\documentclass[12pt]{article}

\textwidth=17cm
\textheight=22.5cm

\topmargin -1.5cm
\oddsidemargin -0.3cm
%\evensidemargin -0.8cm

\parskip 0.3cm

\pagestyle{plain}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Indranil's LaTeX macros
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Equation environments
\def\bea{\begin{eqnarray}}
\def\eea{\end{eqnarray}}
\def\ben{\begin{equation}}
\def\een{\end{equation}}
\def\benu{\begin{enumerate}}
\def\enu{\end{enumerate}}
%
% Special letters 
\def\la{{\langle\, }}
\def\ra{{\,\rangle }}
\def\J{J_{ij}}
\def\j{J_i(0)}
\def\c{\alpha=0}
\def\ptl{\partial}
\def\d{\delta}
\def\a{\alpha}
\def\G{\Gamma}
\def\om{\omega}
\def\tr{{\rm Tr }}
\begin{document}
\begin{center}
Solution to Problem Set 4
\end{center}

We consider a system of Ising spins $(S_i = \pm 1)$ with the partition 
function $Z[J]$ defined as,
\ben
Z[J]= \tr e^{- ( \J S_i S_j + J_i S_i ) } = e^{-W[J]}.
\een
$W[J]$ is the free energy of the system. We will follow the convention that 
when not mentioned explicitly, all repeated indices are to be summed.

The average magnetization is defined as,
\ben
\frac{\ptl W[J]}{\ptl J_i } = \frac{1}{Z}\tr S_i e^{- (\J S_i S_j 
+ J_i S_i ) } = \la S_i \ra = m_i.
\een

The free energy with fixed magnetization $\G [m]$ is defined by the Legendre
tranform,
\ben
\G [m] = W[J] - J_i m_i.
\een
Using equations (2) and (3) we find that,
\ben
\frac{\ptl \G [m]}{\ptl m_i} = \frac{ \ptl W[J]}{\ptl J_k}\frac{\ptl J_k}
{\ptl m_i} - J_i - m_k \frac{\ptl J_k}{\ptl m_i} = - J_i.
\een

The term $\J$ in equation (1) defines interaction between the spins. We want 
to treat this term perturbatively. For this we multiply $\J$ by $\a$, and 
expand relevant quantities in successive orders in $\a$. Here
$\a$ is a parameter which will be set equal to one, like the charge of the 
electron in DFT. The purpose of this exercise is to demonstrate that since 
magnetization is the interesting quantity for the spin system, it is 
meaningful to make a perturbative expansion of the thermodynamic quantity 
which keeps magnetization fixed, namely $\G [m]$. Mathematically, as $\a$ is
tuned from zero to one, from the point of view of $\G [m]$, $J_i(\a)$ has 
the role of a Lagrange multiplier which varies with $\a$ to satisfy the 
constraint $\la S_i \ra = m_i$. This is the same principle 
that motivates DFT. In DFT, density of the electrons is the important quantity
that must be kept fixed. The equivalent of $J_i$ in DFT is the effective 
potential $V(r)$, while the zeroth order approximation to it, which is 
equivalent to $J_i(0)$ is the Kohn-Sham potential.

\benu

\item{{\bf Zeroth Order}}

For $\a =0$ we are dealing with free Ising spins. The partition function 
factorizes and is given by,
\ben
Z_0[J]= \prod_i \left( e^{- J_i} + e^{J_i} \right) = \prod_i \left( 2 \cosh
(J_i) \right).
\een
The free energy $W_0[J]$ is,
\ben
W_0[J] = - \ln Z_0 = - \sum_i \ln \left[ 2 \cosh (J_i) \right].
\een
Since, $m_i =\la S_i \ra$ for all values of $\a$, in particular this is true 
for $\a=0$. Then, using equation (2), the relation between $m_i$ and 
$\j$ is given by,
\ben
m_i = - \tanh (\j).
\een
We can use the above equation to express $\j$ in terms of $m_i$. Using the 
definition of $\G[m]$ we can write,
\ben
\G_0[m] = \sum_i \frac{1+m_i}{2}\ln (\frac{1+m_i}{2}) +\sum_i \frac{1-m_i}
{2}\ln (\frac{1-m_i}{2})
\een
This is nothing but the entropy of free Ising spins with average magnetization
$m_i$ at each site. 

