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We have a molecule that is emitting/absorbing photons. We know the Hamiltonian and that there are several levels. We count the emitted photons at different angles and frequencies. We can also do scattering with a beam of photons. From the absorption data, can we determine if the system occupies each energy level with a thermal probability distribution? I.e. I mean that an energy eigenstate $i$ with energy $E_i$ is occupied with probability proportional to $e^{-\beta E_i}$. How do we do this experimentally?

As a simple example, how can we tell if a hydrogen atom is in a thermal state? Can we generalize this to bigger molecules?

sebastianspiegel
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Interesting question. I would have thought that if you were aware of the exact number of energy states and the populations thereof, you could apply boltzmann statistics to each of the levels in order to fit an appropriate temperature to the population in each state. This temperature, if comparable amongst the included levels, would therefore require that the system is in thermodynamic equilibrium. The temperature defining the populations of the states in this way is called the excitation temperature.

stars83clouds
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The only temperature that would make sense in your question is the temperature of the thermal distribution of the photons (if the photon gas is thermalized). So let us assume we have a photon gas at equilibrium. Its chemical potential is $\mu=0$ because photons have no ground state.

Consider now the atom. One of the characteristics of an atom is that the energy level are discrete and the energy gap between two level is finite. For instance in a harmonic oscillator, the gap is equal to $\hbar \omega$. To emit or absorb a photon, the atom changes its energy by a non-zero amount of energy. Therefore, the atom acts as a system with finite chemical potential.

Since the chemical potentials of the atom and of the photon gas cannot be equal, equilibrium between them is not possible.

Tom-Tom
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