# Superradiance and the assumption of indiscernable atom field coupling

1. Mar 15, 2013

### McLaren Rulez

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
$$\mid\psi\rangle=\frac{1}{\sqrt{N}}\sum_{j}\mid g_{1}g_{2}..e_{j}..g_{N}\rangle$$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!

2. Mar 15, 2013

### Cthugha

This is a bit like asking, why we need to assume indistinguishability in the double slit experiment and cannot just have a look at two individual slits.

If you have an indistinguishable situation, that also means your atoms will coherently radiate in phase with each other. That makes a huge difference. For independently radiating atoms, you will get an intensity proportional to N (the number of atoms). Basically you just take the field of each individual atom, square it and sum up the intensities.

For indistinguishable atoms you cannot neglect phase and you cannot sort one field to one atom. So you have to sum up all the fields (or probability amplitudes if you want a quantum optics treatment). This sum will be proportional to N. Now you need to square afterwards to get the intensity, which will obviously go as N^2.

3. Mar 16, 2013