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How does the amplitude on a EM wave graph correspond to the number of photons?

  1. Aug 20, 2012 #1
    Hello all! I am trying to understand ER on a more intuitive level. I can see the relationship between energy and frequency. The relationship between amplitude and photon number is less clear. So far I have E = hf. I understand that the intensity of light is a function of the number of photons. Can this be correlated to amplitude? If so, then how? Thank you.
     
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  3. Aug 21, 2012 #2

    Drakkith

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    Staff: Mentor

    What amplitude are you referring to?
     
  4. Aug 21, 2012 #3
    The intensity of a light beam depends on the number of photons in the light beam; The energy depends on the energy [frequency] of the individual photons.


    More here:
    http://en.wikipedia.org/wiki/Photoelectric_effect
    Photoelectric effect
     
  5. Aug 21, 2012 #4

    Physics Monkey

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    curiousatlarg,

    I assume by amplitude you mean the magnitude of the electric field, let's call it E. In fact, E is related to the average number of photons via [itex] N \sim E^2[/itex]. Please note also the very important word average. A macroscopic electric field does not correspond to an exact number of photons but instead to something called a coherent state of photons (we can discuss this later if you want).

    My claim above can be understood as follows. Let's think about a EM wave of fixed frequency. If you accept that the energy of one photon is [itex] h f [/itex] then the energy of N photons is [itex] N h f[/itex] and the rate of energy flow, which is proportional to the intensity, is [itex] I \sim c N h f[/itex] ([itex] c [/itex] is the speed of light i.e. how fast the photons are going). Now you also know that the intensity is given by [itex] I \sim E^2 [/itex] in a classical EM wave, so by equating these two forms you find that [itex] E^2 \sim N[/itex] up to lots of factors that you can work out or look up somewhere.

    Hope this helps.
     
  6. Aug 23, 2012 #5
    Thanks. I was referring to both the electric and the magnetic fields. What you gave me so far is what I was looking for. I will work with it some.
     
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