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Electric field of single photon in a cavity

  1. Aug 29, 2011 #1
    1. The problem statement, all variables and given/known data

    Assume that there is a single blue photon in a cavity of the size of 10
    micrometers. The total energy of oscillating electric fields due to this photon is
    one half of the photon energy.
    A. Estimate the magnitude of the electric fields associated with this single
    photon. Compare your result to the typical electrical field experienced by an
    orbiting electron in a hydrogen atom.
    B. At what cavity size does the electrical field in the cavity become compa-
    rable to that in a hydrogen atom?



    2. Relevant equations

    Energy of a photon [itex] U = h\nu [/itex]

    Energy stored in electric field E in volume V: [itex] U = \frac{V}{2}\epsilon_{0}E^{2}[/itex]



    3. The attempt at a solution

    So I just need to check this over because something doesn't seem right.

    If we equate the energy of a photon with the classical energy of an EM wave we get:

    [itex] U = \frac{V}{2}\epsilon_{0}E^{2} + \frac{V}{2}\mu_{0}H^{2}[/itex]

    However, the question states that the energy stored in the electric field is equal to half the photon energy. I.e:

    [itex] \frac{V}{2}\epsilon_{0}E^{2} = \frac{h\nu}{2} = \frac{hc}{2\lambda} [/itex]

    This makes sense as you;d expect the energy of a photon/EM wave to be divided equally between it's electric and magnetic components.

    Solving the above equality for the electric field strength E gives:

    [itex] E = \sqrt{\frac{2hc}{\epsilon_{0}\lambda L^{3}}} [/itex]

    Where we've assumed the box is cubic. Plugging in L = 10um and taking 450 nm to be the wavelength of a blue photon gives the answer: 9974.8 V/m for the electric field.

    As an approximation we take a classical estimate of the electric field of the hydrogen atom, using Coulombs law with the Bohr radius of 0.53 A. This gives an electric field of [itex]5.13x10^11[/itex] V/m. Which strikes me as a bit large.:yuck:

    The difference between these electric fields is [itex]2x10^8[/itex]. Keeping everything else constant that means that

    [tex] \frac{1}{L^{3/2}} = 2x10^{8} [/tex]

    Which means the size of the cavity has to shrink down to 3 um? Which seems a bit small but the relationship is cubic.


    Does this make sense. It seems a little off, maybe too simplistic an approach.
     
  2. jcsd
  3. Aug 30, 2011 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The factor 2 should not be there.


    It is right.

    You meant the ratio instead of difference did you not?

    The ratio can not be equal to length on some power. Write the equation for the ratio of the lengths.

    Anyway, your derivation is correct up to this last stage, good work!

    ehild
     
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