# Ising model and binary alloys

The Ising model can be also used to described systems composed of different kinds of particles, like binary alloys. What we now want to see is that, similarly to what happened in Ising model and fluids for the lattice gas, the Hamiltonian of a binary alloy can be mapped into the Hamiltonian of an Ising model.

Consider a lattice with coordination number $z$ , and suppose that on every site of this lattice there can be an atom of two possible different elements, say $A$ and $B$ , and that they can "move" (i.e. exchange their positions) on the lattice. Let us call $-E_{AA}$ and $-E_{BB}$ (writing explicitly the negative sign) the interaction energies between neighbouring atoms of the same elements, respectively $A$ and $B$ ; similarly, we call $-E_{AB}$ the interaction energy of neighbouring atoms of different kinds. Let us also call:

• $N_{AA}$ , $N_{BB}$ the number of $A$ -$A$ and $B$ -$B$ bonds
• $N_{AB}$ the number of $A$ -$B$ bonds
• $N_{A}$ , $N_{B}$ the numbers of $A$ and $B$ atoms
• $N=N_{A}+N_{B}$ the total number of atoms

This way, the energy of the system will be:

$E=-E_{AA}N_{AA}-E_{AB}N_{AB}-E_{BB}N_{BB}$ However, $N_{AA}$ , $N_{BB}$ and $N_{AB}$ are not independent. Let us in fact consider all the $A$ atoms of our system: we have that every $A$ -$A$ bond contributes to $N_{A}$ with two atoms and every $A$ -$B$ bond with a single $A$ atom. If we add these numbers we have $z$ times the total number of $A$ atoms, i.e.:
$2N_{AA}+N_{AB}=zN_{A}$ and similarly:
$2N_{BB}+N_{AB}=zN_{B}$ The fact that these sums are $z$ times equal to $N_{A}$ or $N_{B}$ is better understood if explicitly verified in simple cases. Check for example that this is true in the following two-dimensional case (assuming periodic boundary conditions):
${\begin{array}{ccccc}A&B&A&A&B\\B&A&A&B&B\\A&B&A&A&B\\B&A&B&B&B\\B&A&A&B&A\end{array}}$ If we now solve these last two equations, expressing everything in terms of $N_{A}$ and $N_{AA}$ we get:
$N_{B}=N-N_{A}\qquad N_{BB}=N_{AA}+(N-2N_{A}){\frac {z}{2}}\qquad N_{AB}=zN_{A}-2N_{AA}$ and thus the energy of the system can be rewritten as:
$-E=N_{AA}(E_{AA}+E_{BB}-2E_{AB})+zN_{A}(E_{AB}-E_{BB})+{\frac {zN}{2}}E_{BB}$ Now, in order to establish a correspondence with the Ising model, similarly to what we have done for the lattice gas, we can define a site variable $n_{i}$ which represents if an $A$ or $B$ atom is occupying the $i$ -th site. We define $n_{i}$ such that $n_{i}=1$ when an $A$ atom is present while $n_{i}=0$ when a $B$ atom is present in the site; this way, we can map this system into an Ising model setting (just like before):
$n_{i}={\frac {1+S_{i}}{2}}$ Therefore:
$N_{A}=\sum _{i}n_{i}=\sum _{i}{\frac {1+S_{i}}{2}}\quad \qquad N_{AA}=\sum _{\left\langle i,j\right\rangle }{\frac {1+S_{i}}{2}}{\frac {1+S_{j}}{2}}$ Computing the sums just like we have done for the lattice gas, we have:
$N_{A}={\frac {N}{2}}+{\frac {1}{2}}\sum _{i}S_{i}\quad \qquad N_{AA}={\frac {1}{4}}\sum _{\left\langle i,j\right\rangle }S_{i}S_{j}+{\frac {z}{2}}\sum _{i}S_{i}+{\frac {zN}{8}}$ Thus, the energy of the system can be rewritten as:
$-E=J\sum _{\left\langle i,j\right\rangle }S_{i}S_{j}+h\sum _{i}S_{i}+CN$ where:
$J={\frac {1}{4}}(E_{AA}+E_{BB}-2E_{AB})\quad \qquad h={\frac {z}{4}}(E_{AA}-E_{BB})$ $C={\frac {z}{8}}(E_{AA}+E_{BB}+2E_{AB})$ The energy $E$ has again the same form of the Hamiltonian of the Ising model, apart from the irrelevant constant shift $CN$ . Therefore, the partition functions of the Ising model and the binary alloy are perfectly equivalent.

1. This happens, for example, in $\beta$ -brasses: at temperatures lower than approximately $733^{\circ }{\text{C}}$ the atoms are arranged in a body-centered cubic lattice, with zinc atoms occupying the center of the copper cubes; if the temprature is raised then zincs and coppers freely exchange.
2. From this definition we can guess that from now on we can do exactly what we have seen for the lattice gas.