Statistical Physics of Fields By Mehran Kardar SOLUTION MANUAL

SOLUTION MANUAL

Problems & Solutions

for

Statistical Physics of Fields

Updated July 2008 by Mehran Kardar Department of Physics Massachusetts Institute of Technology Cambridge, Massachusetts 02139, USA

Table of Contents 1.

Collective Behavior, From Particles to Fields . . . . . . . . . . . . . . . .

1

2.

Statistical Fields

3.

Fluctuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4.

The Scaling Hypothesis

5.

Perturbative Renormalization Group

6.

Lattice Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

7.

Series Expansions

. . . . . . . . . . . . . . . . . . . . . . . . . .

106

8.

Beyond Spin Waves . . . . . . . . . . . . . . . . . . . . . . . . . .

132

. . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 . . . . . . . . . . . . . . . . . . . . . . . . . 55 . . . . . . . . . . . . . . . . . . . 63

Solutions to problems from chapter 1- Collective Behavior, From Particles to Fields 1. The binary alloy: A binary alloy (as in β brass) consists of NA atoms of type A, and NB atoms of type B. The atoms form a simple cubic lattice, each interacting only with its six nearest neighbors. Assume an attractive energy of −J (J > 0) between like neighbors

A − A and B − B, but a repulsive energy of +J for an A − B pair.

(a) What is the minimum energy configuration, or the state of the system at zero temperature?

• The minimum energy configuration has as little A-B bonds as possible. Thus, at zero

temperature atoms A and B phase separate, e.g. as indicated below.

A

B

(b) Estimate the total interaction energy assuming that the atoms are randomly distributed among the N sites; i.e. each site is occupied independently with probabilities pA = NA /N and pB = NB /N .

• In a mixed state, the average energy is obtained from E = (number of bonds) × (average bond energy)

= 3N · −Jp2A − Jp2B + 2JpA pB 2

NA − NB . = −3JN N (c) Estimate the mixing entropy of the alloy with the same approximation. Assume NA , NB ≫ 1.

• From the number of ways of randomly mixing NA and NB particles, we obtain the mixing entropy of

S = kB ln

N! NA !NB !

.

Using Stirling’s approximation for large N (ln N ! ≈ N ln N − N ), the above expression can be written as

S ≈ kB (N ln N − NA ln NA − NB ln NB ) = −N kB (pA ln pA + pB ln pB ) .

1

(d) Using the above, obtain a free energy function F (x), where x = (NA −NB )/N . Expand F (x) to the fourth order in x, and show that the requirement of convexity of F breaks down below a critical temperature Tc . For the remainder of this problem use the expansion obtained in (d) in place of the full function F (x).

• In terms of x = pA − pB , the free energy can be written as F = E − TS 2

= −3JN x + N kB T

1+x 2

ln

1+x 2

+

x2 +

1−x 2

ln

1−x 2

.

Expanding about x = 0 to fourth order, gives F ≃ −N kB T ln 2 + N

kB T − 3J 2

N kB T 4 x .

12

Clearly, the second derivative of F , ∂ 2F = N (kB T − 6J) + N kB T x2 , 2 ∂x becomes negative for T small enough. Upon decreasing the temperature, F becomes concave first at x = 0, at a critical temperature Tc = 6J/kB . (e) Sketch F (x) for T > Tc , T = Tc , and T < Tc . For T < Tc there is a range of compositions x < |xsp (T )| where F (x) is not convex and hence the composition is locally unstable. Find xsp (T ).

• The function F (x) is concave if ∂ 2 F /∂x2 < 0, i.e. if 2

x <

6J −1 . kB T

This occurs for T < Tc , at the spinodal line given by xsp (T ) =

r 2

6J − 1, kB T

F(x)/NJ T=0 x sp(T) T<Tc

T=Tc T>Tc +1

-1

x

as indicated by the dashed line in the figure below. (f) The alloy globally minimizes its free energy by separating into A rich and B rich phases of compositions ±xeq (T ), where xeq (T ) minimizes the function F (x). Find xeq (T ).

• Setting the first derivative of dF (x) /dx = N x (kB T − 6J) + kB T x2 /3 , to zero yields

the equilibrium value of

 r  6J  ±√3 − 1 for T < Tc . xeq (T ) = kB T   0 for T > Tc (g) In the (T, x) plane sketch the phase separation boundary ±xeq (T ); and the so called

spinodal line ±xsp (T ). (The spinodal line indicates onset of metastability and hysteresis

effects.)

• The spinodal and equilibrium curves are indicated in the figure above. In the interval

between the two curves, the system is locally stable, but globally unstable. The formation of ordered regions in this regime requires nucleation, and is very slow. The dashed area is locally unstable, and the system easily phase separates to regions rich in A and B. ********

2. The Ising model of magnetism:

The local environment of an electron in a crystal

sometimes forces its spin to stay parallel or anti-parallel to a given lattice direction. As a model of magnetism in such materials we denote the direction of the spin by a single 3

T Tc x eq(T) x sp(T)

unstable

metastable

metastable

1

1

x

variable σi = ±1 (an Ising spin). The energy of a configuration {σi } of spins is then given

by

N

X 1 X Jij σi σj − h σi H= 2 i,j=1

;

i

where h is an external magnetic field, and Jij is the interaction energy between spins at sites i and j. (a) For N spins we make the drastic approximation that the interaction between all spins is the same, and Jij = −J/N (the equivalent neighbor model). Show that the energy can now PN be written as E(M, h) = −N [Jm2 /2+hm], with a magnetization m = i=1 σi /N = M/N .

P • For Jij = −J/N , the energy of each configuration is only a function of m = i σi /N ,

given by

N N X J X E (M, h) = − σi σj − h σi 2N i,j=1 i=1  ! N !

N N X X X J = −N σi /N  σj /N  − N h σi /N 2 i=1 j=1 i=1

J 2 m + hm . = −N 2

4

P (b) Show that the partition function Z(h, T ) = {σi } exp(−βH) can be re-written as P Z = M exp[−βF (m, h)]; with F (m, h) easily calculated by analogy to problem (1). For

the remainder of the problem work only with F (m, h) expanded to 4th order in m.

• Since the energy depends only on the number of up spins N+ , and not on their configuration, we have

Z (h, T ) =

X

exp (−βH)

{σi }

=

N X

N+ =0

=

N X

N+ =0

=

N X

N+ =0

=

(number of configurations with N+ fixed) · exp [−βE (M, h)]

X

N! exp [−βE (M, h)] N+ ! (N − N+ )!

exp −β E (M, h) − kB T ln

N! N+ ! (N − N+ )!

exp [−βF (m, h)] .

M

By analogy to the previous problem (N+ ↔ NA , m ↔ x, J/2 ↔ 3J),

F (m, h) 1 kB T 4 = −kB T ln 2 − hm + (kB T − J) m2 + m + O m5 .

N 2 12 (c) By saddle point integration show that the actual free energy F (h, T ) = −kT ln Z(h, T )

is given by F (h, T ) = min[F (m, h)]m . When is the saddle point method valid? Note that F (m, h) is an analytic function but not convex for T < Tc , while the true free energy F (h, T ) is convex but becomes non-analytic due to the minimization.

• Let m∗ (h, T ) minimize F (m, h), i.e. min [F (m, h)]m = F (m∗ , h). Since there are N terms in the sum for Z, we have the bounds

exp (−βF (m∗ , h)) ≤ Z ≤ N exp (−βF (m∗ , h)) , or, taking the logarithm and dividing by −βN , F (h, T ) F (m∗ , h) ln N F (m∗ , h) ≥ ≥ + .

N N N N 5

Since F is extensive, we have therefore F (h, T ) F (m∗ , h) = N N in the N → ∞ limit. (d) For h = 0 find the critical temperature Tc below which spontaneous magnetization appears; and calculate the magnetization m(T ) in the low temperature phase.

• From the definition of the actual free energy, the magnetization is given by m=− i.e.

1 1 dF (m, h) =− m=− N dh N

1 ∂F (h, T ) , N ∂h

∂F (m, h) ∂F (m, h) ∂m + ∂h ∂m ∂h

Thus, if m∗ minimizes F (m, h), i.e. if ∂F (m, h)/∂m|m∗ = 0, then m=−

.

1 ∂F (m, h) = m∗ .

N ∂h m∗

For h = 0, m∗2 =

3 (J − kB T ) , kB T

yielding Tc = and

J , kB

 s  3 (J − kB T )  ± m= kB T   0

if T < Tc

.

if T > Tc

(e) Calculate the singular (non-analytic) behavior of the response functions C=

∂m ∂E , and χ = . ∂T h=0 ∂h h=0

• The hear capacity is given by C=

∗2

∂E N J ∂m =− ∂T h=0,m=m∗ 2 ∂T 6

  3N JTc 2T 2 =  0

if T < Tc if T > Tc

,

i.e. α = 0, indicating a discontinuity. To calculate the susceptibility, we use h = (kB T − J) m +

kB T 3 m .

3

Taking a derivative with respect to h, 1 = kB T − J + kB T m2 which gives

 1    2kB (Tc − T ) ∂m = χ= 1 ∂h h=0    kB (T − Tc )

∂m , ∂h if T < Tc . if T > Tc

From the above expression we obtain γ± = 1, and A+ /A− = 2. ******** 3. The lattice–gas model: Consider a gas of particles subject to a Hamiltonian H=

N X p~i 2 i=1

2m

+

1X V(~ri − ~rj ), in a volume V. 2 i,j

(a) Show that the grand partition function Ξ can be written as  

Z N ∞ N βµ X 1 e X Y 3 − β V(~ri − ~rj ) . d ~ r exp Ξ= i N ! λ3 2 i,j i=1 N=0

• The grand partition function is calculated as Ξ=

∞ X eNβµ

N=0

=

N!

ZN

Z Y ∞ N X d3 pi d3 ri eNβµ

N=0

N!

i=1

h3

e−βH

  !Z N N Z 3 Y X Y d pi −βp2i /2m β Vij  d3 ri exp − e = 3 N! h 2 i,j i=1 i=1 N=0  

N Z Y ∞ N X X 1 eNβ 3 − β Vij  , d r exp = i N ! λ3 2 i,j i=1 ∞ X eNβµ

N=0

7

where λ−1 =

√

2πmkB T /h.

(b) The volume V is now subdivided into N = V /a3 cells of volume a3 , with the spacing a

chosen small enough so that each cell α is either empty or occupied by one particle; i.e. the

cell occupation number nα is restricted to 0 or 1 (α = 1, 2, · · · , N ). After approximating R PN the integrals d3~r by sums a3 α=1 , show that Ξ≈

X

{nα =0,1}

βµ 3 e a λ3

P

α

nα

 N X β nα nβ V(~rα − ~rβ ) . exp − 2 

α,β=1

• Since Z Y N

i=1



d3 ri exp −

βX 2 i,j



Vij  ≈ a3N

′ X

  N  β X  exp − nα nβ V (~rα − ~rβ ) · N !,  2  α,β=1

where the primed sum is over the configurations {nα = 0, 1} with fixed N , and N=

N X

nα ,

α=1

we have X

Ξ≈

{nα =0,1}

  P βµ 3 nα N   X α β e a n n V (~ r − ~ r ) . exp − α β α β   2 λ3 α,β=1

(c) By setting nα = (1 + σα )/2 and approximating the potential by V(~rα − ~rβ ) = −J/N ,

show that this model is identical to the one studied in problem (2). What does this imply about the behavior of this imperfect gas?

• With nα = (1 + σα ) /2, and V (~rα − ~rβ ) = −J/N , Ξ=

X

exp

{nα =0,1}

Setting m ≡

 

βµ + 3 ln



N a X 1 + σ α

λ

α=1

2



 N X βJ 1 + σα 1 + σβ + .  2N 2 2 α,β=1

3 a J 1 ′ ′ µ + ln + σ /N , h = α α 2 β λ 2 , and J = J/4, the grand partition

P

function is written

Ξ = const.

X

{nα =0,1}

exp N β J ′ m2 /2 + h′ m .

8

The phase diagram of the lattice-gas can thus be mapped onto the phase diagram of the Ising model of problem 2. In particular, at a chemical potential µ such that h′ = 0, there is a continuous “condensation” transition at a critical temperature Tc = J/4kB . (Note that m=

X

σα /N =

α

X α

(2nα − 1) /N = 2a3 ρ − 1,

where ρ = N/V is the density of the gas.)

• The manifest equivalence between these three systems is a straightforward consequence

of their mapping onto the same (Ising) Hamiltonian. However, there is a more subtle equivalence relating the critical behavior of systems that cannot be so easily mapped onto each other due to the Universality Principle. ******** 4. Surfactant condensation:

N surfactant molecules are added to the surface of water

over an area A. They are subject to a Hamiltonian H=

N X p~i 2 i=1

2m

+

1X V(~ri − ~rj ), 2 i,j

where ~ri and p~i are two dimensional vectors indicating the position and momentum of particle i. (a) Write down the expression for the partition function Z(N, T, A) in terms of integrals over ~ri and p~i , and perform the integrals over the momenta.

• The partition function is obtained by integrating the Boltzmann weight over phase space, as

Z(N, T, A) =



~i d ~qi i=1 d p exp −β 2N N !h

N X p2

Z Y N



Z QN

2

2

i

i=1

2m

−β

X i<j



V(~qi − ~qj ) ,

with β = 1/(kB T ). The integrals over momenta are simple Gaussians, yielding

Z(N, T, A) =

1 1 N ! λ2N

i=1

d2 ~qi exp −β

X i<j



V(~qi − ~qj ) ,

√ where as usual λ = h/ 2πmkB T denotes the thermal wavelength.