\item{{\bf First Order}}

After introducing the parameter $\a$, the partition function has the form
\[
Z[J,\a] = \tr e^{ - ( \a \J S_i S_j + J_i S_i )}.
\]
We get,
\ben
\frac{\ptl W}{\ptl \a} = - \frac{1}{Z} \frac{\ptl Z}{\ptl \a} = \J \la S_i S_j
\ra,
\een
and
\ben
\frac{\ptl \G}{\ptl \a} = \frac{\ptl W}{\ptl \a} + \frac{\ptl W}{\ptl J_i}
\frac{\ptl J_i}{\ptl \a} - m_i \frac{\ptl J_i}{\ptl \a} = \frac{\ptl W}
{\ptl \a}.
\een
Then, $(\frac{\ptl \G}{\ptl \a})_{\c} = \J \la S_i S_j \ra_0$, where 
$\la \cdots \ra_0$ means average with Boltzman weight for free spins. We 
note that for $i \neq j$, $\la S_i S_j \ra_0 = \la S_i \ra_0 \la S_j \ra_0 = 
m_i m_j$. While for $i=j$ , $\la S_i^2 \ra = 1$ (for Ising spins). Thus, 
ignoring a constant term, we have,
\ben
 \G_1[m] = \left(\frac{\ptl \G}{\ptl \a}\right)_{\c} = \sum_{i \neq j} 
\J m_i m_j.
\een
We identify this term, together with the zeroth order term,  as the free 
energy of the system within Weiss mean field approximation.

\item{{\bf Second Order}}

To calculate the second order term we have to work a little more. But the 
strategy is similar to what we did for the first order. We note that,
\ben
\frac{\ptl^2 W}{\ptl \a^2} = \J J_{kl} \left[ \la S_i S_j \ra \la S_k S_l \ra
- \la S_i S_j S_k S_l \ra \right],
\een
and that
\bea
\frac{\ptl^2 \G}{\ptl \a^2} &=& \frac{\ptl^2 W}{\ptl \a^2} + \frac{\ptl}{\ptl
J_i}\left( \frac{\ptl W}{\ptl \a} \right) \frac{\ptl J_i}{\ptl \a} 
\nonumber \\
&=& \J J_{kl} \left[ \la S_i S_j \ra \la S_k S_l \ra - \la S_i S_j S_k S_l 
\ra \right] \nonumber \\
&+& \J \left[ \la S_i S_j \ra \la S_k \ra - \la S_i S_j S_k \ra 
\right] \frac{\ptl J_k}{\ptl \a}.
\eea
In the above equation we know, in principle, how to calculate all the terms
for $\a =0$, except for the term $\ptl J_k /\ptl \a$. We will calculate this 
term in two ways. The first method, though longer, sheds light on the 
mathematics of the formalism. For this we note that $J_i(\a)$ varies to 
satisfy the constraint $m_i = \la S_i \ra$. Thus $\ptl \la S_i \ra/ \ptl \a 
=0$, and this condition should give how $J_i(\a)$ varies. We get,
\ben
\left( \frac{\ptl \la S_i \ra}{\ptl \a} \right)_{\c} = J_{kl} \left[ \la S_i 
\ra_0 \la S_k S_l \ra_0 - \la S_i S_k S_l \ra_0 \right] + \left[ \la S_i \ra_0
\la S_j \ra_0 - \la S_i S_j \ra_0 \right]\left( \frac{\ptl J_j}{\ptl \a} 
\right)_{\c} = 0.
\een
We note that $\la S_i \ra_0 \la S_j \ra_0 - \la S_i S_j \ra_0 = (m_i^2 -1)
\d_{ij} $. Then, from the above equation we have,
\[
\left( \frac{\ptl J_i}{\ptl \a}\right)_{\c} = - \frac{1}{(m_i^2 -1)}
\sum_{kl} J_{kl} \left[ \la S_i \ra_0 \la S_k S_l \ra_0 - \la S_i S_k S_l 
\ra_0 \right].
\]
In the above equation, the expression on the right hand side has contribution
only for $i=k \neq l$ and for $i=l \neq k$. Thus,
\bea
\left( \frac{\ptl J_i}{\ptl \a}\right)_{\c} &=& - \frac{1}{(m_i^2 -1)}
\sum_{i \neq j} 2 \J (m_i^2 -1)m_j \nonumber \\
&=& -2 \sum_{i \neq j} \J m_j.
\eea
We can also derive the above equation very easily using equations (4) and 
(11). Thus,
\bea
\left( \frac{\ptl J_i}{\ptl \a}\right)_{\c} &=& - \left[ \frac{\ptl}{\ptl \a}
\left( \frac{\ptl \G}{\ptl m_i} \right) \right]_{\c} = - \frac{\ptl}{\ptl m_i}
\left( \frac{\ptl \G}{\ptl \a}\right)_{\c} \nonumber \\
&=& - \frac{\ptl}{\ptl m_i} \sum_{i \neq j} \J m_i m_j \nonumber \\
&=& -2 \sum_{i \neq j} \J m_j. \nonumber
\eea
The above is nothing but an application of Maxwell relation. Note that we 
can write $\frac{\ptl}{\ptl \a} \frac{\ptl}{\ptl m_i} = \frac{\ptl}{\ptl m_i}
\frac{\ptl}{\ptl \a}$ because $m_i$ is independent of $\a$.

