PHYSICS 210A : STATISTICAL PHYSICS
FINAL EXAMINATION SOLUTIONS
(1) A classical three-dimensional gas consists of nonrelativistic particles of mass m which
can exist in either of two spin polarization states: σ = ±1. The energy for a gas particle is
p2
εσ (p) = 2m
− µ0 Hσ, where µ0 is the magnetic moment. The gas is in contact with a surface
which has K adsorption sites. The adsorption sites are themselves magnetic, each of which
has a spin polarization τ = ±1. The energy of an empty adsorption site is −µ0 Hτ , and that
of an occupied adsorption site is −Jτ σ − µ0 H(τ + σ), where σ is the spin polarization of
the adsorbate particle. Note that, for simplicity, we are taking the adsorption sites and the
gas particles to have the same magnetic moment.
(a) Find the grand potential Ωg (T, V, H, µg ) for the gas. (Take care not to confuse the
chemical potential µ and the magnetic moment µ0 !)
(b) Enumerate all the possible energy states of an adsorption site, and indicate their
corresponding energies and degeneracies.
(c) Find the grand potential Ωa (T, K, H, µa ) for the surface.
(d) Assuming the gas and the surface are in thermal and particle equilibrium, compute
the fraction ν0 of adsorption sites which are empty.
(e) Compute the fraction ν ′ of adsorption sites which are both occupied and whose spin
polarization is aligned with the adsorbate particle’s polarization. (I.e. sites with τ =
σ = 1 or τ = σ = −1.)
Solution :
(a) Multiplying the usual expression by e±µg /kB T and summing over the two spin polarizations, we have
µg /kB T
.
Ωg (T, V, H, µg ) = −2V λ−3
T kB T cosh(µ0 H/kB T ) e
(b) There are six possible energy states, corresponding to two states for the adsorption site
polarization and three possibilities for an adsorbate particle (none, spin up, or spin down).
The corresponding energies are shown in Tab. 1.
(c) The grand potential for the surface is Ξ = ξ K , where ξ is the contribution from a single
adsorption site:
h
i
ξ = 2 cosh(µ0 H/kB T ) + 2 e−J/kB T + 2 eJ/kB T cosh(2µ0 H/kB T ) eµa /kB T .
Thus,
Ωa (T, K, J, µa ) = −KkB T ln 2 cosh(µ0 H/kB T )+ 2e−J/kB T +2eJ/kB T cosh(2µ0 H/kB T ) eµa /kB T .
1
σ
−
−
↑
↑
↓
↓
τ
↑
↓
↑
↓
↑
↓
E
−µ0 H
µ0 H
−J − 2µ0 H
J
J
−J + 2µ0 H
N
0
0
1
1
1
1
Table 1: Configurations and their energies and occupancies for problem 1b.
(d) The fraction of empty adsorption sites is
ν0 =
=
2 cosh(µ0 H/kB T )
h
i
2 cosh(µ0 H/kB T ) + 2 e−J/kB T + 2 eJ/kB T cosh(2µ0 H/kB T ) z
2 cosh2 (µ0 H/kB T )
h
i
,
2 cosh2 (µ0 H/kB T ) + e−J/kB T + eJ/kB T cosh(2µ0 H/kB T ) nλ3T
where n is the gas density, which we relate to the fugacity z = eµg /kB T = eµa /kB T by
n = 2zλ−3
T cosh(µ0 H/kB T ).
(e) The fraction of adsorption sites whose spins are aligned with their adsorbate particle is
ν′ =
=
2z eJ/kB T cosh(2µ0 H/kB T )
h
i
2 cosh(µ0 H/kB T ) + 2 e−J/kB T + 2 eJ/kB T cosh(2µ0 H/kB T ) z
cosh(2µ0 H/kB T ) eJ/kB T nλ3T
h
i
.
2 cosh2 (µ0 H/kB T ) + e−J/kB T + eJ/kB T cosh(2µ0 H/kB T ) nλ3T
(2) The density of states for a particle moving in three space dimensions is
g(ε) =
ε (ε2 + ∆2 )
Θ(ε) .
Ω ∆4
(a) Find the single particle dispersion ε(k).
(b) Assuming the particles obey photon statistics find their density n(T ).
(c) Assuming the particles are bosons, find the Bose condensation temperature Tc (n).
(d) Assuming the particles are fermions, find the Fermi energy εF (n).
2
Solution :
(a) We have
k2 dk
d3k
=
g(ε) dε =
(2π)3
2π 2 dε
3 k
d
=d
6π 2
⇒
1 4
4ε
+ 12 ∆2 ε2
Ω∆4
!
.
Note that Ω has units of volume. Integrating, we have
1 4
4ε
∆4 Ωk3
=0,
6π 2
+ 21 ∆2 ε2 −
with solution
ε(k) = ∆
s
In the limit k → 0, one finds ε(k) = ∆
(b) The photon density is
1+
2Ωk3
3π 2
1/2
−1.
Ω 1/2 3/2
k .