The inter–particle potential V(~r) is infinite for separations |~r | < a, and attractive for R∞ |~r | > a such that a 2πrdrV(r) = −u0 .

9

(b) Estimate the total non–excluded area available in the positional phase space of the system of N particles.

• To estimate the joint phase space of particles with excluded areas, add them to the

system one by one. The first one can occupy the whole area A, while the second can explore only A − 2Ω, where Ω = πa2 . Neglecting three body effects (i.e. in the dilute limit), the area available to the third particle is (A − 2Ω), and similarly (A − nΩ) for the

n-th particle. Hence the joint excluded volume in this dilute limit is

A(A − Ω)(A − 2Ω) · · · (A − (N − 1)Ω) ≈ (A − N Ω/2)N , where the last approximation is obtained by pairing terms m and (N − m), and ignoring

order of Ω2 contributions to their product.

(c) Estimate the total potential energy of the system, assuming a uniform density n = N/A. Using this potential energy for all configurations allowed in the previous part, write down an approximation for Z. • Assuming a uniform density n = N/A, an average attractive potential energy, Ū , is estimated as

Z 1 1X V attr. (~qi − ~qj ) = d2~r1 d2~r2 n(~r1 )n(~r2 )V attr. (~r1 − ~r2 ) Ū = 2 i,j 2 Z n2 N2 ≈ A d2~r V attr. (~r ) ≡ − u0 .

2 2A Combining the previous results gives

βu0 N 2 1 1 N .

(A − N Ω/2) exp Z(N, T, A) ≈ N ! λ2N 2A

(d) The surface tension of water without surfactants is σ0 , approximately independent of temperature. Calculate the surface tension σ(n, T ) in the presence of surfactants.

• Since the work done is changing the surface area is dW = σdA, we have dF = −T dS +

σdA + µdN , where F = −kB T ln Z is the free energy. Hence, the contribution of the surfactants to the surface tension of the film is σs = −

N kB T u0 N 2 ∂ ln Z =− + , ∂A T,N A − N Ω/2 2A2 10

which is a two-dimensional variant of the familiar van der Waals equation. Adding the (constant) contribution in the absence of surfactants gives N kB T ∂ ln Z u0 N 2 =− + . ∂A T,N A − N Ω/2 2A2

σ(n, T ) = σ0 −

(e) Show that below a certain temperature, Tc , the expression for σ is manifestly incorrect.

What do you think happens at low temperatures?

• Thermodynamic stability requires δσδA ≥ 0, i.e. σ must be a monotonically increasing

function of A at any temperature. This is the case at high temperatures where the first term in the equation for σs dominates, but breaks down at low temperatures when the term from the attractive interactions becomes significant. The critical temperature is obtained by the usual conditions of ∂σs /∂A = ∂ 2 σs /∂A2 = 0, i.e. from  N kB T u0 N 2 ∂σs   = − =0   ∂A (A − N Ω/2)2 A3 T , 2 2  ∂ σ 2N k T 3u N s B 0   =− + =0  ∂A2 T (A − N Ω/2)3 A4

The two equations are simultaneously satisfied for Ac = 3N Ω/2, at a temperature Tc =

8u0 .

27kB Ω

As in the van der Waals gas, at temperatures below Tc , the surfactants separate into a high density (liquid) and a low density (gas) phase. (f) Compute the heat capacities, CA and write down an expression for Cσ without explicit evaluation, due to the surfactants.

• The contribution of the surfactants to the energy of the film is given by Es = −

∂ ln Z kB T u0 N 2 = 2N × − . ∂β 2 2A

The first term is due to the kinetic energy of the surfactants, while the second arises from their (mean-field) attraction. The heat capacities are then calculated as CA =

∂E dQ = = N kB , dT A ∂T A 11

and Cσ =

∂E ∂A dQ = −σ . dT σ ∂T σ ∂T σ ********

5. Critical behavior of a gas: The pressure P of a gas is related to its density n = N/V , and temperature T by the truncated expansion b c P = kB T n − n 2 + n 3 2 6

,

where b and c are assumed to be positive, temperature independent constants. (a) Locate the critical temperature Tc below which this equation must be invalid, and the corresponding density nc and pressure Pc of the critical point. Hence find the ratio kB Tc nc /Pc .

• Mechanical stability of the gas requires that any spontaneous change in volume should be opposed by a compensating change in pressure. This corresponds to δP δV < 0, and since

δn = −(N/V 2 )δV , any equation of state must have a pressure that is an increasing func-

tion of density. The transition point between pressure isotherms that are monotonically increasing functions of n, and those that are not (hence manifestly incorrect) is obtained by the usual conditions of dP/dn = 0 and d2 P/dn2 = 0. Starting from the cubic equation of state, we thus obtain

dP c =kB Tc − bnc + n2c = 0 dn 2 . d2 P = − b + cnc = 0 dn2 From the second equation we obtain nc = b/c, which substituted in the first equation gives kB Tc = b2 /(2c). From the equation of state we then find Pc = b3 /(6c2 ), and the dimensionless ratio of

kB Tc nc = 3.

Pc

(b) Calculate the isothermal compressibility κT = − V1 ∂V ∂P T , and sketch its behavior as a

function of T for n = nc .

• Using V = N/n, we get −1

−1 1 ∂P 1 ∂V = = n kB T − bn + cn2 /2 . κT (n) = − V ∂P T n ∂n T 12

For n = nc , κT (nc ) ∝ (T − Tc )−1 , and diverges at Tc . (c) On the critical isotherm give an expression for (P − Pc ) as a function of (n − nc ).

• Using the coordinates of the critical point computed above, we find b3 b2 b c + n − n2 + n3 2 6c 2c 2 6

2 b 2 b b3 c 3 n − 3 n + 3 2n − 3 = 6 c c c c 3 = (n − nc ) .

6

P − Pc = −

(d) The instability in the isotherms for T < Tc is avoided by phase separation into a liquid of density n+ and gas of density n− . For temperatures close to Tc , these densities behave

as n± ≈ nc (1 ± δ). Using a Maxwell construction, or otherwise, find an implicit equation

for δ(T ), and indicate its behavior for (Tc − T ) → 0. (Hint: Along an isotherm, variations of chemical potential obey dµ = dP/n.)

• According to the Gibbs–Duhem relation, the variations of the intensive variables are

related by SdT − V dP + N dµ = 0, and thus along an isotherm (dT = 0) dµ = dP/n =

∂P/∂n|T dn/n. Since the liquid and gas states are in coexistence they should have the same chemical potential. Integrating the above expression for dµ from n− to n+ leads to the so-called Maxwell construction, which reads 0 = µ(n+ ) − µ(n− ) =

Z n+ n−

dP = n

Z nc (1+δ) nc (1−δ)

dn

kB T − bn + cn2 /2 n

.

Performing the integrals gives the equation

c 2 1+δ 1+δ 2 2 0 = kB T ln − bnc (2δ) + nc (1 + δ) − (1 − δ) = kB T ln − 2kB Tc δ, 1−δ 4 1−δ

where for the find expression, we have used nc = b/c and kB Tc = b2 /(2c). The implicit equation for δ is thus T δ= ln 2Tc

1+δ 1−δ

≈

T δ − δ3 + · · · .

Tc

The leading behavior as (Tc − T ) → 0 is obtained by keeping up to the cubic term, and

given by

δ≈

r

1−

13

Tc .

T

(e) Now consider a gas obeying Dieterici’s equation of state:

a , P (v − b) = kB T exp − kB T v where v = V /N . Find the ratio P v/kB T at its critical point.

• The critical point is the point of inflection, described by ∂P = 0, ∂v Tc ,N

∂ 2P = 0. ∂v 2 Tc ,N

pressure

and

P

liq.

1

Q hot

2 T

gas

V

P-dP 4 Q cold

T-dT

3

volume

The first derivative of P is

∂ kB T kB T a a a 1 ∂P = = exp − exp − − ∂v Tc ,N ∂v v − b kB T v v−b kB T v kB T v 2 v−b

a 1 =P , − 2 kB T v v−b while a second derivative gives

∂ a 1 ∂ 2P = P − ∂v 2 Tc ,N ∂v kB T v 2 v−b

∂P a 2a 1 1 = −P − − . ∂v kB T v 2 v−b kB T v 3 (v − b)2 Therefore vc and Tc are determined by 1 a − = 0, 2 kB Tc vc vc − b

and 14

1 2a − = 0, 3 kB Tc vc (vc − b)2

with the solutions vc = 2b,

and

kB Tc =

a .

4b

The critical pressure is

kB Tc a a Pc = = 2 e−2 , exp − vc − b kB Tc vc 4b resulting in the ratio

Pc vc = 2e−2 ≈ 0.27. kB Tc

Note that for the van der Waals gas 3 Pc vc = = 0.375, kB Tc 8 while for some actual gases

Pc vc = 0.230, kB Tc water

and

Pc vc kB Tc

= 0.291.

Argon

(f) Calculate the isothermal compressibility κT for v = vc as a function of T − Tc for the

Dieterici gas.

• The isothermal compressibility is defined by κT ≡ −

1 ∂v , v ∂P T,N

and from part (a), given by ∂P =P ∂v Tc ,N

a 1 − 2 kB T v v−b

.

Expanding this expression, at v = vc , in terms of t ≡ kB T − kB Tc (for T > Tc ), yields

Pc 4bt 1 2Pc a ∂P ≈− − =− t, ≈ Pc 2 ∂v Tc ,N (a/4b + t) 4b b b a vc kB Tc and thus κT =

kB Tc 1 be2 = .

2Pc t 2kB (T − Tc )

Note that expanding any analytic equation of state will yield the same simple pole for the divergence of the compressibility.

15

(g) On the Dieterici critical isotherm expand the pressure to the lowest non-zero order in (v − vc ).

• Perform a Taylor–series expansion along the critical isotherm T = Tc , as P (v, Tc ) = Pc +

1 ∂ 2P ∂P 1 ∂ 3P 2 (v − vc ) + (v − v ) + (v − vc )3 + · · · . c ∂v Tc ,vc 2! ∂v 2 Tc ,vc 3! ∂v 3 Tc ,vc

The first two terms are zero at the critical point, and

∂ 3P ∂ 2a 1 = −Pc − ∂v 3 Tc ,vc ∂v kB Tc v 3 (v − b)2

2 6a − = −Pc kB Tc vc4 (vc − b)3 Pc = − 3.

2b Substituting this into the Taylor expansion for P (v, Tc ), results in

(v − vc )3 , P (v, Tc ) = Pc 1 − 12b3 which is equivalent to P 2 −1= Pc 3

v −1 vc

3

.

******** 6. Magnetic thin films:

A crystalline film (simple cubic) is obtained by depositing a

finite number of layers n. Each atom has a three component (Heisenberg) spin, and they interact through the Hamiltonian −βH =

n X X

α=1 <i,j>

JH ~siα · ~sjα +

n−1 XX

α=1

i

JV ~siα · ~siα+1 .

(The unit vector ~siα indicates the spin at site i in the αth layer. The subscript < i, j > indicates that the spin at i interacts with its 4 nearest-neighbors, indexed by j on the square lattice on the same layer.) A mean–field approximation is obtained from the variational density ρ0 ∝ exp (−βH0 ), with the trial Hamiltonian −βH0 =

n X X

α=1

16

i

~h α · ~s α . i

(Note that the most general single–site Hamiltonian may include the higher order terms α α Lα s.) c1 ,···,cp sc1 · · · sc1 , where sc indicates component c of the vector ~ n o (a) Calculate the partition function Z0 ~h α , and βF0 = − ln Z0 .

• The partition function Z0 is obtained by integrating over all angles as Z0 =

Y Z

~α α d~siα eh ·~si

i,α

=

Y

2π

i,α

=

Z 1

hα x

dxe

−1

4π sinh hα hα

N

=

Y Z 2π

=

dφ

Z π

hα cos θ

d cos θe

i,α

0

Y

ehα − e−hα 2π hα

i,α

0

,

where N is the number of sites in each layer. (b) Obtain the magnetizations mα = h~siα i0 , and hβH0 i0 , in terms of the Langevin function L(h) = coth(h) − 1/h.

• From the partition function, we easily obtain the magnetizations as m ~ α = h~siα i0 =

∂ sinh hα 1 ∂ ln Z0 = ĥ α ln = ĥ α [coth hα − 1/hα ] = ĥ α L(hα ). α ~ N ∂h ∂hα hα

(c) Calculate hβHi0 , with the (reasonable) assumption that all the variational fields n o ~h α are parallel.

• For the zeroth order weight the spins at different sites are independent random variables, and

hβHi0 = JH

n X X

α=1 <i,j>

h~siα i0 ·

~sjα 0 + JV

n−1 XX

α=1

i

h~siα i0 · ~siα+1 0 .

Noting that h~siα i0 = mα ĥ with all spins on average pointing in the same direction, and

that each spin has 4 neighbors in the plane on a square lattice, we obtain n n−1 X 4N X hβHi0 = JH mα mα + JV N mα mα+1 2 α=1 α=1

=N

n X

α=1

2JH m2α + JV mα mα+1 ,

with mn+1 = 0.

17

(d) The exact free energy, βF = − ln Z, satisfies the Gibbs inequality (see below), βF ≤

βF0 + hβH − βH0 i0 . Give the self-consistent equations for the magnetizations {mα } that

optimize βH0 . How would you solve these equations numerically?

