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What happens to a bound electron when

  1. Aug 6, 2004 #1
    What happens to a bound electron when a photon comes along but doesn't have quite enough energy to make it go up a level? What happens to the photon? Quantum mechanical and simple answers welcome.
     
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  3. Aug 6, 2004 #2

    Tom Mattson

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    In that case, the photon will scatter and the atom will recoil.
     
  4. Aug 6, 2004 #3
    That works in a gas, but what about in water or in a solid, like glass?
     
  5. Aug 6, 2004 #4

    Tom Mattson

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    It works everywhere. The photon imparts momentum to an atom when it scatters. If the atom is tightly bound in a lattice, then some of that momentum gets imparted to the rest of the atoms in the lattice. Imagine a chain link fence with a golf ball at each link. Then imagine shooting one of the golf balls with a BB gun. The BB imparts some momentum to the struck ball, and the whole fence will start jiggling.
     
  6. Aug 7, 2004 #5
    Ok, so if a beam goes straight in glass, rather than completely scatter in all directions, it is simply because of its momentum. Conservation of momentum makes scattered photons keep travelling along the same path. Right?
     
  7. Aug 7, 2004 #6
    If I could help you somehow. I think when a photon hits an electron around a nucleus the distribution of an electron cloud is polarized. As is in classical theory, this polarization causes new electromagnetic field which probably means new photon is emitted. According to a book of optical physics in my mother language, elastically scattered electromagnetic field and newly created field interfere to create new electromagnetic field which is described by classical electromagnetic theory in medum, I mean that the wavelength or phase of it, I am not sure, is changed. Because of the sequence of this phenomenon repeated successively, new field with different wavelength is created so that snell's law is satisfied.
     
    Last edited: Aug 7, 2004
  8. Aug 9, 2004 #7

    Claude Bile

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    Also, if you 'pump' a crystalline medium hard enough with an intense source such as a laser, odd things can begin to happen.

    Depending on the symmetry of the crystal lattice, the polarisation that is induced by the electromagnetic wave is not necessarily linear.

    For example, a commonly used second order non-linear process is frequency doubling and is often used in lasers to acheive shorter wavelengths. For example an Nd:YAG laser of wavelength 1064 nm is often frequency doubled to produce green light at 532 nm.

    In other words, two photons incident on a single electron can cause it to emit a single photon at double the frequency under the right conditions.

    The interaction between light and matter is a very rich topic, from both a classical and quantum perspective. Really, this is only the tip of the iceberg.

    Claude.
     
  9. Aug 11, 2004 #8
    When light reflects off of matter, do the photons just bounce off, or are they absorbed by the electrons and then re-emitted?
     
  10. Aug 12, 2004 #9
    They are absorbed and re-emitted.
     
  11. Aug 12, 2004 #10
    Tom,
    Recoil implies momentum transfer. I'm into wave packets just now and would appreciate a bit of help.
    With this result in mind, is the scatter 100% elastic?
    What is the vector pattern of that scatter?
    Is there a very small energy barrier to absorption or is it an exact match absorption?
    When absorption is possible, does the cross section of the photon match the space over which the electron orbits the atom or does the photon only need to match the cross section of the electron itself?
    Thanks!
    Vince
     
  12. Aug 12, 2004 #11
    Wouldnt this mean that a very very small percentage of photons are reflected off of the suface of everyday things? Since the electron is the smallest part of the atom...the probability of it being hit by a photon is not very likely.
     
  13. Aug 12, 2004 #12

    ZapperZ

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    Gonzolo: this is a clear example on why there is no such thing as a SIMPLE question! :) The simpler the question, the more complicated and complex the answer can get. This is because the question leaves open the nature of the system of the bound electron, i.e. under what condition is this going on? Is the bound electron part of a solid structure, is it merely part of a free atom, or is it bound to an electron partner as in a Cooper pair? Each one of these will give you different answers to your question.

    I will not go into the question of optical conductivity/transmission/reflection/etc that seems to keep appearing every so often, because I've described the major mechanism for this a few times already (Integral and a few others can verify this). Suffice to say that, especially for metals, it isn't due to atomic transition and has more to do with the conduction electrons/plasmon states/phonon modes.

