# Thermal States of Light

1. Nov 5, 2007

### 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...

2. Nov 5, 2007

### 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))

3. Nov 6, 2007

### 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...