Answer Key
Statistical Mechanics II: Test 3 Review Solutions
-
University:
Massachusetts Institute of Technology -
Course:
8.334 | Statistical Mechanics II: Statistical Physics of Fields Academic year:
2024
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157
Pages:
32
Author:
GahKeewayi5n
8.334: Statistical Mechanics II
Spring 2014
Test 3
Review Problems & Solutions
The test is ‘closed book,’ but if you wish you may bring a one-sided sheet of formulas.
The intent of this sheet is as a reminder of important formulas and definitions, and not as
a compact transcription of the answers provided here. The test will be composed entirely
from a subset of the following problems as well as those in problem sets 5 and 6.
Thus if you are familiar and comfortable with these problems, there will be no surprises!
********
1. Continuous spins: In the standard O(n) model, n component unit vectors are placed
on the sites of a lattice. The nearest neighbor spins are then connected by a bond JSsi · Ssj .
In fact, if we are only interested in universal properties, any generalized interaction f (Ssi ·Ssj )
leads to the same critical behavior. By analogy with the Ising model, a suitable choice is
exp [f (Ssi · Ssj )] = 1 + (nt)Ssi · Ssj ,
resulting in the so called loop model.
(a) Construct a high temperature expansion of the loop model (for the partition function
Z) in the parameter t, on a two-dimensional hexagonal (honeycomb) lattice.
• The partition function for the loop model has the form
Z=
�
{Dsi }
I
(ij)
[1 + (nt)si · sj ] ,
that we can expand in powers of the parameter t. If the total number of nearest neighbor
bonds on the lattice is NB , the above product generates 2NB possible terms. Each term
may be represented by a graph on the lattice, in which a bond joining spins i and j is
included if the factor si · sj appears in the term considered. Moreover, each included bond
carries a factor of nt. As in the Ising model, the integral over the variables {si } leaves only
graphs with an even number of bonds emanating from each site, because
�
ds sα =
�
ds sα sβ sγ = · · · = 0.
In a honeycomb lattice, as plotted below, there are only 1, 2, or 3 bonds emerging
from each site. Thus the only contributing graphs are those with two bonds at each site,
which, as any bond can only appear once, are closed self-avoiding loops.
1 1
6
2
5
3
4
While the honeycomb lattice has the advantage of not allowing intersections of loops at a
site, the universal results are equally applicable to other lattices.
J
We shall rescale all integrals over spin by the n-dimensional solid angle, such that
ds = 1. Since sα sα = 1, it immediately follows that
Z
�
δαβ
ds sα sβ =
,
n
resulting in
1
sα s′′α .
n
A sequence of such integrals forces the components of the spins around any loop to be the
same, and there is a factor n when integrating over the last spin in the loop, for instance
Z
�
δαβ δβγ δγδ δδη δην δαν
n
= 6.
{Dsi } (s1α s2α )(s2β s3β )(s3γ s4γ )(s4δ s5δ )(s5η s6η )(s6ν s1ν ) =
6
n
n
�
Z
′
ds′ (sα sα
)(sβ′ sβ′′ ) =
Since each bond carrier a factor of nt, each loop finally contributes a factor n × tℓ , where
ℓ is the number of bonds in the loop. The partition function may then be written as
nNℓ tNb ,
Z=
self−avoiding loops
where the sum runs over distinct disconnected or self-avoiding loops collections with a
bond fugacity t, and Nℓ , Nb are the number of loops, and the number of bonds in the
graph, respectively. Note that, as we are only interested in the critical behavior of the
model, any global analytic prefactor is unimportant.
(b) Show that the limit n → 0 describes the configurations of a single self-avoiding polymer
on the lattice.
• While Z = 1, at exactly n = 0, one may obtain non-trivial information by considering
the limit n → 0. The leading term (O(n1 )) when n → 0 picks out just those configurations
with a single self-avoiding loop, i.e. Nℓ = 1.
The correlation function can also be calculated graphically from
Z
�
I
1
Gαβ (n − m) = (snα smβ ) =
[1 + (nt)si · sj ] .
{Dsi }snα smβ
Z
(ij)
2 m
n
After disregarding any global prefactor, and taking the limit n → 0, the only surviving
graph consists of a single line going from n to m, and the index of all the spins along
the line is fixed to be the same. All other possible graphs disappear in the limit n → 0.
Therefore, we are left with a sum over self-avoiding walks that go from n to m, each
carrying a factor tℓ , where ℓ indicates the length of the walk. If we denote by Wℓ (R) the
number of self-avoiding walks of length ℓ whose end-to-end distance is R, we can write
that
X
Wℓ (R)tℓ = lim G(R).
n→0
ℓ
As in the case of phantom random walks, we expect that for small t, small paths
dominate the behavior of the correlation function. As t increases, larger paths dominate
the sum, and, ultimately, we will find a singularity at a particular tc , at which arbitrarily
long paths become possible.
Although we presented the mapping of self-avoiding walks to the n → 0 limit of
the O(n) model for a honeycomb lattice, the critical behavior should be universal, and
therefore independent of this lattice choice. What is more, various scaling properties of
self-avoiding walks can be deduced from the O(n) model with n → 0. Let us, for instance,
characterize the mean square end-to-end distance of a self-avoiding walk, defined as
� 2�
1 X 2
R =
R Wℓ (R),
Wℓ
R
where Wℓ =
L
R
Wℓ (R) is the total number of self-avoiding walks of length ℓ.
The singular part of the correlation function decays with separation R as G ∝
|R|
, up to the correlation length ξ, which diverges as ξ ∝ (tc − t)−ν . Hence,
−(d−2+η)
X
R
R2 G(R) ∝ ξ d+2−(d−2+η) = (tc − t)−ν(4−η) = (tc − t)−γ−2ν .
3 We noted above that G(t, R) is the generating function of Wℓ (R), in the sense that
L
L
ℓ
ℓ
ℓ Wℓ (R)t = G(t, R). Similarly
ℓ Wℓ t is the generating function of Wℓ , and is re
lated to the susceptibility χ, by
X
Wℓ tℓ =
X
R
ℓ
G(R) = χ ∝ (tc − t)−γ .
To obtain the singular behavior of Wℓ from its generating function, we perform a Taylor
expansion of (tc − t)−γ , as
X
l
Wℓ t =
ℓ
tc−γ
−γ
t
1−
tc
which results in
Wℓ =
=
tc−γ
X
ℓ
Γ(1 − γ)
Γ(1 + ℓ)Γ(1 − γ − ℓ)
t
tc
ℓ
,
Γ(1 − γ)
t−ℓ−γ
.
Γ(1 + ℓ)Γ(1 − γ − ℓ) c
After using that Γ(p)Γ(1 − p) = π/ sin pπ, considering ℓ → ∞, and the asymptotic expres
sion of the gamma function, we obtain
Γ(γ + ℓ) −ℓ
t ∝ ℓγ−1 tc−ℓ ,
Γ(1 + ℓ) c
L
L
and, similarly one can estimate R R2 Wℓ (R) from R R2 G(R), yielding
Wℓ ∝
� 2 � ℓ2ν+γ−1 t−ℓ
c
R ∝
= ℓ2ν .
ℓγ−1 t−ℓ
c
Setting n = 0 in the results of the ǫ-expansion for the O(n) model, for instance, gives the
exponent ν = 1/2 + ǫ/16 + O(ǫ2 ), characterizing the mean square end-to-end distance of
a self-avoiding polymer as a function of its length ℓ, rather than ν0 = 1/2 which describes
the scaling of phantom random walks. Because of self-avoidance, the (polymeric) walk is
swollen, giving a larger exponent ν. The results of the first order expansion for ǫ = 1, 2,
and 3, in d = 3, 2, and 1 are 0.56, 0.625, and 0.69, to be compared to 0.59, 3/4 (exact),
and 1 (exact).
********
2. Potts model I: Consider Potts spins si = (1, 2, · · · , q), interacting via the Hamiltonian
L
−βH = K δsi ,sj .
