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\heading{E5721: Ising antiferromagnet}

\auname{Alona Petrushansky}

The problem:

MM 

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

(1) Explain the claim: for zero field (h=0), Ising antiferromagnet is the same as Ising ferromagnet, where M_{s} is the order parameter. 

(2) Given h and \epsilon_{0}, find the coupled mean-field equations for M_{A} and M_{B}. Write the expression for M_{s}(T) for T\sim T_{c}, based on the familiar solution of the ferromagnetic case.

(3) Find the critical temperature {T_{c}} for h=0, and also for small h. Hints: for h=0 use the same procedure of expanding \mbox{arctanh}(x) as in the ferromagnetic case; for small h you may use the most extreme simplification that does not give a trivial solution.

(4) Find the critical magnetic field h_{c} above which the system no longer acts as an antiferromagnet at zero temperature. 

(5) Find an expression for the susceptibility \chi(T), expressed as a function of the staggered magnetization M_{s}(T).

(6) In the region of {T\sim T_{c}} give a linear approximation for 1/\chi as a function of the temperature T

MM 

The solution:



(1) The Hamiltonian is:

\mathcal{H}=-\epsilon\sum s_{i}s_{j}-\underset{i}{h\sum}s_{i}

For zero magnetic field and low temperature the arrangement of spins depends on the interaction term \epsilon s_{i}s_{j} of the spins. By regard the lattice as composed of two sublattices A and B, where each spin A have neighbors spins B we have:

for ferromagnet \epsilon>0, we have minimum energy when \left\langle s_{A}\right\rangle s_{B}>0, as for antiferromagnet, \epsilon<0 and therefore minimum energy will obtain when\left\langle s_{A}\right\rangle s_{B}<0, from this claim we can see that

\mathcal{H}^{AF}(h=0)=\mathcal{H}^{F}(h=0)



(2) The Hamiltonian for sub lattice A is of the form:

\mathcal{H}=-\epsilon c\left<\sigma_{B}\right>\underset{i}{\sum}\sigma_{A_{i}}-\underset{i}{h\sum}\sigma_{A_{i}}=-\underset{i}{\sum}\sigma_{A_{i}}(\underset{\overline{h}}{\underbrace{\epsilon c\left<\sigma_{B}\right>+h}})=-\underset{i}{\sum}\sigma_{A_{i}}\overline{h}

the Hamiltonian of one spin from sub lattice A is of form:

\mathcal{H_{\text{i}}}=-\epsilon c\left<\sigma_{B}\right>\sigma_{A_{i}}-h\sigma_{A_{i}}=-\sigma_{A_{i}}(\underset{\overline{h}}{\underbrace{\epsilon c\left<\sigma_{B}\right>+h}})=-\sigma_{A_{i}}\overline{h}

for summation of index i over spin up(+) and down(-) we have

Z=\underset{i}{\sum}e^{-\beta H_{i}}=2cosh(\beta\overline{h})

M_{A/B}=\left<\sigma_{A/B}\right>=\frac{1}{Z}\underset{i=\pm1}{\sum}\sigma_{A/B}^{i}e^{\beta\overline{h}\sigma_{A/B}^{i}}=\frac{2}{Z}sinh(\beta\overline{h})=tanh(\beta\overline{h})

using c=2^{d}=4as the number of close neighbors depending of the dimension and we

define T_{c}=\epsilon c and get (in next question (3) we prove this definition):

\left\{ \begin{array}{c}
M_{A}=tanh(\beta h-\beta\epsilon cM_{B})\\
M_{B}=tanh(\beta h-\beta\epsilon cM_{A})\end{array}\right.

For T\simeq T_{c}, we assume M_{A}\simeq M_{B}\ll1 and therefore can expand

arctgh(x)\simeq x+\frac{1}{3}x^{3}and with a little algebra we get:

Tarctgh(M_{A})\simeq T(M_{A}+\frac{1}{3}M_{A}^{3})\simeq TM_{A}+\frac{1}{3}T_{c}M_{A}^{3}=h-T_{c}M_{B}

Tarctgh(M_{B})\simeq T(M_{B}+\frac{1}{3}M_{B}^{3})\simeq TM_{B}+\frac{1}{3}T_{c}M_{B}^{3}=h-T_{c}M_{A}

By adding and subtracting the equations we get: 

