E5721: Ising antiferromagnet

E5721: Ising antiferromagnet
Submitted by: Alona Petrushansky
The problem:
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 = 12 (MA + MB ) is the averaged
magnetization per spin, and Ms = 21 (MA − MB ) is the staggered magnetization
(1) Explain the claim: for zero eld (h = 0), Ising antiferromagnet is the same as Ising ferromagnet,
where Ms is the order parameter.
(2) Given h and 0 , nd the coupled mean-eld equations for MA and MB . Write the expression for
Ms (T ) for T ∼ Tc , based on the familiar solution of the ferromagnetic case.
(3) Find the critical temperature Tc for h = 0, and also for small h. Hints: for h = 0 use the same
procedure of expanding arctanh(x) as in the ferromagnetic case; for small h you may use the most
extreme simplication that does not give a trivial solution.
(4) Find the critical magnetic eld hc above which the system no longer acts as an antiferromagnet
at zero temperature.
(5) Find an expression for the susceptibility χ(T ), expressed as a function of the staggered magnetization Ms (T ).
(6) In the region of T ∼ Tc give a linear approximation for 1/χ as a function of the temperature T
The solution:
(1) The Hamiltonian is:
H = −
X
si sj − h
(1)
X
si
i
For zero magnetic eld and low temperature the arrangement of spins depends on the interaction
term si sj 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 > 0, we have minimum energy when hsA i sB > 0, as for antiferromagnet, < 0
and therefore minimum energy will obtain whenhsA i sB < 0, from this claim we can see that
(2)
HAF (h = 0) = HF (h = 0)
(2) The Hamiltonian for sub lattice A is of the form:
H = −c hσB i
X
i
σ Ai − h
X
i
X
X
σAi = − σAi (c hσB i + h) = − σAi h
|
{z
}
i
h
the Hamiltonian of one spin from sub lattice A is of form:
1
i
(3)
Hi = −c hσB i σAi − hσAi = −σAi (c hσB i + h) = −σAi h
|
{z
}
(4)
h
for summation of index i over spin up(+) and down(-) we have
Z=
(5)
X
e−βHi = 2cosh(βh)
i
1 X i
2
βhσ i
MA/B = σA/B =
σA/B e A/B = sinh(βh) = tanh(βh)
Z
Z
(6)
i=±1
using c = 2d = 4as the number of close neighbors depending of the dimension and we
dene Tc = c and get (in next question (3) we prove this denition):
(
MA = tanh(βh − βcMB )
(7)
MB = tanh(βh − βcMA )
For T ' Tc , we assume MA ' MB 1 and therefore can expand
arctgh(x) ' x + 13 x3 and with a little algebra we get:
1
1
T arctgh(MA ) ' T (MA + MA3 ) ' T MA + Tc MA3 = h − Tc MB
3
3
(8)
1
1
T arctgh(MB ) ' T (MB + MB3 ) ' T MB + Tc MB3 = h − Tc MA
3
3
(9)
By adding and subtracting the equations we get:
2h − Tc (MA + MB ) = T (MA + MB ) +
Tc (MA + MB ) = T (MA − MB ) +
Tc
(MA3 + MB3 )
3
Tc
(MA3 − MB3 )
3
(10)
(11)
From Ms = 12 (MA − MB ) and M = 12 (MA + MB ) we get MA = Ms + M and
MB = M − Ms , which gives:
MA3 = (Ms + M )3 = Ms3 + 3Ms2 M + 3Ms M 2 + M 3
(12)
MB3 = (M + Ms )3 = M 3 − 3M 2 Ms + 3M Ms2 − Ms3
(13)
adding and subtracting the equations gives:
2
MA3 + MB3 = 2M 3 + 6Ms2 M
(14)
MA3 − MB3 = 2Ms3 + 6M 2 Ms
(15)
by substitution of Eq. 14, 15 and M, Ms to Eq. 10, 11 we nally get
(
a). h = M (T + Tc ) +
b). 0 = Ms (T − Tc ) +
Tc
3
2
3 (M + 3Ms M )
Tc
3
2
3 (Ms + 3M Ms )
(16)
By using this equations we can nd Ms for h = 0 and T ' Tc :
for T ' Tc and h = 0 → M 1and therefore we can assume that
M 2 ' 0, by substracting this terms to Equ. b. we get
r
Ms =
3
Tc − T
T
(17)
(3)
when T = Tc the staggered magnitization equal to zero, Ms = 0, as can be seen
from the last result (Eq. 17) this result conrms our denition Tc = c.
