Boltzmann distribution and probability of finding the system with specific energy

Question

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 $\ln\left(\Omega_\text{S}(E_i)\right)$?

Comments

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?

Answer 1

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.

Answer 2

I don't know whether I understand your question.
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 $E_i$.