Ising model and fluids

As we have alreadt stated, the Ising model can also be used to describe systems different from magnets. The most important example is the correspondence that can be established between an Ising model and a fluid through a lattice gas model. We will now show how this equivalence can be defined; in order to do so, we will proceed in two steps: we will first show how an Ising model is equivalent to a lattice gas, and then show (qualitatively) that this lattice gas model is equivalent to the classical model for a fluid. Before doing so, we briefly review the formalism used in classical statistical mechanics in order to describe fluids (see The canonical ensemble at work, for example).

Let us then consider a system of $N$$N$ particles subjected to some generic potentials, so that its Hamiltonian can be written as:

$${\mathcal {H}}=\sum _{i=1}^{N}\left({\frac {{\vec {p}}_{i}{}^{2}}{2m}}+U_{1}({\vec {r}}_{i})\right)+{\frac {1}{2}}\sum _{i\neq j}U_{2}({\vec {r}}_{i},{\vec {r}}_{j})+{\frac {1}{3!}}\sum _{i\neq j\neq k}U_{3}({\vec {r}}_{i},{\vec {r}}_{j},{\vec {r}}_{k})+\cdots$$

$${\mathcal {H}}=\sum _{i=1}^{N}\left({\frac {{\vec {p}}_{i}{}^{2}}{2m}}+U_{1}({\vec {r}}_{i})\right)+{\frac {1}{2}}\sum _{i\neq j}U_{2}({\vec {r}}_{i},{\vec {r}}_{j})+{\frac {1}{3!}}\sum _{i\neq j\neq k}U_{3}({\vec {r}}_{i},{\vec {r}}_{j},{\vec {r}}_{k})+\cdots$$

where $U_{n}$$U_{n}$ is an $n$$n$-body potential, which generally depends only on the distances between the particles if it involves two or more of them. Working in the grand canonical ensemble, the grand partition function will be:

$${\mathcal {Z}}=\operatorname {Tr} e^{-\beta ({\mathcal {H}}-\mu N)}$$

$${\mathcal {Z}}=\operatorname {Tr} e^{-\beta ({\mathcal {H}}-\mu N)}$$

where:

$$\operatorname {Tr} =\sum _{N=0}^{\infty }{\frac {1}{N!}}\int \prod _{i=1}^{N}{\frac {1}{h^{dN}}}d^{d}{\vec {r}}_{i}d^{d}{\vec {p}}_{i}$$

$$\operatorname {Tr} =\sum _{N=0}^{\infty }{\frac {1}{N!}}\int \prod _{i=1}^{N}{\frac {1}{h^{dN}}}d^{d}{\vec {r}}_{i}d^{d}{\vec {p}}_{i}$$

($h$$h$ is the Planck constant, and $d$$d$ the dimensionality of the system). It is convenient to separate the contributes due to the kinetic and configurational terms:

$${\mathcal {Z}}=\sum _{N=0}^{\infty }{\frac {1}{N!}}\left(\int \prod _{i=1}^{N}{\frac {d^{d}{\vec {p}}_{i}}{h^{dN}}}e^{-\beta {\frac {{\vec {p}}_{i}{}^{2}}{2m}}}\right)\left(\int \prod _{i=1}^{N}d^{d}{\vec {r}}_{i}e^{-\beta (\lbrace U\rbrace -\mu N)}\right)$$

$${\mathcal {Z}}=\sum _{N=0}^{\infty }{\frac {1}{N!}}\left(\int \prod _{i=1}^{N}{\frac {d^{d}{\vec {p}}_{i}}{h^{dN}}}e^{-\beta {\frac {{\vec {p}}_{i}{}^{2}}{2m}}}\right)\left(\int \prod _{i=1}^{N}d^{d}{\vec {r}}_{i}e^{-\beta (\lbrace U\rbrace -\mu N)}\right)$$

where with $\lbrace U\rbrace$$\lbrace U\rbrace$ we mean all the possible potentials acting on the system. However (see, again, The canonical ensemble at work) we have:

