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Light and photon, confused

  1. Feb 14, 2005 #1
    I am confused about a few things....

    As I understand it, a ray of light is merely a stream of photons.

    Every photon that falls onto the detector will be detected and hence, a frequency will be registered. The frequency of light registered by a detector is actually the photon's frequency.

    Photon is a wavepacket.

    Questions:

    1.) How is it that in classical theories regarding light, we can picture successfully (to a certain extent), light in terms of a sinusoidal wave (infinite sine wave), when, light is only a stream of photons (particles) ?

    2.) In the classical theory of light, how did they detect the crest and trough of the infinite sinusoidal wave? In QM, we talk about photons. So, how does the concept of crest and trough in classical theory related to photon/QM?

    3) Photon is created when an electron drops down from one level to a lower level of energy, in an atom. If this is so, we should get a photon with a definite energy and frequency. Why is it that a photon is a wavepacket?
     
  2. jcsd
  3. Feb 14, 2005 #2

    Andrew Mason

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    How is it that we can successfully picture a tsunami in terms of a sinusoidal wave, when it is really a movement of water molecules?

    Light behaves like a wave and like a particle, but is neither. It is a photon. Waves and particles are macroscopic concepts and are only models that help us to understand and predict the behaviour of light and other phenomena. The wave-like properties of a photon are more evident when dealing with large numbers of them. When dealing with single photons, as occur in the interaction between photons and atoms, photons exhibit more particle-like properties.

    AM
     
  4. Feb 14, 2005 #3

    vanesch

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    The relationship between a classical EM wave, and a photon description is rather subtle (although very well defined). In fact, to a classical EM wave correspond special states which are called "coherent states" and which are superpositions of the vacuum, single-photon, two-photon etc... states.
    A single photon state does not really correspond to a classical EM wave, although in many respects you can get away with that simplified view. But when you talk about "crests and trough" in the EM wave, that's one of the aspects where you NEED the true description as a coherent state. In fact, you need that each time you make reference to the absolute phase of a classical EM wave. A single photon has no absolute phase.
    But things like relative phases, or intensities are dealt with with the simpler (but wrong) picture of 1 photon states as EM waves.


    [/quote]

    Because you can only get such an emission when there is an interaction between the atom and the EM field (which is NOT taken into account when you calculate the levels of the atom alone). This interaction will blur what you think to be a discrete energy level, and as such, your transition doesn't define a single value for energy with 52 decimal places, because it is now a transition between two blurred, approximate energy levels.
    As such, you have some bandwidth to make a wave packet with.
    Indeed, if the energies WERE absolutely discrete, upto 52 decimal places, then these would be STATIONARY STATES, and you wouldn't get any transition at all in finite time.

    cheers,
    Patrick.
     
  5. Feb 14, 2005 #4

    dextercioby

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    Light is both a stream of photons and a wave.Depends on how you look at it.

    Duality means that you cannot tell for sure the nature of light judging only one experiment.You disprove the wave character when discussing photoelectric effect and Compton effect,but turn to the macroscopical phenomena:diffraction is pictured nicely,when one thinks of light as a wave,but it's very rigurously described with photons,too...

    That's because,generally,quantum theories tend to generalize (to extend the domain to microscopcal world) classical ones and not viceversa...

    Because LIGHT IS BOTH...? :wink:

    They don't,to my knowledge.They measure light intensity...Which of course depends on those crests and troughs.

    No,we talk about classical em.field...In QED,we talk about photons...

    It really doesn't,once the quantization is performed,the connection is lost.If there still were a connection,we would not have been able to explain phoelectrical effect.

    The photon is a (quantum) particle.Period.There can be uniparticle photon states with definite helicity and momentum (hence frequency).


    Daniel.
     
  6. Feb 15, 2005 #5
    Light intensity in quantum is interpreted as the number of photons. Which means that what makes a crest a crest or a trough a trough is the relative number of photons.

    How is it that the periodic repetition of crest and trough let us calculate the wavelength and frequency of the light, when crest and trough is only the number of photons? And how is it that the wavelength and frequency calculated from the crests and troughs equals to the wavelength and frequency of a single photon?
     
  7. Feb 15, 2005 #6

    Doc Al

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    Do not think of a "crest" of an EM wave as a place with many photons and a "trough" as a place with few photons. Instead, correlate the number of photons to the average intensity of the EM wave (the power per unit area) which is proportional to the square of the amplitude (amplitude equals the height of the crest). The greater the intensity of the EM wave, the greater the number of photons.

    For an EM wave of a single frequency, the wavelength of the EM wave is the wavelength associated with the photons.
     
  8. Feb 15, 2005 #7

    So, is it that, all experiments performed on light, did not in any way, directly show the existence of crest nor trough? Because they found, wavelength and frequency and other wave behaviour of light, eg, from Young's experiment etc, that leads to scientists making an analogy between light and mechanical waves and water waves, where there are crests and troughs?
    Further, we can only detect or calculate the average intensity etc., but not the instantaneous values.
     
  9. Feb 20, 2005 #8

    Claude Bile

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    1) They are solutions to Maxwell's equations. (Actually a sinusoidal wave is not a solution in 3 dimensions, although infinite plane waves are). Sine waves are easily reconginsed mathematical functions.

    2) You can't directly measure the phase of an electromagnetic wave, one needs to use an interferometer. Only coherent light such as that from a laser has a definite 'crest' and 'trough'. These manifest as points of constructive interferance (bright fringes) and destructive interference (dark fringes). It dosen't make sense to measure absolute phase, only relative phase between two beams.

    3) The Uncertainty Principle says that in order to know energy and hence frequency with absolute precision, the electron must have resided in the upper state for an infinite amount of time. There is always some uncertainty when measuring frequency (Hence the wave-packet).

    Claude.
     
  10. Feb 21, 2005 #9

    jtbell

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    In practice, it's rather difficult to measure directly the instantaneous electric and magnetic fields in a light wave, which are oscillating at a frequency of circa 10^15 Hz and whose crests and troughs are flying past at 3 x 10^8 m/sec. :yuck:

    With microwaves, at least, it's possible to set up standing waves and detect the spatial maxima and minima in the amplitude of the electric field. It's a fairly common undergraduate-level lab experiment. With a fast-enough oscilloscope one can probably even observe the oscillation of the electric field more or less directly.
     
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