Midterm 2 Solutions

Problem 1

Consider a magnetic system described by an Ising model $(S_{i}=\pm 1)$ on a square lattice with $\ H=\sum _{\langle ij\rangle }J_{ij}s_{i}s_{j}+\sum _{\langle ij\rangle }K_{ij}s_{i}s_{j}$. Here, the ferromagnetic interaction $\ J_{ij}<0$ couples only nearest-neighbor spins. In contrast,the antiferromagnetic interaction $\ K_{ij}>0$ couples only the the second neighbors, i.e. spins that are "across the square" ( diagonally placed relative to each other).

a)

Sketch the suare lattice and determine the number of first neighbors $\mathbb {Z} _{FM}$ which interact ferromagnetically with the given spin, and the number of second neighbors $\mathbb {Z} _{AFM}$ which interact antiferromagnetically with the given spin. (5points)

Image of square lattice with ferromagnetic $J_{i,j}<0$ coupling nearest neighbor spins. 
Anti-ferromagnetic $K_{i,j}<0$ coupling only second nearest neighbors. 
$Z_{FM}=4$
$Z_{AFM}=4$

b)

Use the mean-field approximation, where $\ S_{i}S_{j}\rightarrow S_{i}\langle S_{j}\rangle$. Assume that the magnitude of $\ J_{ij}$ is appreciably larger then the magnitude of $\ K_{ij}$; then the system will order ferromagnetically at low temperatures, so that $\langle S_{i}\rangle =m$ for any site. Derive the self consistency condition determining the magnetization "m". (15points)

The point of the mean-field approximation is to simplify a many body problem. To consider all the interactions taking place in a material would require huge amounts of computing power and resources. The mean-field approximation accomplishes this by considering only the average of the interactions.

The first step in this problem is to right down our equation for the energy when $\ J_{ij}$ is "appreciably larger than $\ K_{ij}$ .

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ H = \sum_{ij} J_{ij} S_{i} S_{j}}

We also know:

$\langle S_{i}\rangle =m$

This is due to the fact that we are considering many neighbors, so as z gets very large then we get the following:

$\ 1/z\sum _{j=1}^{z}S_{j}\rightarrow \langle S_{i}\rangle =m$

Now if we sum over many random variables we get the following:

$\sum _{j=1}^{z}S_{j}\rightarrow z\langle S_{i}\rangle$

Now we can simplify our original expression as the following:

$\ H=J\sum _{i}^{N}S_{i}zm$

Pulling out the constants we get:

$\ H=Jzm\sum _{i=1}^{N}S_{i}$

Since we have a problem of non interacting spins in an internal field we can consider and solve for only one spin. This then gives us the following:

$\ H=JzmS_{1}$

$\ S_{1}=\pm 1$

Now we need to consider the probability for the spin to point either up or down. This is given as the following:

$\ P(S_{1}=\pm 1)=1/z(e^{\pm Jzm/kt})$

We can also define z as:

$\ z=e^{\pm Jzm/kt}$

Now we can define the magnetization as the spin times the probability of the spin:

$\ m=\sum _{S_{1}=\pm 1}S_{1}P(S_{1})$

Rewriting this equation we get:

$\ m={\frac {e^{Jzm/kt}-e^{-Jzm/kt}}{e^{Jzm/kt}+e^{-Jzm/kt}}}$

This is the same as the tanh, so we get:

$\ m=\tanh({\frac {Jzm}{kt}})$

Making $\ \beta =1/kt$

We get:

$\ m=\tanh(\beta Jzm)$

Problem 2

a)

b) Since Coulomb interactions are long range by $e^{2}/r$ high densities cause the formation of solids such as crystals.

f) Where $T\rightarrow \infty$ Fermi-Dirac Statistics and all quantum effects can be ignored.