$$\int {\frac {d^{d}{\vec {p}}}{h^{d}}}e^{-\beta {\frac {{\vec {p}}{}^{2}}{2m}}}=\Lambda (T)^{-d}\quad \qquad \Lambda (T)={\frac {h}{\sqrt {2\pi mk_{B}T}}}$$

$$\int {\frac {d^{d}{\vec {p}}}{h^{d}}}e^{-\beta {\frac {{\vec {p}}{}^{2}}{2m}}}=\Lambda (T)^{-d}\quad \qquad \Lambda (T)={\frac {h}{\sqrt {2\pi mk_{B}T}}}$$

where $\Lambda (T)$$\Lambda (T)$ is the thermal wavelength. Therefore:

$${\mathcal {Z}}=\sum _{N=0}^{\infty }{\frac {1}{N!}}\left({\frac {e^{\beta \mu }}{\Lambda (T)^{d}}}\right)^{N}Q_{N}$$

$${\mathcal {Z}}=\sum _{N=0}^{\infty }{\frac {1}{N!}}\left({\frac {e^{\beta \mu }}{\Lambda (T)^{d}}}\right)^{N}Q_{N}$$

where:

$$Q_{N}=\int \prod _{i=1}^{N}d^{d}{\vec {r}}_{i}e^{-\beta \lbrace U\rbrace }$$

$$Q_{N}=\int \prod _{i=1}^{N}d^{d}{\vec {r}}_{i}e^{-\beta \lbrace U\rbrace }$$

The grand free energy is thus:

$${\mathcal {F}}(T,\mu ,\lbrace U\rbrace )=-k_{B}T\ln {\mathcal {Z}}$$

$${\mathcal {F}}(T,\mu ,\lbrace U\rbrace )=-k_{B}T\ln {\mathcal {Z}}$$

and as usual the thermodynamic limit of $F$$F$ is defined as:

$$f(T,\mu ,\lbrace U\rbrace )=\lim _{V\to \infty }{\frac {\mathcal {F}}{V}}$$

$$f(T,\mu ,\lbrace U\rbrace )=\lim _{V\to \infty }{\frac {\mathcal {F}}{V}}$$

with the constraint:

$$\rho =\lim _{V\to \infty }{\frac {\left\langle N\right\rangle }{V}}={\text{const.}}$$

$$\rho =\lim _{V\to \infty }{\frac {\left\langle N\right\rangle }{V}}={\text{const.}}$$

Ising model and lattice gas

The basic idea of the lattice gas model is to describe a fluid where the particles are located on the sites of a lattice instead of continuously occupying any position in space; it is a sort of "discretization" of the classical description of fluids. The correspondence with the Ising model is established relating the local density of a fluid with the local magnetization density of an Ising model.

Let us therefore consider a $d$$d$-dimensional lattice with coordination number $z$$z$. Each site of the lattice can be occupied by a particle, so if we call $n_{i}$$n_{i}$ the occupation number of a site we will have either $n_{i}=0$$n_{i}=0$ or $n_{i}=1$$n_{i}=1$, and the total number of particles will be:

$$N=\sum _{i=1}^{N_{\text{s}}}n_{i}$$

$$N=\sum _{i=1}^{N_{\text{s}}}n_{i}$$

where ${\textstyle N_{\text{s}}}$${\textstyle N_{\text{s}}}$ is the number of sites. In analogy with the continuum case we can guess a Hamiltonian of the form:

$${\mathcal {H}}=\sum _{i=1}^{N_{\text{s}}}U_{1}(i)n_{i}+{\frac {1}{2}}\sum _{i\neq j}U_{2}(i,j)n_{i}n_{j}+\cdots$$

$${\mathcal {H}}=\sum _{i=1}^{N_{\text{s}}}U_{1}(i)n_{i}+{\frac {1}{2}}\sum _{i\neq j}U_{2}(i,j)n_{i}n_{j}+\cdots$$

where the interaction potential $U_{2}$$U_2$ is symmetric, namely $U_{2}(i,j)=U_{2}(j,i)$$U_{2}(i,j)=U_{2}(j,i)$. We have written only the configurational term, because we know that the kinetic part always contributes to the grand partition function with the thermal wavelength. This way, in the grand canonical ensemble we have:

$${\mathcal {H}}-\mu N=\sum _{i=1}^{N_{\text{s}}}\left(U_{1}(i)-\mu \right)n_{i}+{\frac {1}{2}}\sum _{i\neq j}U_{2}(i,j)n_{i}n_{j}+\cdots$$

$${\mathcal {H}}-\mu N=\sum _{i=1}^{N_{\text{s}}}\left(U_{1}(i)-\mu \right)n_{i}+{\frac {1}{2}}\sum _{i\neq j}U_{2}(i,j)n_{i}n_{j}+\cdots$$

Considering now an Ising model defined on the same lattice, we can establish a correspondence with the lattice gas model defining:

$$n_{i}={\frac {1}{2}}(1+S_{i})$$

$$n_{i}={\frac {1}{2}}(1+S_{i})$$

where $S_{i}$$S_{i}$ is the spin variable of the Ising model defined on the same lattice: in fact, doing so we have $n_{i}=0$$n_{i}=0$ when $S_{i}=-1$$S_{i}=-1$ and $n_{i}=1$$n_{i}=1$ when $S_{i}=1$$S_{i}=1$. What we now want to show is that substituting $n_{i}=(1+S_{i})/2$$n_{i}=(1+S_{i})/2$ into ${\mathcal {H}}-\mu N$${\mathcal {H}}-\mu N$ we obtain a Hamiltonian that leads to the grand partition function of the Ising model, so that the equivalence between the two models is made explicit. From now on we will neglect any potential that couples more than two particles; therefore, the first term of expression of ${\mathcal {H}}-\mu N$${\mathcal {H}}-\mu N$ becomes:

$$\sum _{i=1}^{N_{\text{s}}}\left(U_{1}(i)-\mu \right){\frac {1}{2}}(1+S_{i})={\frac {1}{2}}\sum _{i=1}^{N_{\text{s}}}\left(U_{1}(i)-\mu \right)+{\frac {1}{2}}\sum _{i=1}^{N_{\text{s}}}\left(U_{1}(i)-\mu \right)S_{i}$$

$$\sum _{i=1}^{N_{\text{s}}}\left(U_{1}(i)-\mu \right){\frac {1}{2}}(1+S_{i})={\frac {1}{2}}\sum _{i=1}^{N_{\text{s}}}\left(U_{1}(i)-\mu \right)+{\frac {1}{2}}\sum _{i=1}^{N_{\text{s}}}\left(U_{1}(i)-\mu \right)S_{i}$$

while the second term:

$${\frac {1}{8}}\sum _{i\neq j}U_{2}(i,j)(1+S_{i})(1+S_{j})={\frac {1}{8}}\sum _{i\neq j}U_{2}(i,j)+{\frac {1}{4}}\sum _{i\neq j}U_{2}(i,j)S_{i}+{\frac {1}{8}}\sum _{i\neq j}U_{2}(i,j)S_{i}S_{j}$$

$${\frac {1}{8}}\sum _{i\neq j}U_{2}(i,j)(1+S_{i})(1+S_{j})={\frac {1}{8}}\sum _{i\neq j}U_{2}(i,j)+{\frac {1}{4}}\sum _{i\neq j}U_{2}(i,j)S_{i}+{\frac {1}{8}}\sum _{i\neq j}U_{2}(i,j)S_{i}S_{j}$$

