First, I have read this question:What is meant by the term "single particle state"

There is an analysis going on in my book (Mandle F. Statistical Physics) that has brought me in a confusing point.

In chapter 7 of the book, there is an analysis of the classical ideal gas. It is proved that $$Z_{\text{total}}={1 \over N!}(Z_1)^N $$ and that

$$Z_1 =Z_{\text{tr}} \, Z_{\text{int}}$$

where $Z_{\text{total}}$ is the total partition function of the system and $Z_{\text{tr}}$ and $Z_{\text{int}}$ are the transactional and internal partition functions of a sub system in canonical ensemble respectively.

Also, we know from chapter 2 that the partition function is in general of the form:

$$Z_1 ^{general} =\sum_r g(\epsilon_r) \exp\left[-\epsilon_r \over kT \right] \,. \qquad (1)$$

When defining the classical ideal gas we have that there exist energies

$$\epsilon_1 \leq \epsilon_2 \leq \epsilon_3 \ldots \leq \epsilon_r$$

corresponding to a discrete set of quantum states noted with $1,2,3 \ldots r$ where only a *unique* molecule can exist. Then we determine the state of the whole gas by defining a catalog of the molecules of each particular state defining $n_r$ as the occupation number of the state $r$.

Question :

In chapter 9 one finds the expression of the partition function of an ideal quantum gas:

$$Z_{\text{tot}} = \sum_{n_1 , n_2 \ldots} \exp\left(-\sum_r{n_r \epsilon_r} \over kT\right)~~~~~~~~(2) $$

and it is:

$$E(n_1 ,n_2 ,…)=\sum_r n_r \epsilon_r$$

and

$$N=\sum_r n_r \, .$$

So, why this difference between (1) and (2) at the expontential? Why not use

the sum of the occupation numbers in (1) too? Although, if I take the relations for granted I can prove some things, I don't understand the reason the partition functions are different in this analysis. I mean, to be clear, why is there a sum at the exponent of (2) and not (1) or vice-versa? If it is something about indistinguishable particles in QM, or that the $n_r$ number in the second relation isn't considered to be constant but in one for some reason it is, can someone elaborate?

Also, it seems to me I don't quite understand the meaning of unique molecule,maybe it's something that refers to the single-particle state which I understand is what one study in the quantum gas in difference with the classical where one study a particle to determine the statistical behaviour of the system.

Thank you.

## Best Answer

Okay, this is actually pretty straightforward, but I don't know where to start.

## Review: What's a partition function?

Let's step back and derive what we're talking about: what is a partition function? So we have a system which takes on a set of energy levels with degeneracies $\left \{(E_i, g_i) \right \}.$

We know that your system $s$ is in contact with a reservoir $r$, but together they are sealed up in a microcanonical ensemble with $S = S_s + S_r$, $U = U_s + U_r$. Now that reservoir is big and complicated so its internal degrees of freedom over the (to it) smallish changes in $U_r$ can be linearized as $S_r(U - U_s) = S_r(U) - U_s/T$, where $T$ is its (effectively constant) thermodynamic temperature $T^{-1} = \left(\frac{\partial S_r}{\partial U_r}\right)_{N_r,~V_r}$. Therefore the overall entropy of the reservoir system in state $i$ is $S_r(i) = S_0 - E_i/T$ for some $S_0$. But we know that the definition of entropy is $S = k_B \ln W$ where $W$ is a multiplicity of the state, so counting in the degeneracy, the total multiplicity of the state is simply: $$W_i = g_i ~ W_r(i) = g_i ~ e^{S_0/k_B - E_i / (k_B T)} $$and the probability is therefore $$p_i = \frac{W_i}{\sum_k W_k} =\frac {g_i ~ e^{-E_i / (k_B T)}} Z$$ for some constant $Z$ independent of the state index $i$, incorporating both $\sum_k W_k$ and $e^{S_0/k_B}$. Since the probabilities sum to 1, we can say that:$$Z = \sum_i g_i ~ \exp\left(\frac{-E_i} {k_B T}\right). $$If the system is continuous then we need a density-of-states $g(E)$ so that the number of states with energies between $E$ and $E + dE$ is roughly $g(E) ~ dE$, then we convert the above to an integral.

## From particles to complex systems

Okay, now that we're both on the same page about what it

is, what happens if your system hasa bunch of parts? Then each $i$ now labels aconfiguration of the parts. It potentially gets complicated! The first easy thing to do is to ditch the degeneracies $g_i$ and instead store all of their energies in amultiset: this is a set which can hold the same number multiple times. That might be confusing so let's procede formally a different way.Let's talk now about a set $C = \left\{c_i\right\}$ where the $c_i$ is some mathematical object telling me

the configuration of the state $i$, and we'll assume that this is distinct for each $i$, and now we have to transition from a set of $E_i$ to a function $E(c_i)$ which gives the energy of a configuration of the parts. As a side effect now $g_i = 1$ for each $i$ since each configuration is treated independently, but the same result holds:$$Z = \sum_i \exp\left(\frac{-E(c_i)} {k_B T}\right).$$## Non-interacting systems

If you're with me so far, there's just one more step! What is the

formof $c_i$ and $E(c)$?Well for a system of $N$ identical noninteracting particles, we have the single-particle energies $E_i$ from before, and the total energy is the sum of the energies for which the states are occupied. That is, the ideal form for $c_i$ becomes

an occupation function, $c_i = \left\{n_{i,k}\right\}$ which tells us, in configuration $i$, how many particles are in the state with energy $E_k$. Then the energy of the state is:$$E(c_i) = \sum_k n_{i,k} ~ E_k,$$hence,$$Z = \sum_i \exp\left(\frac{-\sum_k n_{i,k} ~ E_k} {k_B T}\right)$$Sothatis where the sum up-top comes from: we now have a complicatedmulti-particle statebut as long as the particles themselves arenoninteractingwe can use thesum of single-particle energiesto get the overall energy.