Mean field problems : Exercises

Introduction

The problems below demonstrate how we can calculate the partition function for the Ising model by exploiting the mean field approximation. The essence of this approach was covered in the videos the problems below show how we can vary the number of interactions that we include explicitly and thus progressively refine our estimate when using this technique.

Example problems

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Problem 1

The mean field Hamiltonian for the 1D-closed Ising model that was introduced in the video had one spin incorporated explicitly. The interaction between this spin and its neighbours was then incorporated using the mean field approximation. Write out a mean field Hamiltonian that models the interaction between a pair of spins explicitly and then then models the interaction with the remaining spins using the mean field approximation

The Hamiltonian for our pair of spins in a mean field is given by: \[ E = -Js_1s_2 - (H + J\langle s\rangle)(s_1 +s_2) \] The $s_1$ and $s_2$ in this expression are the values of the two spin variables that we are incorporating explicitly. The interaction between these two spins is explicitly modelled in the first term. The factor of $J\langle s \rangle (s_1 + s_2)$ ensures that the remaining spin-spin interactions are incoroporated using the mean field approximation.

Calculate the partition function for the Hamiltonian that you introduced in the previous part and derive an expression that can be solved self-consistently to determine the average spin on a lattice site.

The partition function for the Hamiltonian introduced in the previous part is given by: \[ \begin{aligned} Z_c^{mfa} & = \sum_{s_1=0}^1 \sum_{s_2=0}^1 \dots \sum_{s_{N-1}=0}^1 \sum_{s_N=0}^1 \exp\left( \beta \sum_{i=1}^{N/2} Jz(s_{2i-1}) z(s_{2i}) + (H+J\langle s \rangle)[z(s_{2i-1})+z(s_{2i})] \right) \\ & = \sum_{s_1=0}^1 \sum_{s_2=0}^1 \dots \sum_{s_{N-1}=0}^1 \sum_{s_N=0}^1 \prod_{i=1}^{N/2} e^{\beta Jz(s_{2i-1})z(s_{2i})}e^{\beta(H+J\langle s \rangle)[z(s_{2i-1})+z(s_{2i})]} \\ & = \sum_{s_1=0}^1 \sum_{s_2=0}^1e^{\beta Jz(s_{1})z(s_{2})}e^{\beta(H+J\langle s \rangle)[z(s_{1})+z(s_{2})]} \dots \sum_{s_{N-1}=0}^1 \sum_{s_N=0}^1 e^{\beta Jz(s_{N-1})z(s_{N})}e^{\beta(H+J\langle s \rangle)[z(s_{N-1})+z(s_{N})]} \\ & = \left(e^{\beta J}e^{2\beta(H+J\langle s \rangle)} + 2e^{-\beta J} + e^{\beta J}e^{-2\beta(H+J\langle s \rangle)} \right) \dots \left(e^{\beta J}e^{2\beta(H+J\langle s \rangle)} + 2e^{-\beta J} + e^{\beta J}e^{-2\beta(H+J\langle s \rangle)} \right) \\ & = \left(e^{\beta J}e^{2\beta(H+J\langle s \rangle)} + 2e^{-\beta J} + e^{\beta J}e^{-2\beta(H+J\langle s \rangle)} \right)^{N/2} \\ & = \left(2e^{\beta J}\cosh(2\beta(H+J\langle s \rangle)) + 2e^{-\beta J} \right)^{N/2} \end{aligned} \] To calculate the average spin, $\langle s \rangle$, in a system with $N$ sites we use the fact that this quantity is equal to the average magnetisation, $\langle M \rangle$, as shown below: $$ \langle s \rangle = \frac{\langle M \rangle}{N} $$ We can calculate the average magnetization by taking suitable derivatves of the logarithm of the partition function as shown below: \[ \begin{aligned} \langle s \rangle & = \frac{1}{N \beta} \left( \frac{\partial \ln(Z_c)}{\partial H}\right)_T \\ & = \frac{1}{N \beta} \left\{ \frac{\partial }{\partial H} \frac{N}{2} \ln\left( 2e^{\beta J}\cosh(2\beta[H+J\langle s \rangle)] + 2e^{-\beta J} \right) \right\}_T \\ & = \frac{N}{2N \beta} \left[ \frac{4\beta e^{\beta J} \sinh[2\beta(H+J\langle s\rangle)] }{ 2e^{\beta J}\cosh[2\beta(H+J\langle s \rangle)] + 2e^{-\beta J}} \right] \\ & = \frac{\sinh[2\beta(H+J\langle s\rangle)] }{ \cosh[2\beta(H+J\langle s \rangle)] + e^{-2\beta J}} \end{aligned} \]

Show that the inverse temperature, $\beta$, at which the order-disorder transition takes place for a closed-1D Ising model in which the interaction between pairs of spins is modelled explicitly and the interactions with the remaining sites are incorporated using the mean field approximation is given by: \[ \beta = \frac{1+e^{-2\beta J}}{2\beta J} \] You should do this calculation in the absence of a field $H=0$.