(where in the second summation has been obtained relabelling indices and using the symmetry of $U_{2}$$U_2$). If we now suppose that $U_{2}$$U_2$ is a nearest-neighbour interaction, i.e.:

$$U_{2}(i,j)={\begin{cases}U_{2}&i{\text{ and }}j{\text{ are nearest neighbours}}\\0&{\text{otherwise}}\end{cases}}$$

$$U_{2}(i,j)={\begin{cases}U_{2}&i{\text{ and }}j{\text{ are nearest neighbours}}\\0&{\text{otherwise}}\end{cases}}$$

then:

$${\frac {1}{2}}\sum _{i\neq j}U_{2}(i,j)n_{i}n_{j}={\frac {1}{8}}U_{2}zN_{\text{s}}+{\frac {1}{4}}U_{2}z\sum _{i=1}^{N_{\text{s}}}S_{i}+{\frac {1}{8}}U_{2}\sum _{\left\langle i,j\right\rangle }S_{i}S_{j}$$

$${\frac {1}{2}}\sum _{i\neq j}U_{2}(i,j)n_{i}n_{j}={\frac {1}{8}}U_{2}zN_{\text{s}}+{\frac {1}{4}}U_{2}z\sum _{i=1}^{N_{\text{s}}}S_{i}+{\frac {1}{8}}U_{2}\sum _{\left\langle i,j\right\rangle }S_{i}S_{j}$$

If we now set $U_{1}=0$$U_{1}=0$ (i.e. there are no external fields acting on the fluid) we have:

$${\frac {1}{2}}\sum _{i=1}^{N_{\text{s}}}\left(U_{1}(i)-\mu \right)n_{i}=-{\frac {1}{2}}\mu N_{\text{s}}-{\frac {1}{2}}\mu \sum _{i=1}^{N_{\text{s}}}S_{i}$$

$${\frac {1}{2}}\sum _{i=1}^{N_{\text{s}}}\left(U_{1}(i)-\mu \right)n_{i}=-{\frac {1}{2}}\mu N_{\text{s}}-{\frac {1}{2}}\mu \sum _{i=1}^{N_{\text{s}}}S_{i}$$

Therefore:

$${\mathcal {H}}-\mu N=-{\frac {\mu }{2}}N_{\text{s}}+{\frac {z}{8}}U_{2}N_{\text{s}}+\left(-{\frac {\mu }{2}}+{\frac {z}{4}}U_{2}\right)\sum _{i=1}^{N_{\text{s}}}S_{i}+{\frac {U_{2}}{8}}\sum _{\left\langle i,j\right\rangle }S_{i}S_{j}$$

$${\mathcal {H}}-\mu N=-{\frac {\mu }{2}}N_{\text{s}}+{\frac {z}{8}}U_{2}N_{\text{s}}+\left(-{\frac {\mu }{2}}+{\frac {z}{4}}U_{2}\right)\sum _{i=1}^{N_{\text{s}}}S_{i}+{\frac {U_{2}}{8}}\sum _{\left\langle i,j\right\rangle }S_{i}S_{j}$$

For the sake of simplicity, we set:

$$E_{0}=-{\frac {\mu }{2}}N_{\text{s}}+{\frac {z}{8}}U_{2}N_{\text{s}}\quad \quad -H=-{\frac {\mu }{2}}+{\frac {z}{4}}U_{2}\quad \quad -J={\frac {U_{2}}{8}}$$

$$E_{0}=-{\frac {\mu }{2}}N_{\text{s}}+{\frac {z}{8}}U_{2}N_{\text{s}}\quad \quad -H=-{\frac {\mu }{2}}+{\frac {z}{4}}U_{2}\quad \quad -J={\frac {U_{2}}{8}}$$

so that:

$${\mathcal {H}}-\mu N=E_{0}-H\sum _{i=1}^{N_{\text{s}}}S_{i}-J\sum _{\left\langle i,j\right\rangle }S_{i}S_{j}$$

$${\mathcal {H}}-\mu N=E_{0}-H\sum _{i=1}^{N_{\text{s}}}S_{i}-J\sum _{\left\langle i,j\right\rangle }S_{i}S_{j}$$