(a) To treat this problem graphically at high temperatures, the Boltzmann weight for each
bond is written as
(
J
exp Kδsi ,sj = C(K) [1 + T (K)g(si, sj )] ,
with g(s, s′) = qδs,s′ − 1. Find C(K) and T (K).
4 • To determine the two unknowns C(K) and T (K), we can use the expressions
eK = C [1 + T (q − 1)]
1 = C [1 − T ]
if
si = sj
if
si =
6 sj
,
from which we obtain
T (K) =
eK − 1
,
eK + q − 1
and
C(K) =
eK + q − 1
.
q
(b) Show that
q
X
′
g(s, s ) = 0 ,
s=1
q
X
g(s1 , s)g(s, s2) = qg(s1 , s2 ) , and
q
X
s,s′
s=1
g(s, s′)g(s′ , s) = q 2 (q − 1).
• Moreover, it is easy to check that
q
X
s=1
q
X
s=1
q
g(s, s′) = q − 1 − (q − 1) = 0,
q
X
[ 2
]
q δs1 s δs2 s − q(δs1 s + δs2 s ) + 1 = q (qδs1 s2 − 1) = qg(s1 , s2 ),
g(s1 , s)g(s, s2) =
X
s=1
q
X
[ 2
]
g(s, s )g(s, s ) =
q δss′ δss′ − 2qδss′ + 1 = q 3 − 2q 2 + q 2 = q 2 (q − 1).
′
s,s′ =1
′
s,s′ =1
(c) Use the above results to calculate the free energy, and the correlation function
(g(sm , sn )) for a one–dimensional chain.
• The factor T (K) will be our high temperature expansion parameter. Each bond conL
tributes a factor T g(si , sj ) and, since s g(s, s′) = 0, there can not be only one bond per
any site. As in the Ising case considered in lectures, each bond can only be considered
once, and the only graphs that survive have no dangling bonds. As a result, for a onedimensional chain, with for instance open boundary conditions, it is impossible to draw
any acceptable graph, and we obtain
Z=
XI
{si } (ij)
(
JN −1
C(K) [1 + T (K)g(si, sj )] = C(K)N −1 q N = q eK + q − 1
.
Ignoring the boundary effects, i.e., that there are N − 1 bonds in the chain, the free energy
per site is obtained as
(
J
βF
−
= ln eK + q − 1 .
N
5 With the same method, we can also calculate the correlation function (g(snsm )). To get a
nonzero contribution, we have to consider a graph that directly connects these two sites.
Assuming that n > m, this gives
(g(sn sm )) =
=
{si }
(ij)
X
C(K)N
T (K)n−m
g(sn sm )g(sm , sm+1 ) · · · g(sn−1 , sn )
Z
N
=
I
C(K)N X
g(snsm )
[1 + T (K)g(si, sj )]
Z
{si }
C(K)
T (K)n−m q n−m+1 (q − 1)q N −(n−m)−1 = T n−m (q − 1)
Z
where we have used the relationships obtained in (b).
(d) Calculate the partition function on the square lattice to order of T 4 . Also calculate
the first term in the low–temperature expansion of this problem.
• The first term in the high temperature series for a square lattice comes from a square of
4 bonds. There are a total of N such squares. Therefore,
Z=
XI
{si } (ij)
[
]
C(K) [1 + T (K)g(si, sj )] = C(K)2N q N 1 + N T (K)4 (q − 1) + · · · .
Note that any closed loop involving ℓ bonds without intersections contributes T ℓ q ℓ (q − 1).
On the other hand, at low temperatures, the energy is minimized by the spins all
being in one of the q possible states. The lowest energy excitation is a single spin in a
different state, resulting in an energy cost of K × 4 with a degeneracy factor N × (q − 1),
resulting in
[
]
Z = qe2NK 1 + N (q − 1)e−4K + · · · .
(e) By comparing the first terms in low- and high–temperature series, find a duality rule
for Potts models. Don’t worry about higher order graphs, they will work out! Assuming
a single transition temperature, find the value of Kc (q).
• Comparing these expansions, we find the following duality condition for the Potts model
e−K̃ = T (K) =
eK − 1
.
eK + q − 1
This duality rule maps the low temperature expansion to a high temperature series, or vice
versa. It also maps pairs of points, K̃ ⇔ K, since we can rewrite the above relationship
in a symmetric way
( K˜
J(
J
e − 1 eK − 1 = q,
6 and consequently, if there is a single singular point Kc , it must be self-dual point,
˜ c,
Kc = K
√
Kc = ln ( q + 1) .
=⇒
(f) How do the higher order terms in the high–temperature series for the Potts model
differ from those of the Ising model? What is the fundamental difference that sets apart
the graphs for q = 2? (This is ultimately the reason why only the Ising model is solvable.)
• As mentioned in lectures, the Potts model with q = 2 can be mapped to the Ising model
by noticing that δss′ = (1 + ss′ )/2. However, higher order terms in the high-temperature
series of the Potts model involve, in general, graphs with three or more bonds emanating
from each site. These configurations do not correspond to a random walk, not even a
constrained one as introduced in class for the 2d-Ising model on a square lattice. The
quantity
q
X
s1 =1
g(s1 , s2 )g(s1 , s3 )g(s1 , s4 ) = q 3 δs2 s3 δs2 s4 − q 2 (δs2 s3 + δs2 s4 + δs3 s4 ) + 2q,
is always zero when q = 2 (as can be easily checked for any possible state of the spins s2 , s3
and s4 ), but is in general different from zero for q > 2. This is the fundamental difference
that ultimately sets apart the case q = 2. Note that the corresponding diagrams in the
low temperature expansion involve adjacent regions in 3 (or more) distinct states.
********
3. Potts model II:
An alternative expansion is obtained by starting with
exp [Kδ(si , sj )] = 1 + v(K)δ(si , sj ),
where v(K) = eK − 1. In this case, the sum over spins does not remove any graphs, and
all choices of distributing bonds at random on the lattice are acceptable.
L
(a) Including a magnetic field h i δsi ,1 , show that the partition function takes the form
Z(q, K, h) =
X
all graphs
I
clusters c in graph
[
(
)]
c
c
v nb × q − 1 + ehns ,
where nbc and nsc are the numbers of bonds and sites in cluster c. This is known as the
random cluster expansion.
• Including a symmetry breaking fiela along direction 1, the partition function
Z=
XI
[1 + v(K)δ(si , sj )]
I
i
{si } (ij)
7
ehδsi ,1 , can be expanded in powers of v(K) as follows. As usual, if there is a total number NB
of nearest neighbor bonds on the lattice, the product over bonds generates 2NB possible
terms. Each term may be represented by a graph on the lattice, in which a bond joining
sites i and j is included if the factor vδ(si , sj ) appears in the term considered. Each
included bond carries a factor v(K), as well as a delta function enforcing the equality of
the spins on the sites which it connects. In general, these bonds form clusters of different
sizes and shapes, and within each cluster, the delta functions force the spins at each vertex
L
c
to be the same. The sum {si } therefore gives a factor of (q − 1) + ehns for each cluster c,
where ncs is the number of point in the cluster. The partition function may then be written
as
[
(
)]
I
X
c
s
v(K)nb q − 1 + ehnc ,
Z(q, v, h) =
all graphs
clusters in graph
nbc
where
is the number of bonds in cluster c, and the sum runs over all distinct cluster
collections. Note that an isolated site is also included in this definition of a cluster. While
the Potts model was originally defined for integer q, using this expansion, we can evaluate
Z for all values of q.
(b) Show that the limit q → 1 describes a percolation problem, in which bonds are randomly
distributed on the lattice with probability p = v/(v +1). What is the percolation threshold
on the square lattice?
• In the problem of bond percolation, bonds are independently distributed on the lattice,
with a probability p of being present. The weight for a given configuration of occupied
and absent bonds bonds is therefore
ncb
I
p
zN
.
