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\begin{document}

\heading{E5642: Long-range Ising model} 
\auname{Submitted by Evgeniy Kogan}

{\bf The problem:}\\

Consider a cluster of ${N}$ spins ${s_{i}=\pm 1}$.
The interaction
between {\em any} two spins is ${-\epsilon s_{i} s_{j}}$, with
${\epsilon>0}$.
The interaction of each spin with the external
magnetic field ${H}$ is ${-H s_{i}}$. The total magnetization is
defined as ${\mathsf{m}=\sum s_{i}}$. The inverse temperature is
${\beta}$.

\begin {itemize}

\item[(a)]

Show that the partition function can be written as
${ Z \left ( \beta, H \right) = \sum_{ \mathsf{m}}g \left ( \mathsf{m}\right) \exp \left (\frac{1}{2} B m^{2} + h \mathsf{m}\right) }$.
Express
${g\left(\mathsf{m}\right)}$ and ${B}$ and ${h}$ using
${\left(N,\epsilon, H, \beta\right)}$.

\item[(b)]
Assume that ${B=\frac{b}{N}}$, define the magnetization as
${M=\frac{\mathsf{m}}{N}}$, and write the partition function as
${Z\left(b,h\right)=\sum_{M} \exp\left(-N*A\left(M\right)\right)}$.

Write the expressions for ${A\left(M\right)}$ and for its
derivatives ${A'\left(M\right)}$ and ${A''\left(M\right)}$.

\item[(c)]
Determine the critical temperature ${T_{c}}$, and write an equation
for the mean field value of ${M}$. Make a qualitative plot of
${A\left(M\right)}$ below and above the critical temperature.

\item[(d)]
Write an approximation for ${A\left(M\right)}$ up to order ${M^{4}}$.
On the basis of this expression determine the temperature range
where mean filed theory cannot be trusted. Hint: you have to
estimate the variance ${\langle M^{2} \rangle}$ in the Gaussian
approximation.
What happens with this condition in the thermodynamic
limit (${N\rightarrow\infty}$)?

\item[(e)]
What is the susceptibility for ${T>T_c}$.

\item[(f)]
What is the heat capacity.

\end {itemize}


\Dn


{\bf The solution:}\\

\begin {itemize}

\item[(a)]

Hamiltonian of a system:
\beq
H = -\epsilon \sum_{<i,j>} s_i s_j - H \sum_{i=1}^{N} s_i
\eeq
where ${<i,j>}$ means summation over all i and j from 1 to N except duplcations (for example i=1,j=2 and i=2,j=1 it's duplication) and cases when i=j.\\
First term in Hamiltonian describes interactions of spin $s_i$ with all other spins $s_j$
In view of mean field approximation we assume that each spin $s_j$ has average value $<s>$,
thus Hamiltonian takes following form:
\beq
H = -\epsilon \frac{N}{2} \sum_{i=1}^{N} s_i \avg{s} - H \sum_{i=1}^{N} s_i
\eeq
Where factor $\frac{1}{2}$ appears to exclude duplications and N comming from summation over j.
Average spin value given by:
\beq
\avg{s} = \frac{1}{N}\sum_{i=1}^{N} s_i = \frac{m}{N}
\eeq
Thus we can rewrite Hamiltonian in following form:
\beq
H = -\frac{1}{2} \epsilon m^2 - H m
\label{eq:HAM}
\eeq

Let us find function g(m) which gives number of possible spin's configuration for given m. \\
We use follwoing notations: \\
$N_{\uparrow}$ - total number of "up" spins\\
$N_{\downarrow}$ - total number of "down" spins\\
$m = N_{\uparrow} - N_{\downarrow}$\\
$N_{\uparrow} = \frac{N + m}{2}$\\
$N_{\downarrow} = \frac{N - m}{2}$\\
\beq
g(m)=\frac{N!}{N_{\uparrow}!N_{\downarrow}!}=\frac{N!}{\left[\frac{1}{2}(N+m)\right]!\left[\frac{1}{2}(N-m)\right]!}
\label{eq:GM}
\eeq

