PHYSICS 140B : STATISTICAL PHYSICS
PRACTICE MIDTERM SOLUTIONS
Consider a four-state ferromagnetic Ising model with the Hamiltonian
X
X
X
Si Sj − H
Si Sj − J 2
Si ,
Ĥ = −J1
hhijii
hiji
i
where the first sum is over all nearest neighbor pairs and the second
sum is over all next
nearest neighbor pairs. The spin variables S i take values in the set − 32 , − 21 , + 12 , + 23 .
(a) Making the mean field Ansatz Si = m + (Si − m), where m = hSi i is presumed
independent of i, derive the mean field Hamiltonian ĤMF . You may denote z1 as the
number of nearest neighbors and z2 as the number of next nearest neighbors of any
site on the lattice.
[15 points]
Solution : As usual, we neglect fluctuations and obtain
X
ˆ m+H
Si ,
ĤMF = 21 N Jˆ(0) m2 − J(0)
i
where
ˆ =
J(0)
X
J(R) = z1 J1 + z2 J2 .
R
(b) Find the mean field free energy F (m, T, H).
[15 points]
Solution : The free energy is obtained from the partition function,
N
ˆ m+H)S
ˆ m2
−βF
− 12 N β J(0)
β(J(0)
Z=e
=e
Tr e
S
=e
ˆ m2
− 12 N β J(0)
"
ˆ
ˆ
#N
J(0) m + H
3(J (0) m + H)
2 cosh
+ 2 cosh
.
2kB T
2kB T
Thus,
F (m, T, H) =
2
1
ˆ
2 N J(0) m −N kB T
"
ˆ
ˆ
#
J(0) m + H
3(J (0) m + H)
ln 2 cosh
.
+2 cosh
2kB T
2kB T
ˆ
(c) Adimensionalize, writing θ = kB T /Jˆ(0) and h = H/J(0).
Find the dimensionless free
ˆ
energy per site f = F/N J(0).
[15 points]
Solution : We have
"
#
m
+
h
3(m
+
h)
f (m, θ, h) = 21 m2 − θ ln 2 cosh
+ 2 cosh
.
2θ
2θ
1
(d) What is the self-consistent mean field equation for m?
[15 points]
∂f
= 0 we obtain the mean field equation
Solution : Setting ∂m
m=
sinh
2 cosh
m+h
+ 3 sinh 3(m+h)
2θ
2θ
m+h
+ 2 cosh 3(m+h)
2θ
2θ
.
(e) Find the critical temperature θ c .
[15 points]
Solution : We set h = 0 and expand the RHS of the above equation to lowest order
in m. This yields
m
+ 3 sinh 3m
sinh 2θ
2θ
m=
m
2 cosh 2θ
+ 2 cosh 3m
2θ
5m
=
+ O(m3 ) .
4θ
The critical temperature occurs when the slope of the RHS matches the slope of the
LHS, which occurs at θc = 54 .
(f) For θ > θc , find m(h, θ) assuming |h| 1.
[15 points]
Solution : For θ > θc , if |h| 1 then |m| 1 and we can again expand, obtaining
Thus,
5(m + h)
5h
m'
=⇒
m(h) =
.
4θ
4θ − 5
(g) What is the mean field result for |Si | ? Interpret your result in the θ → ∞ and
θ → 0 limits. Hint : We don’t neglect fluctuations from the same site.
[10 points]
Solution : We have
· e3(m+h)2/θ + 12 · e−(m+h)2/θ + 12 · e(m+h)2/θ
2 cosh m+h
+ 2 cosh 3(m+h)
2θ
2θ
+ 3 cosh 3(m+h)
cosh m+h
2θ
2θ
=
.
2 cosh m+h
+ 2 cosh 3(m+h)
2θ
2θ
Note that as θ → ∞ we have |S| → 1, since all four states are equally probable
and two of them have |S| = 12 and the other two have |S| = 32 . As θ → 0 the ground
state configurations are selected. These are two completely
polarized states, one with
Si = + 32 ∀ i and the other with Si = − 32 ∀ i. Thus |S| → 23 in this limit.
|S| =
3
2
· e−3(m+h)2/θ +
3
2
2