JohnnyGui said:
My intention in post #126 with writing it as ##n(\epsilon_j)## instead of ##n_j## is because energy is being considered continuous so that one can not dedicate a certain particle number ##n_j## to a certain discrete energy level ϵj\epsilon_j.
You're confusing yourself with this too. The difference between continuous and discrete is irrelevant to the issue that you're struggling with.
JohnnyGui said:
##e^{-\alpha} = \frac{N}{Z}##.
Ah, so now you're confusing ##e^{- \alpha}## with ##e^{- \alpha - \beta E_j}##. And I see that you've been mixing those together for the last few posts. And you have managed to confuse me as well in the process.
I'm not going to try an disentangle all of that. Instead, let's start with the final distribution function, as given in the slides (from the summary, the next to last slide):
$$
f(E_j) = \frac{N_j}{g_j} = e^{\frac{- \left( E_j - E_F \right)}{k_B T}}
$$
What is this? Well, since it's a fraction ##N_j / g_j##, it's clearly a
number of particles--the number in a single state of energy ##E_j## where there is a number ##g_j## of degenerate states with that energy. (This easily simplifies to the non-degenerate case ##g_j = 1##.)
But what about that exponential of a negative number on the right? Well, the numerator of that number (the argument of the exponential) is ##- \left( E_j - E_F \right)##. In other words, it's minus the
difference between the energy of the state, ##E_j##, and the Fermi energy. But if there are no chemical reactions taking place--i.e., there is only a single type of particle present, or the temperature is too low for any reactions between multiple particle types to take place--then ##E_j < E_F##! Which means that the argument of the exponential is minus a negative number, i.e., a positive number.
In other words, this is telling us that the average number of particles in a single state of energy ##E_j## is more than one if ##E_j## is below the Fermi energy. If ##E_j## exactly
equals the Fermi energy, then there is exactly one particle in that state (on average); and if ##E_j## is greater than the Fermi energy, there is on average less than one particle in that state.
Now, "average number of particles" is easier to interpret if the states are discrete. But the formula itself works just fine if the states are continuous. If you struggle with how to interpret "average number of particles" for continuous states, then you can, as I said in what you quoted from post #99, just divide ##f(E_j)## by ##N##, the total number of particles, to get a formula that looks like this
$$
\frac{f(E_j)}{N} = \frac{N_j}{N g_j} = \frac{1}{N} e^{\frac{- \left( E_j - E_F \right)}{k_B T}}
$$
which has a simple interpretation as the statistical probability that a randomly selected particle will have energy ##E_j##.
Now, you keep on separating that exponential factor into two pieces. You write it in terms of ##\alpha## and ##\beta##, but we know what those actually turn out to equal so there's no need to obfuscate. We can just write:
$$
f(E_j) = \frac{N_j}{g_j} = e^{\frac{E_F}{k_B T}} e^{\frac{- E_j}{k_B T}}
$$
And if you want, you can go back and dig up that ##e^{\frac{E_F}{k_B T}} = N / Z## and rewrite this as
$$
f(E_j) = \frac{N_j}{g_j} = \frac{N}{Z} e^{\frac{- E_j}{k_B T}}
$$
which, again, clearly has an interpretation as "number of particles" (because ##N## appears explicitly on the right and we take the exponential of a negative number to get a fraction, with the partition function as an additional divisor). And the "divide by ##N##" form of this would be
$$
\frac{f(E_j)}{N} = \frac{N_j}{N g_j} = \frac{1}{Z} e^{\frac{- E_j}{k_B T}}
$$
which, again, clearly has an interpretation as "fraction of particles".
I really don't care which one of these you want to look at; they are all valid formulas and they all have well-defined meanings. But I think you need to take care to be clear about
which of them you are talking about. Otherwise you will just confuse yourself, and, as I noted above, you will end up confusing others (e.g., me) as well.