W (graph) = (1 − p)
1−p
clusters in graph
The prefactor of (1 − p)zN is merely the weight of the configuration with no bonds. The
above weights clearly become identical to those appearing in the random cluster expansion
8 of the Potts model for q = 1 (and h = 0). Clearly, we have to set p = v/(v + 1), and
neglect an overall factor of (1 +v)N , which is analytic in v, and does not affect any singular
behavior. The partition function itself is trivial in this limit as Z(1, v, h) = (1 + v)zN ehN .
On the other hand, we can obtain information on the number of clusters by considering
the limt of q → 1 from
�
X
X
s
∂ ln Z(q, v) ��
=
[probablility of graph]
e−hnc .
�
∂q
q=1
all graphs
clusters in graph
Various properties of interest to percolation can then be calculated from the above
generating function. This mapping enables us to extract the scaling laws at the percolation
point, which is a continuous geometrical phase transition. The analog of the critical tem
perature is played by the percolation threshold pc , which we can calculate using duality as
pc = 1/2 (after noting that v ∗ = 1).
An alternative way of obtaining this threshold is to find a duality rule for the percola
tion problem itself: One can similarly think of the problem in terms of empty bonds with
a corresponding probability q. As p plays the role of temperature, there is a mapping of
low p to high q or vice versa, and such that q = 1 − p. The self-dual point is then obtained
by setting p∗ = 1 − p∗ , resulting in p∗ = 1/2.
(c) Show that in the limit q → 0, only a single connected cluster contributes to leading
order. The enumeration of all such clusters is known as listing branched lattice animals.
• The parition function Z(q, v, h) goes to zero at q = 0, but again infomration about
geometrical lattice structure can be obtain by taking the limit q → 0 in an appropriate
fashion. In particular, if we set v = q a x, then
X
Z(q, v = xq a , h = 0) =
xNb q Nc +aNb ,
all graphs
where Nb and Nc are the total number of bonds and clusters. The leading dependence
on q as q → 0 comes from graphs with the lowest number of Nc + aNb , and depends on
the value of a. For 0 < a < 1, these are the spanning trees, which connect all sites of the
lattice (hence Nc = 1) and that enclose no loops (hence Nb = N − 1). Such spanning trees
have a power of xa(N −1) q aN−a+1 , and all other graphs have higher powers of q. For a = 0
one can add bonds to the spanning cluster (creating loops) without changing the power,
as long as all sites remain connected in a single cluster. These have a relation to a problem
referred to as branched lattice animals.
********
4. Potts duality: Consider Potts spins, si = (1, 2, · · · , q), placed on the sites of a square
lattice of N sites, interacting with their nearest-neighbors through a Hamiltonian
X
−βH = K
δsi ,sj .
9 (a) By comparing the first terms of high and low temperature series, or by any other
method, show that the partition function has the property
]
[
]2N
[
Z(K) = qe2N K Ξ e−K = q −N eK + q − 1
Ξ
eK − 1
,
eK + (q − 1)
for some function Ξ, and hence locate the critical point Kc (q).
• The low temperature series takes the form
[
]
]
[
Z = qe2N K 1 + N (q − 1)e−4K + · · · ≡ qe2N K Ξ e−K ,
while at high temperatures
eK + q − 1
Z=
q
2N
q
N
1 + N (q − 1)
[
]2N
≡ q −N eK + q − 1
Ξ
eK − 1
eK + q − 1
4
+···
eK − 1
.
eK + q − 1
Both of the above series for Ξ are in fact the same, leading to the duality condition
e−K̃ =
eK − 1
,
eK + q − 1
and a critical (self-dual) point of
˜ c,
Kc = K
√
Kc = ln ( q + 1) .
=⇒
(b) Starting from the duality expression for Z(K), derive a similar relation for the internal
energy U (K) = (βH) = −∂ ln Z/∂ ln K. Use this to calculate the exact value of U at the
critical point.
• The duality relation for the partition function gives
[
]
[
]
ln Z(K) = ln q + 2N K + ln Ξ e−K = −N ln q + 2N ln eK + q − 1 + ln Ξ
eK − 1
.
eK + q − 1
The internal energy U (K) is then obtained from
−
]
[
U (K)
∂
=
ln Z(K) = 2N − e−K ln Ξ′ e−K
K
∂K
eK
qeK
eK − 1
′
.
= 2N K
+
ln
Ξ
eK + q − 1
e + q − 1 (eK + q − 1)2
10 ln Ξ′ is the derivative of ln Ξ with respect to its argument, whose value is not known in
general. However, at the critical point Kc , the arguments of ln Ξ′ from the high and low
√
temperature forms of the above expression are the same. Substituting eKc = 1 + q, we
obtain
ln Ξ′c
2N
ln Ξc′
q−1
2N −
√ = √ +
√ , =⇒ ln Ξ′c = √ N,
1+ q
q
1+ q
q
and,
U (Kc )
−
=N
Kc
q−1
2− √
,
q+q
=⇒
U (Kc ) = N Kc
√
q+1
√ .
q
********
5. Anisotropic Random Walks: Consider the ensemble of all random walks on a square
ℓ
lattice starting at the origin (0,0). Each walk has a weight of txℓx × ty y , where ℓx and ℓy
are the number of steps taken along the x and y directions respectively.
(a) Calculate the total weight W (x, y), of all walks terminating at (x, y). Show that W is
well defined only for t = (tx + ty )/2 < tc = 1/4.
• Defining (0, 0| W (ℓ) |x, y) to be the weight of all walks of ℓ steps terminating at (x, y),
we can follow the steps in sec.VI.F of the lecture notes. In the anisotropic case, Eq.(VI.47)
(applied ℓ times) is trivially recast into
X
(x, y| T ℓ |qx , qy ) =
(x, y| T ℓ |x′ , y ′ ) (x′ , y ′ | qx , qy )
x′ ,y ′
ℓ
= (2tx cos qx + 2ty cos qy ) (x, y| qx , qy ) ,
√
L
where (x, y| qx , qy ) = eiqx x+iqy y / N . Since W (x, y) = ℓ (0, 0| W (ℓ) |x, y), its Fourier
transform is calculated as
XX
(0, 0| T ℓ |x, y) (x, y| qx , qy )
W (qx , qy ) =
ℓ
=
X
x,y
ℓ
(2tx cos qx + 2ty cos qy ) =
ℓ
1
.
1 − (2tx cos qx + 2ty cos qy )
Finally, Fourier transforming back gives
� π 2
� π
Z
Z
e−iqx x−iqy y
d2 q
d q
−iqx x−iqy y
,
q
)
e
=
.
W (x, y) =
W
(q
x y
2
2
−π (2π)
−π (2π) 1 − (2tx cos qx + 2ty cos qy )
Note that the summation of the series is legitimate (for all q’s) only for 2tx + 2ty < 1, i.e.
for t̄ = (tx + ty ) /2 < tc = 1/4.
(b) What is the shape of a curve W (x, y) = constant, for large x and y, and close to the
transition?
11 • For x and y large, the main contributions to the above integral come from small q’s. To
second order in qx and qy , the denominator of the integrand reads
Then, with qi′ ≡
√
1 − 2 (tx + ty ) + tx qx2 + ty qy2 .
ti qi , we have
′
∞
d2 q ′
e−iq ·v
W (x, y) ≈
,
√
2
tx ty 1 − 2 (tx + ty ) + q′2
−∞ (2π)
y
x
√
√
where we have extended the limits of integration to infinity, and v =
,
. As the
t
�
Z
x
ty
denominator is rotationally invariant, the integral depends only on the magnitude of the
vector v. In other words, W (x, y) is constant along ellipses
x2
y2
+
= constant.
tx
ty
(c) How does the average number of steps, (ℓ) = (ℓx + ℓy ), diverge as t approaches tc ?
