I am having a hard time applying the grand canonical theory to a simple example. I expose my understanding of the matter, the problem, my attempt of solution, the solution and my question on this solutions; I apologize for the lengthy question and will be very grateful to whoever feels like going through it!
I also add an answer including some ideas and a second solution.
Preliminars
I follow Kubo, Statistical Mechanics, but having a look around the notation should be standard. An open system is in contact with a reservoir fixing temperature $T$ and chemical potential $\mu$. A microstate of the open system is denoted by $s$; the grand canonical partition function is
$$Z_G(T,V,\mu) = \sum_se^{-\beta E_s + \beta \mu N_s}$$
where $s$ denotes each available microstate of the system, $N_s$ the number of particles in that microstate, and $E_s$ the energy of that microstate. This can be related to the canonical partition function $Z$: let $l$ denote a microstate for fixed number of particles, then
$$Z_G = \sum_{N_s=0}^{N_{tot}}\left(\sum_le^{-\beta E_l} \right)e^{\beta\mu N_s} = \sum_{N_s=0}^{N_{tot}}Z(T,V,N_s)\, e^{\beta\mu N_s}$$
This is useful: $Z$ is relatively hard to compute due to the fixed number of particles condition, but $\sum_{N_s=0}^{N_{tot}}$ allows to get rid of this condition. We consider the single particle properties:
- $i$ runs over the single particle possible microstates
- $\epsilon_i$ denotes the energy of the state $i$, that is the energy that a single particle has when happens to be in the state $i$
- $n_i=$ is the occupation number of the state $i$, that is the number of particles that happen to be in the state $i$. For fermions
$n_i=0,1$; for bosons $n_i=0,1,2,…$.
A microstate of the whole system $s$ is then specified by the sequence of occupation numbers $n_1, n_2, …$, and
$$N_s=\sum_i n_i, \quad E_s = \sum_i \epsilon_i n_i $$
The canonical partition function is
$$Z=\sum_l e^{-\beta E_l} = \underbrace{\sum_{n_1}\sum_{n_2}…\sum_{n_i}…}_{\text{with the condition } N_s=\sum_i n_i} e^{-\beta E_s}$$
Plugging this in the grand canonical equation
$$ Z_G = \sum_{N_s=0}^{N_{tot}} \underbrace{\sum_{n_1}\sum_{n_2}…\sum_{n_i}…}_{\text{with the condition } N_s=\sum_i n_i} e^{-\beta E_s} e^{\beta \mu N_s} = \underbrace{\sum_{n_1}\sum_{n_2}…\sum_{n_i}…}_{\text{on all possible values}} e^{-\beta \sum_i \epsilon_i n_i } e^{\beta \mu \sum_i n_i} = \prod _i\sum_{n_i}e^{-\beta(\epsilon_i-\mu)n_i}$$
We define the single state grand canonical partition function
$$z_{G,i}=\sum_{n_i}e^{-\beta(\epsilon_i-\mu)n_i}$$
$$Z_G=\prod_i z_{G,i}$$
The problem
We consider a gas in contact with a solid surface (e.g. argon on
graphene or molecular nitrogen on iron, as in the Haber-Bosch
synthesis). The gas molecules can be adsorbed at $N$ specific adsorption
sites while one site can only bind one molecule. The energies of the
bound and unbound state are $\epsilon$ and 0, respectively. The gas acts as a reservoir fixing $T$ and $\mu$.
How I would procede
- The system role is played by the $N$ adsorption sites
- The single particle role is played by one adsorption site
- The site admits two states, empty $i=0$ and full $i=1$
- The corresponding energies are $\epsilon_0=0$ and $\epsilon_1=\epsilon$
- The occupation numbers are $n_0$ = number of empty sites, $n_1$ = number of full sites. They both run from 0 to the total number of available sites, $n_i=0,1,…,N$
- A microstate of the system is determined by $n_0$ and $n_1$ such that $E_s=\sum_i\epsilon_i n_i = n_1 \epsilon$ and $N_s=\sum_i n_i = n_0+n_1=N$.
The grand canonical partition function should then read ($x_i:=e^{-\beta(\epsilon_i-\mu)}$)
$$Z_G=\prod_{i=0}^1\sum_{n_i=0}^N x_i^{n_i} = \prod_{i=0}^{1}\frac{1-x_i^{N+1}}{1-x_i}$$
Which is wrong (evaluating the product).
What may be wrong
- Assigning to the sites the role of "single particle" the total number of particles is fixed, namely $N$, why it should be allowed to change
Question (see answer attempt)
What is wrong with this approach?