2h-T_{c}(M_{A}+M_{B})=T(M_{A}+M_{B})+\frac{T_{c}}{3}(M_{A}^{3}+M_{B}^{3})

T_{c}(M_{A}+M_{B})=T(M_{A}-M_{B})+\frac{T_{c}}{3}(M_{A}^{3}-M_{B}^{3})

From M_{s}=\frac{1}{2}(M_{A}-M_{B}) and M=\frac{1}{2}(M_{A}+M_{B}) we get M_{A}=M_{s}+M and 

M_{B}=M-M_{s}, which gives:

M_{A}^{3}=(M_{s}+M)^{3}=M_{s}^{3}+3M_{s}^{2}M+3M_{s}M^{2}+M^{3}

M_{B}^{3}=(M+M_{s})^{3}=M^{3}-3M^{2}M_{s}+3MM_{s}^{2}-M_{s}^{3}

adding and subtracting the equations gives: 

M_{A}^{3}+M_{B}^{3}=2M^{3}+6M_{s}^{2}M

M_{A}^{3}-M_{B}^{3}=2M_{s}^{3}+6M^{2}M_{s}

by substitution of Eq. 14, 15 and M,\: M_{s} to Eq. 10, 11 we finally get 

\left\{ \begin{array}{c}
a).\: h=M(T+T_{c})+\frac{T_{c}}{3}(M^{3}+3M_{s}^{2}M)\\
b).\:0=M_{s}(T-T_{c})+\frac{T_{c}}{3}(M_{s}^{3}+3M^{2}M_{s})\end{array}\right.

By using this equations we can find M_{s} for h=0 and T\simeq T_{c}: 

for T\simeq T_{c}and h=0 \rightarrow M\ll1and therefore we can assume that

M^{2}\simeq0, by substracting this terms to Equ. b. we get

M_{s}=\sqrt{3\frac{T_{c}-T}{T}}



(3) 

when T=T_{c} the staggered magnitization equal to zero, M_{s}=0, as can be seen 

from the last result (Eq. 17) this result confirms our definition T_{c}=\epsilon c.

As for h\neq0:

from Eq. 16.b we have:

M_{s}(\underset{\tilde{T_{C}}}{\underbrace{T-T_{c}(1-M^{2}}}))+\frac{1}{3}T_{c}M_{s}^{3}=0

\tilde{T_{c}}=T_{c}(1-M^{2})=c\epsilon(1-M^{2})

for T\simeq T_{c}from Eq. 16.a we get:

h=2T_{c}M+\frac{1}{3}T_{c}M^{3}+M_{s}^{2}T_{c}M

for T\simeq T_{c} we can assume that M,\: M_{s}\ll1 therefore M_{s}^{2}\rightarrow0 and M^{3}\rightarrow0 

With these assumptions from Eq. 20 we get

M=\frac{h}{2T_{c}}\Rightarrow\tilde{T_{c}}=T_{c}(1-\frac{h^{2}}{4T_{c}^{2}})





(4) In antiferromagnet M_{s}=0,\:(T=0), but if the magnetic field h\neq0 at T=0 

we can steel use previous algebraic definitions like in ferromagnetic case:

\left\{ \begin{array}{c}
(T-T_{c})M_{s}+\frac{1}{3}T_{c}(3M^{2}M_{s}+M^{3})=0\\
(T+T_{c})M+\frac{1}{3}T_{c}(3M_{s}^{2}M+M^{3})=h\end{array}\right.

In antiferromagnetic case we have: M_{s}=0, so h=MT_{c},\: T_{c}=\epsilon_{c}

At T=0 there is a definition of magnetization M=mN, in our case we have two sabblatices 

so in order do not count nearest neighbors twice:

N\longrightarrow\frac{N}{2}

so h=T_{c}m\frac{N}{2},\:but in case of antiferromagnet that have \uparrow\downarrow constraction

m=-1, so h=-\frac{N}{2}T_{c}=-\frac{N}{2}\epsilon_{c}=h_{c} and this is an identification of ground state:

E(\uparrow\downarrow\uparrow\downarrow)=-\frac{N}{2}\epsilon_{c}, so for ddestroying the antiferromagnetic order needed h<h_{c}.