• The weight of all walks of length ℓ, irrespective of their end point location, is
X
ℓ
ℓ
(0, 0| W (ℓ) |x, y) = (0, 0| T ℓ |qx = 0, qy = 0) = (2tx + 2ty ) = (4t̄ ) .
x,y
Therefore,
"
#
ℓ
X
ℓ
(4t̄
)
∂
∂
1
4t̄
ℓ
(ℓ) = Lℓ
=
4t̄
ln
(4t̄
)
=
4t̄
ln
=
,
ℓ
∂
(4t̄
)
∂
(4t̄
)
1
−
4t̄
1
−
4t̄
(4t̄
)
ℓ
ℓ
L
i.e.
(ℓ) =
t̄
,
tc − t̄
diverges linearly close to the singular value of t̄.
********
6. Anisotropic Ising Model: Consider the anisotropic Ising model on a square lattice with
a Hamiltonian
X(
J
−βH =
Kx σx,y σx+1,y + Ky σx,y σx,y+1 ;
x,y
i.e. with bonds of different strengths along the x and y directions.
(a) By following the method presented in the text, calculate the free energy for this model.
You do not have to write down every step of the derivation. Just sketch the steps that
need to be modified due to anisotropy; and calculate the final answer for ln Z/N .
12 • The Hamiltonian
−βH =
X
(Kx σx,y σx+1,y + Ky σx,y σx,y+1 ) ,
x,y
leads to
Z=
X
N
(2 cosh Kx cosh Ky ) txℓx tyℓy ,
where ti = tanh Ki , and the sum runs over all closed graphs. By extension of the isotropic
case,
X tℓx tℓyy
ln Z
x
f=
= ln (2 cosh Kx cosh Ky ) +
(0| W ∗ (ℓx , ℓy ) |0) ,
N
ℓx + ℓy
ℓx ,ℓy
where
′
1X
number of crossings
(0| W (ℓx , ℓy ) |0) =
(−1)
,
2
and the primed sum runs over all directed (ℓx , ℓy )-steps walks from (0, 0) to (0, 0) with
no U-turns. As in the isotropic case, this is evaluated by taking the trace of powers
of the 4N × 4N matrix described by Eq.(VI.66), which is block-diagonalized by Fourier
transformation. However, unlike the isotropic case, in which each element is multiplied by
t, here they are multiplied by tx and ty , respectively, resulting in
�
Z
[
1
d2 q
∗]
f = ln (2 cosh Kx cosh Ky ) +
tr ln 1 − T (q) ,
2
2
(2π)
∗
where
[
[
∗]
∗]
tr ln 1 − T (q) = ln det 1 − T (q)
[(
J(
J
(
J
(
J
]
= ln 1 + t2x 1 + t2y − 2tx 1 − t2y cos qx − 2ty 1 − t2x cos qy
cosh 2Kx cosh 2Ky − sinh 2Kx cos qx − sinh 2Ky cos qy
= ln
,
cosh 2 Kx cosh 2 Ky
resulting in
1
f = ln 2 +
2
�
Z
d2 q
2
(2π)
ln (cosh 2Kx cosh 2Ky − sinh 2Kx cos qx − sinh 2Ky cos qy ) .
(b) Find the critical boundary in the (Kx , Ky ) plane from the singularity of the free energy.
Show that it coincides with the condition Kx = K̃y , where K̃ indicates the standard dual
interaction to K.
• The argument of the logarithm is minimal at qx = qy = 0, and equal to
cosh 2Kx cosh 2Ky − sinh 2Kx − sinh 2Ky
)2
J
1 ( Kx J
=
e
cosh 2Ky − 1 − e−Kx cosh 2Ky + 1 .
2
13 Therefore, the critical line is given by
e2Kx =
cosh 2Ky + 1
= coth Ky .
cosh 2Ky − 1
Note that this condition can be rewritten as
sinh 2Kx =
1
1
(coth Ky − tanh Ky ) =
,
2
sinh 2Ky
˜ y , where the dual interactions, K̃
i.e. the critical boundary can be described as Kx = K
and K, are related by sinh 2K sinh 2K̃ = 1.
(c) Find the singular part of ln Z/N , and comment on how anisotropy affects critical
behavior in the exponent and amplitude ratios.
• The singular part of ln Z/N for the anisotropic case can be written as
fS =
1
2
�
Z
(
)
2
X
2
J
J
qi
d q
eKx cosh 2Ky − 1 − e−Kx cosh 2Ky + 1 +
sinh 2Ki .
2 ln
2
(2π)
2
i=x,y
In order to rewrite this expression in a form closer to that of the singular part of the free
energy in the isotropic case, let
qi =
�
2
q′ ,
sinh 2Ki i
and
δt = eKx
J
J
cosh 2Ky − 1 − e−Kx cosh 2Ky + 1
(δt goes linearly through zero as (Kx , Ky ) follows a curve which intersects the critical
boundary). Then
1
fS = J
sinh 2Kx sinh 2Ky
Z
�
d2 q ′
2
(2π)
(
J
ln δt2 + q ′2 .
Thus, upon approaching the critical boundary (sinh 2Kx sinh 2Ky = 1), the singular part
of the anisotropic free energy coincides more and more precisely with the isotropic one,
and the exponents and amplitude ratios are unchanged by the anisotropy. (The amplitudes
themselves obviously depend on the locatio
********
14 +
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
- - - - h1 - - - - - - - -
-
-
-
+ + + +
+ + + + - - - - - - - - - - - - -
-
1
2
.
.
.
hN
N
7. Müller–Hartmann Zittartz estimate of the interfacial energy of the d = 2 Ising model
on a square lattice:
(a) Consider an interface on the square lattice with periodic boundary conditions in one
direction. Ignoring islands and overhangs, the configurations can be labelled by heights
hn for 1 ≤ n ≤ L. Show that for an ansiotropic Ising model of interactions (Kx , KY ), the
energy of an interface along the x-direction is
X
−βH = −2Ky L − 2Kx
|hn+1 − hn | .
n
• For each unsatisfied (+−) bond, the energy is increased by 2Ki from the ground state
energy, with i = x if the unsatisfied bond is vertical, and i = y if the latter is horizontal.
Ignoring islands and overhangs, the number of horizontal bond of the interface is L, while
L
the number of vertical bonds is n |hn+1 − hn |, yielding
−βH = −2Ky L − 2Kx
L
X
n=1
|hn+1 − hn | .
(b) Write down a column–to–column transfer matrix (h|T |h′ ), and diagonalize it.
• We can define
(h| T |h′ ) ≡ exp (−2Ky − 2Kx |h′ − h|) ,
or, in matrix form,
1
T = e−2Ky e−2Kx
···
e−2Kx
1
e−4Kx
e−2Kx
···
e−HKx
−2( H
2 −1)Kx
··· e
15
e−HKx
−2( H
2 +1)Kx
e
e−2( 2 −1)Kx
e−HKx
H
· · · e−2Kx
· · · e−4Kx where H is the vertical size of the lattice. In the H → ∞ limit, T is easily diagonalized since
each line can be obtained from the previous line by a single column shift. The eigenvectors
of such matrices are composed by the complex roots of unity (this is equivalent to the
statement that a translationally invariant system is diagonal in Fourier modes). To the
eigenvector
( 2π
)
2π
2π
2π
ei k , ei k ·2 , ei k ·3 , · · · , ei k ·(H+1) ,
is associated the eigenvalue
−2Ky
λk = e
H+1
X
T1n ei
2π
k ·(n−1)
.
n=1
Note that there are H + 1 eigenvectors, corresponding to k = 1, · · · , H + 1.
(c) Obtain the interface free energy using the result in (b), or by any other method.
• One way of obtaining the free energy is to evaluate the largest eigenvalue of T . Since all
elements of T are positive, the eigenvector (1, 1, · · · , 1) has the largest eigenvalue
H/2
H+1
X
X
λ1 = e−2Ky
T1n = e−2Ky 1 + 2
e−2Kx n
n=1
−2Ky
=e
n=1
H/2
2
X
n=0
e−2Kx n − 1 = e−2Ky coth Kx ,
in the H → ∞ limit. Then, F = −LkB T ln λ1 .
