Ising model : Exercises

Introduction

The questions below require you to find partition functions using the transfer matrix technique that was introduced in the video on finding the partition function for the 1D-closed Ising model. The problems below are difficult so don't get disillusioned if you cannot do them. They are here mostly to show you the scope of this technique.

Example problems

Click on the problems to reveal the solution

Problem 1

In the video on the closed, 1D Ising model we calculated the canonical partition function for a system of $N$ spins using the transfer matrix method. We are now going to use this method to calculate the ensemble average for the $i$th of our spin variables, $\langle S_i \rangle$, instead. Explain why I can calculate this quantity using the following equation: \[ \langle S_i \rangle = \frac{1}{Z_c} \sum_{s_1=0}^1 \sum_{s_2=0}^1 \sum_{s_3=0}^1 \dots \sum_{s_{N}=0}^1 z(s_i) \exp\left[ \beta J \sum_{j=1}^{N} z(s_j)z(s_{j+1}) + \beta H \sum_{j=1}^N z(s_j) \right] \] where as always $z(0)=-1$, $z(1)=1$ and $s_{N+1}=s_1$.

We note that the probability, $P_i$, of being in microstate $i$ (if we are in the canonical ensemble) is equal to: \[ P_i = \frac{ e^{-\beta E_i}}{Z_c} \] The average of a quantity, $A$, is then defined as: \[ \langle A \rangle = \sum_i A_i P_i \] where the sum runs over all the possible microstates $A_i$ the system can adopt. $A_i$ is the value of the quantity $A$ when we are in microstate $i$ and $P_i$ is the probability (see above) of being in that microstate. Now remember when we write out a sum over all microstates for an Ising system we consider write one summation sign (over the two states each spin can adopt) for each spin in the system. By doing so we thus enumerate over all possible states. If we do the summation above using the value of the $i$th spin as our property $A$ we get the expression in the question above.

Given that I can express the value of the spin variable using the Pauli spin matrix: \[ \sigma = \left( \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right) \] how might I rewrite the summation in the previous question using the transfer matrix approach.

From the first question we have: \[ \langle S_i \rangle = \frac{1}{Z_c} \sum_{s_1=0}^1 \sum_{s_2=0}^1 \sum_{s_3=0}^1 \dots \sum_{s_{N}=0}^1 z(s_i) \exp\left[ \beta J \sum_{j=1}^{N} z(s_j)z(s_{j+1}) + \beta H \sum_{j=1}^N z(s_j) \right] \] which we can rewrite as follows: \[ \begin{aligned} \langle S_i \rangle = & \frac{1}{Z_c} \sum_{s_1=0}^1 \sum_{s_2=0}^1 \sum_{s_3=0}^1 \dots \sum_{s_{N}=0}^1 z(s_i) \exp\left[ \beta J \sum_{j=1}^{N} z(s_j)z(s_{j+1}) +\frac{H}{2}[z(s_j)+z(s_{j+1})] \right] \\ = & \frac{1}{Z_c} \sum_{s_1=0}^1 \sum_{s_2=0}^1 \sum_{s_3=0}^1 \dots \sum_{s_{N}=0}^1 z(s_i) \prod_{j=1}^N e^{ \frac{\beta H z(s_j)}{2}} e^{\beta J z(s_j)z(s_{j+1})} e^{\frac{\beta H z(s_{j+1})}{2}} \\ = & \frac{1}{Z_c} \sum_{s_1=0}^1 \sum_{s_2=0}^1 e^{\frac{\beta H z(s_1)}{2}} e^{\beta J z(s_1)z(s_{2})} e^{\frac{\beta H z(s_{2})}{2}} \sum_{s_3=0}^1 e^{\frac{\beta H z(s_2)}{2}} e^{\beta J z(s_2)z(s_{3})} e^{\frac{\beta H z(s_{3})}{2}} \dots \\ & \dots \sum_{s_i=0}^1 z(s_i) e^{\frac{\beta H z(s_i)}{2}} e^{\beta J z(s_i)z(s_{i+1})} e^{\frac{\beta H z(s_{i+1})}{2}} \dots \\ & \dots \sum_{s_N=0}^1 e^{\frac{\beta H z(s_{N-1})}{2}} e^{\beta J z(s_{N-1})z(s_{N})} e^{\frac{\beta H z(s_{N})}{2}} e^{\frac{\beta H z(s_N)}{2}} e^{\beta J z(s_N)z(s_{1})} e^{\frac{\beta H z(s_{1})}{2}} \end{aligned} \] We now introduce the transfer matrix: \[ \mathbf{P} = \left( \begin{matrix} e^{\beta(J+H)} & e^{-\beta J} \\ e^{-\beta J} & e^{\beta(J-H)} \end{matrix} \right) \] We can use this matrix and the matrix defined in the question to rewrite the above sum as: \[ \langle S_i \rangle = \frac{1}{Z_c} \textrm{Tr}(\mathbf{P}^{i-1} \sigma \mathbf{P}^{N+1-i}) \] (Hopefully you are beginning to be able to see how we can rewrite these sums as products of matrices easily now. If you are still struggling, however, try to do all this explicitly for the case of 3 spins to convince yourself that the above works)

We next recall that the trace of a product of matrices is invariant to cyclic permutations in the order in which the matrix products are calculated. This is is important as we can use this fact to rewrite the above as \[ \langle S_i \rangle = \frac{\textrm{Tr}(\sigma \mathbf{P}^N) }{Z_c} = \frac{\textrm{Tr}(\sigma \mathbf{P}^N) }{\textrm{Tr}(\mathbf{P}^N)} \] We get to the final result here by using the fact that the canonical partition function, $Z_c$, is equal to the trace of the transfer matrix, $\mathbf{P}$.