Now we can evaluate the right hand side of equation (13). The first term, 
$\la S_i S_j \ra_0 \la S_k S_l \ra_0 - \la S_i S_j S_k S_l \ra_0$ is nonzero
for $i=l \neq j =k$. There are two such terms. Also, it is nonzero for 
$i=k \neq j \neq l$. There are four such terms. Thus,
\ben
\J J_{kl} \left[ \la S_i S_j \ra_0 \la S_k S_l \ra_0 - \la S_i S_j S_k S_l 
\ra_0 \right] = 2\sum_{i \neq j} \J^2 \left( m_i^2 m_j^2 -1 \right)
+ 4 \sum_{i \neq j \neq l} \J J_{jl} \left( m_j^2 -1 \right) m_i m_l.
\een
The second term in equation (13) gives,
\bea
\lefteqn{\J \left[ \la S_i S_j \ra_0 \la S_k \ra_0 - \la S_i S_j S_k \ra_0 
\right] \left(\frac{\ptl J_k}{\ptl \a}\right)_{\c}} \\
&&= -4 \sum_{i \neq j} \J m_i \left(m_j^2 -1 \right) \sum_{k \neq j} J_{jk}
m_k \nonumber \\
&& = -4 \sum_{i \neq j \neq l}\J J_{jl} \left( m_j^2 -1 \right) m_i m_l
-4 \sum_{i \neq j} \J^2 m_i^2 \left(m_j^2 -1 \right) .
\eea
Using equations (13), (16) and (18) we finally get,
\bea
\G_2[m] &=& \left(\frac{\ptl^2 \G}{\ptl \a^2}\right)_{\c} \nonumber \\
&=& 2\sum_{i \neq j} \J^2 \left( m_i^2 m_j^2 -1 \right)
-4 \sum_{i \neq j} \J^2 m_i^2 \left(m_j^2 -1 \right) \nonumber \\
&=& -2 \sum_{i \neq j} \J^2 \left(1- m_i^2 \right) \left( 1- m_j^2 \right).
\eea
This is the famous Onsager correction to the mean field free energy.
\enu



\end{document}