As for h6= 0 :
from Eq. 16.b we have:
1
Ms (T − Tc (1 − M 2 )) + Tc Ms3 = 0
{z
}
|
3
(18)
T˜c = Tc (1 − M 2 ) = c(1 − M 2 )
(19)
T˜C
for T ' Tc from Eq. 16.a we get:
1
h = 2Tc M + Tc M 3 + Ms2 Tc M
3
(20)
for T ' Tc we can assume that M, Ms 1 therefore Ms2 → 0 and M 3 → 0
With these assumptions from Eq. 20 we get
M=
h2
h
⇒ T˜c = Tc (1 −
)
2Tc
4Tc2
(21)
3
(4) In antiferromagnet Ms = 0, (T = 0), but if the magnetic eld h 6= 0 at T = 0
we can steel use previous algebraic denitions like in ferromagnetic case:
(T − Tc )Ms + 13 Tc (3M 2 Ms + M 3 ) = 0
(T + Tc )M + 31 Tc (3Ms2 M + M 3 ) = h
(22)
In antiferromagnetic case we have: Ms = 0, so h = M Tc , Tc = c
At T = 0 there is a denition of magnetization M = mN , in our case we have two sabblatices
so in order do not count nearest neighbors twice:
N −→ N2
so h = Tc m N2 , but in case of antiferromagnet that have ↑↓ constraction
m = −1, so h = − N2 Tc = − N2 c = hc and this is an identication of ground state:
E(↑↓↑↓) = − N2 c , so for ddestroying the antiferromagnetic order needed h < hc .
(5) we can dene MA = MA0 + δMA and MB = MB0 + δMB
(23)
M = MA + MB = MA0 + δMA + MB0 + δMB
by using the approximation:
tanh(x) = tanh(x0 + δx) ' tanh(x0 ) +
δx
cosh2 (x0 )
(24)
and for small magnetic eld we can dene δx = hβ − cβδMB we get:
MA = tanh(hβ − cβ(MB0 + δMB )) ' tanh(−cβMB0 ) +
hβ − cβδMB
cosh2 (−cβMB0 )
(25)
hβ − cβδMA
cosh2 (−cβMA0 )
(26)
and for δx = hβ − cβδMA we get:
MB = tanh(hβ − cβ(MA0 + δMA )) ' tanh(−cβMA0 ) +
At temperature that much lower from the critical temperature we can assume that
MA0 = −MB0 , and by attaching MA and MB we get an equation for the magnetism M :
M = M A + MB =
2hβ − cβ(δMA + δMB )
cosh2 (cβMA0 )
2Ms = MA − MB = MA0 + δMA − MB0 − δMB = 2MA0 + (δMA − δMB ) ' 2MA0
for δMA − δMB → 0, and we get:
4
(27)
M'
2hβ − cβ(δMA + δMB )
cosh2 (cβMs )
(28)
at low T we have very small M . therefore we can assume that δMA + δMB = δM ' M
by substitution and isolating M we have:
M=
1
T +T
2 c
h ,
cosh2 TTMs
1
β
c
∂M
∂h h→0
χ=
where Tc = c, T =
=
T cosh2
1
T Ms
Tc
+ 21 Tc
(6)
let dene linear approximation of suseptability in case of T ≈ Tc at two regions
T ≥ Tc and T ≤ Tc :
for h = 0: χ =
1
Tc +T cosh2 ( TcTMs )
a) for T ≥ Tc
Here Ms = 0 because the order parameter of antiferromagnet dened by a magnetization
of lattices A and B, that in region of T>Tc have zero values, so
Ms = 21 (MA − MB ) = 0,where MA = MB = 0, than
cosh2
χ=
1
χ
Tc Ms
T
=1
1
Tc +T
= T + Tc
b) for T ≤ Tc ,
q
In this region order parameter of antiferromagnet dened by Ms = 3 TcT−T
c
we can use the approximation cosh2 x ' 1 + 12 x2
cosh2
Tc Ms
T
'1+
Tc 2
T
3 TcT−T
c
2
' 1 + x2
' 1 + 3 TcT−T
c
the last approximation we can do because Tc ∼ T and therefore
χ=
1
T 2M2
Tc +T 1+ c 2 s
T
1
χ
∼
1
Tc +T 1+ 3 TcT−T
=
1
4Tc −2T
c
= 4Tc − T
5
Tc 2
T
→1