Use the result from the previous calculation and the fact that: \[ \mathbf{S} = \mathbf{V}^{-1} \sigma \mathbf{V} = \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \] where $a$, $b$, $c$ and $d$ are parameters to calculate $\langle S_i \rangle$

We begin by recalling that we can write $\mathbf{P}^N$ as: \[ \mathbf{P}^N = \mathbf{V}\mathbf{\Lambda^N} \mathbf{V}^{-1} \] where $\mathbf{V}$ is the matrix of right eigenvectors and $\mathbf{\Lambda}$ is the diagonal matrix of eigenvalues. Substituting this into the equation for $\langle S_i \rangle$ we derived in the previous part gives: \[ \begin{aligned} \langle S_i \rangle & = \frac{ \textrm{Tr}\left[ \sigma \mathbf{V} \left( \begin{matrix} \lambda_1^N & 0 \\ 0 & \lambda_2^N \end{matrix} \right) \mathbf{V}^{-1} \right]}{ \textrm{Tr}\left[ \mathbf{V} \left( \begin{matrix} \lambda_1^N & 0 \\ 0 & \lambda_2^N \end{matrix} \right) \mathbf{V}^{-1} \right] } = \frac{ \textrm{Tr}\left[ \mathbf{V}^{-1} \sigma \mathbf{V} \left( \begin{matrix} \lambda_1^N & 0 \\ 0 & \lambda_2^N \end{matrix} \right) \right]}{ \textrm{Tr}\left[ \mathbf{V}^{-1} \mathbf{V} \left( \begin{matrix} \lambda_1^N & 0 \\ 0 & \lambda_2^N \end{matrix} \right) \right] } = \frac{ \textrm{Tr}\left[ \mathbf{V}^{-1} \sigma \mathbf{V} \left( \begin{matrix} \lambda_1^N & 0 \\ 0 & \lambda_2^N \end{matrix} \right) \right]}{ \textrm{Tr}\left[ \mathbf{I} \left( \begin{matrix} \lambda_1^N & 0 \\ 0 & \lambda_2^N \end{matrix} \right) \right] } \\ & = \frac{ \textrm{Tr}\left[ \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} \lambda_1^N & 0 \\ 0 & \lambda_2^N \end{matrix} \right) \right]}{ \lambda_1^N + \lambda_2^N } = \frac{\textrm{Tr} \left( \begin{matrix} a \lambda_1^N & b \lambda_2^N \\ c\lambda_1^N & d \lambda_2^N \end{matrix} \right)}{\lambda_1^N + \lambda_2^N} = \frac{a\lambda_1^N + d\lambda_2^N}{\lambda_1^N + \lambda_2^N} = \frac{a \left[ 1 + d \left(\frac{\lambda_2}{\lambda_1}\right)^N \right]}{ \left[ 1 + \left(\frac{\lambda_2}{\lambda_1}\right)^N \right]} \end{aligned} \] Notice how we have used the fact that the trace of a product of matrices is invariant to cyclic permutations in the order of the matrix multiplications in the above.

Now suppose that I want to calculate whether or not (on average) the spin on the $i$th site and the $(i+r)$th site are arranged parallel or antiparallel, $\langle s_i s_{i+r} \rangle$. Write an expression for this quantity using summation notation and show that when this is rewritten using transfer matrices you obtain. \[ \langle s_i s_{i+r} \rangle = \frac{ \textrm{Tr}(\sigma \mathbf{P}^r\sigma \mathbf{P}^{N-r} )}{\textrm{Tr}(\mathbf{P}^N)} \]

