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Light quantisation

  1. Nov 27, 2006 #1
    A hydrogen filled bulb emits discrete light wavelength, but why is it that a normal bulb emits the entire spectrum, i.e. continuous spectrum? I'm sure the gas in the bulb also emits discrete spectrum, since there's nothing so special about hydrogen.
     
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  3. Nov 27, 2006 #2

    jtbell

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    A normal incandescent light bulb doesn't use an excited gas. The light is thermal radiation from a hot metal filament. It's the mechanism that ideally produces a black-body spectrum, but I don't know how closely the spectrum from an incandescent bulb matches a black body.
     
  4. Nov 28, 2006 #3

    dextercioby

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    Certainly worse than the sun's photosphere which has roughly 5000K.

    Daniel.
     
  5. Nov 28, 2006 #4
    There should be emission lines in the spectrum, although they are probably relatively faint, from the reflection of the black body radiation by the gas particles inside the bulb.

    The black body radiation itself is also discrete. This was the first discovery for quantum mechanics with Planck's formula: [itex]E=h\nu[/itex], which solved the "Ultraviolet catastrophe" problem.

    Hope this helped.
     
    Last edited: Nov 28, 2006
  6. Nov 28, 2006 #5

    vanesch

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    Uh ! :bugeye: The blackbody radiation is not discrete (unless you're talking about box quantization).
     
  7. Nov 28, 2006 #6
    What is the deal with the E=hv then? I thought the whole point was that because the energy is quantised you can avoid the Ultraviolet catastrophe since the wavelength can not become vanishingly small since the energy is constrained to being integer multiples of h and if the energy is constrained in this way then so must the wavelength, and hence the frequency, and therefore the black body spectrum.

    Sorry if I added confusion. In hindsight I probably should have looked it up before responding... seeing as I am a self taught noob.
     
  8. Nov 28, 2006 #7

    vanesch

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    The frequency v doesn't need to be an integer!
    What you have, indeed, is that for each individual frequency, upon measurement, you will find an integer number of photons. However, this doesn't mean that the values of the frequencies are quantized (as long one is working in free space and not within a cavity).
     
  9. Nov 28, 2006 #8
    I see, I think I have totally misunderstood the "Ultraviolet catastrophe" problem if a photon is free to have any frequency it wishes.

    If the black body spectrum is continuous then at least /some/ photons should be emitted with very small wavelengths, and although their energies are not infinite, they must be quite large, right? So I guess that the intensity of the emission must just drop off quickly enough for it to not matter... i.e. these photons are so rare that their total contribution to the output energy of a black body radiator is small.

    In which case, why was it not possible to do a similar thing with the classical theory? Making the intensity drop off as a function of wavelength such that the intensity is zero for the zero wavelength isn't really that hard to concieve of...

    If the energy is only dependent on the frequency and the number of photons, then surely the intensity and the frequency of a classical wave could be used in place as an approximation? Since "number of photons" becomes the intensity of the radiation if we deal with large enough numbers... I struggle to see why having a number of photons with frequencies instead of a continuous wave with intensity and frequency dodges the problem.

    Sorry to hijack this thread... I have looked around a bit on the net, but I still don't understand any better.
     
  10. Nov 28, 2006 #9

    jtbell

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    Rayleigh and Jeans tried to derive the black-body radiation spectrum using only classical electromagnetic wave theory. The spectral equation that they came up with failed to agree with experiment.

    You can find a summary of the derivation of the Rayleigh-Jeans equation at

    http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html

    It's also discussed at greater length in many "introductory modern physics" textbooks.
     
  11. Nov 28, 2006 #10

    vanesch

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    The problem is that one wanted to obtain thermodynamic equilibrium. That is to say that one considered the degrees of freedom of the classical EM field and those of a "matter system at temperature T", and one wanted to find the configuration that is statistically most plausible. Now, one can derive that this comes down to identical amounts of energy per "degree of freedom" (the so-called equi-partition theorem in statistical mechanics), and when you give the SAME amount of energy to every classical radiation mode, then you get more and more energy per frequency interval, simply because there are more and more modes (in box normalisation) per frequency interval. This is the classical UV catastrophy.

    One way to avoid this, is to require discrete amounts of energy per mode which are proportional to the frequency. This changes the probabilities of having certain configurations in just the right way to obtain thermodynamic equilibrium. That's an artificial trick, but it is the trick that Planck used. Doing that, he found that he got exactly the right (continuous) distribution that was experimentally observed from a black body. It turns out that these discrete amounts of energy fit perfectly with the photons of QED (but that came only much later).