We thus have that the grand partition function of the lattice gas can be written as:

$${\begin{aligned}{\mathcal {Z}}_{\text{l.g.}}=\operatorname {Tr} e^{-\beta ({\mathcal {H}}-\mu N)}=\operatorname {Tr} e^{-\beta \left(-E_{0}+H\sum _{i}S_{i}+J\sum _{\left\langle i,j\right\rangle }S_{i}S_{j}\right)}=\\=e^{-\beta E_{0}}\operatorname {Tr} e^{-\beta \left(H\sum _{i}S_{i}+J\sum _{\left\langle i,j\right\rangle }S_{i}S_{j}\right)}\end{aligned}}$$

$${\begin{aligned}{\mathcal {Z}}_{\text{l.g.}}=\operatorname {Tr} e^{-\beta ({\mathcal {H}}-\mu N)}=\operatorname {Tr} e^{-\beta \left(-E_{0}+H\sum _{i}S_{i}+J\sum _{\left\langle i,j\right\rangle }S_{i}S_{j}\right)}=\\=e^{-\beta E_{0}}\operatorname {Tr} e^{-\beta \left(H\sum _{i}S_{i}+J\sum _{\left\langle i,j\right\rangle }S_{i}S_{j}\right)}\end{aligned}}$$

and so:

$$\Rightarrow \quad {\mathcal {Z}}_{\text{l.g.}}=e^{-\beta E_{0}}{\mathcal {Z}}_{\text{I.m.}}$$

$$\Rightarrow \quad {\mathcal {Z}}_{\text{l.g.}}=e^{-\beta E_{0}}{\mathcal {Z}}_{\text{I.m.}}$$

where ${\mathcal {Z}}_{\text{I.m.}}$${\mathcal {Z}}_{\text{I.m.}}$ is the grand partition function of the Ising model. Therefore, we see that the grand partition functions of both the Ising model and the lattice gas are equal, except for the irrelevant constant rescaling factor $e^{-\beta E_{0}}$$e^{-\beta E_{0}}$. This way we can see explicitly that the two models are perfectly equivalent.

Lattice gas and continuous fluids

We now want to show (although not really rigorously) that the lattice gas model can be derived from the "classical" model of a fluid. Consider the configurational sum:

$$Q_{N}=\int \prod _{i=1}^{N}d^{d}{\vec {r}}_{i}e^{-\beta \lbrace U\rbrace }$$

$$Q_{N}=\int \prod _{i=1}^{N}d^{d}{\vec {r}}_{i}e^{-\beta \lbrace U\rbrace }$$

We can approximate it dividing our system in $N_{\text{c}}$$N_{\text{c}}$ cells of linear dimension $a$$a$, such that the probability to find more than one particle inside one cell is negligible; in other words, if our particles interact via a potential that has a hard core component we can take $a$$a$ of the order of the hard core radius. This way the integral in $Q_{N}$$Q_{N}$ can be rewritten as:

$$\int \prod _{i=1}^{N}d^{d}{\vec {r}}_{i}\approx a^{dN}\sum _{\alpha =1}^{N_{\text{c}}}$$

$$\int \prod _{i=1}^{N}d^{d}{\vec {r}}_{i}\approx a^{dN}\sum _{\alpha =1}^{N_{\text{c}}}$$

where $i$$i$ labels the particles while $\alpha$$\alpha$ labels the cells. Since we have focused our attention towards the cells, we can substitute the interaction $U_{2}({\vec {r}}_{i},{\vec {r}}_{j})$$U_{2}({\vec {r}}_{i},{\vec {r}}_{j})$ between particles with an interaction between occupied cells:

$$U_{2}(\alpha ,\beta )=U_{2}({\vec {r}}_{i},{\vec {r}}_{j})\quad {\text{if }}{\vec {r}}_{i}\in \alpha {\text{ and }}{\vec {r}}_{j}\in \beta$$

$$U_{2}(\alpha ,\beta )=U_{2}({\vec {r}}_{i},{\vec {r}}_{j})\quad {\text{if }}{\vec {r}}_{i}\in \alpha {\text{ and }}{\vec {r}}_{j}\in \beta$$