• According to the Gibbs inequality βF ≤ N Ψ({mα }), with Ψ({mα }) =

X α

− ln

4π sinh hα hα

+ hα mα − 2JH m2α − JV mα mα+1

,

with mα = L(hα ). The best variational choice for {hα } is obtained by minimizing Ψ, and

thus

dhα dhα ∂Ψ = −L(hα ) + mα + hα − 4JH mα − JV (mα+1 + mα−1 ) = 0. ∂mα dmα dmα After canceling the first two terms, the above equation can be re-written as hα = L−1 (mα ) = 4JH mα + JV (mα+1 + mα−1 ) , which is the effective field acting on a site in layer α. (e) Find the critical temperature, and the behavior of the magnetization in the bulk by considering the limit n → ∞. (Note that limm→0 L−1 (m) = 3m + 9m3 /5 + O(m5 ).)

• When examining the bulk, we can drop the index α, and using the expansion of L−1 (m)

we obtain

9 3m + m3 + · · · = (4JH + JV )m.

5 The condition for criticality is the equality of the linear terms, i.e. 4JH + 2JV = 3. Setting JH = JˆH /kB T , and JV = JˆV /kB T , this gives kB Tc = (4JˆH + 2JˆV )/3. For T < Tc , we have 9 3 m =3 5

Tc −1 T

≡ 3t,

=⇒

p m = ± 5t/3,

i.e. the usual saddle-point exponent of β = 1/2 is recovered.

(f) By linearizing the self-consistent equations, show that the critical temperature of film depends on the number of layers n, as kTc (n ≫ 1) ≈ kTc (∞) − JV π 2 /(3n2 ).

• For a finite number of layers, the linearized recursion relations are

L−1 (mα ) = 3mα + O(m3 ) = 4JH mα + JV (mα+1 + mα−1 ) .

18

These equations have to be supplemented by boundary conditions m0 = mn+1 = 0. The solution to the linearized equations takes the form

απ , mα = µ sin n+1 which gives the profile of magnetization at the critical point. It can be checked that this is a solution to the linearized equations, by noting that

απ απ (α + 1)π (α − 1)π 3µ sin + · · · = 4JH µ sin + JV µ sin + sin n+1 n+1 n+1 n+1

απ π απ + JV µ 2 sin cos . = 4JH µ sin n+1 n+1 n+1 Thus the linear terms match is 4JH + 2JV cos

π n+1

= 3,

giving the critical temperature of the finite system as

π π 2 Jˆv ˆ kB Tc (n) = JH + 2JV cos /3 ≈ kB Tc (∞) − + ···, n+1 3n2 where the last expression is obtained by expanding the cosine for large n. (g) Derive a continuum form of the self-consistent equations, and keep terms to cubic order in m. Show that the resulting non-linear differential equation has a solution of the form m(x) = mbulk tanh(kx). What circumstances are described by this solution?

• The continuum limit is obtained by replacing the discrete layer number α with a continuous coordinate z, such that mα → m(x), and

(mα+1 + mα−1 ) ≈ 2m(x) +

d2 m + ···. dx2

In this limit the discrete set of equations are replaced with the differential equation (4JH + 2JV ) m(x) + JV

d2 m 9 + · · · = 3m + m3 + · · · .

2 dx 5

To get the profile of a large system close to criticality the leading terms included above are sufficient. It is in general not possible to give closed form solutions to non-linear equations.

However, it is easy to check that m(x) = mbulk tanh(kx) satisfies this equation, using

dm = mbulk k 1 − tanh2 kx , dx

and

d2 m 2 2 = −2m k tanh(kx) 1 − tanh kx . bulk dx2 19

Substituting in the differential equation gives

(4JH + 2JV ) mbulk tanh(kx) − 2mbulk k 2 JV tanh kx − tanh3 kx 9 = 3mbulk tanh(kx) + m3bulk tanh3 kx.

5 Since the equality must hold irrespective of x, we must equate the coefficients of tanh kx and tanh3 kx at each point. From the terms of order tanh kx, we get 3 − (4JH + 2JV ) (4JH + 2JV ) − 2k 2 JV = 0, =⇒ k 2 = , 2JV while at order of tanh3 kx, we get 10 2 5 m2bulk = k JV = [3 − (4JH + 2JV )] .

9 9 The latter agrees with the bulk magnetization obtained earlier, validating the consistency of this solution. This profile corresponds to a domain wall in the system, located at x = 0, separating a down magnetized phase at x → −∞, and an up magnetized phase at x → +∞.

(h) How can the above solution be modified to describe a semi–infinite system? Obtain the critical behaviors of the healing length λ ∼ 1/k.

• As a first approximation to a semi–infinite system, we can take one half of the above solution, say for x > 0. The magnetization is small (almost zero) at the surface, and recovers to the bulk value over a distance λ ∼ 1/k ∝ t−1/2 . Note that the divergence of the healing length is precisely the same as the bulk correlation length. (i) Show that the magnetization of the surface layer vanishes as |T − Tc |.

• A more accurate answer is obtained by noting that in the discrete system, the coupling to neighboring layers comes from mα−1 + mα+1 . For the equation at the surface layer, we have m−1 + m1 , and we must set m−1 = 0. Thus a more accurate solution is actually shifted by one lattice spacing, as m(x ≥ 0) = mbulk tanh[k(x + a)] for a lattice spacing a.

Expanding the tanh gives m0 = msurface ≈ mbulk ka ∝ t.

The result in (f) illustrates a quite general result that the transition temperature of a finite system of size L, approaches its asymptotic (infinite–size) limit from below, as Tc (L) = TC (∞) − A/L1/ν , where ν is the exponent controlling the divergence of the correlation length. However, some liquid crystal films appeared to violate this behavior.

In fact, in these films the couplings are stronger on the surface layers, which thus order before the bulk. For a discussion of the dependence of Tc on the number of layers in this case, see H. Li, M. Paczuski, M. Kardar, and K. Huang, Phys. Rev. B 44, 8274 (1991). ******** 20

Solutions to problems from chapter 2- Statistical Fields 1. Cubic invariants:

When the order parameter m, goes to zero discontinuously, the

phase transition is said to be first order (discontinuous). A common example occurs in systems where symmetry considerations do not exclude a cubic term in the Landau free energy, as in βH =

Z

K t d x (∇m)2 + m2 + cm3 + um4 2 2 d

(K, c, u > 0).

(a) By plotting the energy density Ψ(m), for uniform m at various values of t, show that as t is reduced there is a discontinuous jump to m 6= 0 for a positive t in the saddle–point approximation.

• To simplify the algebra, let us rewrite the energy density Ψ(m), for uniform m, in terms

of the rescaled quantity

mr =

u m. c

In this way, we can eliminate the constant parameters c, and u, to get the expression of the energy density as Ψr (mr ) =

1 tr m2r + m3r + m4r , 2

where we have defined 4 c Ψ, Ψr = u3

and

tr =

u c2

t.

To obtain the extrema of Ψr , we set the first derivative with respect to mr to zero, i.e.

dΨr (mr ) = mr tr + 3mr + 4m2r = 0. dmr

The trivial solution of this equation is m∗r = 0. But if tr ≤ 9/16, the derivative vanishes √ also at m∗r = (−3 ± 9 − 16tr )/8. Provided that tr > 0, m∗r = 0 is a minimum of the function Ψr (mr ). In addition, if tr < 9/16, Ψr (mr ) has another minimum at √ 3 + 9 − 16tr ∗ , mr = − 8

and a maximum, located in between the two minima, at √ −3 + 9 − 16tr ∗ . mr = 8 21

Fr

2

1

1

2

3

4

3 4

mr

The accompanying figure depicts the behavior of Ψr (mr ) for different values of tr .

1. For tr > 9/16, there is only one minimum m∗r = 0 .

2. For 0 < tr < tr < 9/16, there are two minima, but Ψr (m∗r ) > Ψr (0) = 0.

3. For 0 < tr = tr , Ψr (m∗r ) = Ψr (0) = 0.

4. For 0 < tr < tr , Ψr (m∗r ) < Ψr (0) = 0.

The discontinuous transition occurs when the local minimum at m∗r < 0 becomes the absolute minimum. There is a corresponding jump of mr , from m∗r = 0 to m∗r = mr , where mr = m∗r (tr = tr ). (b) By writing down the two conditions that m and t must satisfy at the transition, solve for m and t.

• To determine mr and tr , we have to simultaneously solve the equations dΨr (mr ) = 0, dmr

and

Ψr (mr ) = Ψr (0) = 0.

Excluding the trivial solution m∗r = 0, from   tr + 3mr + 4m2r = 0 ,  tr + mr + m2r = 0 2

we obtain tr = −mr = 1/2, or in the original units, t=

c2 , 2u

and 22

m=−

c .

2u

ξ

ξ max

t

t

(c) Recall that the correlation length ξ is related to the curvature of Ψ(m) at its minimum

by Kξ −2 = ∂ 2 Ψ/∂m2 |eq. . Plot ξ as a function of t.

•

Likewise, the equilibrium value of m = meq in the original units equals to

meq =

   

0 c 3 + p9 − 16ut/c2   − u 8

for

c2 , t>t= 2u

for

t < t.

The correlation length ξ, is related to the curvature of Ψ(m) at its equilibrium minimum by Kξ −2 =

∂ 2Ψ = t + 6cmeq + 12um2eq , ∂m2 meq

which is equal to  1/2 K     t ξ= 1/2   K   − 2t + 3cmeq

if

t > t,

if

t < t.

(To arrive to the last expression, we have useddΨ(m)/dm|m=meq = 0.) ξmax = ξ(t) = 23

√

2Ku . c

A plot of ξ as a function of t is presented here. Note that the correlation length ξ, is finite at the discontinuous phase transition, attaining a maximum value of ******** 2. Tricritical point: By tuning an additional parameter, a second order transition can be made first order. The special point separating the two types of transitions is known as a tricritical point, and can be studied by examining the Landau–Ginzburg Hamiltonian

Z K t 2 d 2 4 6 βH = d x (∇m) + m + um + vm − hm , 2 2 where u can be positive or negative. For u < 0, a positive v is necessary to ensure stability. (a) By sketching the energy density Ψ(m), for various t, show that in the saddle–point approximation there is a first-order transition for u < 0 and h = 0.

• If we consider h = 0, the energy density Ψ(m), for uniform m, is Ψ(m) =

t 2 m + um4 + vm6 .

2

As in the previous problem, to obtain the extrema of Ψ, let us set the first derivative with respect to m to zero. Again, provided that t > 0, Ψ(m) has a minimum at m∗ = 0. But the derivative also vanishes for other nonzero values of m as long as certain conditions are satisfied. In order to find them, we have to solve the following equation t + 4um2 + 6vm4 = 0, from which, ∗2

m

√

u =− ± 3v

4u2 − 6tv .

6v

Thus, we have real and positive solutions provided that u < 0,

and

2u2 t< .

3v

Under these conditions Ψ(m) has another two minima at ∗2

|u| + = 3v

∗2

|u| = − 3v

m and two maxima at m

√

√

24

4u2 − 6tv , 6v 4u2 − 6tv , 6v

F

2

1

1

2

3 3 4 4

m

as depicted in the accompanying figure.

The different behaviors of the function Ψ(m) are as follows: 1. For t > 2u2 /3v, there is only one minimum m∗ = 0.

2. For 0 < t < t < 2u2 /3v, there are three minima, but Ψ(±m∗ ) > Ψ(0) = 0.

3. For 0 < t = t, Ψ(±m∗ ) = Ψ(0) = 0.

4. For 0 < t < t, Ψ(±m∗ ) < Ψ(0) = 0.

There is a thus discontinuous phase transition for u < 0, and t = t(u). (b) Calculate t and the discontinuity m at this transition.

• To determine t, and m = m∗ (t = t), we again have to simultaneously solve the equations dΨ(m) = 0, dm2 or equivalently,

Ψ(m2 ) = Ψ(0) = 0,

and

 t   + 2um2 + 3vm4 = 0 2 ,   t + um2 + vm4 = 0 2

from which we obtain t=

u2 , 2v

and

m2 = − 25

u |u| = .

2v 2v

(c) For h = 0 and v > 0, plot the phase boundary in the (u, t) plane, identifying the phases, and order of the phase transitions.

• In the (u, t) plane, the line t = u2 /2v for u < 0, is a first-order phase transition boundary. In addition, the line t = 0 for u > 0, defines a second-order phase transition

boundary, as indicated in the accompanying figure.

t

first order boundary

disordered phase

ordered phase

u

second order boundary

tricritical point

(d) The special point u = t = 0, separating first– and second–order phase boundaries, is a tricritical point. For u = 0, calculate the tricritical exponents β, δ, γ, and α, governing the singularities in magnetization, susceptibility, and heat capacity. (Recall: C ∝ t−α ;

m(h = 0) ∝ tβ ; χ ∝ t−γ ; and m(t = 0) ∝ h1/δ .)

• For u = 0, let us calculate the tricritical exponents α, β, γ, and δ. In order to calculate α and β, we set h = 0, so that

Ψ(m) = Thus from

t 2 m + vm6 .

2

∂Ψ = m t + 6vm4 = 0, ∂m m 26

we obtain,

resulting in,

   0 m = t 1/4   − 6v

for

t > t = 0,

for

t<0

m(h = 0) ∝ tβ ,

with

β=

,

1 .

4

The corresponding free energy density scales as Ψ(m) ∼ m 6 ∝ (−t)3/2 .

