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Oscillating E Field Interpretation of Light

  1. Jul 12, 2014 #1
    I don't know how to explain the classical view of an EM wave quantum mechanically. First off, when you see the graphical representation, there is an E field (x-axis) and a B field (y-axis) oscillating in parallel. Parallel to both of those is the direction of motion (z-axis).

    http://missionscience.nasa.gov/images/ems/emsAnatomy_mainContent_EMwave.png

    So the z-axis is in units of meters I'm guessing, but what about the x and y axes? Are they just the intensity of the field, so coulombs/newtons and teslas? If so, wouldn't varying field strengths just translate to more/less photon density? Are photons popping into and out of existence periodically?

    Also, by this model, the strength of the E or B field depends on the time you absorb it, given its sinusoidal nature. Could someone clear this up for me?

    Thanks a bunch!!
     
  2. jcsd
  3. Jul 12, 2014 #2

    Drakkith

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    Photons are the quantized transfer of energy from an EM wave to matter. They do not pop into and out of existence with the variation in field strength.

    The X and Y axis are a measure of the intensity of the field strength. They are perpendicular to each other and to the direction of propagation, not parallel. Also, remember that the graph you see is exactly that. A graph. It is not what an EM wave "looks like". It just shows us the relation between the intensity of the different field vectors over time.
     
  4. Jul 12, 2014 #3
    Sorry, meant perpendicular.

    The intensity is what is confusing me. This graph could represent one photon, could it not? So why is the photon's energy varying with time when in quantum mechanics you have a definite energy of E=hf?
     
  5. Jul 12, 2014 #4

    Drakkith

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    The graph doesn't represent the energy of anything. The intensity of the fields are measured as a force, not as energy.
     
  6. Jul 12, 2014 #5
    Ahh, so does is this wave represent the same thing as the probability wave described by the schrodinger WF? Are they one and the same?
     
  7. Jul 12, 2014 #6
    Also something that confuses me is the magnetic field. We know that magnetism is just relativistic electric fields, right? This makes me think that there is a "magnetism" type parallel for all of the fundamental particles, like electrons.
     
  8. Jul 12, 2014 #7

    jtbell

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    No. If you have one photon (or a "few" of them), I don't think it is meaningful to talk about them in terms of classical E and B fields. When you have a bazillion photons, then you can describe their collective effect in terms of a classical electromagnetic wave of E and B fields. However, in that case you cannot divvy up those E and B fields and say that this chunk "belongs" to one photon and that chunk "belongs" to another photon.

    As far as I know, the only safe way to make a connection between photons and the E and B fields in a classical real-world electromagnetic wave (that corresponds to a very large number of photons) is via the total energy that they carry. A sinusoidal electromagnetic wave with amplitude (maximum value) ##E_{max}## of its electric field, impinging perpendicularly on a screen in vacuum, carries energy per unit area per unit time of
    $$\frac{1}{2}\epsilon_0 c E_{max}^2$$
    (This includes the contribution from the wave's magnetic field.) If this wave is monochromatic with frequency ##\nu##, then this corresponds to a number of photons per unit area per unit time:
    $$\frac {\epsilon_0 c E_{max}^2} {2h \nu}$$

    I don't think that is the case. I don't think the photons associated with the wave arrive in "bunches", as this picture would seem to suggest. However, I will gladly defer to someone who knows more about QED and quantum optics than I remember from graduate school.
     
    Last edited: Jul 12, 2014
  9. Jul 13, 2014 #8
    It just seems that the quantum and classical representations are incompatible, but I know I'm missing something.

    What does the wave represent in the classical view shown in my first post. If the amplitude is increased, then that would represent a higher number of photons no doubt. What is the wavelength? Is it the same as the De Broglie wavelength? Is the De Broglie wavelength the same thing as the Schrodinger probability wave?
     
  10. Jul 13, 2014 #9

    Drakkith

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    The graph in your first post is a representation of the amplitude and direction of the electric and magnetic field vectors over time. That's it.

    Increasing the amplitude while keeping the wavelength/frequency the same would require an increase in photons absorbed over time. If you double the intensity, the energy deposited by the wave in the same time period quadruples, which means that the number of photons absorbed is 4 times what it was before.

    Since the graph represents a classical wave, the wavelength is exactly what it sounds like. It is the distance the wave travels between oscillation peaks. This is not the same as either the De Broglie or Schrodinger probability waves, but I believe they can be reduced to a classical wave when dealing with photons. I'm not sure honestly. Perhaps someone with more knowledge of quantum physics can help.

    However, I can tell you that quantum physics ends up reducing to classical physics at the macro scale, so they are indeed compatible in that sense.
     
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