# In quantum optics if you measure the energy of a thermal state does it collapse to a Fock state (Photon number state)?

As I understand it a thermal state is a state of light such that the light is in a superposition of many Fock states (Photon number states) with the probability of the light being found in a particular Fock state (i.e. having a certain photon number) following a Boltzmann distribution.

So surely if you had light in a thermal state in an experiment and measured it, it would collapse into a single Fock state, is this correct?

I understand creating Fock states experimentally is of some interest at the moment and it seems like it would be trivial if this was the case.

It should be noted that the current approaches to creating Fock states involve the strong coupling of an atom or quantum dot with a cavity and passing some squeezed state $a|\alpha\rangle = \alpha |\alpha\rangle$ into the system; here, the nonlinearities introduced by the atom-cavity interactions reject states that aren't particular Fock states. This is often taught in a first course in cavity QED as an application of strong interactions between a two-level system coupled with a cavity, but there has been some recent work in applications of these systems (c.f. Vuckovic, et al 2007.) to other problems.