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:

> 
> 1. $i$ runs over the *single particle* possible microstates
> 2. $\epsilon\_i$ denotes the energy of the state $i$, that is the energy that a single particle has when happens to be in the state $i$
> 3. $n\_i=$ 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](http://userpage.fu-berlin.de/krinne/klausur_problems_solution.pdf) and [this page](http://mtdevans.com/projects/physics-problems/absorption-of-molecules-onto-a-surface/) 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?**
> 
> 
>