Is the energy density normalized differently in the quantum case?by Hypersphere Tags: case, density, differently, energy, normalized, quantum 

#1
Dec713, 11:26 PM

P: 178

Hi all,
This is all in the context of interaction between (twolevel) atoms and an electromagnetic field, basically the WignerWeisskopf model. In particular, I tried to derive the value of the atomfield 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
Dec1013, 11:36 AM

P: 178

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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