Now we can write partition function:
\beq
Z = \sum_{m=-N}^{N} g(m) \exp{(-\beta H(m))} = \sum_{m=-N}^{N} g(m) \exp{(\frac{1}{2}Bm^2 + hm)}
\label{eq:PART}
\eeq
Where ${B \equiv \beta \epsilon}$ \ and \ ${h \equiv \beta H}$ 

\item[(b)]

We assume ${B=\frac{b}{N}}$ and define the magnetization as
${M=\frac{\mathsf{m}}{N}}$ we can rewrite \\
partition function (\ref{eq:PART}):
\beq
Z = \sum_{M=-1}^{1} \exp{\left[ -N \left(-\frac{1}{2}bM^2 - hM - \frac{ln(g(M))}{N} \right) \right]}
  = \sum_{M=-1}^{1} \exp{(-N*A(M))}
\label{eq:PART1}
\eeq
Where ${A(M) \equiv -\frac{1}{2}bM^2 - hM - \frac{ln(g(M))}{N}}$.\\

First we would like to simplify expression (\ref{eq:GM}) for ln(g(m)) and after Stirling's approximation ${ln (x!) \approx x ln(x)-x}$ and little bit algebra we have:
\beq
ln(g(m)) \approx N\left(ln(2)-\frac{M}{2}ln\left(\frac{1+M}{1-M}\right)-\frac{1}{2}ln(1-M^2)\right)
\eeq

Thus A(M) gets following form:
\beq
A(M) = -\frac{1}{2}bM^2 - hM - ln(2)+\frac{M}{2}ln\left(\frac{1+M}{1-M}\right)+\frac{1}{2}ln(1-M^2)
\label{eq:AM}
\eeq


First and second derivatives of $A(M)$ with respect to M:
\beq
A'(M) = -bM - h + \frac{1}{2}ln\left(\frac{1+M}{1-M}\right) = -bM - h + arctanh(M)
\label{eq:FDRV}
\eeq
\beq
A''(M) = -b + \frac{1}{1-M^2}
\eeq
Here relation ${arctanh(x) = \frac{1}{2}ln\left(\frac{1+x}{1-x}\right)}$ has been used.

\item[(c)]

In order to find minima of $A(M)$ we equal first derivative (\ref{eq:FDRV}) to zero:
\beq
A'(M) = -bM - h + arctanh(M) = 0
\eeq
From here we deduce:
\beq
M = tanh(bM+h)
\label{eq:TAN}
\eeq

We solve graphically this equation with zero external magnetic field  ${(h \to 0)}$:

\begin{figure}[h]
	\begin{center}
		\begin{center}
		\includegraphics[width=0.4\hsize]{LessThanCrTemp}
		\includegraphics[width=0.4\hsize]{GraterThanCrTemp}
		\end{center}
	\end{center}
	\caption{Graphs of $f_1(M)=M$ and $f_2(M)=tanh(bM)$ for different values of b.}
	\label{fig:F1}
\end{figure}

Parameter $b$ influences on curvature of $tanh(bM)$ (red line on graph) and 
starting from some critical value of $b$ the equation has two additional solutions
(intersection points on a graph).
It happens when slope of blue line is equal to slope of red line near point $M=0$.
Expanding $tanh(bM)$ by Taylor expansion about point $M=0$ we get:
\beq
tanh(bM) \approx  bM + ...
\eeq

By comparing between slope $b$ of $tanh$ and slope $1$ of line we get critical temperature:
\beq
T_c = \frac{\epsilon N}{k_B}
\eeq

Using (\ref{eq:PART}) and (\ref{eq:PART1}) we can rewrite $A(M)$ in following terms:
\beq
A(M) = \beta (-\frac{1}{2}\epsilon \frac{m^2}{N} - H \frac{m}{N} - \frac{1}{\beta} \frac{ln(g(m))}{N})= \frac{\beta}{N} (E(m)-TS(m)) = \frac{\beta}{N} F(m) 
\eeq
Where $E(m)$ is energy of given magnitization defined by (\ref{eq:HAM}), ${S(m) \equiv ln(g(m))}$ is entropy for given magnitization and F(m) is nothing else Helmholtz free energy for given magnitization.

