Just to clarify the terminology:
In a cavity (which is easier because of the somewhat delicate "infinite-volume limit" in QFT) you have
$$\hat{H}=\sum_{\lambda} \sum_{\vec{k}} |\vec{k}| \hat{N}(\vec{k},\lambda).$$
A complete set of energy eigenstates of the free em. field thus is given by the corresponding Fock basis,
$$|\{N(\vec{k},\lambda) \}_{\vec{k,\lambda}} \rangle=\prod_{\vec{k},\lambda} \frac{1}{\sqrt{N(\vec{k},\lambda)!}} \hat{a}^{\dagger N(\vec{k},\lambda)} |\Omega \rangle.$$
The energy eigenvalues are
$$E[\{N \}]=\sum_{\lambda,\vec{p}} |\vec{k}| N(\vec{k},\lambda),$$
i.e., each occupied single-photon state contributes ##E_{\vec{k}}=|\vec{k}|=\omega##.
The thermal state is given by the statistical operator,
$$\hat{\rho}=\frac{1}{Z} \exp(-\hat{H}/T)$$
with the partition sum
$$Z=\prod_{\vec{k}} \left [\frac{1}{1-\exp(-|\vec{k}|/T)} \right]^2,$$
where each factor comes from summing over ##N(\vec{k},\lambda) \in \{0,1,2,\ldots\}##. With the two cases ##\lambda=\pm 1## giving the same factor (that's why each factor is squared in the product).
A single-mode coherent state is given by
$$|\alpha,\vec{k},\lambda \rangle=\sum_{n=0}^{\infty} \exp(-|\alpha|^2/2) \frac{\alpha^n}{n!} \hat{a}^{n}(\vec{k},\lambda) |\Omega \rangle.$$
The photon numbers are Poisson distributed,
$$P[N(\vec{k},\lambda)=n]=\exp(-|\alpha|^2) \frac{|\alpha|^{2n}}{n!}$$
with
$$\langle n \rangle =\langle n^2 \rangle-\langle n \rangle^2=|\alpha|^2$$
The mean energy is
$$\langle E \rangle=|\alpha|^2 \omega, \Delta E=|\alpha| \omega.$$
What is now the precise question, we try to answer in this thread, concerning either of these states, loosely mentions in this thread?