The tricritical exponent α characterizes the non-analytic behavior of the heat capacity C ∼ (∂ 2 Ψ/∂T 2 )|h=0,m , and since t ∝ (T − Tc ), C∼

∂ 2Ψ ∝ t−α , ∂t2 h=0,m

with

α=

1 .

2

To calculate the tricritical exponent δ, we set t = 0 while keeping h 6= 0, so that Ψ(m) = vm6 − hm.

Thus from

∂Ψ = 6vm5 − h = 0, ∂m m

we obtain, m ∝ h1/δ ,

with

δ = 5.

Finally, for h 6= 0 and t 6= 0, ∂Ψ = tm + 6vm5 − h = 0, ∂m m so that the susceptibility scales as χ=

∂m ∝ |t|−1 , ∂h h=0

i.e. with the exponents γ± = 1.

for both

t<0

and

t > 0,

********

3. Transverse susceptibility: An n–component magnetization field m(x) ~ is coupled to an R d external field ~h through a term − d x ~h · m(x) ~ in the Hamiltonian βH. If βH for ~h = 0 27

is invariant under rotations of m(x); ~ then the free energy density (f = − ln Z/V ) only depends on the absolute value of ~h; i.e. f (~h) = f (h), where h = |~h|.

R (a) Show that mα = h dd xmα (x)i/V = −hα f ′ (h)/h.

• The magnetic work is the product of the magnetic field and the magnetization density,

and appears as the argument of the exponential weight in the (Gibbs) canonical ensemble.

R We can thus can “lower” the magnetization M = dd xmα (x) “inside the average” by

taking derivatives of the (Gibbs) partition function with respect to hα , as R R d ′ −βH

Z ′ Dm (x) d x m (x ) e 1 1 α R dd xmα (x) = mα = −βH V V Dm (x) e Z 1 1 1 ∂ 1 ∂ ∂f = Dm (x) e−βH = ln Z = − .

V Z β ∂hα βV ∂hα ∂hα

For an otherwise rotationally symmetric system, the (Gibbs) free energy depends only on the magnitude of h, and using

we obtain

p ∂ hβ hβ hα 1 2δαβ hβ ∂h = = = p , ∂hα ∂hα 2 hβ hβ h mα = −

df ∂h hα ∂f =− = −f ′ . ∂hα dh ∂hα h

(b) Relate the susceptibility tensor χαβ = ∂mα /∂hβ , to f ′′ (h), m, ~ and ~h.

• The susceptibility tensor is now obtained as

∂mα ∂ hα ′ hα ∂f ′ ∂hα 1 ′ ∂h−1 χαβ = = − f (h) = − hα f ′ − f − ∂hβ ∂hβ h ∂hβ h ∂hβ h ∂hβ ′

hα hβ ′′ hα hβ f − f . = − δαβ − 2 h h h2 In order to express f ′ in terms of the magnetization, we take the magnitude of the result of part (a), m = |f ′ (h)| = −f ′ (h) , from which we obtain

hα hβ m hα hβ dm χαβ = δαβ − + . h2 h h2 dh (c) Show that the transverse and longitudinal susceptibilities are given by χt = m/h and χℓ = −f ′′ (h); where m is the magnitude of m. ~ 28

• Since the matrix δαβ − hα hβ /h2 removes the projection of any vector along the magnetic field, we conclude

 dm   χℓ = −f ′′ (h) = dh . m   χt = h Alternatively, we can choose the coordinate system such that hi = hδi1 (i = 1, . . . , d), to

get

 dm h1 h1 m h1 h1 ′′   − 2 f (h) =  χℓ = χ11 = δ11 − h2 h h dh

.  h h m h2 h2 ′′ m   χt = χ22 = δ11 − 2 2 − 2 f (h) = h2 h h h

(d) Conclude that χt diverges as ~h → 0, whenever there is a spontaneous magnetization.

Is there any similar a priori reason for χℓ to diverge?

• Provided that limh→0 m 6= 0, the transverse susceptibility clearly diverges for h → 0.

There is no similar reason, on the other hand, for the longitudinal susceptibility to diverge.

In the saddle point approximation of the Landau–Ginzburg model, for example, we have tm + 4um3 + h = 0, implying (since 4um2 = −t at h = 0, for t < 0) that χℓ |h=0 =

dh dm

−1

h=0

= (t − 3t)

−1

,

i .e.

χℓ =

1 , 2 |t|

at zero magnetic field, in the ordered phase (t < 0).

NOTE: Another, more pictorial approach to this problem is as follows. Since the Hamiltonian is invariant under rotations about h, m must be parallel to h, i.e. mα =

hα ϕ (h) , h

where ϕ is some function of the magnitude of the magnetic field. For simplicity, let h = he1 , with e1 a unit vector, implying that m = me1 = ϕ (h) e1 .

The longitudinal susceptibility is then calculated as χℓ =

dm ∂m1 = = ϕ′ (h) . ∂h1 h=he1 dh 29

To find the transverse susceptibility, we first note that if the system is perturbed by a small external magnetic field δhe2 , the change in m1 is, by symmetry, the same for δh > 0 and δh < 0, implying

m1 (he1 + δhe2 ) = m1 (he1 ) + O δh2 .

Hence

∂m1 = 0. ∂h2 h=he1

Furthermore, since m and h are parallel, m1 (he1 + δhe2 ) m2 (he1 + δhe2 ) = , h δh from which m2 (he1 + δhe2 ) = yielding χt =

m1 (he1 ) δh + O δh3 , h

m ∂m2 = . ∂h2 h=he1 h ********

4. Superfluid He4 –He3 mixtures:

The superfluid He4 order parameter is a complex

number ψ(x) . In the presence of a concentration c(x) of He3 impurities, the system has the following Landau–Ginzburg energy βH[ψ, c] =

Z

K t c(x)2 2 2 4 6 2 d x |∇ψ| + |ψ| + u |ψ| + v |ψ| + − γc(x)|ψ| , 2 2 2σ 2 d

with positive K, u and v. (a) Integrate out the He3 concentrations to find the effective Hamiltonian, βHeff [ψ], for the superfluid order parameter, given by Z Z Z = Dψ exp (−βHeff [ψ]) ≡ DψDc exp (−βH[ψ, c]) .

• The integration over c(x) is merely a Gaussian integral, which at each point gives Z

c(x)2 dc(x) exp − + γc(x)|ψ|2 2 2σ

30

1 2 2 4 ∝ exp + γ σ |ψ| .

2

This modifies the quartic term in the Hamiltonian, resulting in −βHeff [ψ] =

Z

t γ 2σ2 K 4 6 2 2 |ψ| + v |ψ| . |∇ψ| + |ψ| + u − d x 2 2 2 d

(b) Obtain the phase diagram for βHeff [ψ] using a saddle point approximation. Find the limiting value of σ ∗ above which the phase transition becomes discontinuous.

• As long as the quartic term is positive, in the saddle point approximation there is a

continuous phase transition at t = 0, separating a disordered phase for t > 0, and a √ superfluid phase for t < 0. The fourth order term becomes negative for σ > σ ∗ = 2u/γ, and the phase transition is then discontinuous. (c) The discontinuous transition is accompanied by a jump in the magnitude of ψ. How does this jump vanish as σ → σ ∗ ?

• The saddle point solution minimizes the above energy, leading to tψ + 4ũψ 3 + 6vψ 5 = 0,

where ũ = u − γ 2 σ 2 /2 ∝ (σ ∗ − σ). At the first order boundary, the resulting energy must equal to that of the disordered phase with ψ = 0, resulting in a second equation t 2 ψ + ũψ 4 + vψ 6 = 0.

2 Eliminating t between the above two equations gives ψ2 = −

ũ ∝ (σ − σ ∗ ) , 2v

i.e. the discontinuity in ψ along the boundary vanishes as

√

σ − σ∗.

(d) Show that the discontinuous transition is accompanied by a jump in He3 concentration.

• From the initial Gaussian Hamiltonian for c(x), it is easy to see that hc(x)i = γσ 2 |ψ|2 .

Hence a discontinuity in ψ is necessarily accompanied by a discontinuity in the density c(x). (Note that the ensemble for He3 is grand canonical, thus allowing this change in density.) (e) Sketch the phase boundary in the (t, σ) coordinates, and indicate how its two segments join at σ ∗ .

31

t

Normal Fluid

s*

s

Superfluid

2

• Using the above value of Ψ in one of the equations determining the transition point,

one finds

2 σ2γ 2 γ4 2 2 4 σ − σ ∗2 , t = −4 u − Ψ − 6vΨ = 2 8v

i.e. the discontinuous line joins the continuous portion quadratically, as illustrated in the figure (f) Going back to the original joint probability for the fields c(x) and Ψ(x), show that hc(x) − γσ 2 |Ψ(x)|2 i = 0.

• By grouping the quadratic forms involving c(x), the joint probability can be written as −

e−β H = e

R

d

d x

2 2 t K 2 |∇Ψ| + 2 |Ψ| +

2 2 u− σ 2γ

(c(x)−γσ2 |Ψ(x)|2 ) |Ψ| +v|Ψ| +

4

6

2σ 2

2

.

We can change variables to c′ (x) = c(x) − γσ 2 |Ψ(x)|2 , whose average is clearly zero by symmetry, and hence

hc(x)i = γσ 2 h|Ψ(x)|2 i. (g) Show that hc(x)c(y)i = γ 2 σ 4 h|Ψ(x)|2 |Ψ(y)|2 i, for x 6= y.

• Since the field c′ (x) is not coupled to any other field at a different location in space, hc′ (x)c′ (y)i = 0, and hc′ (x)|Ψ(y)2 |i = 0. Hence, hc(x)c(y)i = h c′ (x) + γσ 2 |Ψ(x)|2

c′ (y) + γσ 2 |Ψ(y)|2 i = γ 2 σ 4 h|Ψ(x)|2 |Ψ(y)|2 i.

(h) Qualitatively discuss how hc(x)c(0)i decays with x = |x| in the disordered phase.

32

• In the disordered phase hΨ(x)Ψ(0)i decays exponentially, as e−x/ξ . From the result of

the previous part, we expect

2x . hc(x)c(0)i ∼ γ σ exp − ξ 2 4

(i) Qualitatively discuss how hc(x)c(0)i decays to its asymptotic value in the ordered phase.

• In the ordered phase, hΨ(x)Ψ(0)i ∼ 1/xd−2 due to the Goldstone mode. Hence in d = 3

dimensions,

hc(x)c(0)i ∼ 1/x2 . ******** 5. Crumpled surfaces: The configurations of a crumpled sheet of paper can be described by a vector field ~r(x), denoting the position in three dimensional space, ~r = (r1 , r2 , r3 ), of the point at location x = (x1 , x2 ) on the flat sheet. The energy of each configuration is assumed to be invariant under translations and rotations of the sheet of paper. (a) Show that the two lowest order (in derivatives) terms in the quadratic part of a Landau– Ginzburg Hamiltonian for this system are: βH0 [~r] =

X Z

α=1,2

•

t K 2 2 d x ∂α~r · ∂α~r + ∂α~r · ∂α~r .

2 2 2

Since the energy in the sheet is invariant under translations, it cannot depend on

the vector ~r, but only on its gradients such as ∂α~r, and ∂α ∂β ~r. To ensure rotational invariance, each component ri must be contracted (paired and summed over) with another ri . Similarly to ensure isotropy in x = (x1 , x2 ), each derivative ∂α must be paired with another one. Taking advantage of contractions is a good way to find invariants. Thus using the summation notation, and ensuring proper contractions, the lowest order terms in the effective Hamiltonian are

Z t K 2 βH0 [~r] = d x ∂α ri ∂α ri + ∂α ∂α ri ∂β ∂β ri .

2 2 (b) Write down the lowest order terms (there are two) that appear at the quartic level.

33

• The same procedure allows us to construct 4-th order terms in ~r: There should be two

ri , two rj and derivatives that are contracted. The two possible terms are ′

βH [~r] =

Z

d2 x [u∂α ri ∂α ri ∂β rj ∂β rj + v∂α ri ∂β ri ∂α rj ∂β rj ] .

(c) Discuss what happens when t changes sign, assuming that quartic terms provide the required stability (and K > 0).

• The above theory has the same structure as the Landau-Ginzburg Hamiltonian, which a more complicated order parameter (∂α ri in place of mi ). As in the Landau-Ginzburg

model, when t become negative, the order parameter is no longer zero, but the weight is maximized for a finite value of ∂α ri . This implies that the sheet of paper gets stretched along a specific direction. This is known as the crumpling (or uncrumpling) transition.

The sheet of paper is in a crumpled state for t > 0, while for t < 0, it opens up in some arbitrary direction. ********

34

Solutions to problems from chapter 3 - Fluctuations 1. Spin waves:

In the XY model of n = 2 magnetism, a unit vector ~s = (sx , sy ) (with

s2x + s2y = 1) is placed on each site of a d–dimensional lattice. There is an interaction that tends to keep nearest–neighbors parallel, i.e. a Hamiltonian X

−βH = K

<ij>

~si · ~sj

.

The notation < ij > is conventionally used to indicate summing over all nearest–neighbor pairs (i, j). (a) Rewrite the partition function Z =

RQ

si exp(−βH), as an integral over the set of i d~

angles {θi } between the spins {~si } and some arbitrary axis.

• The partition function is Z=

Z Y i



d2~si exp K

X hiji



~si · ~sj  δ ~si 2 − 1 .

Since ~si · ~sj = cos (θi − θj ), and d2~si = dsi dθi si = dθi , we obtain Z=

Z Y i



dθi exp K

X hiji



cos (θi − θj ) .