To understand $A(M)$ function behavior under and above ${T_c}$ with and without magnetic field
we plot following qualitative graphs:

\begin{figure}[h]
	\begin{center}
		\subfigure[] {\label{fig:fig-a} \includegraphics[width=0.4\hsize]{MeanField(T_grater_T_c)}}
		\subfigure[] {\label{fig:fig-b} \includegraphics[width=0.4\hsize]{MeanField(T_eq_T_c)}}
		\subfigure[] {\label{fig:fig-c} \includegraphics[width=0.4\hsize]{MeanField(T_less_T_c)}}
		\subfigure[] {\label{fig:fig-d} \includegraphics[width=0.4\hsize]{MeanField(T_less_T_c_H_not_zero)}}		
	\end{center}
	\caption{$A(M)$ for different values of temerature and external magnetic field.}
\end{figure}

\newpage

We note that above ${T_c}$ in average magnitization of a system is zero.
Under ${T_c}$ and without external magnetic field system can be
in one of two possible magnitizations are different from zero.
Magnitudes both of them are equal, because without external magnetic field
system is symmetric and there is no preference to be in one of possible magnitizations.
If we add external magnetic field spins get preference direction (direction of magnetic field)
symmetry is broken and we see in graph that one minima is above the other.

\item[(d)]
We can expand $A(M)$ defined by (\ref{eq:AM}) about $M=0$ up to order $M^4$:
\beq
A(M) = -ln(2)-hM+\frac{1}{2}(1-b)M^2+\frac{1}{12}M^4
\eeq

We can rewrite $1-b$ by using previous definitions ${b=BN= \beta \epsilon N}$,
${T_c = \frac{\epsilon N}{k_B}}$ and ${\beta = 1/(k_BT)}$ and we get:
\beq
1-b = \frac{T-T_c}{T} \equiv t
\eeq

and thus A(M) becomes:
\beq
A(M) = const-hM + \frac{1}{2}t M^2+\frac{1}{12}M^4
\label{eq:AM1}
\eeq

\newpage

\underline{Case of $T>T_c$ ($t>0$)}\\

We can neglect term of $M^4$ in (\ref{eq:AM1}) in case it's much smaller than contribution from $M^2$ term:
\beq
tM^2 \gg M^4 \quad \Rightarrow \quad t \gg M^2
\label{eq:C1}
\eeq

In the absence of external magnetic field from (\ref{eq:PART1}) and (\ref{eq:AM1}) we get that partition function is:
\beq
Z \sim \sum \exp{(-\frac{1}{2}NtM^2)} = \sum \exp{(-\frac{M^2}{2\sigma^2})}
\label{eq:GAP}
\eeq
"Width" or variance of Gaussian is $\sigma^2$, where ${ \sigma \equiv \frac{1}{\sqrt{Nt}} }$.

In mean field approximation we neglect fluctuations, therefore ${ \avg{M^2}=\avg{M}^2=M^2}$.\\
Gaussian approximation (\ref{eq:GAP}) is valid if disperssion of $M^2$ is of order $\sigma^2$ or smaller.
\beq
M^2 \sim \sigma^2 = \frac{1}{Nt}
\label{eq:MM}
\eeq

By substitution (\ref{eq:MM}) into (\ref{eq:C1}) we get following condition:
\beq
t \gg \frac{1}{Nt} \quad \Rightarrow \quad t \gg \frac{1}{\sqrt{N}}
\label{eq:C2}
\eeq

\underline{Case of $T<T_c$ ($t<0$)}\\

We define ${t'\equiv-t}$.\\
When $T<T_c$ function $A(M)$ has two minima (see figure \ref{fig:fig-c}), which in the absence of external magnetic field are:
\beq
 M_{1,2} = \pm \sqrt{3 \left( \frac{T_c-T}{T} \right)} = \pm \sqrt{3t'}
\eeq

