
  12. Jan 1, 2008 #11

    olgranpappy

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    Because if the classical vector potential is identically zero (which it is for spontanious emmision) then there is no purturbation acting on the atom and it will never make a transition out of the higher state.

    As I said in my previous post, one can actually get the right answer semi-classically, but a proper explanation relies on the quantized electromagnetic field.

    Cheers.
     
  13. Jan 1, 2008 #12
    I understand quantum case... but what about classical wave?
    What did you mean when introduced classical wave? Was that just light of large enough energy? If so, can we reduce its energy using, for example, filters, which absorb 99% of light and transmitt 1% og light? If the energy of classical wave is very small (for example E = hw/100) is it still classical wave or at some small level of energy (E of the order hw) classical wave becomes a photon?l
     
    Last edited by a moderator: Jan 1, 2008
  14. Jan 1, 2008 #13
    Semi-classical model of atomic transitions is a pretty good one, but I have a question.

    Let spontaneous emission already started, and electron is in superposition
    state 2p - probability, for example 90% and
    state 1s - probability 10%
    In such situation superposition of the two states can be considered as classical antena.

    But what about the VERY BEGINNING of the spontaneous emission process when electron exist in PURE 2p state (probability 100%) and there is NO superposition of two states and NO classical antena.

    In this model we cannot explain HOW SPONTANEOUS EMISSION CAN BE STARTED.
    Of course, if it already started for some misterious reason, your model can easily explain how it can continue. But the question is HOW it can be started from PURE 2p state without superposition?
     
  15. Jan 1, 2008 #14
    I wonder if this is like asking if a pencil will ever fall over if you balance it perfectly on its nose. Do we really have to be able to answer this kind of question? Is it even possible to set up an experiment where a hydrogen atom is prepared in a pure p state with no mixture of any other state?

    Have you considered that if you could isolate an atom by itself in the pure p state, maybe it WOULD stay that way forever? Or at least for a very long time. It almost seems like more of a philosophical diversion than a question of physics.

    The other point to remember is: a semi-classical calculation and the "correct" QFT calculation both seem to give the same result. Yet people object to the description that goes along with the semi-classical calculation. I think we should remember that the description that accompanies the "correct" QFT explanation (Copenhagen interpretation?) also contains some pretty disturbing aspects. Who can really say that one is preferable to the other?
     
  16. Jan 1, 2008 #15

    olgranpappy

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    Okay. okay. You can't ever completely isolate a system. so what? So maybe it would decay and maybe it wouldn't but since the completely utterly isolated atom can never be prepared then the answer doesn't matter. In fact, the question obviously can not ever be answered at all and for this reason appears to be somewhat inappropriate and red-herring-esque.

    Even thought such a perfectly isolated pure 2p atomic state can not be prepared in practice, it can be prepared in theory. And the well-known standard QFT gives an unambiguous answer to the question of whether or not such an atom will decay. And the answer is, "yes, it will, by spontanious emmission". And one can easily calculate the emmision rate via Fermi's golden rule.
     
  17. Jan 1, 2008 #16
    I think it is possible. For example, atom in EM field that correspond to the transition 2p-1s would slowly oscillate (Rabi oscillations) between 2p and 1s states. If we turn off the EM field at the time when the atom is 100% in 2p, it will remain in 2p for a while.

    In semi-classical model, if there is NO superposition it SHOULD stay in the state 2p forever.
    In QFT model it should not.
    So, semi-classical approach may be considered as a kind of 'philosophical diversion'

    QFT calculations give better precision.
    QFT is correct NOT because some GURU told us that is correct, but because better precision... so, the judge in this matter is EXPERIMENT

    If we need precise result, QFT is preferable.
    If we need quick estimate, semi-classics is preferable...

    IHMO
     
  18. Jan 1, 2008 #17
    In atomic trap several deep cooled atoms may stay for many hours. I think, if we put only one atom in such trap, it can be considered as isolated system for our purposes. Because lifetime of the transition 2p-1s is of the order 10^(-6) sec, and lifetime of the atom in the trap is of the order 10^4 sec ...
     
  19. Jan 1, 2008 #18

    jtbell

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    They're waves of electric and magnetic fields that satisfy Maxwell's equations. When an electromagnetic wave hits a boundary between two media, you apply the boundary conditions on the E and B fields and find that some of the wave is reflected and some is refracted (transmitted), If you know enough about the properties of the media, you can calculate reflection and transmission coefficients.

    Light is what it is, whatever it is. I assume that its fundamental nature, whatever it is, is the same regardless of intensity. We have two descriptions of light which allow us to calculate how light behaves:

    1. The classical description in terms of electric and magnetic fields using Maxwell's equations.

    2. The quantum description in terms of photons, using quantum electrodynamics (QED).

    As far as I know, the quantum description "works" (gives good predictions for experimental results) wherever it has been tested. The classical description "works" only when the light is "strong" (corresponding to many many photons in the quantum description).

    Even when the light is "strong", the two descriptions give different predictions, but the differences are so tiny that they cannot be detected, except perhaps with very carefully designed experiments. The classical description is much easier to work with, so we use it in practice unless the quantum effects are significant.

    When we switch between the two descriptions as we make the light stronger or weaker, that doesn't mean that light has changed its fundamental nature. We're simply re-evaluating the tradeoff between absolute correctness and practical convenience.

    Because the quantum description currently "works" for a larger range of phenomena, I think it is likely to be closer to the "true nature" of light (whatever it is), than the classical description is. But this is just my opinion. Future discoveries (experiments or theoretical work) may change this.
     
  20. Jan 2, 2008 #19
    gr8 post.. i had many confusions with what a 'photon' actually was.. however.. i'd be thankful if u cud answer a few queries..

    Q] Why do we consider that photon has mass '0' but still explain mainy of it's properties based on collisions with other particles? How can a particle having '0' mass collide with another particle? Or, is collision purely an electrostatic phenomena? If it is, why do we care to call photon as a particle?

    thx..
     
  21. Jan 2, 2008 #20

    malawi_glenn

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    Collision is not collision in classical sense, collision is exchange of virtual gauge bosons, the force transmitters. When a photon and an electron "collide", it is not like two balls hit eachother, because non of them have spatial extension, but they iteract via gauge boson exhange.
    (quarks and leptons are point like particle)
     
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