The average we have been asked to calculate can be calculated using the following summation \[ \begin{aligned} \langle S_i S_{i+r} \rangle = & \frac{1}{Z_c} \sum_{s_1=0}^1 \sum_{s_2=0}^1 \sum_{s_3=0}^1 \dots \sum_{s_{N}=0}^1 z(s_i) z(s_{i+r}) \exp\left[ \beta J \sum_{k=1}^{N} z(s_k)z(s_{k+1}) + \beta H \sum_{k=1}^N z(s_k) \right] \\ = & \frac{1}{Z_c} \sum_{s_1=0}^1 \sum_{s_2=0}^1 \sum_{s_3=0}^1 \dots \sum_{s_{N}=0}^1 z(s_i) z(s_{i+r}) \exp\left[ \beta J \sum_{k=1}^{N} z(s_k)z(s_{k+1}) +\frac{H}{2}[z(s_k)+z(s_{k+1})] \right] \\ = & \frac{1}{Z_c} \sum_{s_1=0}^1 \sum_{s_2=0}^1 \sum_{s_3=0}^1 \dots \sum_{s_{N}=0}^1 z(s_i) z(s_{i+r}) \prod_{k=1}^N e^{ \frac{\beta H z(s_k)}{2}} e^{\beta J z(s_k)z(s_{k+1})} e^{\frac{\beta H z(s_{k+1})}{2}} \\ = & \frac{1}{Z_c} \sum_{s_1=0}^1 \sum_{s_2=0}^1 e^{ \frac{\beta H z(s_1)}{2}} e^{\beta J z(s_1)z(s_{2})} e^{\frac{\beta H z(s_{2})}{2}} \sum_{s_3=0}^1 e^{ \frac{\beta H z(s_2)}{2}} e^{\beta J z(s_2)z(s_{3})} e^{\frac{\beta H z(s_{3})}{2}} \dots \\ & \dots \sum_{s_i=0}^1 z(s_i) e^{ \frac{\beta H z(s_i)}{2}} e^{\beta J z(s_i)z(s_{i+1})} e^{\frac{\beta H z(s_{i+1})}{2}} \dots \sum_{s_{i+r}=0}^1 e^{ \frac{\beta H z(s_{i+r})}{2}} e^{\beta J z(s_{i+r})z(s_{i+r+1})} e^{\frac{\beta H z(s_{i+r+1})}{2}} \dots \\ & \dots \sum_{s_N=0}^1 e^{ \frac{\beta H z(s_{N-1})}{2}} e^{\beta J z(s_{N-1})z(s_{N})} e^{\frac{\beta H z(s_{N})}{2}} e^{ \frac{\beta H z(s_N)}{2}} e^{\beta J z(s_N)z(s_{1})} e^{\frac{\beta H z(s_{1})}{2}} \end{aligned} \] Where here we once again have $z(0)=-1$ and $z(1)=1$. When we introduce the transfer matrix, $\mathbf{P}$, from above we see that this can be rewritten as: \[ \langle S_i S_j \rangle = \frac{1}{Z_c} \textrm{Tr}( \mathbf{P}^{i-1} \sigma \mathbf{P}^r \sigma \mathbf{P}^{N-r-i+1}) = \frac{1}{Z_c} \textrm{Tr}( \sigma \mathbf{P}^r\sigma \mathbf{P}^{N-r} ) \] where we again use the fact that we can permute the order of the matrices when we calculate the trace of their product. To get the result in the question you use the fact that $Z_c = \textrm{Tr}(\mathbf{P}^N)$.

Use the result from the previous question to express the ensemble average $\langle S_i S_j \rangle$ in terms of the parameters $a$, $b$, $c$ and $d$ and the eigenvalues of the transfer matrix $\lambda_1$ and $\lambda_2$

At the end of the previous question we arrrived at: \[ \langle S_i S_j \rangle = \frac{1}{Z_c} \textrm{Tr}( \sigma \mathbf{P}^r\sigma \mathbf{P}^{N-r} ) \] Lets now consider the numerator in the above expression only. Using our usual tricks with the eigendecomposition we can rewrite this as: \[ \begin{aligned} \textrm{Tr}\left[ \sigma \mathbf{P}^r\sigma \mathbf{P}^{N-r} \right] & = \textrm{Tr}\left[ \sigma \mathbf{V}\mathbf{\Lambda}^{r}\mathbf{V}^{-1} \sigma \mathbf{V}\mathbf{\Lambda}^{N-r} \mathbf{V}^{-1} \right] = \textrm{Tr}\left[\mathbf{V}^{-1} \sigma \mathbf{V}\mathbf{\Lambda}^{r}\mathbf{V}^{-1} \sigma \mathbf{V}\mathbf{\Lambda}^{N-r} \right] \\ & = \textrm{Tr}\left[ \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} \lambda_1^r & 0 \\ 0 & \lambda_2^r \end{matrix} \right) \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} \lambda_1^{N-r} & 0 \\ 0 & \lambda_2^{N-r} \end{matrix} \right) \right] \\ & = \textrm{Tr}\left[ \left( \begin{matrix} a \lambda_1^r & b \lambda_2^r \\ c \lambda_1^r & d \lambda_2^r \end{matrix} \right) \left( \begin{matrix} a \lambda_1^{N-r} & b \lambda_2^{N-r} \\ c \lambda_1^{N-r} & d \lambda_2^{N-r} \end{matrix} \right) \right] \\ & = \textrm{Tr}\left[ \left( \begin{matrix} a^2 \lambda_1^N + bc \lambda_1^{N-r}\lambda_2^r & ab\lambda_1^r\lambda_2^{N-r} + bd \lambda_2^N \\ ac\lambda_1^N + cd \lambda_1^{N-r}\lambda_2^r & bc \lambda_1^r \lambda_2^{N-r} + d^2 \lambda_2^N \end{matrix} \right) \right] \\ & = a^2 \lambda_1^N + bc \lambda_1^{N-r}\lambda_2^r + bc \lambda_1^r \lambda_2^{N-r} + d^2 \lambda_2^N \\ & = \lambda_1^N \left[ a^2 + d^2 \left(\frac{\lambda_2}{\lambda_1}\right)^N \right] + bc \lambda_1^N\left[ \left(\frac{\lambda_2}{\lambda_1}\right)^r + \left(\frac{\lambda_2}{\lambda_1}\right)^{N-r} \right] \end{aligned} \] Inserting this into the question at the start (and using the value we know for for $Z_c$) we thus have: \[ \langle S_i S_j \rangle = \frac{1}{Z_c} \textrm{Tr}( \sigma \mathbf{P}^r\sigma \mathbf{P}^{N-r} ) = \frac{\left[ a^2 + d^2 \left(\frac{\lambda_2}{\lambda_1}\right)^N \right] + bc \left[ \left(\frac{\lambda_2}{\lambda_1}\right)^r + \left(\frac{\lambda_2}{\lambda_1}\right)^{N-r} \right]}{\left[ 1 + \left(\frac{\lambda_2}{\lambda_1}\right)^N\right]} \]