As we have done before we consider only two-particle interactions and suppose that the relative potential is short-ranged, namely $U_{2}$$U_2$ acts only between nearest neighbours and in that case is equal to a constant value $U_{2}$$U_2$; therefore:

$$\sum _{\alpha \neq \beta }U_{2}(\alpha ,\beta )n_{\alpha }n_{\beta }=U_{2}\sum _{\left\langle \alpha ,\beta \right\rangle }n_{\alpha }n_{\beta }$$

$$\sum _{\alpha \neq \beta }U_{2}(\alpha ,\beta )n_{\alpha }n_{\beta }=U_{2}\sum _{\left\langle \alpha ,\beta \right\rangle }n_{\alpha }n_{\beta }$$

where of course $n_{\alpha }$$n_{\alpha }$ is the occupation number of the $\alpha$$\alpha$-th cell. Now, for each specified configuration of the occupation numbers $n_{\alpha }$$n_{\alpha }$ there can be $N!$$N!$ possible configurations for the positions ${\vec {r}}_{i}$${\vec {r}}_{i}$: in fact, the configuration $\lbrace n_{\alpha }\rbrace$$\lbrace n_{\alpha }\rbrace$ of the occupied cells only tells us where we can find a particle, but not which specific particle is in a given cell; furthermore, the system is left unchanged if we interchange the particles. We can thus write:

$$Q_{N}=N!a^{dN}\sum _{\lbrace n_{\alpha }=0,1\rbrace }'e^{-\beta U(\lbrace n_{\alpha }\rbrace )}$$

$$Q_{N}=N!a^{dN}\sum _{\lbrace n_{\alpha }=0,1\rbrace }'e^{-\beta U(\lbrace n_{\alpha }\rbrace )}$$

where ${\textstyle \sum '}$${\textstyle \sum '}$ means that the sum on $n_{\alpha }$$n_{\alpha }$ must be subject to the constraint that the total number of particle be fixed and equal to $N$$N$:

$$\sum _{\lbrace n_{\alpha }=0,1\rbrace }'=\sum _{\lbrace n_{\alpha }=0,1\rbrace }\quad \quad {\text{with }}\sum _{\alpha }n_{\alpha }=N$$

$$\sum _{\lbrace n_{\alpha }=0,1\rbrace }'=\sum _{\lbrace n_{\alpha }=0,1\rbrace }\quad \quad {\text{with }}\sum _{\alpha }n_{\alpha }=N$$

Therefore:

$${\mathcal {Z}}_{\text{c.f.}}=\sum _{N=0}^{\infty }{\frac {1}{N!}}\left({\frac {e^{\beta \mu }}{\Lambda (T)^{d}}}\right)^{N}Q_{N}=\sum _{N=0}^{\infty }\left[e^{\beta \mu }\left({\frac {a}{\Lambda (T)}}\right)^{d}\right]^{N}\sum _{\lbrace n_{\alpha }=0,1\rbrace }'e^{-\beta U(\lbrace n_{\alpha }\rbrace )}$$

$${\mathcal {Z}}_{\text{c.f.}}=\sum _{N=0}^{\infty }{\frac {1}{N!}}\left({\frac {e^{\beta \mu }}{\Lambda (T)^{d}}}\right)^{N}Q_{N}=\sum _{N=0}^{\infty }\left[e^{\beta \mu }\left({\frac {a}{\Lambda (T)}}\right)^{d}\right]^{N}\sum _{\lbrace n_{\alpha }=0,1\rbrace }'e^{-\beta U(\lbrace n_{\alpha }\rbrace )}$$

where "c.f." stands for "continuum fluid". The presence of the sum ${\textstyle \sum _{N=0}^{\infty }}$${\textstyle \sum _{N=0}^{\infty }}$ strongly simplifies the calculations: in fact, if $f(n_{\alpha })$$f(n_{\alpha })$ is a generic function, then:

$$\sum _{N=0}^{\infty }\sum _{\lbrace n_{\alpha }\rbrace }'f(n_{\alpha })=\underbrace {\sum _{\lbrace n_{\alpha }\rbrace }f(n_{\alpha })} _{\sum _{\alpha }n_{\alpha }=0}+\underbrace {\sum _{\lbrace n_{\alpha }\rbrace }f(n_{\alpha })} _{\sum _{\alpha }n_{\alpha }=1}+\cdots +\underbrace {\sum _{\lbrace n_{\alpha }\rbrace }f(n_{\alpha })} _{\sum _{\alpha }n_{\alpha }=\infty }=\sum _{n_{\alpha }}f(n_{\alpha })$$

