- #1
McLaren Rulez
- 292
- 3
Hi,
If we have a system on N atoms confined to dimensions much smaller than the wavelength corresponding to the transition, we see superradiant decay. Now, in these cases, we always assume that the excitation is symmetrically or antisymmetrically distributed. For instance, if we have one excitation among N atoms, an example of an initial state is
[tex]
\mid\psi\rangle=\frac{1}{\sqrt{N}}\sum_{j}\mid g_{1}g_{2}..e_{j}..g_{N}\rangle
[/tex]The claim is that the correct treatment of superradiance needs to assume that the atomic ensemble couples to the field in an indiscernible way. In other words, there is no way to know which atom emitted the photon. Why is this assumption necessary and why does it correctly describe superradiance?
Why can I not say, for instance, that the kth atom is excited and all others are in the ground state? Remember, we are not doing an actual experiment so for our purposes, all the atoms are just dipoles and no dimensions have been specified for the atoms.
Thank you!
If we have a system on N atoms confined to dimensions much smaller than the wavelength corresponding to the transition, we see superradiant decay. Now, in these cases, we always assume that the excitation is symmetrically or antisymmetrically distributed. For instance, if we have one excitation among N atoms, an example of an initial state is
[tex]
\mid\psi\rangle=\frac{1}{\sqrt{N}}\sum_{j}\mid g_{1}g_{2}..e_{j}..g_{N}\rangle
[/tex]The claim is that the correct treatment of superradiance needs to assume that the atomic ensemble couples to the field in an indiscernible way. In other words, there is no way to know which atom emitted the photon. Why is this assumption necessary and why does it correctly describe superradiance?
Why can I not say, for instance, that the kth atom is excited and all others are in the ground state? Remember, we are not doing an actual experiment so for our purposes, all the atoms are just dipoles and no dimensions have been specified for the atoms.
Thank you!