Problem 2

Consider a 1-D open chain Ising model with $N$ spins in an external magnetic field $H$. Each spin will interact with its neighbours and with the external field. Write an expression that can be used calculate the energy of a microstate of this system. Remember in this model, at variance with the closed Ising model, the 1st spin does not interact with the $N$th spin.

The energy of a microstate can be calculated using: \[ E = -\sum_{i=1}^{N-1} J s_i s_{i+1} - \sum_{i=1}^N H s_i = -\frac{H}{2}s_1 - \sum_{i=1}^{N-1} \left[J s_i s_{i+1} + \frac{H}{2}(s_i + s_{i+1}) \right] - \frac{H}{2}s_N \] The important difference from the 1D-closed Ising model is that the first sum in the above goes from $1$ to $N-1$ rather than from $1$ to $N$ as it did in the closed case. The reason for this is that $s_{N+1} \ne s_1$ as it was for the closed Ising model. Now $s_{N+1}$ is a spin variable that just does not exist as the spins are now on a line and not on a circle.

Use the summation notation that we used to solve the closed, 1D Ising model to show that the canonical partition function for open, 1D Ising model is given by: \[ \begin{aligned} Z_c & \sum_{s_1=0}^1 e^{\frac{H\beta}{2}z(s_1)} \sum_{s_{2}=0}^1 e^{\frac{\beta H}{2}z(s_1)} e^{\beta J z(s_1) z(s_{2})}e^{\frac{\beta H}{2}z(s_{2})} \dots \sum_{s_{N-1}=0}^1 e^{\frac{\beta H}{2}z(s_{N-2})} e^{\beta J z(s_{N-2}) z(s_{N-1})}e^{\frac{\beta H}{2}z(s_{N-1})} \\ & \sum_{s_N=0}^1 e^{\frac{\beta H}{2}z(s_{N-1})} e^{\beta J z(s_{N-1}) z(s_{N})}e^{\frac{\beta H}{2}z(s_{N})} e^{\frac{H\beta}{2}z(s_N)} \end{aligned} \] Use the arguments the derivation we performed for the closed, 1D Ising model to show that the above summations can be written as a product of matrices as follows: \[ Z_c = \mathbf{s}^T \mathbf{P}^{N-1}\mathbf{s} \] where $\mathbf{P}$ is the transfer matrix that we introduced when solving the closed, 1D Ising model and where $\mathbf{s}$ is: \[ \mathbf{s} = \left( \begin{matrix} e^{\frac{\beta H}{2}} \\ e^{-\frac{\beta H}{2}} \end{matrix} \right) \] $\mathbf{s}^T$ is the (row vector) transpose of $\mathbf{s}$. (N.B. One way to do this is would be to enumerate all the microstates of an open-1D ising model with two spins, work out the energy of each and the contribution they make to the partition function. You could then compare the result you get by summing all these contributions to the result that you get by evaluating the above matrix product with $N=2$)