(5) we can define M_{A}=M_{A0}+\delta M_{A} and M_{B}=M_{B0}+\delta M_{B}

M=M_{A}+M_{B}=M_{A0}+\delta M_{A}+M_{B0}+\delta M_{B}

by using the approximation:

tanh(x)=tanh(x_{0}+\delta x)\simeq tanh(x_{0})+\frac{\delta x}{cosh^{2}(x_{0})}

and for small magnetic field we can define \delta x=h\beta-\epsilon c\beta\delta M_{B} we get:

M_{A}=tanh(h\beta-\epsilon c\beta(M_{B0}+\delta M_{B}))\simeq tanh(-\epsilon c\beta M_{B0})+\frac{h\beta-\epsilon c\beta\delta M_{B}}{cosh^{2}(-\epsilon c\beta M_{B0})}

and for \delta x=h\beta-\epsilon c\beta\delta M_{A} we get:

M_{B}=tanh(h\beta-\epsilon c\beta(M_{A0}+\delta M_{A}))\simeq tanh(-\epsilon c\beta M_{A0})+\frac{h\beta-\epsilon c\beta\delta M_{A}}{cosh^{2}(-\epsilon c\beta M_{A0})}

At temperature that much lower from the critical temperature we can assume that

M_{A0}=-M_{B0}, and by attaching M_{A}and M_{B}we get an equation for the magnetism M:

M=M_{A}+M_{B}=\frac{2h\beta-\epsilon c\beta(\delta M_{A}+\delta M_{B})}{cosh^{2}(\epsilon c\beta M_{A0})}

2M_{s}=M_{A}-M_{B}=M_{A0}+\delta M_{A}-M_{B0}-\delta M_{B}=2M_{A0}+(\delta M_{A}-\delta M_{B})\simeq2M_{A0}

for \delta M_{A}-\delta M_{B}\rightarrow0, and we get:

M\simeq\frac{2h\beta-\epsilon c\beta(\delta M_{A}+\delta M_{B})}{cosh^{2}(\epsilon c\beta M_{s})}

at low T we have very small M. therefore we can assume that \delta M_{A}+\delta M_{B}=\delta M\simeq M

by substitution and isolating M we have:

M=\frac{h}{\frac{1}{2}T_{c}+T\cosh^{2}\left(\frac{TM_{s}}{T_{c}}\right)}, where T_{c}=\epsilon c,\,\,\,\, T=\frac{1}{\beta} 

\chi=\left(\frac{\partial M}{\partial h}\right)_{h\rightarrow0}=\frac{1}{T\cosh^{2}\left(\frac{TM_{s}}{T_{c}}\right)+\frac{1}{2}T_{c}}



(6) 

let define linear approximation of suseptability in case of T\approx T_{c} at two regions 

T\geq T_{c} and T\leq T_{c}:

for h=0: \chi=\frac{1}{T_{c}+T\cosh^{2}\left(\frac{T_{c}M_{s}}{T}\right)}

a) for T\geq T_{c} 

Here {M_{s}=0} because the order parameter of antiferromagnet defined by a magnetization 

of lattices A and B, that in region of T>T_{c} have zero values, so 

M_{s}=\frac{1}{2}(M_{A}-M_{B})=0,where M_{A}=M_{B}=0, than

\cosh^{2}\left(\frac{T_{c}M_{s}}{T}\right)=1

\chi=\frac{1}{T_{c}+T}

\frac{1}{\chi}=T+T_{c}

b) for {T\leq T_{c}}, 

In this region order parameter of antiferromagnet defined by {M_{s}=\sqrt{3\frac{T_{c}-T}{T_{c}}}} 

we can use the approximation \cosh^{2}x\simeq\left(1+\frac{1}{2}x^{2}\right)^{2}\simeq1+x^{2} 

\cosh^{2}\left(\frac{T_{c}M_{s}}{T}\right)\simeq1+\left(\frac{T_{c}}{T}\right)^{2}\left(3\frac{T_{c}-T}{T_{c}}\right)\simeq1+\left(3\frac{T_{c}-T}{T_{c}}\right)

the last approximation we can do because T_{c}\sim T and therefore\left(\frac{T_{c}}{T}\right)^{2}\rightarrow1

\chi=\frac{1}{T_{c}+T\left(1+\frac{T_{c}^{2}M_{s}^{2}}{T^{2}}\right)}\sim\frac{1}{T_{c}+T\left(1+\left(3\frac{T_{c}-T}{T_{c}}\right)\right)}=\frac{1}{4T_{c}-2T}

\frac{1}{\chi}=4T_{c}-T


