How Do Thermal States Affect Single-Mode Light Cavities?

[guruoleg]

The Hamiltonian (ignores vacuum energy), H = \hbar\omega_{p}a_{p}^+a_{p}, represents some cavity at temperature T. For simplicity assume the cavity only supports a single mode.
H = \hbar\omega_{m} a_{m}^+ a_{m}

1) Given that in thermal equilibrium the probability of a system to have energy E is proportional to ~exp(-E/kT), give an expression for \rho_{m}(n) where:
\hat{\rho}_{m} = \sum_{n}\rho_{m}(n) | n>_{m} _{m}<n|
The constant is independent of n
2) Find this constant by imposing condition Tr( \rho_{m}) = 1

For the first part since the constant is independent of n and the Hamilitonian is governed by \hbar\omega a^+a where a^+a equals n so in order to isolate the n component from the Hamiltonian and write an expression \rho I think we should start by evaluating <m|\rho|n> where we will get a summation expression that needs to be evaluated. This is what I think we should do and I tried it but it is not taking me anywhere..I am afraid that I don't even know what I am doing. I just need a small hint to get started and then I will be all set...

 

[Galileo]

You know that the probability that the system has energy E is proportional to exp(-E/kT).
What are possible values for E and what is the probability of finding the system to have energy a particular energy E? (In terms of rho(n))

 

[guruoleg]

Thanks Galileo, I believe I have gotten the answer since E (non-vacuum state) is:

\hbar\omega n

We could just assume that \rho (n) = C*exp(- \hbar\omega n) and then it is easy street from there on...