Once again in what follows we have $z(0)=-1$ and $Z(1)=1$. We can write the sum over microstates that we have to perform in order to evaluate the canonical partition function for this system as: \[ \begin{aligned} Z_c = & \sum_{s_1=0}^1 \sum_{s_2=0}^1 \dots \sum_{s_{N-1}=0}^1 e^{\left( \frac{\beta H}{2}z(s_1) + \sum_{i=1}^{N-1} \left[\beta J z(s_i) z(s_{i+1}) + \frac{\beta H}{2}(z(s_i) + z(s_{i+1})) \right]\right)} \sum_{s_N=0}^1 e^{\frac{H\beta}{2}z(s_N)} \\ = & \sum_{s_1=0}^1 \sum_{s_2=0}^1 \dots \sum_{s_{N-1}=0}^1 e^{\frac{H\beta}{2}z(s_1)} \left\{ \prod_{i=1}^{N-1} e^{\frac{\beta H}{2}z(s_i)} e^{\beta J z(s_i) z(s_{i+1})}e^{\frac{\beta H}{2}z(s_{i+1})} \right\} \sum_{s_N=0}^1 e^{\frac{H\beta}{2}z(s_N)} \\ = & \sum_{s_1=0}^1 e^{\frac{H\beta}{2}z(s_1)} \sum_{s_{2}=0}^1 e^{\frac{\beta H}{2}z(s_1)} e^{\beta J z(s_1) z(s_{2})}e^{\frac{\beta H}{2}z(s_{2})} \dots \sum_{s_{N-1}=0}^1 e^{\frac{\beta H}{2}z(s_{N-2})} e^{\beta J z(s_{N-2}) z(s_{N-1})}e^{\frac{\beta H}{2}z(s_{N-1})} \\ & \sum_{s_N=0}^1 e^{\frac{\beta H}{2}z(s_{N-1})} e^{\beta J z(s_{N-1}) z(s_{N})}e^{\frac{\beta H}{2}z(s_{N})} e^{\frac{H\beta}{2}z(s_N)} \end{aligned} \] To show that this sum can be written in the matrix product notation we note first note that terms like $$ e^{\frac{\beta H}{2}z(s_i)} e^{\beta J z(s_i) z(s_{i+1})}e^{\frac{\beta H}{2}z(s_{i+1})} $$ appeared when we derived the partition function for the closed 1D-Ising model. If you remember in that derivation we rewrote the summations involving these objects as matrix products. Notice here though that in the first product we do (the rightmost one in my convention) is different to the one we had when we were solving for the closed, 1D Ising model because the geometry of the lattice the spins are on is different in this case. In this, open, case the first three exponentials after the product are the objects that we turn into the transfer matrix $\mathbf{P}$. The last exponential, however, is simply $e^{\frac{H\beta}{2}z(s_N)}$ which can take on only two values $e^{+\frac{H\beta}{2}}$ and $e^{-\frac{H\beta}{2}}$, which is why we write the bit after the $\sum_{s_N=0}^1$ as a product of a vector and a matrix. Notice next that we have an $e^{\frac{\beta H}{2}z(s_i)} e^{\beta J z(s_i) z(s_{i+1})}e^{\frac{\beta H}{2}z(s_{i+1})}$ after all but one of the $N$ summation signs. Whenever we see this we replace it by a transfer matrix $\mathbf{P}$. Hence, we can rewrite everything from the summation over $s_2$ onwards as $\mathbf{P}^{n-1}\mathbf{s}$. Hopefully, you can see why this then is multiplied by the row vector transpose of $\mathbf{s}$ as I am not sure that I can explain it clearly or even whether this paragraph of text explains clearly how we have rewritten this as matrix products. Ah well... if you have learnt anything from this experience learn this - matrices are awesome! Just look at how simple the equation above now looks: \[ Z_c = \mathbf{s}^T \mathbf{P}^{N-1}\mathbf{s} \] Cool huh!

Show that when $H = 0$ the eigenvectors of the transfer matrix, $\mathbf{P}$, are equal to: \[ v_1 = \frac{1}{\sqrt{2}} \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) \qquad \textrm{and} \qquad v_2 = \frac{1}{\sqrt{2}} \left( \begin{matrix} 1 \\ -1 \end{matrix} \right) \] Use the eigenvalues of this matrix (we already calculated these when we considered the closed 1D-Ising model) and all the other tricks that we used for calculating powers of matrices to show that, when we evaluate the canonical partition function (in the absence of a magnetic field) using the transfer matrix method for an open 1D-Ising model, we get the following: $$ Z_c = 2^N\cosh^{N-1}(\beta J) $$

The eigenvalues we obtained for the transfer matrix $\mathbf{P}$ in the derivation for the 1D-Ising model were: \[ \lambda = e^{\beta J} \cosh(\beta H) \pm \sqrt{ e^{2\beta J}\sinh^2(\beta H) + e^{-2\beta J} } \] In the absence of a magentic field, $H=0$, these reduce to $\lambda_1=2\cosh(\beta J)$ and $\lambda_2=2\sinh(\beta J)$.

To find the partition function for the open, 1D-Ising model we also need to calculate the eigenvectors of this matrix. Remember that we calculate an eigenvector by rearranging the equation that defines our eigensystem of equations as follows: \[ \mathbf{P}\mathbf{v} = \lambda \mathbf{v} \qquad \rightarrow \qquad \left( \mathbf{P} - \lambda \mathbf{I} \right) \mathbf{v} = 0 \] Lets now write out the matrices explicitly and find the first eigenvector (the one with eigenvalue $2\cosh(\beta J)=e^{\beta J} + e^{-\beta J}$). \[ \begin{aligned} 0= \left( \mathbf{P} - \lambda \mathbf{I} \right) \mathbf{v} & = \left\{ \left( \begin{matrix} e^{\beta J} & e^{-\beta J} \\ e^{-\beta J} & e^{\beta J} \end{matrix} \right) - \left( e^{\beta J} + e^{-\beta J} \right) \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) \right\} \left( \begin{matrix} v_1 \\ v_2 \end{matrix} \right) \\ & = \left( \begin{matrix} -e^{-\beta J} & e^{-\beta J} \\ e^{-\beta J} & -e^{-\beta J} \end{matrix} \right) \left( \begin{matrix} v_1 \\ v_2 \end{matrix} \right) =0 \end{aligned} \] Multiplying the top row of the matrix by the vector gives us $-e^{-\beta J}v_1 + e^{-\beta J}v_2 = 0$ which we can rearrange to give $v_1=v_2$. Now to get the final values of the components of the eigenvector we require that the vector $\mathbf{v}$ be normalised. In other words we require $v_1^2 + v_2^2 = 1$. We can rewrite this requirement as: $2v_1^2 =1$ as we know that $v_1=v_2$ and hence arrive at $v_1=v_2 = \frac{1}{\sqrt{2}}$.