Alternatively, we can directly sum the partition function, as
Z = e−2Ky L
X
{hn }
= e−2Ky 2
yielding
exp −2Kx
X
d≥0
L
X
n=1
|hn+1 − hn |
= e−2Ky L
L
"
X
d
exp (−2Kx |d|)
#L
(
JL
e−2Kx d − 1 = e−2Ky coth Kx ,
F = −LkB T [ln (coth Kx ) − 2Ky ] .
(d) Find the condition between Kx and Ky for which the interfacial free energy vanishes.
Does this correspond to the critical boundary of the original 2d Ising model?
• The interfacial free energy vanishes for
coth Kx = e2Ky ,
16 which coincides with the result from an earlier problem. This illustrates that long wave
length fluctuations, such as interfaces, are responsible for destroying order at criticality.
********
8. Anisotropic Landau Theory:
d–dimensions.
Consider an n–component magnetization field m(x)
S
in
(a) Using the previous problems on anisotropy as a guide, generalize the standard Landau–
Ginzburg Hamiltonian to include the effects of spacial anisotropy.
• Requiring different coupling constants in the different spatial directions, along with
rotational invariance in spin space, leads to the following leading terms of the Hamiltonian,
−βH =
Z
�
#
d
X
t
K
∂
m
S
∂
m
S
i
2
4
dd x m
S (x) +
·
+ um
S (x) .
∂x
2
2
∂x
i
i
i=1
"
(b) Are such anisotropies “relevant?”
• Clearly, the apparent anisotropy can be eliminated by the rescaling
x′i
=
�
r
K
xi .
Ki
In terms of the primed space variables, the Hamiltonian is isotropic. In particular, the
universal features are identical in the anisotropic and isotropic cases, and the anisotropy
is thus “irrelevant” (provided all Ki are non-vanishing).
(c) In La2 CuO4 , the Cu atoms are arranged on the sites of a square lattice in planes,
and the planes are then stacked together. Each Cu atom carries a “spin”, which we
assume to be classical, and can point along any direction in space. There is a very strong
antiferromagnetic interaction in each plane. There is also a very weak interplane interaction
that prefers to align successive layers. Sketch the low–temperature magnetic phase, and
indicate to what universality class the order–disorder transition belongs.
• For classical spins, this combination of antiferromagnetic and ferromagnetic couplings is
equivalent to a purely ferromagnetic (anisotropic) system, since we can redefine (e.g. in
the partition function) all the spins on one of the two sublattices with an opposite sign.
Therefore, the critical behavior belongs to the d = 3, n = 3 universality class.
Nevertheless, there is a range of temperatures for which the in-plane correlation length
is large compared to the lattice spacing, while the interplane correlation length is of the
order of the lattice spacing. The behavior of the system is then well described by a d = 2,
n = 3 theory.
********
17 9. Anisotropic nonlinear σ model: Consider unit n-component spins, Ss(x) = (s1 , · · · , sn )
L
with α s2α = 1, subject to a Hamiltonian
βH =
�
Z
1
2
2
d x
(∇Ss ) + gs1 .
2T
d
For g = 0, renormalization group equations are obtained through rescaling distances by
a factor b = eℓ , and spins by a factor ζ = bys with ys = − (n−1)
4π T , and lead to the flow
equation
dT
(n − 2) 2
= −ǫT +
T + O(T 3 ),
dℓ
2π
where ǫ = d − 2.
(a) Find the fixed point, and the thermal eigenvalue yT .
• Setting dT /dℓ to zero, the fixed point is obtained as
T∗ =
2πǫ
+ O(ǫ2 ).
n−2
Linearizing the recursion relation gives
yT = −ǫ +
(n − 2) ∗
T = +ǫ + O(ǫ2 ).
π
(b) Write the renormalization group equation for g in the vicinity of the above fixed point,
and obtain the corresponding eigenvalue yg .
• Rescalings x → bx′ and Ss → ζSs ′ , lead to g → g ′ = bd ζ 2 g, and hence
yg = d + 2ys = d −
n−1 ∗
n−1
1
T =2+ǫ−
ǫ=2−
ǫ + O(ǫ2 ).
2π
n−2
n−2
(c) Sketch the phase diagram as a function of T and g, indicating the phases, and paying
careful attention to the shape of the phase boundary as g → 0.
•
The term proportional to g removes full rotational symmetry and leads to a bicritical
phase diagram as discussed in recitations. The phase for g < 0 has order along direction
1, while g > 0 favors ordering along any one of the (n − 1) directions orthogonal to 1. The
phase boundaries as g → 0 behave as g ∝ (δT )φ , with φ = yg /yt ≈ 2/ǫ + O(1).
********
10. Matrix models:
In some situations, the order parameter is a matrix rather than
a vector. For example, in triangular (Heisenberg) antiferromagnets each triplet of spins
aligns at 120◦ , locally defining a plane. The variations of this plane across the system are
18 g
1
continuous
phase transitions
discontinuous
phase transition
disordered
phase
T
1
described by a 3 × 3 rotation matrix. We can construct a nonlinear σ model to describe a
generalization of this problem as follows. Consider the Hamiltonian
�
Z
[
]
K
βH =
dd x tr ∇M (x) · ∇M T (x)
,
4
where M is a real, N × N orthogonal matrix, and ‘tr’ denotes the trace operation. The
condition of orthogonality is that M M T = M T M = I, where I is the N × N identity
T
matrix, and M T is the transposed matrix, Mij
= Mji . The partition function is obtained
by summing over all matrix functionals, as
Z
�
(
J
Z = DM (x)δ M (x)M T (x) − I e−β H[M (x)] .
(a) Rewrite the Hamiltonian and the orthogonality constraint in terms of the matrix ele
ments Mij (i, j = 1, · · · , N ). Describe the ground state of the system.
• In terms of the matrix elements, the Hamiltonian reads
Z
�
X
K
βH =
dd x
∇Mij · ∇Mij ,
4
i,j
and the orthogonality condition becomes
X
Mik Mjk = δij .
k
19 Since ∇Mij · ∇Mij ≥ 0, any constant (spatially uniform) orthogonal matrix realizes a
ground state.
(b) Define the symmetric and anti-symmetric matrices
J
1(
σ =
M + M T = σT
2
π = 1 (M − M T J = −π T
2
.
Express βH and the orthogonality constraint in terms of the matrices σ and π.
• As M = σ + π and M T = σ − π,
�
�
Z
Z
[
]
K
K
2
2
d
d x tr [∇ (σ + π) · ∇ (σ − π)] =
dd x tr (∇σ) − (∇π) ,
βH =
4
4
where we have used the (easily checked) fact that the trace of the cummutator of matrices
∇σ and ∇π is zero. Similarly, the orthogonality condition is written as
σ 2 − π 2 = I,
where I is the unit matrix.
(c) Consider small fluctuations about the ordered state M (x) = I. Show that σ can be
expanded in powers of π as
1
σ = I − ππ T + · · · .
2
Use the orthogonality constraint to integrate out σ, and obtain an expression for βH to
fourth order in π. Note that there are two distinct types of fourth order terms. Do not
include terms generated by the argument of the delta function. As shown for the nonlinear
σ model in the text, these terms do not effect the results at lowest order.
