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\heading{E5721: Antiferromagnet}
\auname{Dror Moshe Aharoni}


{\bf The problem: Antiferromagnet}

\Dn


Consider Ising model on a 2D lattice with antiferromagnetic interaction
${(\epsilon = -\epsilon_0)}$. You can regard the lattice as composed of
two sublattices A and B, such that ${M=\frac{1}{2}(M_A+M_B)}$ is the 
averaged magnetization per spin and ${M_s=\frac{1}{2}(M_A-M_B)}$ is the
staggered magnetization
  
\Dn
\begin{itemize}
\item[(a)]
Explain the claim: for zero field $(h=0)$, Ising antiferromagnet is the 
\textbf{same} as Ising ferromagnet, where $M_s$ is the order parameter.\\
(write the expression for $M_s(T)$ at $T\approx T_c$, based on the familiar
solution of the ferromagnet case)

\item[(b)]
Given h and $\epsilon_0$, find the coupled mean field equations for $M_A$ and $M_B$.

\item[(c)]
Find the critical temperature ${T_c}$ for $h=0$ and also for small $h$.\\
(Hint for $h=0$: use the same procedure of expanding $arctanh(x)$ as in the ferromagnet case)
(Hint for small $h$: you may use the most extreme simplification that doesn't give the trivial solution)

\item[(d)]
Find the critical magnetic field $h_c$ above which the system no longer acts as an antiferromagnet at zero temperature (T=0).

\item[(e)]
Find an expression for the susceptibility as a function of the staggered magnetization ${\chi (M_s)}$.

\item[(f)]
In the region of ${T \approx T_c}$ give a linear approximation of  
${\frac{1}{\chi}}$ as a function of the temperature $T$
\end{itemize}


\Dn\Dn
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{\bf The solution:}

\Dn

(a) The Hamiltonian as h=0:
\[H_{AF}=\epsilon_0\cdot\sum_{<i,j>}\sigma_i\sigma_j\]
\[=-\epsilon\cdot\sum_{<i,j>}\sigma_i\sigma_j=H_{F}(h=0)\]


\[M_s(T)=\left(3\frac{T_c-T}{T}\right)^\frac{1}{2}\]

\Dn
(b)  The Hamiltonian for lattice A is of the form:

\[
H_A=\epsilon_0\cdot\sum_{<i,j>}\sigma_i\sigma_j - h\sum_{i}\sigma_i
\]

Where  ${<i,j>}$  symbols close neighbors between lattice A to B and ${h=\mu_B\cdot B}$ \\
\Dn
Using ${\gamma}=2^d=4$ as the number of close neighbors depending of the dimension and \\
we define magnetism per spin (of lattice A/B):
\[{<\sigma^{A/B}> = \frac{\tilde{M}_{A/B}}{\mu_B N_{A/B}} = M_{A/B}=\frac{1}{N_{A/B}}\cdot\sum_{i}\sigma^{A/B}_{i}}\]

\Dn

Let us develop ${H}$ using the mean field theory:

\begin{eqnarray*}
H_A=\epsilon_0\gamma\sum_{i}\sigma^{A}_{i}\cdot<\sigma^{B}> - h\sum_{i}\sigma_i\\
   = \epsilon_0\gamma\cdot N_A\cdot M_A\cdot M_B - h\cdot N_A\cdot M_A\\
   = (\epsilon_0\gamma\cdot M_B - h)\cdot N_A\cdot M_A
\end{eqnarray*}
%\vspace{100 mm}
\Dn

The Hamiltonian of one spin from sub lattice A:

\[H^A_i=(\epsilon_0\gamma\cdot M_B - h)\cdot\sigma_i\]
\Dn
\[Z^A_1=\sum_{\sigma^A_i}e^{-\beta\cdot H^A_i}=2cosh(\frac{1}{T}\cdot(\epsilon_0\gamma\cdot M_B - h))\]
\Dn
\[ (I) M_A=-T\cdot\frac{\partial ln(Z^A_i)}{\partial h}=tanh(\frac{1}{T}(h-\epsilon_0\gamma\cdot M_B))\]
\[(II) M_B= tanh(\frac{1}{T}(h-\epsilon_0\gamma\cdot M_A))\]
\Dn