We now turn to the second eigenvalue and eigenvector. By a similar logic to before we have: \[ \begin{aligned} 0= \left( \mathbf{P} - \lambda \mathbf{I} \right) \mathbf{v} & = \left\{ \left( \begin{matrix} e^{\beta J} & e^{-\beta J} \\ e^{-\beta J} & e^{\beta J} \end{matrix} \right) - \left( e^{\beta J} - e^{-\beta J} \right) \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) \right\} \left( \begin{matrix} v_1 \\ v_2 \end{matrix} \right) \\ & = \left( \begin{matrix} e^{-\beta J} & e^{-\beta J} \\ e^{-\beta J} & e^{-\beta J} \end{matrix} \right) \left( \begin{matrix} v_1 \\ v_2 \end{matrix} \right) =0 \end{aligned} \] which brings us neatly to $e^{-\beta J}v_1 + e^{-\beta J}v_2 = 0$ and hence $v_2=-v_1$. The requirement of normalisation once again gives $2v_1^2 =1$ and hence $v_1=\frac{1}{\sqrt{2}}$ while $v_2 = -\frac{1}{\sqrt{2}}$

Lets now turn again to the partition function. When we left off we had: \[ Z_c = \mathbf{s}^T \mathbf{P}^{N-1}\mathbf{s} \] We can exploit the fact that $\mathbf{P}$ is diagonazable and rewrite this as: \[ Z_c = \mathbf{s}^T \mathbf{V} \mathbf{\Lambda}^{N-1} \mathbf{V}^{-1} \mathbf{s} \] The matrix $\mathbf{V}$ is the matrix that has the eigenvectors of $\mathbf{P}$ in its columns. In other words it is: \[ \mathbf{V} = \left( \begin{matrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{matrix} \right) \] Lets invert this matrix (remember we do this by dividing the adjoint matrix by the determinant as below): \[ \mathbf{V}^{-1} = \frac{1}{-\frac{1}{2}-\frac{1}{2}} \left( \begin{matrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{matrix} \right) = \left( \begin{matrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{matrix} \right) \] What do you know $\mathbf{V}^{-1}=\mathbf{V}$. We could have worked this out by recalling something about symmetric matrices; namely, that we can decompose a symmetric diagonalisable matrix as: \[ \mathbf{P} = \mathbf{V} \mathbf{\Lambda} \mathbf{V}^T \] Here $\Lambda$ is still the diagonal element containing the eigenvalues and $\mathbf{V}$ is the matrix containing the eigenvectors.

Right so the matrix multiplication we have to do is as follows: \[ \begin{aligned} Z_c & = \left( \begin{matrix} 1 & 1 \end{matrix} \right) \left( \begin{matrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{matrix} \right) \left( \begin{matrix} \lambda_1^{N-1} & 0 \\ 0 & \lambda_2^{N-1} \end{matrix} \right) \left( \begin{matrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{matrix} \right) \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) \\ & = \left( \begin{matrix} \sqrt{2} & 0 \end{matrix} \right) \left( \begin{matrix} \lambda_1^{N-1} & 0 \\ 0 & \lambda_2^{N-1} \end{matrix} \right) \left( \begin{matrix} \sqrt{2} \\ 0 \end{matrix} \right) \\ & = \left( \begin{matrix} \sqrt{2} & 0 \\ \end{matrix} \right) \left( \begin{matrix} \sqrt{2}\lambda_1^{N-1} \\ 0 \end{matrix} \right) = 2\lambda_1^{N-1} = 2^N\cosh^{N-1}(\beta J) \end{aligned} \] Notice that in the first line of the above we have used $H=0$ for the $\mathbf{s}$ vectors that were introduced in part (b). In the final part I have inserted in the value of the first eigenvalue.

Calculate the ensemble average of the energy $\langle E \rangle$ for an open, 1D Ising model with $H=0$.

The ensemble average of the energy is given by: $$ \langle E \rangle = -\left(\frac{\partial \ln Z_c }{\partial \beta } \right) $$ For the 1D Ising model with $H=0$ we thus have: $$ \langle E \rangle = -(N-1)J\tanh(\beta J) $$

Problem 3

The energy of a microstatestate in the Potts model is given by: \[ E = -J \sum_{i=1}^{N}\delta(\sigma_i,\sigma_{i+1}) \qquad \textrm{where} \qquad \sigma_{N+1}=\sigma_1 \] where $\sigma_i$ is the value of the spin variable on site $i$. This can take values between $1$ and $p$. $\delta(\sigma_i,\sigma_{i+1})$ is the Kronecker delta and is given by: \[ \delta(\sigma_i,\sigma_j) = \begin{cases} 1 & \textrm{if} \quad i = j \\ 0 & \textrm{if} \quad i \ne j \end{cases} \] Which sites interact in the Potts model? When is the interaction energy between two interacting sites in the Potts model non-zero?