• Taking the square root of
σ 2 = I + π 2 = I − ππ T ,
results in
σ=I−
( J
1
ππ T + O π 4 ,
2
(as can easily be checked by calculating the square of I − ππ T /2). We now integrate out
σ, to obtain
� Z
�
�
�
Z
K
1 ( ( T JJ2
2
d
Z = Dπ (x) exp
d x tr (∇π) −
∇ ππ
,
4
4
I
where Dπ (x) = j>i Dπij (x), and π is a matrix with zeros along the diagonal, and
elements below the diagonal given by πij = −πji . Note that we have not included the
20 terms generated by the argument of the delta function. Such term, which ensure that the
measure of integration over π is symmetric, do not contribute to the renormalization of K
at the lowest order. Note also that the fourth order terms are of two distinct types, due
to the non-commutativity of π and ∇π. Indeed,
[ ( T J]2 [ ( 2 J]2
2
∇ ππ
= ∇ π
= [(∇π) π + π∇π]
2
= (∇π) π · (∇π) π + (∇π) π 2 · ∇π + π (∇π) π + π (∇π) π · ∇π,
and, since the trace is unchanged by cyclic permutations,
[
]
[ (
J]2
2
2
tr ∇ ππ T
= 2 tr (π∇π) + π 2 (∇π) .
(d) For an N -vector order parameter there are N − 1 Goldstone modes. Show that an
orthogonal N × N order parameter leads to N (N − 1)/2 such modes.
• The anti-symmetry of π imposes N (N + 1) /2 conditions on the N × N matrix elements,
and thus there are N 2 − N (N + 1) /2 = N (N − 1) /2 independent components (Goldstone modes) for the matrix. Alternatively, the orthogonality of M similarly imposes
N (N + 1) /2 constraints, leading to N (N − 1) /2 degrees of freedom. [Note that in the
analogous calculation for the O (n) model, there is one condition constraining the mag
nitude of the spins to unity; and the remaining n − 1 angular components are Goldstone
modes.]
(e) Consider the quadratic piece of βH. Show that the two point correlation function in
Fourier space is
(πij (q)πkl (q′ )) =
(2π)d δ d (q + q′ )
[δik δjl − δil δjk ] .
Kq 2
• In terms of the Fourier components πij (q), the quadratic part of the Hamiltonian in (c)
has the form
�
Z
KX
dd q 2
q |πij (q)|2 ,
βH0 =
(2π)d
2 i (q),
with q in the shell Λ/b < |q| < Λ.
(f) Calculate the coarse grained expectation value for htr(σ)i>
0 at low temperatures after
′ ′
removing these modes. Identify the scaling factor, M (x ) = M < (x)/ζ, that restores
tr(M ′ ) = tr(σ ′ ) = N .
• As a result of fluctuations of short wavelength modes, tr σ is reduced to
>
htr σi0
>
π2
1
>
= tr I +
+···
≈N+
tr π 2 0
2
2
0
*
+>
*
+>
2
1 X
1 X 2
1
=N−
πij
=N−
N 2 − N πij
=N+
πij πji
2
2
2
i=
i=
6 j
6 j
0
0
!
Z
N − 1 Λ dd q 1
N −1
=N 1−
=N 1−
Id (b) .
d
2
2
2K
Λ/b (2π) Kq
>
0
To restore tr M ′ = tr σ ′ = N , we rescale all components of the matrix by
ζ =1−
N −1
Id (b) .
2K
NOTE: An orthogonal matrix M is invertible (M −1 = M T ), and therefore diagonalizable.
In diagonal form, the transposed matrix is equal to the matrix itself, and so its square is
the identity, implying that each eigenvalue is either +1 or −1. Thus, if M is chosen to be
very close to the identity, all eigenvalues are +1, and tr M = N (as the trace is independent
of the coordinate basis).
˜ Evaluate
(g) Use perturbation theory to calculate the coarse grained coupling constant K.
only the two diagrams that directly renormalize the (∇πij )2 term in βH, and show that
N
K̃ = K +
2
Z
Λ
Λ/b
dd q 1
(2π)d q 2
.
• Distinguishing between the greater and lesser modes, we write the partition function as
Z=
Z
<
>
<
>
> −βH<
0 −βH0 +U [π ,π ]
Dπ Dπ e
22
=
Z
0
<
Dπ < e−δfb −βH0
eU
>
0
, where H0 denotes the quadratic part, and
�
Z
K X
U =−
dd x [(∇πij ) πjk · (∇πkl ) πli + πij (∇πjk ) · (∇πkl ) πli ]
8
i,j,k,l
�
Z d d d
K X
d q1 d q2 d q3
=
[(q1 · q3 + q2 · q3 ) ·
3d
8
(2π)
i,j,k,l
·πij (q1 ) πjk (q2 ) πkl (q3 ) πli (−q1 − q2 − q3 )] .
To first order in U , the following two averages contribute to the renormalization of K:
(i)
�
Z d d d
>
K X
d q1 d q2 d q3
>
>
<
<
(q
)
π
(−q
−
q
−
q
)
(q1 · q3 ) πij
(q1 ) πkl
(q3 )
π
2
1
2
3
jk
li
3d
8
0
(2π)
i,j,k,l
! Z
� Λ d ′
� Λ/b d
Z
X
K
d q
1
d q 2
<
<
=
q
πij
(q) πji
(−q) ,
d
d
′2
8
(2π)
Λ/b (2π) Kq
0
i,j
and
�
�
*
+>
�
Z
K X dd q1 dd q2 dd q3 X >
<
<
πij (q2 ) πli> (−q1 − q2 − q3 )
(q2 · q3 ) πjk
(q2 ) πkl
(q3 )
(ii)
3d
8
(2π)
6�
j,k,l
i=j,l
0
"
#
� Λ d ′
�
Z
Z Λ/b d
X
K
1
d q
d q 2
<
<
=
(N − 1)
q
πjk
(q) πkj
(−q) .
d
d
′2
Kq
8
(2π)
(2π)
0
Λ/b
j,k
Adding up the two contributions results in an effective coupling
K̃
K
K
=
+ N
4
4
8
�
Z
Λ
Λ/b
dd q
1
,
(2π) Kq 2
d
i.e.
˜ = K + N Id (b) .
K
2
(h) Using the result from part (f), show that after matrix rescaling, the RG equation for
K ′ is given by:
"
#
�
Z Λ
d
N
−
2
d
q
1
.
K ′ = bd−2 K −
d 2
2
Λ/b (2π) q
• After coarse-graining, renormalizing the fields, and rescaling,
N
−
1
N
˜ =b
1−
Id (b) K 1 +
Id (b)
K =b ζ K
K
2K
N −2
d−2
=b
K−
Id (b) + O (1/K) ,
2
′
d−2 2
d−2
23 i.e., to lowest non-trivial order,
"
K ′ = bd−2 K −
N −2
2
�
Z
Λ
Λ/b
d
#
d q 1
.
d
(2π) q 2
(i) Obtain the differential RG equation for T = 1/K, by considering b = 1 + δℓ. Sketch
the flows for d < 2 and d = 2. For d = 2 + ǫ, compute Tc and the critical exponent ν.
• Differential recursion relations are obtained for infinitesimal b = 1 + δℓ, as
dK
N −2
′
d−2
K =K+
δℓ = [1 + (d − 2) δℓ] K −
Kd Λ δℓ ,
dℓ
2
leading to
dK
N −2
= (d − 2) K −
Kd Λd−2 .
dℓ
2
To obtain the corresponding equation for T = 1/K, we divide the above relation by −K 2 ,
to get
dT
N −2
= (2 − d) T +
Kd Λd−2 T 2 .
dℓ
2
For d < 2, we have the two usual trivial fixed points: 0 (unstable) and ∞ (stable). The
system is mapped unto higher temperatures by coarse-graining. The same applies for the
case d = 2 and N > 2.
For d > 2, both 0 and ∞ are stable, and a non-trivial unstable fixed point appears at
a finite temperature given by dT /dℓ = 0, i.e.
T∗ =
( 2J
2 (d − 2)
4πǫ
=
+
O
ǫ .
N −2
(N − 2) Kd Λd−2
In the vicinity of the fixed point, the flows are described by
�
{
[
] }
d
dT ��
′
d−2 ∗
δT = 1 +
δℓ
δT
=
1
+
(2
−
d)
+
(N
−
2)
K
Λ
T
δℓ δT
d
dT dℓ �T ∗
= (1 + ǫδℓ) δT.
