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Is the energy density normalized differently in the quantum case?

  1. Dec 7, 2013 #1
    Hi all,

    This is all in the context of interaction between (two-level) atoms and an electromagnetic field, basically the Wigner-Weisskopf model. In particular, I tried to derive the value of the atom-field interaction constant and show that it satisfied
    [tex]|g_\mathbf{k}|^2=\frac{\omega_\mathbf{k}}{2\hbar \epsilon_0 V} \left( d^2 \cos^2 \theta \right)[/tex]
    where [itex]d[/itex] is the dipole moment and [itex]\theta[/itex] is the angle between the dipole moment and the polarization vector.

    These notes claim that the vacuum field amplitude satisfy the normalization
    [tex]\int \epsilon_0 E^2 d^3r = \frac{\hbar \omega}{2}[/tex]
    which does lead to the above form of [itex]|g|^2[/itex], but from classical electrodynamics (eg. eq. (6.106) in Jackson, 3rd ed.) I'm used to defining the energy density of the electric field as
    [tex]u_E=\frac{1}{2} \epsilon_0 E^2 [/tex]

    Now, the notes seem to use a energy density that is [itex]2u_E[/itex]. Is there a good explanation for this, or does it boil down to one of these conventions? Thanks in advance.
  2. jcsd
  3. Dec 10, 2013 #2
    Actually, the author of those notes probably just switched to a complex field
    [tex]E_V=\sqrt{\frac{\epsilon_0}{2}}E + i\frac{B}{\sqrt{2\mu_0}}[/tex]
    in which case the energy density comes out as
    [tex]u=\int |E_V|^2 d^3 r = \int \left( \frac{\epsilon_0}{2}E^2 + \frac{B^2}{2\mu_0} \right) d^3 r[/tex]
    as it should.
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