(b) At low temperatures (K ≫ 1), the angles {θi } vary slowly from site to site. In this case expand −βH to get a quadratic form in {θi }.

• Expanding the cosines to quadratic order gives Z = eNb K

Z Y i



dθi exp −

KX 2

hiji

2



(θi − θj )  ,

where Nb is the total number of bonds. Higher order terms in the expansion may be p neglected for large K, since the integral is dominated by |θi − θj | ≈ 2/K.

(c) For d = 1, consider L sites with periodic boundary conditions (i.e. forming a closed

chain). Find the normal modes θq that diagonalize the quadratic form (by Fourier transformation), and the corresponding eigenvalues K(q). Pay careful attention to whether the modes are real or complex, and to the allowed values of q.

35

• For a chain of L sites, we can change to Fourier modes by setting X

θj =

q

eiqj θ (q) √ .

L

Since θj are real numbers, we must have ∗

θ (−q) = θ (q) , and the allowed q values are restricted, for periodic boundary conditions, by the requirement of ⇒

θj+L = θj ,

qL = 2πn,

L with n = 0, ±1, ±2, . . . , ± .

2

Using θj − θj−1 =

X q

eiqj θ (q) √ 1 − e−iq , L

the one dimensional Hamiltonian, βH = K 2 Fourier components as

2 j (θj − θj−1 ) , can be rewritten in terms of

P

X ei(q+q )j

KX ′ −iq ′ −iq . θ (q) θ (q ) 1−e 1−e βH = 2 L ′ j ′

q,q

Using the identity

i(q+q ′ )j = Lδq,−q ′ , we obtain je

P

βH = K

X q

2

|θ (q)| [1 − cos (q)] .

(d) Generalize the results from the previous part to a d–dimensional simple cubic lattice with periodic boundary conditions.

• In the case of a d dimensional system, the index j is replaced by a vector j 7→ j = (j1 , . . . , jd ) , which describes the lattice. We can then write βH =

2 K XX θj − θj+eα , 2 α j

36

where eα ’s are unit vectors {e1 = (1, 0, · · · , 0) , · · · , ed = (0, · · · , 0, 1)}, generalizing the one

dimensional result to

X X ei(q+q′ )·j

KX ′ −iq′ ·eα −iq·eα θ (q) θ (q ) .

1 − e 1 − e βH = d 2 L ′ α q,q

j

Again, summation over j constrains q and −q′ to be equal, and βH = K

X q

|θ (q)|

2

X α

[1 − cos (qα )] .

(e) Calculate the contribution of these modes to the free energy and heat capacity. (Evaluate the classical partition function, i.e. do not quantize the modes.)

P • With K (q) ≡ 2K α [1 − cos (qα )], Z=

Ys

1 2π 2 , dθ (q) exp − K (q) |θ (q)| = 2 K (q) q q

Z Y

and the corresponding free energy is # 1X ln K (q) , F = −kB T ln Z = −kB T constant − 2 q "

d

or, in the continuum limit (using the fact that the density of states in q space is (L/2π) ), "

1 F = −kB T constant − Ld 2

Z

d

d q d

(2π)

#

ln K (q) .

As K ∼ 1/T at T → ∞, we can write

1 d ′ F = −kB T constant + L ln T , 2 and the heat capacity per site is given by C = −T

kB ∂ 2F 1 · d = .

2 ∂T L 2

This is because there is one degree of freedom (the angle) per site that can store potential energy.

37

(f) Find an expression for h~s0 · ~sx i = ℜhexp[iθx − iθ0 ]i by adding contributions from different Fourier modes. Convince yourself that for |x| → ∞, only q → 0 modes contribute

appreciably to this expression, and hence calculate the asymptotic limit.

• We have

θx − θ0 =

X

θ (q)

q

eiq·x − 1 , Ld/2

and by completing the square for the argument of the exponential in ei(θx −θ0 ) , i.e. for eiq·x − 1 1 2 − K (q) |θ (q)| + iθ (q) , 2 Ld/2 it follows immediately that ( ) ( Z ) 2 E D iq·x d X e − 1 1 d q 1 − cos (q · x) ei(θx −θ0 ) = exp − d = exp − . d L q 2K (q) K (q) (2π) For x larger than 1, the integrand has a peak of height ∼ x2 /2K at q = 0 (as it is

seen by expanding the cosines for small argument). Furthermore, the integrand has a first

node, as q increases, at q ∼ 1/x. From these considerations, we can obtain the leading behavior for large x:

• In d = 1, we have to integrate ∼ x2 /2K over a length ∼ 1/x, and thus

E D |x| i(θx −θ0 ) . ∼ exp − e 2K 2

• In d = 2, we have to integrate ∼ x2 /2K over an area ∼ (1/x) . A better approximation,

at large x, than merely taking the height of the peak, is given by Z Z dd q 1 − cos (q · x) dqdϕq 1 − cos (qx cos ϕ) ≈ 2 d K (q) Kq 2 (2π) (2π) Z Z dqdϕ cos (qx cos ϕ) dqdϕ 1 , = 2 Kq − 2 Kq (2π) (2π) or, doing the angular integration in the first term, Z

dd q 1 − cos (q · x) ≈ d K (q) (2π)

Z 1/|x|

dq 1 + subleading in x, 2π Kq

resulting in D

i(θx −θ0 )

e

E

ln |x| − 1 = |x| 2πK , ∼ exp − 2πK 38

as x → ∞.

3

• In d ≥ 3, we have to integrate ∼ x2 /2K over a volume ∼ (1/x) . Thus, as x → ∞, the x dependence of the integral is removed, and D

i(θx −θ0 )

e

E

→ constant,

implying that correlations don’t disappear at large x.

The results can also be obtained by noting that the fluctuations are important only for small q. Using the expansion of K(q) ≈ Kq 2 /2, then reduces the problem to calculation R of the Coulomb Kernel dd qeiq·x /q 2 , as described in the preceding chapter.

R (g) Calculate the transverse susceptibility from χt ∝ dd xh~s0 · ~sx ic . How does it depend on the system size L? • We have

D

E

( Z = exp −

eiθx

( Z = exp −

ei(θx −θ0 )

dd q 1 − cos (q · x) d K (q) (2π)

)

,

and, similarly, dd q

1 d 2K (q) (2π)

)

.

Hence the connected correlation function E E D D h~sx · ~s0 ic = ei(θx −θ0 ) = ei(θx −θ0 ) − eiθx c

eiθ0 ,

is given by −

h~sx · ~s0 ic = e

R

dd q 1 (2π)d K(q)

(

exp

"Z

#

d

)

d q cos (q · x) −1 . d K (q) (2π)

In d ≥ 3, the x dependent integral vanishes at x → ∞. We can thus expand its exponential,

for large x, obtaining h~sx · ~s0 ic ∼

Z

dd q cos (q · x) ≈ d K (q) (2π)

Z

dd q cos (q · x) 1 1 = Cd (x) ∼ . d d−2 2 Kq K (2π) K |x|

Thus, the transverse susceptibility diverges as χt ∝

Z

dd x h~sx · ~s0 ic ∼

L2 .

K

(h) In d = 2, show that χt only diverges for K larger than a critical value Kc = 1/(4π).

39

• In d = 2, there is no long range order, h~sx i = 0, and −1/(2πK)

h~sx · ~s0 ic = h~sx · ~s0 i ∼ |x|

.

The susceptibility χt ∼

Z L

d2 x |x|

−1/(2πK)

,

thus converges for 1/(2πK) > 2, for K below Kc = 1/(4π). For K > Kc , the susceptibility diverges as χt ∼ L2−2Kc /K . ******** 2. Capillary waves:

A reasonably flat surface in d–dimensions can be described by its

height h, as a function of the remaining (d − 1) coordinates x = (x1 , ...xd−1 ). Convince p R yourself that the generalized “area” is given by A = dd−1 x 1 + (∇h)2 . With a surface tension σ, the Hamiltonian is simply H = σA.

(a) At sufficiently low temperatures, there are only slow variations in h. Expand the energy to quadratic order, and write down the partition function as a functional integral.

• For a surface parametrized by the height function xd = h (x1 , . . . , xd−1 ) , an area element can be calculated as dA =

1 dx1 · · · dxd−1 , cos α

where α is the angle between the dth direction and the normal 1

~n = q 1 + (∇h)

∂h ∂h ,...,− ,1 − ∂x1 ∂xd−1 2

i−1/2 h 2 2 to the surface (n2 = 1). Since, cos α = nd = 1 + (∇h) ≈ 1 − 12 (∇h) , we obtain H = σA ≈ σ

Z

d

d−1

1 2 x 1 + (∇h) , 2

40

and, dropping a multiplicative constant,

Z Z σ 2 d−1 d x (∇h) .

Z = Dh (x) exp −β 2 (b) Use Fourier transformation to diagonalize the quadratic Hamiltonian into its normal modes {hq } (capillary waves).

• After changing variables to the Fourier modes, h (x) =

dd−1 q

Z

d−1

(2π)

h (q) eiq·x ,

the partition function is given by Z=

Z

(

σ Dh (q) exp −β 2

Z

dd−1 q d−1

(2π)

q 2 |h (q)|

2

)

.

(c) What symmetry breaking is responsible for these Goldstone modes?

• By selecting a particular height, the ground state breaks the translation symmetry in the dth direction. The transformation h (x) → h (x) + ξ (x) leaves the energy unchanged

if ξ (x) is constant. By continuity, we can have an arbitrarily small change in the energy by varying ξ (x) arbitrarily slowly.

2 (d) Calculate the height–height correlations h h(x) − h(x′ ) i.

• From

′

h (x) − h (x ) =

we obtain D

′

(h (x) − h (x ))

2

E

=

Z

dd−1 q

Z

dd−1 q

iq·x iq·x′ , h (q) e − e d−1 (2π)

dd−1 q ′

′

iq ·x iq′ ·x′ iq·x iq·x′ . e −e hh (q) h (q )i e −e d−1 d−1 (2π) (2π) ′

The height-height correlations thus behave as D E 2 G (x − x′ ) ≡ (h (x) − h (x′ )) Z 2 dd−1 q 1 − cos [q · (x − x′ )] 2 = = Cd−1 (x − x′ ) . d−1 2 βσ q βσ (2π) (e) Comment on the form of the result (d) in dimensions d = 4, 3, 2, and 1.

41

• We can now discuss the asymptotic behavior of the Coulomb Kernel for large |x − x′ |, either using the results from problem 1(f), or the exact form given in lectures.

• In d ≥ 4, G (x − x′ ) → constant, and the surface is flat.

• In d = 3, G (x − x′ ) ∼ ln |x − x′ |, and we come to the surprising conclusion that there

are no asymptotically flat surfaces in three dimensions. While this is technically correct, since the logarithm grows slowly, very large surfaces are needed to detect appreciable fluctuations.

• In d = 2, G (x − x′ ) ∼ |x − x′ |. This is easy to comprehend, once we realize that the √ interface h(x) is similar to the path x(t) of a random walker, and has similar (x ∼ t)

fluctuations.

2

• In d = 1, G (x − x′ ) ∼ |x − x′ | . The transverse fluctuation of the ‘point’ interface are very big, and the approximations break down as discussed next.

(f) By estimating typical values of ∇h, comment on when it is justified to ignore higher order terms in the expansion for A. 2

• We can estimate (∇h) as D

′

2

(h (x) − h (x )) 2 (x − x′ )

E

∝ |x − x′ |

1−d

.

For dimensions d ≥ dℓ = 1, the typical size of the gradient decreases upon coarse-graining.

The gradient expansion of the area used before is then justified. For dimensions d ≤ dℓ , the whole idea of the gradient expansion fails to be sensible. ******** 3. Gauge fluctuations in superconductors:

The Landau–Ginzburg model of supercon-

ductivity describes a complex superconducting order parameter Ψ(x) = Ψ1 (x) + iΨ2 (x), ~ and the electromagnetic vector potential A(x), which are subject to a Hamiltonian βH =

Z

t K L 2 2 4 ∗ ∗ d x |Ψ| + u|Ψ| + Dµ ΨDµ Ψ + (∇ × A) .

2 2 2 3

The gauge-invariant derivative Dµ ≡ ∂µ − ieAµ (x), introduces the coupling between the two fields. (In terms of Cooper pair parameters, e = e∗ c/h̄, K = h̄2 /2m∗ .)

(a) Show that the above Hamiltonian is invariant under the local gauge symmetry: Ψ(x) 7→ Ψ(x) exp (iθ(x)) ,

and 42

1 Aµ (x) 7→ Aµ (x) + ∂µ θ. e

• Under a local gauge transformation, βH 7−→ ( Z

K t 2 4 (∂µ − ieAµ − i∂µ θ) Ψeiθ (∂µ + ieAµ + i∂µ θ) Ψ∗ e−iθ |Ψ| + u |Ψ| + d3 x 2 2

2 ) 1 L ~ + ∇ × ∇θ ∇×A . + 2 e But this is none other than βH again, since (∂µ − ieAµ − i∂µ θ) Ψeiθ = eiθ (∂µ − ieAµ ) Ψ = eiθ Dµ Ψ, and

1 ∇ × ∇θ = 0. e

~ (b) Show that there is a saddle point solution of the form Ψ(x) = Ψ, and A(x) = 0, and find Ψ for t > 0 and t < 0.

