Computing microstate probabilities based on Boltzmann distribution for chemical systems - Is it rigorous?

Question

One approach to predicting the folded structure of a polymer (DNA, RNA, protein) is to compute the probability that any particular part of the polymer $x_i$ is "paired" with another part of the polymer $x_j$, per molecule of polymer in a solution at equilibrium. This pairing probability is often computed by enumerating all conformations where $x_i$ is paired with $x_j$, summing over the free energy of all such conformations, and then evaluating a Boltzmann distribution.

More precisely, to compute the probability that a polymer adopts a particular conformation $s$, per molecule of polymer in solution at equilibrium, one computes:

$$P(s) = \exp(-G_s/RT)/Z$$

where $G_s$ is the free energy of conformation $s$ and $Z$ is the partition function a.k.a normalization factor needed so that $P$ is a probability distribution.

The question: Is this a rigorous way to compute $P(s)$?

The reason I doubt the rigorousness of the above computation is that I can get widely different results for $P(s)$ by making very slightly different assumptions about the polymer in question.

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Why I believe $P(s)$ is ill-defined:

First, let's take a very simple polymer with just two components and two states: UNPAIRED and PAIRED. Say that the chemical potential of UNPAIRED is $-1$ kcal mol$^{-1}$ and the chemical potential of PAIRED is $-1$ kcal mol$^{-1}$. Then $P($UNPAIRED$)$ is $1/2$, which seems very reasonable.

But, let's say that we add a little bit more complexity to our model for the polymer and split PAIRED into 5 states, which are each very slightly different versions of PAIRED (but they are, in fact, different). These new states, PAIRED$_{i}$ for $i = 1, 2, ... 5$ still have chemical potential $-1$ kcal mol$^{-1}$, and UNPAIRED still has chemical potential $-1$. Now, $P($UNPAIRED$)$ is $1/6$.

Both models accurately describe the polymer, however the latter is slightly more precise. However, it seems that $P(s)$ is sensitive to such changes in the model. To what extent, then, is $P(s)$ rigorous?

Answer 1

$P(s)$ is rigorous. You changed the model so you get different results. In the first case you have two states that have the same energy. The result, as you note, is that the probability of finding the system in a given state is 1/2. In the second model you have six states with the same energy. Again, the result as you note, is that the probability of finding the system in a given state is now 1/6.

Different models, different results.

Answer 2

(I did some of my own research, and came up with some comments:)

My confusion, it turns out, was between energy of the polymer (total internal energy) and Gibb's free energy (thermodynamic free energy). If you split the PAIRED state into 5 smaller substates, each may well have the same internal energy (because they are very similar), but these states have a different value of Gibb's free energy. Because the PAIRED state was split into 5 states, each of these states will contribute some fraction to the "useful work" the solution filled with polymer can perform (free energy), and the sum of the "useful work" done by the 5 new states is equal to the "useful work" the original PAIRED state could perform, i.e. the 5 new states must split the $-1$ kcal mol$^{-1}$ Gibb's free energy contributed by the original PAIRED state.

Redoing my analysis:

With PAIRED and UNPAIRED, each at $-1$ kcal mol$^{-1}$, PAIRED and UNPAIRED occur with probability 1/2.

With PAIRED$_{i}$ for $i = 1, 2, ... 5$, assuming that the energy of PAIRED$_{i}$ is the same as that of PAIRED, the PAIRED$_{i}$ will partition the Gibb's free energy of $-1$ kcal mol$^{-1}$, so UNPAIRED still occurs with probability 1/2, and the PAIRED$_{i}$ will occur with some probability less than 1/2, depending on how the $-1$ kcal mol$^{-1}$ free energy is distributed among the 5 new states.