    But coming back to your original question, after reading the answers you got, I will offer you a DIFFERENT one that hasn't been covered here. In this scenario, the bound electron is part of a solid, or to be more specific, part of an intrinsic semiconductor. So at very low temp (T~0K), you have a filled valence band (filled with bound electrons), and an empty conduction band. These two band are separated by an energy gap (typically 1 to 5 eV). It means that only a photon of energy larger than this energy gap can excite an electron from the valence band to the conduction band. If the energy of the photon is less than the energy gap, then 2 possible things can happen (there may be more, depending on the material): (i) the photon simply pass through the material (ii) the photon may be absorbed by the phonon modes of the semiconductor.

    But here's where it gets interesting, and even for just this one scenario, the answer to your question isn't that simple. If a photon with energy LESS than the energy gap comes in, there is sometime an appreciable probability that an electron CAN in fact be excited to a state IN THE GAP, even though there are no available states there. What happens is this. When the electron is excited, it leaves behind a positive hole in the valence band. What you have now is an electron-hole pair, which is nothing more than an electron in a central force bound by a positive charge. This creates an additional Rydberg-type (hydrogenic) energy states within the gap for that electron. So the electron can occupy an energy state within the "forbidden" gap of the semiconductor. This electron-hole pair is what we call an "exciton".

    Of course, the lifetime for such a pair is very small. However, the existence of the exciton can be easily detected depending on the system and depending on the light source. Today's high-powered laser can easily create these. This brings us to another interesting scenario. If the photon density from the light source is sufficiently high, you can end up with a situation where, after the first photon has created an exciton, a second photon can come in and excite the electron to another state further. If the sum of the two photon's energy is equal or greater than the energy gap, then you can end up with an electron in the conduction band, all done using photons with energy LESS than the band gap! Fancy that! This process is similar to the multi-photon photoemission using photons with energy less than the material's work function (another topic that I've exhaustedly described on here previously).

    So hopefully, by now, you (and anyone else who thinks there's such a thing as a "simple question") realize that there may be no such things as a simple question in physics, especially when you open the question with so many different scenarios. :)

    Zz.
     
    Last edited: Aug 12, 2004
  14. Aug 12, 2004 #13
    Yah, I guess I had an unconcious hint that the question needed a situation ("boundary conditions"). Learning and revising a lot of stuff though. I guess I thought I might be able to figure the answer myself without a little input.

    The question came to mind when I realized I couldn't explain properly why a beam does doesn't scatter all around in glass. Here's my specific problem :

    1: Are photons continuously absorbed and emitted when a beam travels through glass?

    1.a.: If so, why are they always emitted forward rather than in a random direction?

    1.b.: If not, how are the atoms affected without changing energy levels?

    I'm comfortable with optics, n, refratction usually, but for some reason this doesn't click. I just want to understand why light goes STRAIGHT through glass. Both from a collision/quantum point of view or EM .

    shchr seemed to be going in the direction I was looking for. (explaining rectilinear propagation)

    jtolliver' response raises 1.a above (for reflexion)

    ZapperZ, you thought me an exciton was simply a lousy e-hole pair with the e in the gap. I always thought it was somekind of beast. Two-photon absorption! That is sooo neat! (but not quite what I'm looking for for now)
     
  15. Aug 12, 2004 #14

    Claude Bile

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    You have to be careful by what you mean by the terms 'absorbed' and 'emitted' as they refer to specific interactions between a photon and an atom.

    If by absorb you mean that the photon excites the atom into a higher energy state, then re-emitted, then no, the photon is not absorbed and re-emitted. This is due to the simple observational fact that spontaneously emitted photons are isotropically radiated (i.e. they have no preferred direction). For a ray of light (e.g. a laser) travelling through glass, this is clearly not the case as most of the light will emerge from the glass the same direction that it went in.

    Quantum mechanically, an electron is a cloud. This cloud has an intrinsic E-field (due to the charge of the electron). When a photon (i.e. an E-field oscilating at frequency v) is incident on the electron, the electronic E-field also begins to oscillate at frequency v.

    Because the electron cloud is displaced, the atom becomes polarised. Furthermore the polarisation also begins to oscillate at frequency v (in the linear case). Like any wave, this 'polarisation wave' propagates through the medium (it's direction will be different due to refraction). When the polarisation wave reaches the other boundary, the electronic E-field acts as an optical dipole and a photon emerges from the other side.