For the reasons discussed in the video we find the transition temperature by finding the temperature for which: $$ \frac{1}{\beta N} \frac{\partial }{\partial \langle s \rangle } \left( \frac{\partial \ln(Z_c)}{\partial H}\right)_T = 1 $$ When this is done for the partition function introduced in this question we find that: \[ \begin{aligned} \left\{ \frac{\partial}{\partial \langle s \rangle } \frac{\sinh[2\beta J\langle s\rangle] }{ \cosh[2\beta J\langle s \rangle] + e^{-2\beta J}} \right\}_{\langle s\rangle =0} & = 1 \\ \left\{ \frac{2\beta J \cosh[2\beta J\langle s\rangle] }{\cosh[2\beta J\langle s \rangle] + e^{-2\beta J}} - \frac{2\beta J\sinh^2[2\beta J\langle s\rangle] }{(\cosh[2\beta J\langle s \rangle] + e^{-2\beta J})^2} \right\}_{\langle s \rangle = 0 } & = 1 \\ \frac{2\beta J }{1+e^{-2\beta J}} & = 1 \qquad \rightarrow \qquad \beta = \frac{1+e^{-2\beta J}}{2\beta J} \end{aligned} \]

Problem 2

In the 2D-Ising model the spins are arranged on a square lattice - you can think of it like a chess board with one spin on each of the black and white squares. Each spin can take on either the familiar spin-up or spin-down configuration and each spin interacts with its four nearest neighbours. The interaction between any pair of spins is given by $-Js_1s_2$, where $s_1$ and $s_2$ are the spin coordinates of the two interacting sites and can take values of $-1$ and $+1$. When people draw diagrams of the Ising model they generally draw a flat lattice. Really, however, for each spin to have four neighbours system the system must be in a closed geometry so you should think of the spins as arranged on a taurus (donut). <b> Write down a mean field Hamiltonian for the 2D Ising model, which supposes that we model one spin explicitly and the interaction between this spin and its neighbors using the mean field approximation. </b>

\[ E^{mfa} = -\sum_{i=1}^N (H + 4J\langle S \rangle)s_i \]

Now calculate the canonicial partition function for this mean field Hamiltonian for the 2D Ising model. Use the result you obtain to calculate the average spin per site and the temperature at which the order-disorder transition takes place.

The partition function is given by: \[ Z_c^{mfa} = 2^N\cosh^N[\beta(H + 4J\langle s \rangle)] \] The average spin per site is thus: \[ \langle s \rangle = \tanh[\beta(H+4J\langle s\rangle)] \] and the transition temperature is $k_B T = 4J$.

Consider the two-dimensional Ising model on a square lattice once more. This time write down a mean field Hamiltonian for this system that supposes that we model the interaction between a pair of neighbouring spins explicitly and the interaction between these two spins and the surrounding spins using the mean field approximation.

\[ E = -\sum_{i=1}^{N/2} Js_{2i-1}s_{2i} + (H + 3J\langle s \rangle )(s_{2i-1}+s_{2i} ) \]

Now calculate the canonical partition function for this new mean field Hamiltonian. Use the result you obtain to calculate the average spin per site and to derive a self-consistent equation the solution of which tells you temperature at which the order-disorder transition takes place.

The partition function is given by: \[ Z^{mfa} = \left(2e^{\beta J}\cosh(2\beta(H+3J\langle s \rangle)) + 2e^{-\beta J} \right)^{N/2} \] The average spin per site is thus: \langle s \rangle = \frac{\sinh[2\beta(H+3J\langle s\rangle)] }{ \cosh[2\beta(H+3J\langle s \rangle)] + e^{-2\beta J}} The equation for the order-disorder transition is: \[ \beta = \frac{1+e^{-2\beta J}}{6\beta J} \]

Consider the two-dimensional Ising model on a square lattice once more. This time write down a mean field Hamiltonian for this system that supposes that we model the interaction between a cluster of 5 neighbouring spins explicitly and the interaction between these five spins and the surrounding spins using the mean field approximation. Calculate the canonical partition function for this new mean field Hamiltonian.

The Hamiltonian in this case is: \[ E = -\sum_{i=1}^{N/5} \left[ Hs_{5i} + \sum_{j=1}^4 (3J\langle s \rangle + H)s_{5i-j} + Js_{5i-j}s_{5i} \right] \] And the partition function is: \[ Z_N = \left\{e^{\beta H} 2^4 \cosh^4(\beta\left[ 3J\langle s \rangle + H + J \right]) + e^{-\beta H} 2^4 \cosh^4(\beta\left[ 3J\langle s \rangle + H - J \right])\right\}^{N/5} \]

Finally, consider an entire row of spins in the average field of the rest of the lattice. Use the fact that the average value of the spin for a one-dimensional, closed Ising model is given by: $$ \langle s \rangle = \frac{ \sinh(\beta H) }{\sqrt{\sinh^2(\beta H) + e^{-4\beta J}}} $$ To determine a self-consistent equation in $\beta$ that can be used to determine the temperature at which the order-disorder transition takes place in this system.

If we construct the full system by supposing that the spins in a row interact explicitly and then interact with mean field spin spins from adjacent rows we can write the Hamiltonian for the system as: \[ E^{mfa} = - \sum_{i=1}^{N} \sum_{j=1}^{M} J s_{ij}s_{i(j+1)} + Hs_{ij} + 2J\langle s \rangle \] Here we suppose we have $N\times M$ spins in total and that these are arranged in $N$-rows containing $M$ spins each. We can think of the spins on each row as a closed 1D-Ising model in a field of magnitude: $H+2J\langle s\rangle$. Thinking this way and using the equation given in the question we can thus write the average spin as: \[ \langle s \rangle = \frac{ \sinh[\beta (H+2J\langle s \rangle)] }{\sqrt{\sinh^2[\beta(H+2J\langle s \rangle)] + e^{-4\beta J}}} \] which, when $H=0$, becomes: \[ \langle s \rangle = \frac{ \sinh[2\beta J\langle s \rangle)] }{\sqrt{\sinh^2[2\beta J\langle s \rangle] + e^{-4\beta J}}} \] We can thus find (in the usual way) that the transition temperature is the solution of: \[ \beta = \frac{e^{-2\beta J}}{2J} \]

Mean field problems : Exercises

Introduction

Example problems

Problem 1

Problem 2

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