1. In this problem, we are going to show that the mean field theory (MFT) is exact for the infinite-range Ising model. Let's consider the Ising Hamiltonian $$\begin{align} H = -\frac{1}{2}\sum_{ij}J_{ij}S_iS_j \end{align}$$ and assume that \(J_{ij}\) has the same value for any sites \(i,j\). In particular, let \(J_{ij} = J/N\), where \(N\) is the total number of sites.
    1. Show that the Ising model Hamiltonian in this case is $$\begin{align} H = -\frac{J}{2N}\left(\sum_i S_i\right)^2 \end{align}$$
    2. The Hubbard-Stratonovich transformation is easier in this case, because \(\sum_i S_i\) can be treated as one variable. Show that the partition function becomes (apart from a non-essential prefactor \(\sqrt{N/(2\pi JT)}\) $$\begin{align} Z = \sum_{\{S\}}\int_{-\infty}^\infty e^{-\frac{N}{2JT}\varphi^2 + \frac{\varphi}{T}\sum_i S_i}\d\varphi = \int_{-\infty}^\infty e^{-NS[\phi]} \d\varphi \end{align}$$ and write down the expression for \(S[\varphi]\).
    3. Write down the equation for \(\bar{\varphi}\) that minimizes \(S[\varphi]\).
    4. Expand \(S[\varphi]\) around the value \(\bar{\varphi}\) that minimizes it (without explicitly calculating \(\bar{\varphi}\) and \(\d^2S/\d\varphi^2\)), calculate the free energy and show that in the limit \(N\to\infty\) you get $$\begin{align} F = NTS[\bar{\varphi}] \end{align}$$
    1. This is trivial, as we can write $$\begin{align} H = -\frac{1}{2N}J \sum_{ij} S_i S_j = -\frac{J}{2N}\left(\sum_i S_i\right)^2 \end{align}$$
    2. We can write $$\begin{align} e^{\beta J(\sum S)^2/2N} = e^{b^2/4a} \end{align}$$ where we set \(b = \beta \sum S\), and \(a = \frac{N}{2J T}\). Thus $$\begin{align} \int_{-\infty}^\infty e^{-\frac{N}{2JT}\varphi^2 + \frac{\varphi}{T}\sum_i S_i} \d\varphi = e^{\beta J (\sum S)^2 /2N} \end{align}$$ Thus, the partition function is $$\begin{align} Z = \sum_{\{S\}} e^{-\beta H} = \sum_{\{S\}} \int_{-\infty}^\infty e^{-\frac{N}{2JT}\varphi^2 + \frac{\varphi}{T}\sum_i S_i} \d\varphi \end{align}$$ We can now exactly evaluate the sum over \(\{S\}\). This gives us $$\begin{align} Z = \int_{-\infty}^\infty e^{-\frac{N}{2JT}\varphi^2}\left[2\cosh\frac{\varphi}{T}\right]^N \d\varphi \end{align}$$ We can rewrite this as $$\begin{align} Z = \int_{-\infty}^\infty e^{-NS} \d\varphi \end{align}$$ where $$\begin{align} S[\varphi] = \frac{1}{2JT}\varphi^2 - \log\cosh\frac{\varphi}{T} \end{align}$$
    3. Now, we calculate $$\begin{align} \frac{\partial S}{\partial \phi} = \frac{N\varphi}{JT} - \frac{N}{T} \tanh\frac{\varphi}{T} = 0 \end{align}$$ Thus $$\begin{align} \bar{\varphi} = J\tanh\frac{\bar{\varphi}}{T} \end{align}$$
    4. Taylor expanding, we find $$\begin{align} S[\varphi] = S[\bar{\varphi}] + (\varphi - \bar{\varphi})^2S'[\bar{\varphi}] + \frac{1}{2}(\varphi - \bar{\varphi})^2 \frac{\partial^2 S}{\partial\varphi^2} \end{align}$$ Thus $$\begin{align} Z = \int_{-\infty}^\infty e^{-S[\bar{\varphi}]} e^{-\frac{1}{2}\delta\varphi \partial^2S/\partial\varphi^2} \d(\delta\varphi) = e^{-NS[\bar{\varphi}]} \sqrt{\frac{2\pi}{N\partial^2S/\partial\varphi^2}} \end{align}$$ Thus, $$\begin{align} F = -T\log Z = TNS[\bar{\varphi}] + T\log N + \cdots \end{align}$$ The \(N\) part dominates, and we find $$\begin{align} F = NTS[\bar{\varphi}]. \end{align}$$