# PHYSICS 140B : STATISTICAL PHYSICS MIDTERM EXAM SOLUTIONS ```PHYSICS 140B : STATISTICAL PHYSICS
MIDTERM EXAM SOLUTIONS
Consider a four-state ferromagnetic Ising model with the Hamiltonian
X
X
Si Sj − H
Si ,
Ĥ = −J
i
hiji
where the first sum is over all links of a lattice of coordination number z. The spin variables
Si take values in the set {−1 , 0 , 0 , +1}. Note that there are two distinct states, each with
Si = 0, and a total of four possible states on each site. Taking the trace for a single site
means we sum over the four independent states, one with S = +1, two with S = 0, and one
with S = −1.
(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 .
[15 points]
Solution : As usual, we neglect fluctuations and obtain
X
ĤMF = 21 N zJm2 − (zJm + H)
Si .
i
(b) Find the mean field free energy F (m, T, H).
[15 points]
Solution : The free energy is obtained from the partition function,
N
−βF
− 21 N zβJm2
β(zJm+H)S
Z=e
=e
Tr e
S
&quot;
=e
− 21 N zβJm2
2 + 2 cosh
zJm + H
kB T
&quot;
#N
.
Thus,
F (m, T, H) = 12 N zJm2 − N kB T ln 2 + 2 cosh
zJm + H
kB T
#
.
(c) Adimensionalize, writing θ = kB T /zJ and h = H/zJ. Find the dimensionless free
energy per site f = F/N zJ.
[15 points]
Solution : We have
&quot;
#
m
+
h
f (m, θ, h) = 12 m2 − θ ln 2 + 2 cosh
θ
&quot;
#
m
+
h
= 21 m2 − 2θ ln 2 cosh
.
2θ
1
(d) What is the self-consistent mean field equation for m?
[15 points]
∂f
Solution : Setting ∂m
= 0 we obtain the mean field equation
m+h
m = tanh
.
2θ
(e) Find the critical temperature θc . Show that when h = 0 the graphical solution to the
mean field equation depends on whether θ &lt; θc or θ &gt; θc .
[15 points]
Solution : We set h = 0 and ask when the RHS of the above equation has slope
unity. This occurs for θ = θc , where θc = 12 . The graphical solution is depicted in fig
6.4 of the lecture notes.
(f) For θ &gt; θc , find m(h, θ) assuming |h| 1.
[15 points]
Solution : For θ &gt; θc , if |h| 1 then |m| 1 and we can replace the tanh function
by the first term in its Taylor series. Thus,
m'
m+h
2θ
=⇒
m(h) =
h
.
2θ − 1
(g) What is the mean field result for hSi2 i?
Hint : We don’t neglect fluctuations from the same site.
[10 points]
Solution : We have
(−1)2 &middot; e−(m+h)/θ + 2 &middot; (0)2 &middot; e0 + (+1)2 &middot; e(m+h)/θ
e−(m+h)/θ + 2 &middot; e0 + e(m+h)/θ
cosh m+h
θ
.
=
1 + cosh m+h
θ
hS 2 i =
Note that as θ → ∞ we have hS 2 i → 12 , since all four states are equally probable and
two of them have S 2 = +1 and the other two have S 2 = 0. As θ → 0 the ground
state configurations are selected. These are two completely polarized states, one with
Si = +1 ∀ i and the other with Si = −1 ∀ i. Thus hS 2 i → 1 in this limit.
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