• The saddle point solutions are obtained from δH = 0, δΨ∗ and

δH = 0, δAµ

=⇒

=⇒

t K 2 Ψ + 2uΨ |Ψ| − Dµ Dµ Ψ = 0, 2 2

K −ieΨDµ∗ Ψ∗ + ieΨ∗ Dµ Ψ − Lǫαβµ ǫαγδ ∂β ∂γ Aδ = 0.

2

~ = 0, clearly solves these equations. The first equation then The ansatz Ψ (x) = Ψ, A becomes tΨ + 4uΨ Ψ yielding (for u > 0) Ψ = 0 for t > 0, whereas Ψ

2 2

= 0, = −t/4u for t < 0.

(c) For t < 0, calculate the cost of fluctuations by setting

( Ψ(x) = Ψ + φ(x) exp (iθ(x)) , Aµ (x) = aµ (x),

(with ∂µ aµ = 0 in the Coulomb gauge)

and expanding βH to quadratic order in φ, θ, and ~a.

• For simplicity, let us choose Ψ to be real. From the Hamiltonian term

Dµ ΨDµ∗ Ψ∗ = (∂µ − ieaµ ) Ψ + φ eiθ (∂µ + ieaµ ) Ψ + φ e−iθ , 43

we get the following quadratic contribution 2

2

2

2

2

2

Ψ (∇θ) + (∇φ) − 2eΨ aµ ∂µ θ + e2 Ψ |~a| .

The third term in the above expression integrates to zero (as it can be seen by integrating by parts and invoking the Coulomb gauge condition ∂µ aµ = 0). Thus, the quadratic terms read βH

(2)

=

Z

3

d x

(

t K 2 K 2 2 2 + 6uΨ φ2 + (∇φ) + Ψ (∇θ) 2 2 2 K L 2 2 2 + e2 Ψ |~a| + (∇ × ~a) 2 2

)

.

(d) Perform a Fourier transformation, and calculate the expectation values of |φ(q)|2 , |θ(q)|2 , and |~a(q)|2 .

• In terms of Fourier transforms, we obtain (

X t K 2 K 2 2 2 2 (2) + 6uΨ + q |φ (q)| + Ψ q 2 |θ (q)| βH = 2 2 2 q K L 2 2 2 + e2 Ψ |~a (q)| + (q × ~a) 2 2 2

)

.

2

In the Coulomb gauge, q ⊥ ~a (q), and so [q × ~a (q)] = q 2 |~a (q)| . This diagonal form

then yields immediately (for t < 0) D

|φ (q)|

2

E

−1 2 = t + 12uΨ + Kq 2 =

−1

4u 2 = KΨ q 2 =− , Ktq 2 −1 D E

2 2 = |~a (q)| = 2 Ke2 Ψ + Lq 2 D

|θ (q)|

2

E

1 , Kq 2 − 2t 2

Lq 2 − Ke2 t/4u

(~a has 2 components).

Note that the gauge field, “mass-less” in the original theory, acquires a “mass” Ke2 t/4u through its coupling to the order parameter. This is known as the Higgs mechanism. ******** 4. Fluctuations around a tricritical point: As shown in a previous problem, the Hamiltonian βH =

Z

K t 2 2 4 6 d x (∇m) + m + um + vm , 2 2 d

44

with u = 0 and v > 0 describes a tricritical point. (a) Calculate the heat capacity singularity as t → 0 by the saddle point approximation.

• As already calculated in a previous problem, the saddle point minimum of the free

energy m ~ = mêℓ , can be obtained from

yielding,

∂Ψ = m t + 6vm 4 = 0, ∂m m    0 m = t 1/4   − 6v

for

t>t=0

for

t<0

The corresponding free energy density equals to    0 t 2 6 Ψ(m) = m + vm = 1 (−t)3/2  2 − 3 (6v)1/2

.

for

t>0

for

t<0

Therefore, the singular behavior of the heat capacity is given by  0 for  2 ∂ Ψ = C = Cs.p. ∼ −Tc ∂t2 m  Tc (−6vt)−1/2 for 4

t>0 t<0

as sketched in the figure below.

C

t

(b) Include both longitudinal and transverse fluctuations by setting

m(x) ~ = m + φℓ (x) êℓ + 45

n X

α=2

φα t (x)êα ,

.

,

and expanding βH to quadratic order in φ.

• Let us now include both longitudinal and transversal fluctuations by setting m(x) ~ = (m + φℓ (x))êℓ +

n X

φα t (x)êα ,

α=2

where êℓ and êα form an orthonormal set of n vectors. Consequently, the free energy βH is a function of φℓ and φt . Since mêℓ is a minimum, there are no linear terms in the expansion of βH in φ. The contributions of each factor in the free energy to the quadratic term in the expansion are (∇m) ~ 2 =⇒ (∇φℓ )2 +

n X

2 (∇φα t) ,

α=2

(m) ~ 2 =⇒ (φℓ )2 +

n X

2 (φα t) ,

α=2 6

2 3

2

(m) ~ = ((m) ~ ) = (m

+ 2mφℓ + φ2ℓ +

n X

2 3 4 2 4 (φα t ) ) =⇒ 15m (φℓ ) + 3m

n X

2 (φα t) .

α=2

α=2

The expansion of βH to second order now gives βH(φℓ , φα t ) = βH(0, 0) +

Z

φ2ℓ K 2 4 t + 30vm (∇φℓ ) + d x 2 2 ) n α 2 X

K (φ ) t 2 . + t + 6vm4 (∇φα t) + 2 2 α=2 d

We can formally rewrite it as βH(φℓ , φα t ) = βH(0, 0) + βHℓ (φℓ ) +

n X

βHtα (φα t ),

α=2

where βHi (φi ), with i = ℓ, tα , is in general given by K βHi (φi ) = 2

Z

φ2i d x (∇φi ) + 2 , ξi d

2

with the inverse correlation lengths t K ξℓ−2 = −4t   K   

for

t>0

for

t<0

,

46

and

  t −2 ξtα = K  0

for

t>0

for

t<0

.

As shown in the lectures for the critical point of a magnet, for t > 0 there is no difference between longitudinal and transverse components, whereas for t < 0, there is no restoring force for the Goldstone modes φα t due to the rotational symmetry of the ordered state. (c) Calculate the longitudinal and transverse correlation functions.

• Since in the harmonic approximation βH turns out to be a sum of the Hamiltonians of the different fluctuating components φℓ , φα t , these quantities are independent of each other, i.e. hφℓ φα and hφγt φα t i = 0 for α 6= γ. t i = 0, To determine the longitudinal and transverse correlation functions, we first express the free energy in terms of Fourier modes, so that the probability of a particular fluctuation configuration is given by

Y

α 2 K 2 K 2 −2 −2 2 α exp − q + ξℓ |φℓ,q | · exp − q + ξtα |φt, q | .

P({φℓ , φt }) ∝ 2 2 q,α

Thus, as it was also shown in the lectures, the correlation function is hφα (x)φβ (0)i = therefore,

δ eiq·x δα,β X = − α,β Id (x, ξα ), −2 V K q q 2 + ξα K

hφℓ (x)φℓ (0)i = − and β hφα t (x)φt (0)i = −

1 Id (x, ξℓ ), K

δα,β Id (x, ξtα ).

K

(d) Compute the first correction to the saddle point free energy from fluctuations.

• Let us calculate the first correction to the saddle point free energy from fluctuations.

The partition function is

Z Z

K d 2 −2 2 −βH(0,0) d x (∇φ) + ξ φ Dφ(x) exp − Z=e 2 ( ) Z Y X

K = e−βH(0,0) dφq exp − q 2 + ξ −2 φq φ∗q , 2 q q ) ( X Y

1 −1/2 Kq 2 + Kξ −2 = K q 2 + ξ −2 = exp − 2 q q 47

and the free energy density equals to  Z

dd q n  2   ln Kq + t  2 (2π)d βH(0, 0) βf = + Z Z  n−1

V 1 dd q dd q  2 2  ln Kq − 4t + ln Kq  2 (2π)d 2 (2π)d

for

t>0 .

for

t<0

Note that the first term is the saddle point free energy, and that there are n contributions to the free energy from fluctuations. (e) Find the fluctuation correction to the heat capacity.

• As C = −T (d2 f /dT 2 ), the fluctuation corrections to the heat capacity are given by  Z −2 dd q n    Kq 2 + t for t>0  2 d (2π) .

C − Cs.p. ∝ Z d  −2 16 d q  2  Kq − 4t for t<0  2 (2π)d

These integrals change behavior at d = 4. For d > 4, the integrals diverge at large q,

and are dominated by the upper cutoff ∆ ≃ 1/a. That is why fluctuation corrections to the heat capacity add just a constant term on each side of the transition, and the saddle

point solution keeps its qualitative form. On the other hand, for d < 4, the integrals are proportional to the corresponding correlation length ξ 4−d . Due to the divergence of ξ, the fluctuation corrections diverge as Cf l. = C − Cs.p. ∝ K −d/2 |t|d/2−2 .

(f) By comparing the results from parts (a) and (e) for t < 0 obtain a Ginzburg criterion, and the upper critical dimension for validity of mean–field theory at a tricritical point.

• To obtain a Ginzburg criterion, let us consider t < 0. In this region, the saddle point contribution already diverges as Cs.p. ∝ (−vt)−1/2 , so that v 1/2 d−3 Cf l. ∝ (−t) 2 .

Cs.p.

Kd Therefore at t < 0, the saddle point contribution dominates the behavior of this ratio provided that d > 3. For d < 3, the mean field result will continue being dominant far enough from the critical point, i.e. if (−t)

d−3

≫

Kd v

,

or 48

|t| ≫

Kd v

1/(d−3)

.

Otherwise, i.e. if |t| <

Kd v

1/(d−3)

,

the fluctuation contribution to the heat capacity becomes dominant. The upper critical dimension for the tricritical point is then d = 3. (g) A generalized multicritical point is described by replacing the term vm6 with u2n m2n .

Use simple power counting to find the upper critical dimension of this multicritical point.

• If instead of the term vm6 we have a general factor of the form u2n m2n , we can easily generalize our results to m ∝ (−t)1/(2n−2) ,

Ψ(m) ∝ (−t)n/(n−1) ,

Cs.p. ∝ (−t)n/(n−1)−2 .

Moreover, the fluctuation correction to the heat capacity for any value of n is the same as before Cf l. ∝ (−t)d/2−2 .

Hence the upper critical dimension is, in general, determined by the equation n d −2 = − 2, 2 n−1

or

du =

2n . n−1

******** 5. Coupling to a ‘massless’ field: Consider an n-component vector field m(x) ~ coupled to a scalar field A(x), through the effective Hamiltonian βH =

Z

K t 2 L 2 2 2 2 2 2 2 d x (∇m) ~ + m ~ + u(m ~ ) +e m ~ A + (∇A) , 2 2 2 d

with K, L, and u positive. (a) Show that there is a saddle point solution of the form m(x) ~ = mêℓ and A(x) = 0, and find m for t > 0 and t < 0.

• The saddle point solution, assuming uniform m(x) ~ = mêℓ and A(x) = A, is

t 2 4 2 2 2 ln Z = −V min m + um + e m A .

2

49

Since the last term is non-negative, it is minimized for A = 0, while minimizing with respect to m yields m=

(

0 p −t/4u

for t > 0 for t < 0

.

(b) Sketch the heat capacity C = ∂ 2 ln Z/∂t2 , and discuss its singularity as t → 0 in the saddle point approximation.

• From the saddle point free energy ln Z f (t) = − = V

(

0

for t > 0

−t2 /16u

for t < 0

,

we obtain a heat capacity ∂ 2f C∝ 2 = ∂t

(

0

for t > 0

1/8u

for t < 0

.

The heat capacity has a discontinuity at the transition, corresponding to α = 0.

C

0

t

(c) Include fluctuations by setting (

m(x) ~ = m + φℓ (x) êℓ + φt (x)êt , A(x) = a(x),

and expanding βH to quadratic order in φ and a.

• From m(x) ~ = (m + φℓ (x))êℓ + φt (x)êt , we obtain m2 = m2 + 2mφℓ + φ2ℓ + φ2t ,

m4 = m4 + 4m3 φℓ + 6m2 φ2ℓ + 2m2 φ2t + O φ3 .

50

After substituting the above (and A = a) in βH, the linear terms vanish at the minimum, and the second order terms give

Z

Z K t + 12um2 2 t + 4um2 2 K d d 2 2 βH2 = d x (∇φℓ ) + φℓ + d x (∇φt ) + φt 2 2 2 2

Z

2e2 m2 2 L d 2 + d x (∇a) + a + O φ3 .

2 2 (d) Find the correlation lengths ξℓ , and ξt , for the longitudinal and transverse components of φ, for t > 0 and t < 0.

• The longitudinal and transverse correlation lengths are given by  r K   for t > 0  2 t + 12um t −2 , ξℓ = , =⇒ ξℓ = r  K   −K for t < 0 2t

and

ξt−2 =

t + 4um2 , K

=⇒

r   K for t > 0 ξt = . t   ∞ for t < 0

(e) Find the correlation length ξa for the fluctuations of the scalar field a, for t > 0 and t < 0.

• The correlation length for the scalar field is 2e2 m2 , ξa−2 = L

=⇒

ξa =

    

r

∞ for t > 0 −2uL e2 t

.

for t < 0

(f) Calculate the correlation function ha(x)a(0)i for t > 0.