Thus, from
δT ′ = byT δT = (1 + yT δℓ) δT,
we get yT = ǫ, and
ν=
1
.
ǫ
J
(j) Consider a small symmetry breaking term −h dd x tr(M ), added to the Hamiltonian.
Find the renormalization of h, and identify the corresponding exponent yh .
24 • As usual, h renormalizes according to
N −1
d−2
h = b ζh = (1 + dδℓ) 1 −
Kd Λ δℓ h
2K
( 2J
N −1
d−2
Kd Λ
= 1+ d−
δℓ + O δℓ
h.
2K
′
d
From h′ = byh h = (1 + yh δℓ) h, we obtain
yh = d −
( J
N −1
N −1
ǫ
(d − 2) = 2 −
+ O ǫ2 .
Kd Λd−2 = d −
∗
N −2
N −2
2K
Combining RG and symmetry arguments, it can be shown that the 3×3 matrix model
is perturbatively equivalent to the N = 4 vector model at all orders. This would suggest
that stacked triangular antiferromagnets provide a realization of the O(4) universality class;
see P. Azaria, B. Delamotte, and T. Jolicoeur, J. Appl. Phys. 69, 6170 (1991). However,
non-perturbative (topological aspects) appear to remove this equivalence as discussed in
S.V. Isakov, T. Senthil, Y.B. Kim, Phys. Rev. B 72, 174417 (2005).
********
11. The roughening transition:
described by the Hamiltonian
Consider a continuum interface model which in d = 3 is
K
βH0 = −
2
�
Z
d2 x (∇h)2
,
where h(x) is the interface height at location x. For a crystalline facet, the allowed values
of h are multiples of the lattice spacing. In the continuum, this tendency for integer h can
be mimicked by adding a term
−βU = y0
�
Z
d2 x cos (2πh) ,
to the Hamiltonian. Treat −βU as a perturbation, and proceed to construct a renormal
ization group as follows:
(a) Show that
*
�
"
exp i
X
α
qα h(xα )
#+
�
0
X
1
qα qβ C(xα − xβ )
= exp
K
α<β
L
for α qα = 0, and zero otherwise. (C(x) = ln |x|/2π is the Coulomb interaction in two
dimensions.)
25 • The translational invariance of the Hamiltonian constrains (exp [i
L
vanish unless α qα = 0, as implied by the following relation
exp iδ
X
α
qα
�
!*
"
exp i
X
�
#+
qα h (xα )
α
=
0
=
�
*
*
�
(
exp i
"
exp i
X
L
α qα h (xα )])0
qα [h (xα ) + δ]
α
X
qα h (xα )
α
to
)+
��
0
#+
�
.
0
The last equality follows from the symmetry H [h (x) + δ] = H [h (x)]. Using general
properties of Gaussian averages, we can set
�
�
*
"
#+
X
1X
exp i
qα h (xα )
= exp −
qα qβ (h (xα ) h (xβ ))0
2
α
αβ
0
D
E
1X
2
.
= exp
qα qβ (h (xα ) − h (xβ ))
4
0
αβ
Note that the quantity (h (xα ) h (xβ ))0 is ambiguous because of the symmetry h(x) →
L
h(x) + δ. When
α qα = 0, we
D
E can replace this quantity in the above sum with the height
2
difference (h (xα ) − h (xβ ))
which is independent of this symmetry. (The ambiguity,
0
or symmetry, results from the kernel of the quadratic form having a zero eigenvalue, which
means that inverting it requires care.) We can now proceed as usual, and
�
#�
*
"
+
( iq·x
J ( −iq·x
J
�
Z
iq·x
−iq·x
2
α
α
β
β
X
X qα qβ
−e
−e
e
d q e
= exp
exp i
qα h (xα )
2
2
4
Kq
(2π)
α
α,β
0
�
Z
2
X
d q 1 − cos (q · (xα − xβ ))
= exp
qα qβ
2
Kq 2
(2π)
α<β
1 X
= exp
qα qβ C (xα − xβ ) ,
K
α<β
where
C (x) =
�
Z
d2 q 1 − cos (q · x)
1
|x|
=
ln
,
2
q2
2π
a
(2π)
is the Coulomb interaction in two dimensions, with a short distance cutoff a.
(b) Prove that
�
|h(x) − h(y)|
2
�
�
�
d2
= − 2 Gk (x − y)��
,
dk
k=0
26 �
[ (
J]�
where Gk (x − y) = exp ik h(x) − h(y) .
• From the definition of Gk (x − y),
D
E
d2
2
Gk (x − y) = − [h (x) − h (y)] exp [ik (h (x) − h (y))] .
dk 2
Setting k to zero results in the identity
�
D
E
�
d2
2
[h (x) − h (y)] = − 2 Gk (x − y)��
.
dk
k=0
(c) Use the results in (a) to calculate Gk (x − y) in perturbation theory to order of y02 .
(
J
(Hint: Set cos(2πh) = e2iπh + e−2iπh /2. The first order terms vanish according to the
result in (a), while the second order contribution is identical in structure to that of the
Coulomb gas described in this chapter.)
• Following the hint, we write the perturbation as
�
�
Z
Z
[
]
y0
2
−U = y0 d x cos (2πh) =
d2 x e2iπh + e−2iπh .
2
The perturbation expansion for Gk (x − y) = (exp [ik (h (x) − h (y))]) ≡ (Gk (x − y)) is
calculated as
(Gk ) = (Gk )0 − ((Gk U )0 − (Gk )0 (U )0 )
( J
�
� 2� )
1 (�
2
2
+
Gk U 0 − 2 (Gk U )0 (U )0 + 2 (Gk )0 (U )0 − (Gk )0 U 0 + O U 3 .
2
From part (a),
(U )0 = (Gk U )0 = 0,
and
k2
− 2πK
k2
|x − y|
(Gk )0 = exp − C (x − y) =
.
K
a
Furthermore,
� 2�
y2
U 0= 0
2
y2
= 0
2
�
Z
�
Z
d2 x′ d2 x′′ (exp [2iπ (h (x′ ) − h (x′′ ))])0
y2
d x d x (G2π (x − x ))0 = 0
2
2 ′ 2 ′′
′
′′
�
Z
#
2
(2π)
′
′′
d x d x exp −
C (x − x ) ,
K
2 ′ 2 ′′
"
and similarly,
�
�
exp [ik (h (x) − h (y))] U 2 0 =
(
�
2
2 Z
y0
k2
(2π)
2 ′ 2 ′′
=
d x d x exp − C (x − y) −
C (x′ − x′′ )
2
K
K
�
2πk
2πk
[C (x − x′′ ) + C (y − x′ )] .
+
[C (x − x′ ) + C (y − x′′ )] −
K
K
27 Thus, the second order part of Gk (x − y) is
2
Z
�
y02
k
4π 2
′
′′
2 ′ 2 ′′
exp − C (x − y)
d x d x exp −
C (x − x ) ·
4
K
K
�
�
2πk
′
′′
′′
′
· exp
(C (x − x ) + C (y − x ) − C (x − x ) − C (y − x )) − 1 ,
K
and
2
− kK C(x−y)
Gk (x − y) = e
where
�
�
�
Z
( 2πk
)
2
(
J
′′
′
y02
C
x
−x
2 ′ 2 ′′ − 4π
D
4
)
(
K
K
1+
d xd x e
e
− 1 + O y0 ,
4
D = C (x − x′ ) + C (y − x′′ ) − C (x − x′′ ) − C (y − x′ ) .
(d) Write the perturbation result in terms of an effective interaction K, and show that
perturbation theory fails for K larger than a critical Kc .
• The above expression for Gk (x − y) is very similar to that of obtained in dealing with
the renormalization of the Coulomb gas of vortices in the XY model. Following the steps
in the lecture notes, without further calculations, we find
�
(
)
2
�
Z
2
2π ln(r/a)
y
1
2πk
k2
Gk (x − y) = e− K C(x−y) 1 + 0 ×
× C (x − y) × 2π drr 3 e− K
4
2
K
�
�
�
Z
2
2π ln(r/a)
π 3 k2 2
3 −
− kK C(x−y)
K
y C (x − y) drr e
.