2
3π
(
)
Z∞
kB T 4 3
kB T 2
1
1
g(ε)
=
dx x
x +
x
Ω
e −1
∆
∆
eε/kB T − 1
0
0
4
Γ(4) ζ(4) kB T
Γ(2) ζ(2) kB T 2
=
+
,
Ω
∆
Ω
∆
Z∞
n = dε
since
Z∞
xp−1
dx x
= Γ(p) ζ(p) .
e −1
0
Now Γ(4) = 6, Γ(2) = 1, ζ(4) =
π4
90 ,
and ζ(2) =
π2
6 ,
so
1
1 πkB T 2
πkB T 4
n(T ) =
+
.
15 Ω
∆
6Ω
∆
(c) The equation for Tc is n(Tc ) = n, where n(Tc ) is the photon statistics (i.e. µ = 0) density
at T = Tc . Solving the above quadratic equation in T 2 , we find
r
q
i1/2
h
∆
1 + 48
nΩ
−1.
kB Tc = 54
5
π
(d) We have, for fermions at T = 0,
ZεF
1 4
ε + 1 ∆2 ε2
n = dε g(ε) = 4 F 2 4 F ,
Ω∆
0
3
and hence
εF (n) = ∆
q
1 + 4nΩ
1/2
−1.
(3) An O(3) spin model on a regular lattice of coordination number z is described by the
Hamiltonian
Ĥ = −J
X
X 1 + n̂i · n̂j −H ·
n̂i ,
ln
2
i
hiji
where n̂ is a three-dimensional unit vector.
(a) Find the effective field Heff , including the contribution from the mean field from a
site’s neighbors.
(b) Find the mean field Hamiltonian ĤMF . You may drop terms of order (fluctuations)2 ,
but do not drop any constant contributions.
(c) Find the mean field equation for the order parameter m = hn̂i.
(d) Find the mean field critical temperature when H = 0.
(e) Find the mean field critical exponents α, β, γ, and δ.
Solution :
(a) We write n̂i = m + δn̂i , in which case
1 + n̂i · n̂j
1
1 + m2 1
+ m · δn̂i + δn̂j + 2 δn̂i · δn̂j
ln
= ln
2
2
2
2
1+m
m
= ln
· δn̂i + δn̂j + O(δn̂2 ) .
+
2
2
1+m
When we sum over the links hiji, each site will occur z times, hence the internal field is
Hint = zJm/(1 + m2 ), and
zJm
+H .
Heff =
1 + m2
(b) We now sum the above result for the linearized interaction over links hiji, yielding
X
zJm X
1 + m2
·
(
n̂
−
m)
−
H
·
n̂i
−
ĤMF = − 21 N zJ ln
i
2
1 + m2
i
i
(
)
2
X
1
1+m
− Heff ·
n̂i .
− 12 ln
= N zJ 1 −
2
1+m
2
i
4
Figure 1: Mean field free energy f (θ, h = 0, m) for problem 3, shown for θ = 0.2 (blue),
θ = θc = 13 (green), and θ = 0.5 (red).
(c) We have
m = hn̂i =
Z
Z
dn̂ H ·n̂/k T
dn̂
Heff ·n̂/kB T
B
n̂ e
e eff
4π
4π
where the integral is over all solid angle. Since H breaks the continuous O(3) symmetry,
m will lie along the Ĥ direction if H 6= 0. Else m may lie in any direction and below Tc
the O(3) symmetry is spontaneously broken. We are free to choose the ẑ axis to be parallel
to H, in which case

 π
Z
∂
βH
cos
θ
1
 = ctnh Heff − kB T ,
ln 2 dθ sin θ e eff
m=
∂(βHeff )
kB T
Heff
0
where
Heff = H +
zJm
.
1 + m2
(d) When H = 0, we have Heff = zJm/(1+m2 ). We expand the RHS of the above equation
in powers of Heff and identify the linear term. The critical temperature occurs when the
coefficient of the linear term is unity, i.e. it matches the coefficient on the LHS. Note
!
2
1 + u2 + . . .
1
u
1
− 1 = + O(u3 ) .
ctnh u − = ·
2
u
u
u
3
1 + 6 + ...
Therefore, Tc = zJ/3kB .
5
(e) This part is tedious and designed to provide a challenge for the most advanced students. Adimensionalize by defining f = F/N zJ, θ = kB T /zJ, and h = H/zJ as usual.
Then
sinh(heff /θ)
1 + m2
1
1
− ln
f (θ, h, m) = 1 −
− θ ln
,
1 + m2 2
2
heff /θ
m
4
where heff = h + 1+m
2 . We now derive a Landau expansion valid to order m , and in the
1
vicinity of θ = θc = 3 . This is slightly tedious, since m appears throughout the function
f (θ, h, m). I find
f=
1
2
ln 2 +
1
2
1−
1
3θ
m2 +
1
3θ
−
3
4
+
1
180 θ 3
hm
+ ... .
m4 −
3θ
The coefficient of the quartic term is apparently negative at large values of θ, but we only
care about θ ≈ θc here. Thus,
f (θ, h, m) =
1
2
ln 2 + 23 (θ − θc )m2 +
4
3
10 m
− hm + . . . .
This is a standard Landau expansion, hence we get standard mean field exponents:
α=0 ,
β=
1
2
,
γ=1 ,
6
δ=3
.