Why are degenerate states more likely to be filled at a given temperature?

Consider if we have a simple two-level toy model, where the ground state has energy $$E_0 = 0$$ and the excited state has energy $$E_1 = \epsilon$$ and degeneracy $$g$$. The partition function for this system is $$Z = 1 + g e^{-\beta \epsilon}$$ where $$\beta = 1/k_BT$$. The probability of a particle being in any of the excited states at a temperature $$T$$ is given by $$P(\epsilon) = \frac{ge^{-\beta \epsilon}}{1 + ge^{-\beta \epsilon}}$$ which for various $$g$$ looks like

My question is, why does having a higher degeneracy of the excited state mean it is more likely to be populated at a given temperature? What physical principle tells us this should be so?

It's worth considering the high temperature limit, where $$k_{\textrm{B}}T \gg \epsilon$$ (or $$\beta \epsilon \ll 1$$). In this limit, all of the states have effectively the same energy, and since they have the same energy, they have the same probability of occurring. Thus, if $$g$$ is the degeneracy of the excited state, then the probability of being in any one of the states is $$1/(g+1)$$. This means that the probability of the system having energy 0 is exactly $$1/(g+1)$$, and the probability of the system having energy $$\epsilon$$ is $$g/(g+1)$$.

Thus, in making the transition from low temperature$$-$$where only the ground state is populated$$-$$to high temperature$$-$$where all states are populated equally$$-$$the probability for being in any one of the excited states must increase.

Just to make sure this is clear, let me briefly explain why all of the states must be equally likely in the high temperature limit. The (or a) physical context of statistical mechanics is one where we have a system in thermal and/or mechanical contact with a thermal reservoir (a much larger system) of temperature $$T$$. Roughly speaking, the average energy of a particle in the reservoir is $$k_{\textrm{B}}T$$.

We can imagine a collision between a particle in the reservoir with a particle in the system:

• If $$k_{\textrm{B}}T \ll \epsilon$$, then the reservoir particle doesn't have enough energy to bump the system particle up into an excited state, and so at low temperatures, the system stays mostly in the ground state on average.
• If $$k_{\textrm{B}}T \gg \epsilon$$, however, then the reservoir particle has much more energy than either level of the system, and so if it collides with a system particle, it can either change the state or not: the energetics basically allow either case, and so the net result is that it's equally likely that a system particle has either energy.

An 'excited state with degeneracy $$g$$' is actually a set of $$g$$ different states with the same energy level.

The more excited states there are which a particle can transition to, the more likely the particle is to make a transition to one of the excited states.

• Right so to rephrase the question, why is the particle more likely to make the transition? – Kai Mar 11 '19 at 18:12
• The transition rate will depend on the number of final states (Fermi’s Golden Rule). – flaudemus Mar 11 '19 at 21:45

$$P(\epsilon) = \frac{ge^{-b \epsilon}}{z}$$

where, $$z = \sum_\epsilon ge^{-b \epsilon}$$

So, for two states with some difference in energy, if $$g_1e^{-b \epsilon_1}>g_2e^{-b \epsilon_2}$$, then state with energy $$\epsilon_1$$ is more likely to be filled even though $$\epsilon_1$$ might be less than $$\epsilon_2$$

Now, if we take $$K_bT>>1$$, probability difference in the above two states is proportional to:(I have taken the first order aprox for exponential)

$$g_1e^{-b \epsilon_1} - g_2e^{-b \epsilon_2} =g_1 (1 - b\epsilon_1) - g_2 (1 - b\epsilon_2) = (g_1-g_2)(\frac{\epsilon_2}{K_bT} - \frac{\epsilon_1}{K_bT})$$

as T is very high, one can see that in this limitng condition, the effect of $$(g_1 - g_2)$$ is much higher than the energy difference and thus the higher degeneracy cases are filled preferably.