• Since the gauge field has no correlation length for t > 0, we have Z Z dd q iq·x dd q iq·x 1 1 2 ha(x)a(0)i = e |a(q)| = e = − Cd (x). d d 2 (2π) (2π) Lq L The function Cd (x) satisfies the Coulomb relation ∇2 Cd = δ d (x). Integrating the above relation over a sphere of radius x in d-dimensions yields Sd xd−1

dCd = 1, dx

Cd (x) = 51

x2−d + c, Sd (2 − d)

where Sd = 2π d/2 /(d/2 − 1)! is the d-dimensional solid angle, and c is a constant of integration. Up to this unknown constant, we thus get ha(x)a(0)i = −

x2−d .

LSd (2 − d)

(g) Compute the correction to the saddle point free energy ln Z, from fluctuations. (You can leave the answer in the form of integrals involving ξℓ , ξt , and ξa .)

• The fluctuation corrections due to the three terms in βH2 are independent, and adding

them up (after performing the Gaussian integrals) yields

Z

1 ln Z dd q t 2 −2 4 2 − = m + um + ln K q + ξ ℓ V 2 2 (2π)d Z Z 1

dd q dd q 1 −2 2 + ln K q + ξ ln L q 2 + ξa−2 . + t d d 2 (2π) 2 (2π) (h) Find the fluctuation corrections to the heat capacity in (b), again leaving the answer in the form of integrals.

• For t > 0, the fluctuation corrections to the saddle point come from the 2 components of the field m ∼ φ, and the field a, resulting in 2 ln Z(t > 0) = − V 2

Z

1 dd q 2 ln Kq + t + (2π)d 2

Z

dd q 2 ln Lq .

(2π)d

For t < 0, the various terms can be summed to Z

ln Z(t < 0) dd q t2 1 2 − =− + ln Kq − 2t V 16u 2 (2π)d

Z Z 1 1 dd q dd q te2 2 2 . + ln Kq + ln Lq − 2 (2π)d 2 (2π)d 2u After taking two derivatives with respect to t, we obtain  Z  dd q 1    d  (2π) (Kq 2 + t)2 1 ∂ 2 ln Z Z Z = C∝  e4 dd q dd q 1 1 1 V ∂t2   + 2 +  d 2 2 d  8u (2π) Kq − 2t 8u (2π) Lq 2 − te2 2 2u

52

for t > 0 . for t < 0

(i) Discuss the behavior of the integrals appearing above schematically, and state their dependence on the correlation length ξ, and cutoff Λ, in different dimensions.

• The corrections to the heat capacity depend on the integral Id (ξ) ≡

Z

dd q 1 , d (2π) (q 2 + ξ −2 )2

which has dimensions of [q]d−4 . In dimensions d > 4, the integral is dominated by the upper cut-off Λ, and its leading behavior is proportional to Λd−4 . For 2 < d < 4, the integral is convergent as Λ → ∞ for finite ξ, and its most singular part scales as ξ 4−d ,

which diverges as t → 0.

(j) What is the critical dimension for the validity of saddle point results, and how is it modified by the coupling to the scalar field?

• In dimensions d ≤ 4, the fluctuation corrections to the saddle point diverge as ξ → ∞.

The scalar field fluctuations make an additional contribution, which also diverges for d ≤ 4,

and thus does not modify the upper critical dimension. ********

6. Random magnetic fields: Consider the Hamiltonian

Z K t 2 d 2 4 βH = d x (∇m) + m + u m − h(x)m(x) 2 2

,

where m(x) and h(x) are scalar fields, and u > 0. The random magnetic field h(x) results from frozen (quenched) impurities that are independently distributed in space. For simplicity h(x) is assumed to be an independent Gaussian variable at each point x, such that h(x) = 0,

and

h(x)h(x′ ) = ∆δ d (x − x′ ),

(1)

where the over-line indicates (quench) averaging over all values of the random fields. The above equation implies that the Fourier transformed random field h(q) satisfies h(q) = 0,

and

h(q)h(q′ ) = ∆(2π)d δ d (q + q′ ).

(2)

(a) Calculate the quench averaged free energy, fsp = min{Ψ(m)}m , assuming a saddle point solution with uniform magnetization m(x) = m. (Note that with this assumption, the random field disappears as a result of averaging and has no effect at this stage.)

53

• Assuming a uniform solution m(x) = m to the saddle point of the weight, we obtain

ln Z t 2 4 = min m + um − mh(x) . fsp = − V 2 m Since the random field averages to zero, the saddle point magnetization is obtained as in the case without random fields as

0p for t > 0 3 , tm+4um = 0, ⇒ m = ± −t/4u for t > 0

⇒

fsp =

0 for t > 0 . −t2 /16u for t > 0

(b) Include fluctuations by setting m(x) = m + φ(x), and expanding βH to second order in φ.

• To second order, we find βH = V fsp +

Z

t + 12um 2 2 K 2 (∇φ) + φ − h(x)φ(x) . d x 2 2 d

(c) Express the energy cost of the above fluctuations in terms of the Fourier modes φ(q).

• In terms of Fourier modes K βH = V fsp + 2

Z

dd q 2 −2 2 q + ξ |φ(q)| − h(q)φ(−q) , (2π)d

where we have introduced a correlation length ξ such that n K t = −2t ξ2

for t > 0 . for t > 0

(d) Calculate the mean hφ(q)i, and the variance |φ(q)|2 c , where h· · ·i denotes the usual thermal expectation value for a fixed h(q).

• For fixed h(q), the probability for fluctuations in amplitudes of the independent Fourier

mode is given as a product of Gaussian forms

Y K(q 2 + ξ −2 ) 2 |φ(q)| − h(q)φ(−q) .

P[φ(q)] = exp − 2 q Hence we can read off the mean and variance of the fluctuations as hφ(q)i =

h(q) , K (q 2 + ξ −2 )

and 54

|φ(q)|2 c =

1 K (q 2 + ξ −2 )

.

(e) Use the above results, in conjunction with Eq.(2), to calculate the quench averaged scattering line shape S(q) = h|φ(q)|2 i.

• From the definitions of S(q) and the variance, we get 2

S(q) = |hφ(q)i| + h|φ(q)|2 ic =

|h(q)|2

K 2 (q 2 + ξ ∆ 1 = 2 + K (q 2 + ξ −2 ) .

2 2 −2 K (q + ξ )

1

+ −2 2 K (q 2 + ξ −2 ) )

Note that the result is the sum of a Lorentzian and a squared–Lorentzian. This is a common signature of scattering from random systems. (f) Perform the Gaussian integrals over φ(q) to calculate the fluctuation corrections, δf [h(q)], to the free energy.

Reminder :

Z ∞

2 2π K 2 |h| ∗ ∗ dφdφ exp − |φ| + h φ + hφ = exp 2 K 2K −∞ ∗

• Including the Gaussian fluctuations, the partition function is given by −V fsp

Z[h] = e

YZ

q≥0

#

K q 2 + ξ −2 2 ∗ |φ(q)| + h(−q)φ(q) + h(q)φ(q) dφ(q)dφ(q) exp − 2 ∗

"

where φ(q)∗ = φ(−q). Hence, for a given configuration of the field h, the fluctuation– corrected free energy is ( ) 2 X

ln Z[h] 1 |h(q)| f [h] = − − ln K q 2 + ξ −2 + . = fsp + V 2 2K (q 2 + ξ −2 ) q (g) Use Eq.(2) to calculate the corrections due to the fluctuations in the previous part to the quench averaged free energy f . (Leave the corrections in the form of two integrals.)

• Taking the average of the last result over h gives 1 f = fsp − 2

Z

∆ dd q 2 −2 ln K q + ξ + (2π)d 2 55

Z

dd q 1 . d 2 (2π) K (q + ξ −2 )

(h) Estimate the singular t dependence of the integrals obtained in the fluctuation corrections to the free energy.

• The first integral has dimensions of [q]d , and since the relevant t-dependent momentum √ is qξ ∼ ξ −1 ∝ t, the most singular part of this integral behaves as td/2 . The second

integral has dimensions of [q d−2 ], and by similar reasoning leads to a singular form of td/2−1 . (i) Find the upper critical dimension, du , for the validity of saddle point critical behavior.

• The singular form of the saddle point solution scales as t2 , and we ask if the fluctuation corrections can dominate this singularity. The fluctuation correction that results from

random fields (proportional to ∆) is always more important than the standard contribution from thermal fluctuations. The former masks the saddle point singularity if d/2 − 1 ≤ 2, i.e. for dimensions less that the upper critical dimension of du = 6. ******** 7. Long–range interactions: Consider a continuous spin field ~s(x), subject to a long–range ferromagnetic interaction Z

dd xdd y

~s(x) · ~s(y) , |x − y|d+σ

as well as short-range interactions. (a) How is the quadratic term in the Landau-Ginzburg expansion modified by the presence of this long-range interaction? For what values of σ is the long-range interaction dominant?

• The Fourier transform of the above interaction gives a quadratic term proportional to

|q|σ . Such a term is relevant, and dominates the standard q 2 term that arises from shortrange interactions, provided that σ < 2.

(b) By estimating the magnitude of thermally excited Goldstone modes (or otherwise), obtain the lower critical dimension dℓ below which there is no long–range order.

• Goldstone modes are ‘massless’, and governed by a quadratic term proportional to

|q|σ |φ(q)|2 in the Fourier transformed Hamiltonian. Hence |φ(q)|2

real-space

φ(x)

2

∝

Z

dd q

∝ |q|−σ , and in

eiq·x . |q|σ

The above integral diverges with system size for dimensions d ≤ dℓ = σ. Due to increased stiffness of the Goldstone modes, the lower critical dimension is reduced from 2.

56

(c) Find the upper critical dimension du , above which saddle point results provide a correct description of the phase transition.

• From the above analysis, it is apparent that under a change of scale by a factor of b, the naive dimension of the field is ζ = b(d−σ)/2 . The quartic term in the Landau Ginzburg equation thus scales as 0

u → u′ = byu u + O(u2 ),

where

yu0 = d − 4 ×

d−σ = 2σ − d.

2

The upper critical dimension is then identified as du = 2σ, below which the quartic term is relevant, invalidating the results from a Gaussian analysis. ******** 8. Ginzburg criterion along the magnetic field direction: Consider the Hamiltonian

Z K t 2 d 2 2 2 , βH = d x (∇m) ~ + m ~ + u(m ~ ) 2 2 describing an n–component magnetization vector m(x), ~ with u > 0. (a) In the saddle point approximation, the free energy is f = min{Ψ(m)}m . Indicate the resulting phase boundary in the (h, t) plane, and label the phases. (h denotes the magnitude of ~h.) (b) Sketch the form of Ψ(m) for t < 0, on both sides of the phase boundary, and for t > 0 at h = 0. (c) For t and h close to zero, the spontaneous magnetization can be written as m = tβ gm (h/t∆ ). Identify the exponents β and ∆ in the saddle point approximation.

For the remainder of this problem set t = 0. (d) Calculate the transverse and longitudinal susceptibilities at a finite h.

~ t (x)êt , and expanding βH to (e) Include fluctuations by setting m(x) ~ = m + φℓ (x) êℓ + φ

second order in the φs. (êℓ is a unit vector parallel to the average magnetization, and êt is perpendicular to it.) (f) Calculate the longitudinal and transverse correlation lengths.

(g) Calculate the first correction to the free energy from these fluctuations. (The scaling form is sufficient.) (h) Calculate the first correction to magnetization, and to longitudinal susceptibility from the fluctuations.

57

(i) By comparing the saddle point value with the correction due to fluctuations, find the upper critical dimension, du , for the validity of the saddle point result. (j) For d < du obtain a Ginzburg criterion by finding the field hG below which fluctuations are important. (You may ignore the numerical coefficients in hG , but the dependances on K and u are required.) ********

58

Solutions to problems from chapter 4 - The Scaling Hypothesis 1. Scaling in fluids:

Near the liquid–gas critical point, the free energy is assumed to take

the scaling form F/N = t2−α g(δρ/tβ ), where t = |T − Tc |/Tc is the reduced temperature,

and δρ = ρ − ρc measures deviations from the critical point density. The leading singular

behavior of any thermodynamic parameter Q(t, δρ) is of the form tx on approaching the critical point along the isochore ρ = ρc ; or δρy for a path along the isotherm T = Tc . Find the exponents x and y for the following quantities: • Any homogeneous thermodynamic quantity Q(t, δρ) can be written in the scaling form Q(t, δρ) = t

xQ

gQ

δρ tβ

.

Thus, the leading singular behavior of Q is of the form txQ if δρ = 0, i.e. along the critical isochore. In order for any Q to be independent of t along the critical isotherm as t → 0,

the scaling function for a large enough argument should be of the form lim gQ (x) = xxQ /β ,

x→∞

so that Q(0, δρ) ∝ (δρ)yQ ,

with

yQ =

xQ . β

(a) The internal energy per particle hHi/N , and the entropy per particle s = S/N. • Let us assume that the free energy per particle is

δρ F 2−α =t g β , f= N t

∂ ∂ = − T1c ∂t . The entropy is then given by and that T < Tc , so that ∂T

∂f 1 ∂f t1−α s=− gS = = ∂T V T c ∂t ρ Tc

δρ tβ

,

so that xS = 1 − α, and yS = (1 − α)/β. For the internal energy, we have hHi − T s, f= N

or

hHi ∼ Tc s(1 + t) ∼ t1−α gH N

therefore, xH = 1 − α and yH = (1 − α)/β.

59

δρ tβ

,

(b) The heat capacities CV = T ∂s/∂T |V , and CP = T ∂s/∂T |P .