(c) At zero magnetic field ${(h=0)}$ :

\[
M_{A_0}=-M_{B_0}\ \ \  \rightarrow\ \ \  M_{A_0}=tanh\left(\frac{\epsilon_0\gamma M_{A_0}}{T}\right)\]

\Dn

This equation should be solved for $M_{A_0}$.\\

\begin{center}
		\includegraphics{ex5721graph1.png}
\end{center}


By inspection of the plot we observe that for h = 0 the condition for getting a non trivial solution is $\gamma\cdot\epsilon_0 /T > 1$.
Therefore
\[\ (c_1)\ \mathbf{T_c = \gamma\cdot\epsilon_0}\]
\Dn

Lets use the solution of (b) and search for an expression for M and $M_s$:
\Dn
\[\left\{\begin{matrix} (I)\ \frac{1}{T}(h-T_c\cdot M_B)=arctanh(M_A)\approx 
                             M_A+\frac{1}{3}M^3_A\\
                        (II)\ \frac{1}{T}(h-T_c\cdot M_A)=arctanh(M_B)\approx 
                              M_B+\frac{1}{3}M^3_B
					\end{matrix}\right. \]
\[\left\{\begin{matrix} (I)\ h=T_c\cdot M_B+T\cdot M_A+\frac{1}{3}T\cdot M^3_A\\
                        (II)\ h=T_c\cdot M_A+T\cdot M_B+\frac{1}{3}T\cdot M^3_B
					\end{matrix}\right. \]
\[\left\{\begin{matrix} (I)+(II):\ 2h=(T+T_c)\cdot(M_A+M_B)+\frac{1}{3}T(M^3_A+M^3_B)\\
																		\approx (T+T_c)\cdot(M_A+M_B)+\frac{1}{3}T_c(M^3_A+M^3_B)\\
                        (I)-(II):\ 0=T_c\cdot(M_B-M_A)+T\cdot(M_A-M_B)+\frac{T}{3}(M^3_A-M^3_B)
					\end{matrix}\right. \]
After some calculations [and using ${M_A\cdot M_B=m^2-M^2_s}$] we get:
\[\left\{\begin{matrix} (I)+(II):\ h=(T+T_c+T_c\cdot M^2_s)\cdot M+\frac{1}{3}T_c\cdot M^3\\
                        (I)-(II):\ 0=(T-T_c+T\cdot M^3)\cdot M_s+\frac{T}{3}M^3_s
					\end{matrix}\right. \]
Which gives the following solutions:
\[(1)\ M_s=0\ \ \ or\ \ \ 3\cdot \left(\frac{T_c-T}{T}\right)=3M^2+M^2_s\]
\[(2)\ T\approx T_c\ \mapsto\ \frac{h}{T_c}=(2+M^2_s)\cdot M+\frac{1}{3}M^3\]

As expected from the second equation we get M = 0 in the absence of an external field, and from the first equation we get the order parameter $M_s(T)= \left(3\cdot\frac{T_c-T}{T}\right)^\frac{1}{2} $, which satisfies the same equation as in the ferromagnetic problem.\\
\Dn

We will show that if we switch on the magnetic field, $T_c$ is shifted to a lower temperature.\\
Assuming small magnetic field we can neglect $M^3$ and $M_s$ at equation (2) and get:
\[M=\frac{h}{2T^{(0)}_c}\]
Putting this result at equation (1) we get:
\[M^2_s=3\left(\frac{T^{(0)}_c-\frac{h^2}{4T^{(0)}_c}-T}{T^{(0)}_c}\right)\equiv 3\left(\frac{T_c-T}{T}\right)\]
Where we got $T_c=T^{(0)}_c-\frac{h^2}{4T^{(0)}_c}$


\Dn

\begin{center}
		\includegraphics{ex5721graph2.png}
\end{center}

\Dn

(d) We can now express the critical magnetic field from the critical temperature:
\[h^2_c=4\epsilon^2\gamma^2-4\epsilon\gamma T_c(h)\]
\[\Rightarrow\ \ h_c=2\sqrt{\epsilon^2\gamma^2-\epsilon\gamma T_c(h)}\]


