The chemical potential, $\mu$, is equal to the free energy $F$, per photon.
$$F = \mu = u-Ts$$
The energy per photon is $u=h\nu$ and $s$ is the entropy per photon. Say we have a process where an incoming photon ($\nu_{1}$) is absorbed by a dye and then emitted at ($\nu_{2}$). Then at equilibrium,
$$h\nu_{1} - s_{1} = h\nu_{2} - s_{2}$$
What happens at equilibrium if the dye immediately emits two (or more?) photons, so the total number of photons is no longer conserved? Can I still write that the chemical potentials of all the photons are equal as in the equation above?
An attempt at a solution:
The change in Gibbs free energy at equilibrium is $\Delta G=0$. Since this is a homogeneous system, the Gibbs free energy is given by
$$\Delta G = \sum_{i} \mu_{x}x_{i} - \sum_{k} \mu_{y}y_{k} $$
Where $x_{i}$ is the number of particles of product and $y$ is the number of particles reactant. If one photon just becomes one photon then $i=k$, and so
$$\sum_{i} \mu_{y}y_{i} = \sum_{k} \mu_{x}x_{k} \implies \mu_{x}=\mu_{y}$$
and so the chemical potentials match. Now, if one photon becomes two, so $2k=i$, then
$$\sum_{k} \mu_{y}y_{k} = \sum_{2k} \mu_{x}x_{k} \implies 2 \mu_{x}=\mu_{y}$$. This implies that
$$h\nu_{y} - s_{1} = 2 \times (h\nu_{2} - s_{2})$$
at equilibrium.