    What is much less known, is that there are other ways to establish this too. Another way is for instance to postulate a uniform background noise in the classical EM field, and to declare that what we call the EM field, is what is *above* this background noise (as we can only do differential measurements). If you do this, and require a background noise that is relativistically invariant, you also find the correct blackbody curve.
     
  12. Nov 28, 2006 #11
    Nice didn't know that!

    I've been trying to teach myself Quantum optics lately and the explanation of the photoelectric effect is one thing that has puzzled me for some time. Some books (Fox for example) claim that the photoelectric effect can be explained by the semi-classical theory, while others (Gerry/Knight for example) claim that it doesn't explain all aspects. So which is it?

    I know that the semi-classical model gives the correct discrete probability distribution and "instanteneous" ejection of electrons, but another argument of the need for the photon model usually put forward is that in the classical model the energy of the ejected electrons would depend on the intensity, whereas in experiments it is shown not to.

    Does the semi-classical model explain this?
     
  13. Nov 28, 2006 #12
    This seems to be the insight that I was missing. I will research this on my own now that I have a good direction to go in. Thanks very much for your help.
     
  14. Nov 28, 2006 #13

    vanesch

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    There is another thread on this forum somewhere about this

    https://www.physicsforums.com/showthread.php?t=144950

    I'm not an expert on it, but I read Mandel and Wolf (some while ago) and the semiclassical approach can explain quite some features (although an expert on the matter like Zapperz seems to suggest that this is not used in practice). So which details can, and which can't, be explained by the semiclassical approach are not entirely clear to me ; what I know is that the elementary features usually mentionned in textbooks, can be obtained also semiclassically (including the frequency threshold for instance).
     
  15. Dec 12, 2006 #14
    Sorry if I resume this thread.
    vanesh, do you know where is possible to find an abstract about it?
     
  16. Dec 12, 2006 #15
    yes, it does sound very interesting
     
  17. Dec 12, 2006 #16

    vanesch

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    http://prola.aps.org/abstract/PR/v182/i5/p1374_1

    http://prola.aps.org/abstract/PRD/v4/i6/p1597_1

    Now, mind you, it is not because this works in this specific example, that we can do away with quantum optics ! However, it shows you that there cannot be a *proof* based upon the blackbody radiation curve of the exactness of quantum theory, given that there are other ways to establish this particular effect.

    In fact, it turns out that this trick with some kind of background radiation can mimic quite (but not all) some quantum effects. It ruins quite some "proofs" of the necessity of quantum theory that are propagated in textbooks.
     
    Last edited: Dec 12, 2006
  18. Dec 13, 2006 #17
    Last edited: Dec 13, 2006
  19. Dec 18, 2006 #18

    Hans de Vries

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    There is a simple semi-classical argument to explain that an electron in
    a laser bundle can only absorb integer quanta in the form of photons
    as it does in QED:


    According to Aharonov-Bohm the deBroglie frequency of an electron in a
    strong laser field becomes a superposition of multiple frequencies. These
    extra frequencies are simply the same as the sidebands signals in good
    old FM (Frequency Modulated) radio.

    [tex]
    J_0(m_i)f_e\ \ +\ \ J_1(m_i)(f_e + f_p)\ \ +\ \ J_2(m_i)(f_e +
    2f_p)\ \ +\ \ J_3(m_i)(f_e + 3f_p)\ \ + .....
    [/tex]

    Where fe is the frequency of the electron, fp the frequency of the laser
    and the J's are the Bessel functions. These terms correspond to the
    absorption of 0,1,2 and 3 photons and so on. mi is the 'modulation index'
    which is the ratio eV/E, where E is the electron's energy. All the higher
    terms can be neglected in the not so strong fields with a small mi.
    (as is done by the Feynman rules)


    So the fact that an electron can only absorb an integer number of photons
    is actually quite "classical". The only thing which QM adds here seems to be
    that the electron apparently has to choose between any of its superposition
    states.


    Regards, Hans

    P.S: some frequency modulation links:

    http://www.vk1od.net/FM/FM.htm
    http://www.st-andrews.ac.uk/~jcgl/Scots_Guide/RadCom/part12/page1.html
     
    Last edited: Dec 18, 2006
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