Solution A
With no further explanation beyond the fact that the sites are non-interacting the lecturer, this page and this page claim
$$Z_G=z_G^N$$
This $z_G$ is given as
$$z_G=1+e^{-\beta(\epsilon-\mu)}$$
Questions (still open)
- Is the $z_G$ used here the same as the single state grand canonical partition function $z_{G,i}$ defined above?
- Where is $Z_G=z_G^N$ from?
The similar canonical relation $Z=z^N$ for non interacting systems of identical particles goes like this: we start with N distinguishable particles labeled by $j=1,…,N$; $\epsilon_{ij}$ is the $i$-energy level of the $j$-particle. Then
$$Z=\sum_l e^{-\beta E_l} = \sum_{i_1}\sum_{i_2}…\sum_{i_j}…e^{-\beta\sum_{j=1}^N \epsilon_{j i_j}} = \left( \sum_{i_1} e^{-\beta \epsilon_{1i_1}} \right)…\left( \sum_{i_N} e^{-\beta \epsilon_{Ni_N}} \right)$$
The $j$ subscript can be dropped if the particles are identical, so that
$$z_j=z=\sum_i e^{-\beta \epsilon_i}$$
$$Z=\prod_{j=1}^N z = z^N$$
Question (still open)
The subscript can not be dropped in the relation $Z_G=\prod_i
z_{G,i}$, as $z_{G,i}$ is an object strictly related to a state $i$,
so again, how is $Z_G=z_G^N$ obtained?
Best Answer
I'm not sure I will be able to clarify all your doubts, but this is the right approach to tackle this problem.
We consider $N_s$ available adsorption sites, an energy $\epsilon$ for each bound state, chemical potential $\mu$ and temperature $T$.
The grandpartition function $\mathcal Q$ is always expressed as $$ \mathcal Q= \sum_{N=0}^\infty e^{\beta \mu N} Z_N $$ in terms of the $N$-particle partition function. The latter is defined by $$ Z_N=\sum_{\substack{N-\text{particle}\\ \text{states}}} e^{-\beta E(\text{state})}\,. $$ In our case, the energy of a given $N$-particle state, of the ensemble of bound states, is $N\epsilon$ and there are $\binom{N_s}{N}$ such $N$-particle states, since each of the $N$ bound states can be placed by choosing one site among the $N_s$ sites, without repetition: $$ Z_N=\binom{N_s}{N}e^{-\beta \epsilon N}\,. $$ Note that this partition function does not bear a factorized form. Finally, $$ \mathcal Q= \sum_{N=0}^\infty \binom{N_s}{N}e^{\beta (\mu-\epsilon) N} = \sum_{N=0}^{N_s} \binom{N_s}{N}e^{\beta (\mu-\epsilon) N} =(1+e^{\beta(\mu-\epsilon)})^{N_s}\,, $$ where in the last step we used the binomial formula $$ (1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k. $$
EDIT: One can also reason directly using the grandpartition function as follows. Using the occupation number representation $\{n_{\alpha,k}\}$ for noninteracting particles, with $|\alpha, k\rangle$ labelling single-particle states with energy $\epsilon_\alpha$ and $g_\alpha$-fold degeneracy $k=1,2,\ldots,g_{\alpha}$, $$ \mathcal Q= \sum_{N=0}^\infty e^{\beta\mu N} \sum_{\substack{ \{n_{\alpha, k}\}:\\ \sum_{\alpha, k}n_{\alpha,k}=N}} e^{-\beta \sum_{\alpha, k}n_{\alpha,k} \epsilon_\alpha } = \sum_{\{n_{\alpha,k}\}} e^{\beta \sum_{\alpha, k}n_{\alpha,k} (\mu-\epsilon_\alpha) } = \prod_{\alpha,k} \sum_{n_{\alpha, k}}e^{\beta(\mu-\epsilon_\alpha)n_{\alpha,k}}\,. $$ This is a proof that, for noninteracting systems, the grandpartition function always takes the factorized form $$ \mathcal Q = \prod_{\alpha, k} \left( \sum_{n_{\alpha,k}} e^{\beta(\mu-\epsilon_\alpha)n_{\alpha,k}}\right)\,. $$ In the case at hand, the single-particle states all have energy $\epsilon$ and have multiplicity $N_s$; in the above notation $\alpha=1$ and $k=1,2,\ldots,N_s$ so $$ \mathcal Q = \prod_{k=1}^{N_s}(1+e^{\beta(\mu-\epsilon)})=(1+e^{\beta(\mu-\epsilon)})^{N_s}\,. $$