E5721: Antiferromagnet

E5721: Antiferromagnet
Submitted by: Dror Moshe Aharoni
The problem: Antiferromagnet
Consider Ising model on a 2D lattice with antiferromagnetic interaction ( = −0 ). You can regard
the lattice as composed of two sublattices A and B, such that M = 21 (MA + MB ) is the averaged
magnetization per spin and Ms = 12 (MA − MB ) is the staggered magnetization
(a) Explain the claim: for zero field (h = 0), Ising antiferromagnet is the same as Ising ferromagnet, where Ms is the order parameter.
(write the expression for Ms (T ) at T ≈ Tc , based on the familiar solution of the ferromagnet
case)
(b) Given h and 0 , find the coupled mean field equations for MA and MB .
(c) Find the critical temperature Tc 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)
(d) Find the critical magnetic field hc above which the system no longer acts as an antiferromagnet
at zero temperature (T=0).
(e) Find an expression for the susceptibility as a function of the staggered magnetization χ(Ms ).
(f) In the region of T ≈ Tc give a linear approximation of
1
χ
as a function of the temperature T
The solution:
(a) The Hamiltonian as h=0:
X
HAF = 0 ·
σi σj
<i,j>
= − ·
X
σi σj = HF (h = 0)
<i,j>
Ms (T ) =
Tc − T
3
T
1
2
(b) The Hamiltonian for lattice A is of the form:
HA = 0 ·
X
<i,j>
σi σj − h
X
σi
i
Where < i, j > symbols close neighbors between lattice A to B and h = µB · B
Using γ = 2d = 4 as the number of close neighbors depending of the dimension and
we define magnetism per spin (of lattice A/B):
< σ A/B >=
X A/B
M̃A/B
1
= MA/B =
·
σi
µB NA/B
NA/B
i
1
Let us develop H using the mean field theory:
HA = 0 γ
X
σiA · < σ B > −h
i
X
σi
i
= 0 γ · NA · MA · MB − h · NA · MA
= (0 γ · MB − h) · NA · MA
The Hamiltonian of one spin from sub lattice A:
HiA = (0 γ · MB − h) · σi
Z1A =
X
A
e−β·Hi = 2cosh(
σiA
1
· (0 γ · MB − h))
T
∂ln(ZiA )
1
= tanh( (h − 0 γ · MB ))
∂h
T
1
(II)MB = tanh( (h − 0 γ · MA ))
T
(I)MA = −T ·
(c) At zero magnetic field (h = 0) :
MA0 = −MB0
→
MA0 = tanh
0 γMA0
T
This equation should be solved for MA0 .
2
By inspection of the plot we observe that for h = 0 the condition for getting a non trivial solution
is γ · 0 /T > 1. Therefore
(c1 ) Tc = γ · 0
Lets use the solution of (b) and search for an expression for M and Ms :
(I) T1 (h − Tc · MB ) = arctanh(MA ) ≈ MA + 13 MA3
(II) T1 (h − Tc · MA ) = arctanh(MB ) ≈ MB + 13 MB3
(I) h = Tc · MB + T · MA + 31 T · MA3
(II) h = Tc · MA + T · MB + 31 T · MB3

(I) + (II) : 2h = (T + Tc ) · (MA + MB ) + 31 T (MA3 + MB3 )

≈ (T + Tc ) · (MA + MB ) + 13 Tc (MA3 + MB3 )

(I) − (II) : 0 = Tc · (MB − MA ) + T · (MA − MB ) + T3 (MA3 − MB3 )
After some calculations [and using MA · MB = m2 − Ms2 ] we get:
(I) + (II) : h = (T + Tc + Tc · Ms2 ) · M + 13 Tc · M 3
(I) − (II) : 0 = (T − Tc + T · M 3 ) · Ms + T3 Ms3
Which gives the following solutions:
Tc − T
(1) Ms = 0 or 3 ·
= 3M 2 + Ms2
T
(2) T ≈ Tc 7→
h
1
= (2 + Ms2 ) · M + M 3
Tc
3
As expected from the second equation we get M = 0 in the absence of an external field, and from the
1
first equation we get the order parameter Ms (T ) = 3 · TcT−T 2 , which satisfies the same equation
as in the ferromagnetic problem.
We will show that if we switch on the magnetic field, Tc is shifted to a lower temperature.
Assuming small magnetic field we can neglect M 3 and Ms at equation (2) and get:
M=
h
(0)
2Tc
Putting this result at equation (1) we get:

Ms2
(0)
Tc
h2
(0)
4Tc
(0)
Tc
−
= 3
(0)
Where we got Tc = Tc
−
−T

≡3
Tc − T
T
h2
(0)
4Tc
3
(d) We can now express the critical magnetic field from the critical temperature:
h2c = 42 γ 2 − 4γTc (h)
p
⇒ hc = 2 2 γ 2 − γTc (h)
(e) Let us define:
MA = MA0 + δMA ; MB = MB0 + δMB = −MA0 + δMB
M = MA + MB = δMA + δMB
By using the approximation:
tanh(x) = tanh(x0 + δx) ≈ tanh(x0 ) +
δx
cosh2 (x0 )
We get:
1
1
Tc
Tc
MA = tanh − MB + · h = − (MB0 + δMB ) + · h
T
T
T
T
Tc
−Tc · δMB + h
MB 0 +
≈ tanh
T
T · cosh2 (− TTc MB0 )
|
{z
} |
{z
}
M A0
δMA
And in the same way we can get:
δMB =
−Tc · δMA + h
−Tc · δMA + h
=
Tc
2
T · cosh (− T MA0 )
T · cosh2 ( TTc MB0 )
And by attaching the two equations we get an equation for the magnetism M:
− T2c · δM + h
1
M = (δMA + δMB ) =
2
T · cosh2 ( TTc MB0 )
4
After isolating M we have:
M=
h
Tc + T · cosh2 ( TTc MB0 )
1
Ms (h = 0) = Ms (T ) = (MA0 − MB0 ) = −MB0
2
h
→ M=
Tc + T · cosh2 ( TTc Ms (T ))
Let us find the susceptibility:
X=
∂M
∂h
⇒ X=
h→0
1
Tc + T · cosh2 ( TTc Ms (T ))
(f) Let us examine the susceptibility near Tc at two cases:
T > Tc :
at that case the system is not an antiferromagnet yet so Ms (T ) = 0 → cosh( TTc Ms (T )) = 1
which gives us:
XT >Tc =
⇒
1
T + Tc
1
= T + Tc
XT >Tc
T < Tc :
By using the approximation:
1
cosh2 (y) ≈ (1 + y 2 )2 ≈ 1 + y 2
2
XT <Tc =
1
T +T ·
2
Tc
T Ms (T )
+ Tc
A reminder:
5
Ms2 (T )
=
Tc − T
3
T
And after substitution of Ms (T ) and taking into account that where T ≈ Tc →
we get:
XT <Tc =
XT <Tc =
1
XT <Tc
1
T +3·T
Tc −T
T
+ Tc
1
4Tc − 2T
= 4Tc − 2T
for conclusion we got:
1
=
X
T + T c ; T > Tc
4Tc − 2T ; T < Tc
6
Tc
T
≈1