$$\sum _{N=0}^{\infty }\sum _{\lbrace n_{\alpha }\rbrace }'f(n_{\alpha })=\underbrace {\sum _{\lbrace n_{\alpha }\rbrace }f(n_{\alpha })} _{\sum _{\alpha }n_{\alpha }=0}+\underbrace {\sum _{\lbrace n_{\alpha }\rbrace }f(n_{\alpha })} _{\sum _{\alpha }n_{\alpha }=1}+\cdots +\underbrace {\sum _{\lbrace n_{\alpha }\rbrace }f(n_{\alpha })} _{\sum _{\alpha }n_{\alpha }=\infty }=\sum _{n_{\alpha }}f(n_{\alpha })$$

where the last sum in unconstrained. Therefore we can write ${\mathcal {Z}}_{\text{c.f.}}$${\mathcal {Z}}_{\text{c.f.}}$ as:

$${\begin{aligned}{\mathcal {Z}}_{\text{c.f.}}=\sum _{\lbrace n_{\alpha }=0,1\rbrace }\left[e^{\beta \mu }e^{d\log {\frac {a}{\Lambda (T)}}}\right]^{N}e^{-\beta U_{2}\sum _{\left\langle \alpha ,\beta \right\rangle }n_{\alpha }n_{\beta }}=\\=\sum _{\lbrace n_{\alpha }=0,1\rbrace }e^{-\beta \left[U_{2}\sum _{\left\langle \alpha ,\beta \right\rangle }n_{\alpha }n_{\beta }-\left(\mu +{\frac {d}{\beta }}\log {\frac {a}{\Lambda (T)}}\right)N\right]}\end{aligned}}$$

$${\begin{aligned}{\mathcal {Z}}_{\text{c.f.}}=\sum _{\lbrace n_{\alpha }=0,1\rbrace }\left[e^{\beta \mu }e^{d\log {\frac {a}{\Lambda (T)}}}\right]^{N}e^{-\beta U_{2}\sum _{\left\langle \alpha ,\beta \right\rangle }n_{\alpha }n_{\beta }}=\\=\sum _{\lbrace n_{\alpha }=0,1\rbrace }e^{-\beta \left[U_{2}\sum _{\left\langle \alpha ,\beta \right\rangle }n_{\alpha }n_{\beta }-\left(\mu +{\frac {d}{\beta }}\log {\frac {a}{\Lambda (T)}}\right)N\right]}\end{aligned}}$$

namely:

$${\mathcal {Z}}_{\text{c.f.}}=\operatorname {Tr} e^{-\beta ({\mathcal {H}}-{\overline {\mu }}N)}={\mathcal {Z}}_{\text{l.g.}}$$

$${\mathcal {Z}}_{\text{c.f.}}=\operatorname {Tr} e^{-\beta ({\mathcal {H}}-{\overline {\mu }}N)}={\mathcal {Z}}_{\text{l.g.}}$$

where ${\mathcal {Z}}_{\text{l.g. }}$${\mathcal {Z}}_{\text{l.g. }}$ is the grand partition function of the lattice gas model, while:

$${\overline {\mu }}=\mu _{\text{cl.}}+{\frac {d}{\beta }}\log {\frac {a}{\Lambda (T)}}$$

$${\overline {\mu }}=\mu _{\text{cl.}}+{\frac {d}{\beta }}\log {\frac {a}{\Lambda (T)}}$$

and $\mu _{\text{cl.}}$$\mu _{\text{cl.}}$ is the "classical" chemical potential. However, since the chemical potential has the meaning of an energy, any shift in $\mu$$\mu$ by a constant quantity is physically irrelevant, so ${\overline {\mu }}$${\overline {\mu }}$ and $\mu _{\text{cl.}}$$\mu _{\text{cl.}}$ are equivalent. Therefore, the grand partition functions of the continuum fluid and the lattice gas model are the same: we thus see that these two models are equivalent.