# Grand Canonical Molecular adsorption onto a surface

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

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?

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 $$\mathcal Q = \prod_{k=1}^{N_s}(1+e^{\beta(\mu-\epsilon)})=(1+e^{\beta(\mu-\epsilon)})^{N_s}\,.$$

• Thanks, this is precisely solution B in my answer (exchanging the notation for $N$ and $N_s$, see "What should be done" paragraph), confirming the correctness of Kubo's approach. My main doubt remains the origin of the equation $Z_G=z_G^N$ though, widely used in literature. Jan 11, 2018 at 11:15
• @DavideL I added a method to work only with the grandpartition function, maybe this goes in the direction clarifying your doubt. Jan 11, 2018 at 17:10
• Thanks @Brightsun, it definitely does. Understood the factorized form of the grandpartition function $Z_G=\prod_i z_G(i) = \prod_i \sum_{n_i} e^{-\beta(\epsilon_i-\mu)n_i}$ where $i$ labels single particle states with energy $\epsilon_i$ and $n_i$ is the occupation number, my problem was defining who plays the role of the "system" and of the "single particle". For some reason I was focusing on adsorbing sites and hence got $i$ and $n_i$ running over wrong values, while I just understod it's all about gas particles. Jan 12, 2018 at 15:51
• The single particle is, well, a gas particle: it has $i=1,..,N$ possible single-particle states, the adsorbing sites. They all have the same energy $\epsilon_i=\epsilon$ and the possible occupation numbers are $n_i=0,1$. The system is composed by all the $N_s=0,...,N$ captured particles, so that $E_s=\sum_i \epsilon_i n_i = \epsilon \sum_i n_i = \epsilon N_s$. It all works. Jan 12, 2018 at 15:54
• @akozi Yes, rather than the binomial series, I was using the binomial formula for a binomial with an integer power en.wikipedia.org/wiki/Binomial_theorem That's the generalization of $(1+x)^2=1+2x+x^2$ and $(1+x)^3=1+3x+3x^2+x^3$ to any power. Mar 6 at 17:05

The questions took literally hours to be written and during the writing I may have gained a partial understanding of the problem, which I'll try to expose here.

### Question

What is wrong with this approach?

The choice of the system: a fixed number $$N$$ of sites does not make a good gran canonical ensamble and $$Z_G=\prod_i z_{G,i}$$ does not apply.

# On the meaning of $$z_G$$

As given in the solution, $$z_G=1+e^{-\beta(\epsilon-\mu)}$$. Formally this is precisely $$\sum_{n=0}^1e^{-\beta(\epsilon-\mu)n}=\sum_{n=0}^1e^{-\beta \epsilon n +\beta \mu n} \sim \sum_n e^{-\beta E_n + \beta \mu n}$$

This expression resembles the one of the full grand canonical partition function and suggests the following interpretation: it describes a system

• whose microstate is determined by $$n=0,1$$
• whose energy when in the microstate $$n$$ is $$\epsilon n$$
• such that the number of elements in the system when in the microstate $$n$$ is precisely $$n$$

This system is one adsorption site, and the elements in the system are the captured particles. This number is allowed to change (between zero and one), so this is a good gran canonical ensamble. This picture may clarify the situation. On the left the array of $$N$$ adsorption sites is the system and one site is an element; on the right one site is the system and the captured (or not) particle is the element.

It then makes sense to write, for the system on the right side $$z_G=\sum_{n=0}^1 e^{-\beta \epsilon n + \beta \mu n }$$ It still has to be clarified how $$Z_G=z_G^N$$.

## Solution B

Kubo, pag. 92. He denotes by $$N_s$$ the number of full sites, i.e. of captured particles, and calculates the canonical partition function. What should not be done in two wrong albeit attempting ways giving the same result:

Way one $$Z=z^N=\left(\sum_{i=0}^1 e^{-\beta \epsilon_i}\right)^N = (1+e^{-\beta \epsilon})^N$$

Way two

This time we consider as the system the occupied sites: the number $$N_s$$ of occupied sites can vary between $$0$$ and $$N$$ and the energy of the system in the microstate $$s$$ is $$E_s=\epsilon N_s$$, so (using Newton's formula)

$$Z=\sum_s e^{-\beta E_s} = \sum_{N_s=0}^N \underbrace{g_s}_{\text{degeneracy}} e^{-\beta \epsilon N_s} = \sum_{N_s=0}^N \binom{N}{N_s} e^{-\beta \epsilon N_s} = (1+e^{-\beta \epsilon})^N \$$

What should be done is calculating first the canonical partition function for a given value of $$N_s$$ between $$0$$ and $$N$$. For this fixed value the energy of the system is always $$\epsilon N_s$$ with degeneracy $$\binom{N}{N_s}$$: $$Z(N_s)=\sum_{\text{distributions of } N_s \text{ particles in N boxes}}e^{-\beta \epsilon N_s} = \binom{N}{N_s}e^{-\beta \epsilon N_s}$$ The grand canonical partition function follows (using Newton's formula): $$Z_G=\sum_{N_s=0}^N Z(N_s) e^{\beta \mu N_s} = \sum_{N_s=0}^N \binom{N}{N_s} \left( e^{-\beta(\epsilon-\mu)} \right)^{N_s} = (1+e^{-\beta(\epsilon-\mu})^N$$

This could be taken as the sought proof that $$Z_G=z_G^N$$

I've found this way to understand (please someone notify me if this is considered a wrong general formula!):

We know that:

ZG= ZG1 * ZG2 * ...

Lets consider the case with 1 single-particle energy, but with a degeneracy of g1=2. Then the ZG can be written as:

ZG= ZG1 * ZG1 ( since we have to take into consideration every singe particle state).

So it's actually ZG= (ZG1)^g1 .

You can easily expand to r singe-particle states with degeneracy gr each, to :

ZG= ((ZG1)^g1 )* ((ZG2)^g2) * * * ((ZGr)^gr)

• Hi K. Milas, welcome to physics.SE! Please note that this site supports Mathjax, which is an engine that allows you to write formulas with latex-like commands. See MathJax basic tutorial and quick reference for more details. Please note that all users are expected to use this, as it becomes very difficult to understand formulas without it. Thank you. Aug 30, 2018 at 20:43