Grand Canonical Molecular adsorption onto a surface

Question

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?

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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?

Answer 1

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

Answer 2

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?

Answer

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.

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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 $Z_G=z_G^N$.

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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 $Z_G=z_G^N$

Answer 3

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)