PHYSICS 140B : STATISTICAL PHYSICS
HW ASSIGNMENT #5 SOLUTIONS
(1) Consider the S = 1 Ising Hamiltonian
Ĥ = −J
X
Siz Sjz − H
X
Siz + ∆
i
hiji
X
Siz
i
2
,
where Siz ∈ {−1, 0, +1}.
(a) Making the mean field approximation in the first term, but treating the other terms
exactly, find the corresponding mean field Hamiltonian ĤMF .
(b) Defining θ = kB T /zJ, h = H/zJ, and δ = ∆/zJ, find the free energy f = F/N zJ as
a function of θ, δ, h, and m.
(c) Write the mean field equation for this model.
Solution :
(a) The mean field Hamiltonian is
ĤMF = 21 N zJm2 − (H + zJm)
X
Siz + ∆
i
X
Siz
i
2
.
The mean field free energy is then
"
FMF = 21 N zJm2 − N kB T ln 1 + 2 e−∆/kB T
H + zJm
cosh
kB T
#
.
(b) The scaled free energy is
"
m+h
f = 21 m2 − θ ln 1 + 2 e−δ/θ cosh
θ
#
.
(c) The mean field equation is
∂f
=m−
∂m
sinh
1
2
m+h
θ
exp(δ/θ) + cosh
m+h
θ
=0.
Sjz , where the
individual spin polarizations take values Siz ∈ {−S, . . . , +S}. Find an expression for the
mean field value of Tc . Hint: Expand the free energy for small m = hSiz i, and find what
value of T makes the coefficient of the quadratic term vanish.
(2) Consider a spin-S Ising model. The Hamiltonian is Ĥ = −J
1
P
z
hiji Si
Solution :
The mean field Hamiltonian is ĤMF = −zJm
FMF =
2
1
2 N zJm
P
i
Siz , hence the mean field free energy is
− N kB T ln
S
X
ejzJm/kB T .
j=−S
Note that
S
X
ju
e
Su
=e
−ju
e
Su
=e
j=0
j=−S
Thus,
2S
X
1 − e−(2S+1)u
1 − e−u
sinh (S + 12 )u
.
=
sinh( 21 u)
f = 12 m2 − θ ln sinh (2S + 1) m/2θ + θ ln sinh(m/2θ) .
The mean field equation is
∂f
= m − (S + 12 ) ctnh (S + 21 ) m/θ +
∂m
Now use the Laurent expansion for ctnh (u) =
m = 31 S(S + 1)
1
u
+
u
3
−
1
2
u3
45
ctnh (m/2θ) .
+ . . . to obtain
m
+ O(m3 ) ,
θ
from which we identify θc = 31 S(S + 1), or TcMF = S(S + 1)zJ/3kB .
(3) Consider the O(2) model,
Ĥ = − 12
X
Jij n̂i · n̂j − H ·
X
n̂i ,
i
i,j
where n̂i = cos φi x̂ + sin φi ŷ. Consider the case of infinite range interactions, where
Jij = J/N for all i, j, where N is the total number of sites.
(a) Show that
"
#
Z
P
βJ X
N βJ
2
exp
n̂i · n̂j =
d2 m e−N βJm /2 eβJm· i n̂i .
2N
2π
i,j
(b) Using the definition of the modified Bessel function I0 (z),
Z2π
dφ z cos φ
e
,
I0 (z) =
2π
0
show that
−β Ĥ
Z = Tr e
=
Z
d2 m e−N A(m,h)/θ ,
where θ = kB T /J and h = H/J. Find an expression for A(m, h).
2
(c) Find the equation which extremizes A(m, h) as a function of m.
(d) Look up the properties of I0 (z) and write down the first few terms in the Taylor
expansion of A(m, h) for small m and h. Solve for θc .
Solution :
(a) We have
J
Ĥ
=−
kB T
2N kB T
Therefore
−Ĥ/kB T
e
X
i
n̂i
2
−
H X
·
n̂i .
kB T
i
"
#
X 2
1
h X
= exp
n̂i + ·
n̂i
2N θ
θ
i
i
"
X #
Z
N
m+h
N m2
2
=
+
d m exp −
·
n̂i .
2πθ
2θ
θ
i
(b) Integrating the previous expression, we have
Y Z dn̂
−Ĥ/kB T
i −Ĥ[{n̂i }]/kB T
Z = Tr e
=
e
2π
i
Z
h
iN
N
2
=
d2 m e−N m /2θ I0 |m + h|/θ
.
2πθ
Thus, we identify
θ
ln(N/2πθ) .
A(m, h) = 12 m2 − θ ln I0 |m + h|/θ −
N
(c) Extremizing with respect to the vector m, we have
∂A
m + h I1 |m + h|/θ
=0,
=m−
·
∂m
|m + h| I0 |m + h|/θ
where I1 (z) = I0′ (z). Clearly any solution requires that m and h be colinear, hence
I1 (m + h)/θ
.
m=
I0 (m + h)/θ
(d) To find θc , we first set h = 0. We then must solve
m=
I1 (m/θ)
.
I0 (m/θ)
3
The modified Bessel function Iν (z) has the expansion
Iν (z) =
1 ν
2z
∞
X
k=0
1 2 k
4z
k! Γ(k + ν + 1)
.
Thus,
I0 (z) = 1 + 41 z 2 + . . .
I1 (z) = 21 z +
and therefore I1 (z)/I0 (z) = 21 z −
1 3
16 z
1 3
16 z
+ ... ,
+ O(z 5 ), and we read off θc = 12 .
4