Neighboring spins interact in the Pots model and the spins are arranged on a circle so that each spin variable can interact with two neighbors. The interaction between two neighbouring sites is non zero when they have the same value for their spin variables.

Use the summation notation that we have used in exercises 5 and 6 to show that the canonical partition function for a Potts model with $N$ sites that can each be in one of $p$ sites can be written as: \[ Z_c = \sum_{s_1=1}^p \sum_{s_2=1}^p e^{J\beta \delta(\sigma_1,\sigma_2)} \dots \sum_{s_N=1}^p e^{J\beta \delta(\sigma_{N-1},\sigma_N)} e^{J\beta \delta(\sigma_N,\sigma_1)} \]

The canonical partition function is: \[ \begin{aligned} Z_c & = \sum_{s_1=1}^p \sum_{s_2=1}^p \dots \sum_{s_N=1}^p e^{J\beta \sum_{i=1}^N \delta(\sigma_i,\sigma_{i+1})} \\ & = \sum_{s_1=1}^p \sum_{s_2=1}^p \dots \sum_{s_N=1}^p \prod_{i=1}^N e^{J\beta \delta(\sigma_i,\sigma_{i+1})} \\ & = \sum_{s_1=1}^p \sum_{s_2=1}^p e^{J\beta \delta(\sigma_1,\sigma_2)} \dots \sum_{s_N=1}^p e^{J\beta \delta(\sigma_{N-1},\sigma_N)} e^{J\beta \delta(\sigma_N,\sigma_1)} \end{aligned} \]

We are going to find the canonical partition function for the Potts model using a transfer matrix in a manner similar to what we did for the closed and open Ising models. Show that the transfer matrix, $\mathbf{P}$, is given by: \[ \mathbf{P} = \mathbf{D}(\beta) + \mathbf{A} = \left( \begin{matrix} e^{\beta J} - 1 & 0 & 0 & \dots & 0 \\ 0 & e^{\beta J} - 1 & 0 & \dots & 0 \\ 0 & 0 & e^{\beta J} - 1 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & e^{\beta J} - 1 \end{matrix} \right) + \left( \begin{matrix} 1 & 1 & 1 & \dots & 1 \\ 1 & 1 & 1 & \dots & 1 \\ 1 & 1 & 1 & \dots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & 1& \dots & 1 \end{matrix} \right) \]

The formula for the partition function we wrote out in the previous part tells us that should calculate the elements of the transition matrix using $e^{J\beta \delta(\sigma_N,\sigma_1)}$. The transfer matrix is thus: \[ \mathbf{P} = \left( \begin{matrix} e^{\beta J} & 1 & 1 & \dots & 1 \\ 1 & e^{\beta J} & 1 & \dots & 1 \\ 1 & 1 & e^{\beta J} & \dots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & 1& \dots & e^{\beta J} \end{matrix} \right) \] This can be decomposed into the sum of matrices that is required in the question.

To determine the partition function we will use the tricks that we have used for the open and closed Ising models. In other words we are going to take the trace of the transfer matrix raised to the power $N$. Furthermore, we are going to do this by taking the sum of the powers of the eigenvalues as usual. As per usual the first step in doing this is to determine the eigenvalues. Show that for this system the first of these is given by: \[ \lambda_1 = e^{\beta J} + p - 1 \] and that the remaining eigenvalues are all equal to: \[ \lambda_i = e^{\beta J} -1 \]