(e) Let us define:
\[ M_A=M_{A_0}+\delta M_A\ ;\ M_B=M_{B_0}+\delta M_B=-M_{A_0}+\delta M_B\]
\[M=M_A+M_B=\delta M_A+\delta M_B\]

\Dn
\Dn
By using the approximation:

\[tanh(x)=tanh(x_0+\delta x)\approx tanh(x_0)+\frac{\delta x}{cosh^2(x_0)}\]
\Dn
We get:

\[ M_A=tanh\left(-\frac{T_c}{T}M_B+\frac{1}{T}\cdot h\right)=\left(-\frac{T_c}{T}(M_{B_0}+\delta M_B)+\frac{1}{T}\cdot h\right)\]
\[\approx \underbrace{tanh\left(\frac{T_c}{T}M_{B_0}\right)}_{M_{A_0}}+\underbrace{\frac{-T_c\cdot\delta M_B+h}{T\cdot cosh^2(-\frac{T_c}{T}M_{B_0})}}_{\delta M_A}\]

And in the same way we can get:

\[\delta M_B = \frac{-T_c\cdot \delta M_A+h}{T\cdot cosh^2(-\frac{T_c}{T}M_{A_0})}=\frac{-T_c\cdot \delta M_A+h}{T\cdot cosh^2(\frac{T_c}{T}M_{B_0})}\]

And by attaching the two equations we get an equation for the magnetism M:

\[M=\frac{1}{2}(\delta M_A+\delta M_B)= \frac{-\frac{T_c}{2}\cdot \delta M+h}{T\cdot cosh^2(\frac{T_c}{T}M_{B_0})}\]

After isolating ${M}$ we have:

\[M=\frac{h}{T_c+T\cdot cosh^2(\frac{T_c}{T}M_{B_0})}\]
\[M_s(h=0)=M_s(T)=\frac{1}{2}(M_{A_0}-M_{B_0})=-M_{B_0}\]
\[\rightarrow\ \  M=\frac{h}{T_c+T\cdot cosh^2(\frac{T_c}{T}M_s(T))}\]
\Dn

Let us find the susceptibility:

\[X=\left(\frac{\partial M}{\partial h}\right)_{h\rightarrow 0}\]
\Dn
\[\Rightarrow \ \ X=\frac{1}{T_c+T\cdot cosh^2(\frac{T_c}{T}M_s(T))}\]

\Dn

\Dn

(f) Let us examine the susceptibility near ${T_c}$ at two cases:

\[\underline{T\ >\ T_c\ :}\]

at that case the system is not an antiferromagnet yet so ${M_s(T)=0\ \rightarrow \ cosh(\frac{T_c}{T}M_s(T))=1}$\\
which gives us:

\[X_{T>T_c}=\frac{1}{T+T_c}\]
\[\Rightarrow \frac{1}{X_{T>T_c}}=T+T_c\]
\Dn

\[\underline{T\ <\ T_c\ :}\]
\Dn

By using the approximation:

\[cosh^2(y)\approx (1+\frac{1}{2}y^2)^2\approx 1+y^2\]
\Dn
\[X_{T<T_c}=\frac{1}{T+T\cdot\left(\frac{T_c}{T}M_s(T)\right)^2+T_c}\]

A reminder:

\[M^2_s(T)=\left(3\frac{T_c-T}{T}\right)\]

\Dn
And after substitution of $M_s(T)$ and taking into account that where $T\approx T_c\ \rightarrow\ \frac{T_c}{T}\approx 1$\\
we get:

\[X_{T<T_c}=\frac{1}{T+3\cdot T\left(\frac{T_c-T}{T}\right)+T_c}\]
\[X_{T<T_c}=\frac{1}{4T_c-2T}\]
\[\frac{1}{X_{T<T_c}}=4T_c-2T\]
\Dn
for conclusion we got:

\[\frac{1}{X}=\left\{\begin{matrix} T+T_c\ ;\ T>T_c\\
																		4T_c-2T\ ;\ T<T_c
																			 \end{matrix}\right.
																			 \]


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