# Boltzmann distribution and probability of finding the system with specific energy

For sake of simplicity assume classical discrete systems.

If we have a system ($$\text{S}$$) coupled to a reservoir ($$\text{R}$$), then a microstate of the combined (isolated with fixed energy $$E$$) system is:

$$(\underbrace{\mathbf{x}_1, \mathbf{p}_1, \ldots \mathbf{x}_n, \mathbf{p}_n}_{\text{S microstate}}, \underbrace{\mathbf{X}_1, \mathbf{P}_1, \ldots \mathbf{X}_N, \mathbf{P}_N}_\text{R microstate})$$

where $$n, N$$ is the number of particles for the system and reservoir, respectively. For a specific microstate of $$\text{S}$$, lets call it $$\text{S}_i$$, we can have many microstates of $$\text{R}$$, lets say $$M$$, such that:

$$E_i + E_j = E \quad j=1,2 \ldots M$$

Probability of finding the system in $$i$$-th microstate

So to find the probability that the system is in the $$i$$-th microstate, we must divide the number of times it is found on that microstate by the total number of microstates, that is:

$$P(i) = \frac{\Omega_\text{R} (E - E_i)}{\Omega(E)}$$

Manipulating the above expression can lead to the Boltzmann distribution.

Proability of finding the system with energy $$E_i$$

Now, if we wanted to know the probability that the system has energy $$E_i$$ the above expression is not valid because we have not taken into account the other microstates that satisfy the first constraint. This time, we want to know the ratio:

$$P(E_i) = \frac{\Omega_\text{R} (E - E_i) \cdot \Omega_\text{S}(E_i)}{\Omega(E)}$$

Taking the natural logarith on both sides of the above equation leads to:

$$\ln \left(P(E_i)\right)=\ln\left(\Omega_\text{R} (E - E_i)\right) + \ln\left(\Omega_\text{S}(E_i)\right) - \ln\left(\Omega(E)\right)$$

If we assume that:

$$\ln\left(\Omega_\text{R} (E - E_i)\right) + \ln\left(\Omega_\text{S}(E_i)\right) \approx \ln\left(\Omega_\text{R} (E - E_i)\right)$$

then we can continue and show that the probability of finding the system with energy $$E$$ is:

$$P(E_i) = \frac{e^{-\beta E_i}}{Z}$$

Is this approach valid? That is, can we neglect the term $$\ln\left(\Omega_\text{S}(E_i)\right)$$?

In the specific case where $$\Omega_\text{S} (E_i) = 1$$ then there is no difference on what probability we are calculating. That is, the probability of finding the system in the $$i$$-th microstate equals the probability of finding the system with energy $$E_i$$.

If we have multiple microstates corresponding to the same energy $$E_i$$, should we just multiply the last expression for $$P(E_i)$$ by the number of microstates $$n_i$$ that have the same energy $$E_i$$? Are we allowed to do this is because each microstate with energy $$E_i$$ has the same probability of occuring, due to the fact that each microstate with energy $$E_i$$ can be "married" only with the microstates of the reservoir that satisfy the constraint:

$$E_i + E_j = E \quad j=1, 2, \ldots M$$

Is my reasoning correct?

The assumption $$\ln\left(\Omega_\text{R} (E - E_i)\right) + \ln\left(\Omega_\text{S}(E_i)\right) \approx \ln\left(\Omega_\text{R} (E - E_i)\right)$$ is true iff $$\Omega_S(E_i)\ll \Omega_R(E-E_i)$$, which is not obvious.
When we talk about the $$i$$-th microstate, degeneracy is not in consideration. The concept of degeneracy emerges from more than one microstates of the same energe $$E$$. OP confused the index $$i$$ of a microstate with that of an energe level. Strickly speaking, for an arbitrary energy $$E_i$$ of the system, there are $$\Omega_S(E_i)=g_i$$ microstates, each of which has a probability of $$P(E_i,j)=\frac{\Omega_R(E-E_i)}{\Omega(E)},\quad j=1,2,\cdots,g_i,$$ because for each microstate, the reservoir has $$\Omega_R(E-E_i)$$ sates. So the total probability for a system with energe $$E_i$$ is their sum, $$P(E_i)=\sum_j P(E_i,j)=\frac{\Omega_S(E_i)\Omega_R(E-E_i)}{\Omega(E)}.$$ That is, OP's conclusion is right.
Of course we can't neglect $$\Omega_S(E_i)$$, because it's not 1.
So the probability of finding the system with energy $$E_i$$ is $$\Omega_S(E_i)\mathrm e^{-\beta E_i}$$, which is samiliar to Boltzmann distribution but is different.
That's because Boltzmann distribution describes the probability of finding a system at a state whose energy is $$E_i$$, not the probability of finding a system with energy $$E_i$$.