The disctance between two minima is:
\beq
d = M_2-M_1 = 2\sqrt{3t'} \sim \sqrt{t'}
\eeq

In this case Gaussian approximation is valid if dispersion of magnitization is much smaller than distance between
minima. Comparing (\ref{eq:MM}) with last result we have:
\beq
d \sim \sqrt{t'} \gg M \sim \frac{1}{\sqrt{Nt'}} \quad \Rightarrow \quad t' \gg \frac{1}{\sqrt{N}}
\label{eq:C3}
\eeq

We can combine two conditions (\ref{eq:C2}) and (\ref{eq:C3}) and we have:
\beq
|t| \gg \frac{1}{\sqrt{N}}
\eeq

In thermodynamic limit ${N \rightarrow \infty }$ this condition is always satisfied excluding
cases when temperature is in the vicinity of critical temperature.

\newpage

\item[(e)]

Expression (\ref{eq:TAN}) we can be written as ${ arctanh(M) = bM+h }$. We derivate both side with respect to external magnetic field H
and write magnetic susceptibility of the system:
\beq
\chi = \left( \frac{\partial M}{\partial H} \right)_T= N \beta \left( \frac{\partial M}{\partial h} \right)_T = \frac{N}{k_B} \frac{1-M^2}{T-T_c(1-M^2)}
\eeq

At high temperature ${(T > T_c)}$ $M \ll 1$ and we obtain Curie-Weiss law:
\beq
\chi \approx  \frac{N}{k_B} \frac{1}{T-T_c}
\eeq


\item[(f)]

To calculate heat capacity we use definition ${ C_V= \left(\frac{\partial E}{\partial T} \right)_{N,V} }$.
From (\ref{eq:HAM}) we can write in the absence of magnetic field:
\beq
E(M) = -\frac{1}{2}\epsilon M^2 N^2
\eeq

Thus heat capacity is:
\beq
C_V = -\epsilon N^2 M \left(\frac{\partial M}{\partial T} \right)
\label{eq:CV}
\eeq

Expression (\ref{eq:TAN}) we can be written as ${ arctanh(M) = bM }$. We derivate both side with respect to T
(taking into account dependence of b on T) and after a little bit algebra we have:
\beq
\left(\frac{\partial M}{\partial T} \right) = \frac{M/T_c}{T/T_c-(T/T_c)^2\frac{1}{1-M^2}}
\eeq

Substituting this result into (\ref{eq:CV}) we get expression for heat capacity:
\beq
C = \frac{k_B N M^2}{(T/T_c)^2\frac{1}{1-M^2}-T/T_c}
\label{eq:CV1}
\eeq

Heat capacity for $T>T_c$ is always zero (mean value of magnitization is zero).\\
In $T=T_c$ straight line is tangent to $tanh(bM)$ see figure \ref{fig:F1} and
only one possible solution of $M=tanh(bM)$ is $M(T=T_c)=0$, therefore near $T=T_c$
we can expand $tanh$ by Teylor series and using fact that ${T \simeq T_c}$ we get:
\beq
M(T) = tanh(bM) \approx bM-\frac{1}{3}(bM)^3 \quad \Rightarrow \quad M(T) \approx \sqrt{3(1-T/T_c)}
\eeq

We substitute our result into (\ref{eq:CV1}) and take a limit:
\beq
\lim_{T \rightarrow T_c^-} \frac{3 k_B N (1-T/T_c)}{(T/T_c)^2\frac{1}{1-3(1-T/T_c)}-T/T_c} = \frac{3}{2} k_B N
\eeq

When $T \rightarrow 0$ we can expand (\ref{eq:TAN}) about $T=0$:
\beq
M(T) \approx 1-2\exp{(-2T/T_c)} 
\eeq

We again substitute our result into (\ref{eq:CV1}) and get heat capacity for low temperature: 
\beq
C(T) \approx 4Nk_B \left(\frac{T_c}{T}\right)^2 \exp{(-2T_c/T)} 
\eeq


\newpage



\end {itemize}

\end{document}