Remember the eigenvalue equation is: \[ \mathbf{P}\mathbf{v} = \lambda \mathbf{v} \] We can substitute $\mathbf{P}$ into the above to give: \[ (\mathbf{D}(\beta) + \mathbf{A})\mathbf{v} = \lambda \mathbf{v} \quad \rightarrow \quad (\mathbf{D}(\beta) - \lambda \mathbf{I}) \mathbf{v} = -\mathbf{A}\mathbf{v} \] Now notice from above that $\mathbf{A}$ is a square matrix containing all ones. As a consequence the right hand side, $\mathbf{A}\mathbf{v}$ is a vector that has all equal elements. Because $\mathbf{A}$ is a matrix of all ones the elements of the vector $\mathbf{A}\mathbf{v}$ are equal to the sum of the elements of $\mathbf{v}$. With this in mind lets write out the matrices: \[ \left\{ \left( \begin{matrix} e^{\beta J} - 1 & 0 & 0 & \dots & 0 \\ 0 & e^{\beta J} - 1 & 0 & \dots & 0 \\ 0 & 0 & e^{\beta J} - 1 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & e^{\beta J} - 1 \end{matrix} \right) - \lambda \left( \begin{matrix} 1 & 0 & 0 & \dots & 0 \\ 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & 1 \end{matrix} \right) \right\} \left( \begin{matrix} v_1 \\ v_2 \\ v_3 \\ \vdots \\ v_p \end{matrix} \right) = \left( \begin{matrix} -\nu \\ -\nu \\ -\nu \\ \vdots \\ -\nu \end{matrix} \right) \] Where here we have introduced the symbol $\nu=\sum_{i=1}^p v_i$. If we add up the matrices on the left hand side this is: \[ \left( \begin{matrix} e^{\beta J} - 1 - \lambda & 0 & 0 & \dots & 0 \\ 0 & e^{\beta J} -1 - \lambda & 0 & \dots & 0 \\ 0 & 0 & e^{\beta J} - 1 - \lambda & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & e^{\beta J} - 1 - \lambda \end{matrix} \right) \left( \begin{matrix} v_1 \\ v_2 \\ v_3 \\ \vdots \\ v_p \end{matrix} \right) = \left( \begin{matrix} -\nu \\ -\nu \\ -\nu \\ \vdots \\ -\nu \end{matrix} \right) \] Now notice that if take any row in the matrix on the left hand side of the equation and multiply it by the vector. We get something like the following: \[ \left(e^{\beta J} + 1 - \lambda \right)v_i = \nu \] where $i$ is the particular row that we chose to multiply. Now remember that $\nu=\sum_{i=1}^p v_i$, which is important because when this is combined with the above it tells us that for any $\nu \ne 0$: \[ v_1 = v_2 = v_3 = \dots = v_p \] as only then can this equality hold for {\bf all} $v_i$ as it must. Notice we get $\left(e^{\beta J} + 1 - \lambda \right)v_i = \nu$ whenever we multiply any row of the matrix by the vector $\mathbf{v}$. Inserting this all into our matrices gives: \[ \left( \begin{matrix} e^{\beta J} - 1 - \lambda & 0 & 0 & \dots & 0 \\ 0 & e^{\beta J} -1 - \lambda & 0 & \dots & 0 \\ 0 & 0 & e^{\beta J} - 1 - \lambda & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & e^{\beta J} - 1 - \lambda \end{matrix} \right) \left( \begin{matrix} v_1 \\ v_1 \\ v_1 \\ \vdots \\ v_1 \end{matrix} \right) = \left( \begin{matrix} -p v_1 \\ -p v_1 \\ -p v_1 \\ \vdots \\ -p v_1 \end{matrix} \right) \] which in turn implies: \[ \left(e^{\beta J} - 1 - \lambda_1 \right) = -p \quad \rightarrow \quad \lambda_1 = e^{\beta J} - 1 + p \] as the factors of $v_1$ on the right and left hand sides of the equation cancel.

We now note that if $\nu=0$ we must have: \[ \left(e^{\beta J} - 1 - \lambda \right) = 0 \qquad \rightarrow \qquad \lambda = e^{\beta J} - 1 \] This will be the eigenvalue for any eigenvector that has $\nu=0$. $\lambda_1$ will be the eigenvalue corresponding to any eigenvector that has $\nu\ne0$.

To confirm these arguments are correct consider that the trace of the original matrix $\mathbf{P}$ is equal to: \[ \textrm{Tr}(\mathbf{P}) = pe^{\beta J} \] which is equal to: \[ \lambda_1 + (p-1)\lambda = (e^{\beta J} - 1 + p ) + p(e^{\beta J} - 1) = 6e^{\beta J} + p -p = 6e^{\beta J} \] and as we now know the trace of a matrix is equal to the sum of its eigenvalues.

Use the eigenvalues that you determined in the previous question to determine the canonical partition function for a system of $N$ Potts sites. Use the expression you derived to evaluate the average energy. Explain how you would calculate the heat capacity from the average energy.

We calculate the canonical partition function by taking the sum of the powers of the eigenvalues as we did when we examined the closed 1D Ising model. This gives us: \[ Z_c = \lambda_1^N + (p-1)\lambda_2^N = (e^{\beta J} + p -1)^N + (p-1)(e^{\beta J} -1)^N \] We can then calculate the average energy per site from the canonical partition function using: \[ \begin{aligned} \frac{\langle E \rangle}{N} = - \frac{1}{N} \left( \frac{\partial \ln Z_c}{\partial \beta} \right) & = - \frac{1}{N} \frac{\partial }{\partial \beta}\left\{ \ln[ (e^{\beta J} + p -1)^N + (p-1)(e^{\beta J} -1)^N ] \right\} \\ & = - \frac{Je^{\beta J}[(e^{\beta J} + p -1)^{N-1} + (p-1)(e^{\beta J} - 1)^{N-1}] }{(e^{\beta J} + p -1)^N + (p-1)(e^{\beta J} -1)^N} \end{aligned} \] The per-site heat capacity can be calculated from the average energy per site as follows: \[ \begin{aligned} C_v = \frac{1}{N} \left( \frac{\partial \langle E \rangle }{\partial \beta} \right) & = -\frac{\partial }{\partial \beta} \left[ \frac{Je^{\beta J}[(e^{\beta J} + p -1)^{N-1} + (p-1)(e^{\beta J} - 1)^{N-1}] }{(e^{\beta J} + p -1)^N + (p-1)(e^{\beta J} -1)^N} \right] \\ \end{aligned} \] The differentiation here is going to be a nightmare so it is not worth doing

Ising model : Exercises

Introduction

Example problems

Problem 1

Problem 2

Problem 3

Contact Details