=e
1+
K2 0
The second order term can be exponentiated to contribute to an effective coupling constant
Keff , according to
Z
�
π 3 2π/K 2 ∞
1
1
=
− 2a
y0
drr 3−2π/K .
Keff
K
K
a
Clearly, the perturbation theory is inconsistent if the above integral diverges, i.e. if
K>
π
≡ Kc .
2
(e) Recast the perturbation result in part (d) into renormalization group equations for K
and y0 , by changing the “lattice spacing” from a to aeℓ .
• After dividing the integral into two parts, from a to ab and from ab to ∞, respectively,
and rescaling the variable of integration in the second part, in order to retrieve the usual
limits of integration, we have
π3
1
1
=
− 2 a2π/K y02
K
K
Keff
Z
�
ab
drr
a
3−2π/K
π3
− 2 a2π/K × y02 b4−2π/K ×
K
28
Z
�
a
∞
drr 3−2π/K . (To order y02 , we can indifferently write K or K ′ (defined below) in the last term.) In
other words, the coarse-grained system is described by an interaction identical in form,
but parameterized by the renormalized quantities
1
1
π 3 2π/K 2
=
−
a
y0
K
K2
K′
�
Z
ab
drr 3−2π/K ,
a
and
y0′ 2 = b4−2π/K y02 .
With b = eℓ ≈ 1 + ℓ, these RG relations are written as the following differential equations,
which describe the renormalization group flows
( J
dK
= π 3 a4 y02 + O y04
dℓ
.
(
)
dy
0 = 2 − π y + O (y 3 J
0
0
dℓ
K
(f) Using the recursion relations, discuss the phase diagram and phases of this model.
• These RG equations are similar to those of the XY model, with K (here) playing the
role of T in the Coulomb gas. For non-vanishing y0 , K is relevant, and thus flows to larger
and larger values (outside of the perturbative domain) if y0 is also relevant (K > π/2),
suggesting a smooth phase at low temperatures (T ∼ K −1 ). At small values of K, y0 is
irrelevant, and the flows terminate on a fixed line with y0 = 0 and K ≤ π/2, corresponding
to a rough phase at high temperatures.
(g) For large separations
|x − y|, find the magnitude of the discontinuous jump in
�
�
2
|h(x) − h(y)| at the transition.
• We want to calculate the long distance correlations in the vicinity of the transition.
Equivalently, we can compute the coarse-grained correlations. If the system is prepared at
K = π/2− and y0 ≈ 0, under coarse-graining, K → π/2− and y0 → 0, resulting in
2k 2
Gk (x − y) → (Gk )0 = exp −
C (x − y) .
π
From part (b),
D
2
[h (x) − h (y)]
E
�
�
d2
4
2
= C (x − y) = 2 ln |x − y| .
= − 2 Gk (x − y)��
π
dk
π
k=0
On the other hand, if the system is prepared at K = π/2+ , then K → ∞ under the RG
(assuming that the relevance of K holds also away from the perturbative regime), and
D
E
2
[h (x) − h (y)] → 0.
29 D
E
2
Thus, the magnitude of the jump in [h (x) − h (y)] at the transition is
2
ln |x − y| .
π2
********
12. Roughening and duality:
Consider a discretized version of the Hamiltonian in the
previous problem, in which for each site i of a square lattice there is an integer valued
height hi . The Hamiltonian is
K X
βH =
|hi − hj |∞ ,
2
where the “∞” power means that there is no energy cost for Δh = 0; an energy cost of
K/2 for Δh = ±1; and Δh = ±2 or higher are not allowed for neighboring sites. (This is
known as the restricted solid on solid (RSOS) model.)
(a) Construct the dual model either diagrammatically, or by following these steps:
(i) Change from the N site variables hi , to the 2N bond variables nij = hi − hj . Show
that the sum of nij around any plaquette is constrained to be zero.
J 2π
(ii) Impose the constraints by using the identity 0 dθeiθn /2π = δn,0 , for integer n.
(iii) After imposing the constraints, you can sum freely over the bond variables nij to
obtain a dual interaction ṽ(θi − θj ) between dual variables θi on neighboring plaquettes.
• (i) In terms of bond variables nij = hi − hj , the Hamiltonian is written as
KX
∞
−βH = −
|nij | .
2
(ij)
Clearly,
X
any closed loop
nij = hi1 − hi2 + hi2 − hi3 + · · · + hin−1 − hin = 0,
since hi1 = hin for a closed path.
(ii) This constraint, applied to the N plaquettes, reduces the number of degrees of freedom
from an apparent 2N (bonds), to the correct figure N , and the partition function becomes
I
X
Z=
e−βH
δL nα ,0 ,
α
{nij }
�ij�
ij
where the index α labels the N plaquettes, and nα
ij is non-zero and equal to nij only if
the bond (ij) belongs to plaquette α. Expressing the Kronecker delta in its exponential
representation, we get
� 2π
L
X − K L |nij |∞ I Z
dθα iθα �ij� nαij
2
�ij�
Z=
e
e
.
2π
0
α
{nij }
30 (iii) As each bond belongs to two neighboring plaquettes, we can label the bonds by αβ
rather than ij, leading to
!
�
�
Z 2π
X
X
I�
K
dθγ
∞
exp
− |nαβ | + i (θα − θβ ) nαβ
Z=
2
2π
0
γ
{nαβ }
(αβ)
!
�
�
Z 2π
I�
dθγ I X
K
∞
exp
− |nαβ | + i (θα − θβ ) nαβ
.
=
2π
2
0
γ
n
(αβ)
αβ
Note that if all plaquettes are traversed in the same sense, the variable nαβ occurs in
opposite senses (with opposite signs) for the constraint variables θα and θβ on neighboring
plaquettes. We can now sum freely over the bond variables, and since
X
K
K
exp − |n| + i (θα − θβ ) n = 1 + 2e− 2 cos (θα − θβ ) ,
2
n=0,+1,−1
we obtain
Z=
Z
I�
γ
2π
0
dθγ
2π
!
exp
X
[
K
ln 1 + 2e− 2
(αβ)
]
cos (θα − θβ ) .
(b) Show that for large K, the dual problem is just the XY model. Is this conclusion
consistent with the renormalization group results of the previous problem? (Also note the
connection with the loop model.)
• This is the loop gas model , and for K large,
[
]
K
K
ln 1 + 2e− 2 cos (θα − θβ ) ≈ 2e− 2 cos (θα − θβ ) ,
and
Z=
Z
I�
γ
0
2π
dθγ
2π
! L
K
2e− 2
h
i
�
αβ
�
e
cos(θα −θβ )
.
This is none other than the partition function for the XY model, if we identify
K
KXY = 4e− 2 ,
consistent with the results of another problem, in which we found that the low temperature
behavior in the roughening problem corresponds to the high temperature phase in the XY
model, and vice versa.
(c) Does the one dimensional version of this Hamiltonian, i.e. a 2d interface with
−βH = −
KX
|hi − hi+1 |∞
2 i
31
, have a roughening transition?
• In one dimension, we can directly sum the partition function, as
!
!
X
KX
KX
∞
∞
=
exp −
|ni |
Z=
exp −
|hi − hi+1 |
2 i
2 i
{hi }
{ni }
I(
) (
)N
IX
K
∞
=
exp − |ni |
=
1 + 2e−K/2 = 1 + 2e−K/2
,
2
n
i
i
X
i
(ni = hi − hi+1 ). The expression thus obtained is an analytic function of K (for 0 < K <
∞), in the N → ∞ limit, and there is therefore no phase transition at a finite non-zero
temperature.
********
32
Statistical Mechanics II: Test 3 Review Solutions
of 32
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