    Thus, for all intents and purposes the photon has travelled through the medium (albeit with a reduced velocity).

    Claude.
     
  16. Aug 13, 2004 #15

    ZapperZ

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    Unfortunately, as I suspected, this is an optical conductivity question which, as I've said, I've explained several times already.

    Let's look at the specific case of glass, which is a dielectric. There are NO conduction electrons here, so what I described earlier about metals and conduction electrons/plasmon states do not apply. What is more important in the optical conductivity of glass is the phonon states, or modes, that are available in the crystalline structure. This requires that I backtrack and explain what phonon modes are .. (now do you understand why I find this so tedious?)

    If you look into a solid state physics textbook, you'll find, under the chapter on lattice vibrations/harmonic crystals, a mode of oscillation of the lattice ions corresponding to optical phonons. If light, for example, has a frequency equal to one of these modes of vibrations, then there can be coupling between the photons and the phonons (phonons are quantized lattice vibrations). When this occur, at least two things can happen: (i) if you have a well-ordered structure, the vibrations may be transmitted throughout the crystal as heat or (ii) you get "retransmission" of that light in the SAME direction, due conservation of momentum (there will be some scattering of course). If the frequency of light does not match one of these modes, then the lattice vibrations absorbes but cannot sustain such vibrations and it is damped down and the photon is absorbed as heat.

    This is why ANY material you have will have a finite range of transmission. A piece of glass that you find transparent over visible range can be, in fact, opaque in the UV range. It is because the crystal structure doesn't support phonon modes in the UV freq., for example.

    This, btw, is one of the clearest example on why light (or any EM radiation) has an oscillating E-field - the optical phonon modes are excited in a "dipole" fashion. Without it, optical transmission, and all optical conductivity studies, make no sense.

    Zz. [wishing that this will be the last time he has to explain this]
     
  17. Aug 13, 2004 #16

    ZapperZ

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    That's it! I'm convinced that I'm psychic! :)

    In the just released (today) online issue of Science, there is a very good article by J.P. Eisenstein on exitons and the current progress in the study and application of these "beasts". I highly recommend it! [1] It proves to you that I'm not making all of this up as I go along. :)

    Zz.

    1. J.P. Eisenstein, Science, v.305, p.950 (2004)
     
  18. Aug 14, 2004 #17
    Thanks ZapperZ, I suspected this :

    "(ii) you get "retransmission" of that light in the SAME direction, due conservation of momentum (there will be some scattering of course)."

    ...but needed it to be told for confirmation. And I was forgetting the link to phonons, which seems to raise the question of how glass (and water), which is amorphous, can have lattice mode vibrations? I suspect the answer is a detail I am missing.

    BTW, I'm not holding you at gunpoint for an answer! It's not my intention to bore you!

    Claude Bile, thanks. But the missing link (for me) in "Like any wave, this 'polarisation wave' propagates through the medium (it's direction will be different due to refraction)." was conservation of momentum. Some waves travel straight, some radially, I needed to understand why the former in glass. c.o.m. is the short answer. I did check out perturbation theory (from your suggestion in another thread) and it helps a lot.
     
    Last edited by a moderator: Aug 14, 2004
  19. Aug 14, 2004 #18
    When the photon has the energy and wavelength needed to make an absorption possible, what are the requirements to start the absorption process?

    Q1: Does the wavelength of the photon have to match or slightly exceed the 3D spatial size of the electron orbit that is absorbing the photon?

    Q2: Does the wavelength of the phtoton have to match or slightly exceed the 3D spatial size of the electron itself?

    Q3: What is the reason that the electron absorbs the energy of the photon?

    Q4: What happens to the electron when it absorbs the energy of the photon? Does it orbit shape or size change?


    Your advice on any of these would be appreciated.
    ...
     
    Last edited by a moderator: Aug 14, 2004
  20. Aug 15, 2004 #19

    ZapperZ

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    That's a good question. Remember that even in liquid, there has to be some intermolecular forces (H bonding, or van der waal type), or else the material would be a gas. The vibrational energy between these molecules, and even intramolecular vibration itself are relevant here (recall how microwave oven work). So while there isn't a well-defined lattice in a liquid like water (we are ignoring LCD's here), there are still vibrational states involved here.

    Zz.
     
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