• The heat capacity at constant volume

∂S ∂s t−α CV = T =− = gCV ∂T V ∂t ρ Tc

δρ tβ

,

so that xCV = −α and yCV = −α/β.

To calculate the heat capacity at constant pressure, we need to determine first the relation δρ(t) at constant P . For that purpose we will use the thermodynamic identity ∂P

∂δρ ∂t ρ . =− ∂P ∂t P ∂δρ

t

The pressure P is determined as ∂f ∂F = ρ2 ∼ ρ2c t2−α−β gP P =− ∂V ∂δρ

δρ tβ

,

which for δρ ≪ tβ goes like P ∝t

2−α−β

δρ 1+A β , t

and consequently

In the other extreme of δρ ≫ tβ , P ∝ δρ

(2−α−β)/β

1+B

t δρ1/β

,

and

 ∂P  1−α−β    ∂t ∝ t ρ  ∂P   ∝ t2−α−2β  ∂δρ t

 ∂P  (1−α−β)/β    ∂t ∝ δρ ρ  ∂P   ∝ δρ(2−α−2β)/β  ∂δρ t

.

,

where we have again required that P does not depend on δρ when δρ → 0, and on t if

t → 0.

From the previous results, we can now determine ( β−1 t =⇒ δρ ∝ tβ ∂δρ ∝ . ∂t P δρ(β−1)/β =⇒ t ∝ δρ1/β From any of these relationships follows that δρ ∝ tβ , and consequently the entropy is

s ∝ t1−α . The heat capacity at constant pressure is then given by CP ∝ t−α ,

with

xCP = −α 60

and

α yCP = − . β

(c) The isothermal compressibility κT = ∂ρ/∂P |T /ρ, and the thermal expansion coefficient α = ∂V /∂T |P /V .

Check that your results for parts (b) and (c) are consistent with the thermodynamic identity CP − CV = T V α2 /κT .

• The isothermal compressibility and the thermal expansion coefficient can be computed

using some of the relations obtained previously 1 ∂P 1 ∂ρ = κT = ρ ∂P T ρ c ∂ρ T

−1

1 = 3 tα+2β−2 gκ ρc

δρ tβ

,

with xκ = α + 2β − 2, and yκ = (α + 2β − 2)/β. And α=

1 ∂ρ 1 ∂V = ∝ tβ−1 , V ∂T P ρTc ∂t P

with xα = β − 1, and yα = (β − 1)/β. So clearly, these results are consistent with the thermodynamic identity,

(CP − CV )(t, 0) ∝ t−α ,

or

(CP − CV )(0, δρ) ∝ δρ−α/β ,

α2 (t, 0) ∝ t−α , κT

or

α2 (0, δρ) ∝ δρ−α/β . κT

and

(d) Sketch the behavior of the latent heat per particle L, on the coexistence curve for T < Tc , and find its singularity as a function of t.

• The latent heat

L = T (s+ − s− )

is defined at the coexistence line, and as we have seen before

δρ± 1−α .

T s± = t gs tβ The density difference between the two coexisting phases is the order parameter, and vanishes as tβ , as do each of the two deviations δρ+ = ρc − 1/v+ and δρ− = ρc − 1/v−

of the gas and liquid densities from the critical critical value. (More precisely, as seen in (b), δρ|P =constant ∝ tβ .) The argument of g in the above expression is thus evaluated at 61

L

t=0

t

a finite value, and since the latent heat goes to zero on approaching the critical point, we get L ∝ t1−α ,

with

xL = 1 − α.

******** 2. The Ising model: The differential recursion relations for temperature T , and magnetic field h, of the Ising model in d = 1 + ǫ dimensions are  T2 dT   = −ǫT + dℓ 2  dh  =dh . dℓ

,

(a) Sketch the renormalization group flows in the (T, h) plane (for ǫ > 0), marking the fixed points along the h = 0 axis.

• The fixed points of the flow occur along the h = 0 axis, which is mapped to itself under

RG. On this axis, there are three fixed points: (i) T ∗ = 0, is the stable sink for the low temperature phase. (ii) T ∗ → ∞, is the stable sink for the high temperature phase. (iii)

There is a critical fixed point at (T ∗ = 2ǫ, h∗ = 0), which is unstable. All fixed points are unstable in the field direction. (b) Calculate the eigenvalues yt and yh , at the critical fixed point, to order of ǫ.

• Linearizing T = T ∗ + δT , around the critical fixed point yields  dδT (   = − ǫ δT + T ∗ δT = ǫδT yt = + ǫ dℓ , =⇒   dh =(1 + ǫ)h yh =1 + ǫ dℓ 62

.

∞

*

h 0

*

*

T

T*

*

∞

*

−∞

(c) Starting from the relation governing the change of the correlation length ξ under

renormalization, show that ξ(t, h) = t−ν gξ h/|t|∆ (where t = T /Tc − 1), and find the exponents ν and ∆.

• Under rescaling by a factor of b, the correlation length is reduced by b, resulting in the homogeneity relation

ξ(t, h) = bξ(byt t, byh h).

Upon selecting a rescaling factor such that byt t ∼ 1, we obtain

ξ(t, h) = t−ν gξ h/|t|∆ ,

with ν=

1 1 = , yt ǫ

and

∆=

yh 1 = + 1. yt ǫ

(d) Use a hyperscaling relation to find the singular part of the free energy fsing. (t, h), and hence the heat capacity exponent α.

• According to hyperscaling

fsing. (t, h) ∝ ξ(t, h)−d = td/yt gf h/|t|∆ .

Taking two derivatives with respect to t leads to the heat capacity, whose singularity for h = 0 is described by the exponent α = 2 − dν = 2 −

1 1+ǫ = − + 1. ǫ ǫ

63

(e) Find the exponents β and γ for the singular behaviors of the magnetization and susceptibility, respectively.

• The magnetization is obtained from the free energy by m=−

∂f ∼ |t|β , ∂h h=0

with β =

d − yh = 0. yt

(There will be corrections to β at higher orders in ǫ.) The susceptibility is obtained from a derivative of the magnetization, or χ=−

∂ 2f ∼ |t|−γ , 2 ∂h h=0

with γ =

2yh − d 1+ǫ 1 = = + 1. yt ǫ ǫ

(f) Starting the relation between susceptibility and correlations of local magnetizations, calculate the exponent η for the critical correlations (hm(0)m(x)i ∼ |x|−(d−2+η) ).

• The magnetic susceptibility is related to the connected correlation function via χ=

Z

dd x hm(0)m(x)ic .

Close to criticality, the correlations decay as a power law hm(0)m(x)i ∼ |x|−(d−2+η) , which is cut off at the correlation length ξ, resulting in χ ∼ ξ (2−η) ∼ |t|−(2−η)ν .

From the corresponding exponent identity, we find γ = (2 − η)ν,

=⇒

η = 2 − yt γ = 2 − 2yh + d = 2 − d = 1 − ǫ.

(g) How does the correlation length diverge as T → 0 (along h = 0) for d = 1?

• For d = 1, the recursion relation for temperature can be rearranged and integrated, i.e.

1 dT 1 = , 2 T dℓ 2

=⇒ 64

2 = dℓ. d − T

We can integrate the above expression from a low temperature with correlation length ξ(T ) to a high temperature where 1/T ≈ 0, and at which the correlation length is of the order of the lattice spacing, to get

ξ 2 − = ln T a

=⇒

2 ξ(T ) = a exp .

T

******** 3. The nonlinear σ model describes n component unit spins. As we shall demonstrate later, in d = 2 dimensions, the recursion relations for temperature T , and magnetic field h, are  dT (n − 2) 2   = T dℓ 2π   dh =2h . dℓ

,

(a) How does the correlation length diverge as T → 0?

• We can solve for dℓ in the above two recursion relations, and integrate to obtain ℓ as

 1 dT 2π 1 2π    dℓ = , ℓ= −  T (ℓ)−1 = T −1 − (n − 2)ℓ  2 (n − 2) T (n − 2) T T (ℓ)

. , ⇒ 2π   1 h(ℓ) dh 2ℓ  h(ℓ) = he  dℓ = , ℓ = ln 2h 2 h The correlation length is then obtained from (b = eℓ )

ξ(T, h) = eℓ ξ [T (ℓ), h(ℓ)] .

Starting from T and h close to zero, renormalize until a value of T (ℓ∗ ) ∼ 1, which occurs

for ℓ∗ = 2π/[(n − 2)T ], leading to

2π 4π ξ(T, h) = exp g1 h exp . (n − 2)T (n − 2)T (b) Write down the singular form of the free energy as T, h → 0.

• Since d = 2, the singular part of the free energy scales as ξ −2 , i.e.

4π 4π −2 fsing (T, h) ∝ ξ = exp − g2 h exp . (n − 2)T (n − 2)T 65

(c) How does the susceptibility χ, diverge as T → 0 for h = 0?

• The susceptibility is obtained from the second derivative of the free energy as

4π ∂2f ∼ exp . χ= ∂h2 h=0 (n − 2)T ******** 4. Coupled scalars: Consider the Hamiltonian

Z t 2 K L 2 2 d 2 βH = d x m + (∇m) − hm + (∇ φ) + v∇m.∇φ , 2 2 2 coupling two one component fields m and φ. (a) Write βH in terms of the Fourier transforms m(q) and φ(q).

R R dd q • Since all terms are quadratic, and dd xA(x)B(x) = (2π) d A(q)B(−q), we can immediately write down

Z Λ d Lq 4 d q t + Kq 2 2 2 2 |m(q)| + |φ(q)| + vq m(q)φ(−q) − hm(q = 0). βH = d 2 2 0 (2π)

(b) Construct a renormalization group transformation as in the text, by rescaling distances such that q′ = bq; and the fields such that m′ (q′ ) = m̃(q)/z and φ′ (q′ ) = φ̃(q)/y. Do not evaluate the integrals that just contribute a constant additive term.

• For an RG transformation, we subdivide the modes in Fourier space into two parts: One

set with Λ/b < |q| ≤ Λ which we integrate over, and a second set with 0 < |q| ≤ Λ/b which

we keep. Since for this Gaussian Hamiltonian the two sets of modes are not coupled, the integration in trivial, and apart from a constant V δfb , we obtain the same Hamiltonian for

the retained modes (m̃(q) and φ̃(q)) as in the previous part, except that the integration range is now up to Λ/b. The second step of the RG transformation is to remove this difference by the rescaling q = q′ /b. In the third step of RG, we shall renormalize the fields according to m′ (q′ ) = m̃(q′ )/z, and φ′ (q) = φ̃(q′ )/y, to get the final renormalized Hamiltonian ′

βH = V δfb +

Z Λ 0

dd q′ z 2 b−d t + Kz 2 b−d−2 q ′2 ′ ′ 2 Ly 2 b−d−4 q ′4 ′ ′ 2 |m (q )| + |φ (q )| (2π)d 2 2

−d−2 ′2 ′ ′ ′ ′ +vzyb q m (q )φ (−q ) − hzm′ (q′ = 0).

66

The resulting recursion relations are  ′ t = z 2 b−d t       K ′ = z 2 b−d−2 K    L′ = y 2 b−d−4 L      v ′ = zyb−d−2 v     ′ h = zh

.

(c) There is a fixed point such that K ′ = K and L′ = L. Find yt , yh and yv at this fixed point.

• We can choose z 2 = bd+2 such that K ′ = K, and y 2 = bd+4 such that L′ = L. The recursion relations then become  ′ t = b2 t    v ′ = bv    ′ h = b1+d/2 h

,

=⇒

 yt = 2    yv = 1    yh = 1 + d/2

.

(d) The singular part of the free energy has a scaling from f (t, h, v) = t2−α g(h/t∆ , v/tω ) for t, h, v close to zero. Find α, ∆, and ω.

• The RG transformation preserves the partition function, and thus ln Z(t, h, v) = ln Z(t′ , h′ , v ′ ),

=⇒

V f (t, h, v) = V ′ f (t′ , h′ , v ′ ).

Since V ′ = V /bd , we get the homogeneous form f (t, h, v) = b−d f (byt t,yh h,yv v).

Setting byt t = 1 with b = t−1/yt , yields

f (t, h, v) = t−d/yt f 1, h/tyh /yt , v/tyv /yt ≡ t2−α g h/t∆ , v/tω ,

with

 d d   α=2− =2−   yt 2    1 d yh = + ∆=  yt 2 4     y 1  ω= v = yt 2 67

.

(e) There is another fixed point such that t′ = t and L′ = L. What are the relevant operators at this fixed point, and how do they scale?

• We can choose z 2 = bd such that t′ = t, and y 2 = bd+4 such that L′ = L. The recursion relations then become  ′ K = b−2 K    v′ = v    ′ h = bd/2 h

,

 yK = −2 (irrelevant)    yv = 0 (marginal)    yh = d/2 (relevant)

=⇒

********

68

.

Solutions to problems from chapter 5 - Perturbative Renormalization Group 1. Longitudinal susceptibility: While there is no reason for the longitudinal susceptibility to diverge at the mean-field level, it in fact does so due to fluctuations in dimensions d < 4.

This problem is intended to show you the origin of this divergence in perturbation theory.

There are actually a number of subtleties in this calculation which you are instructed to ignore at various steps. You may want to think about why they are justified.

Consider the Landau–Ginzburg Hamiltonian: βH =

Z

K t 2 d x (∇m) ~ 2+ m ~ + u(m ~ 2 )2 2 2 d

,

describing an n–component magnetization vector m(x), ~ in the ordered phase for t < 0.

~ t (x)êt , and expand βH keeping all terms in the expansion. (a) Let m(x) ~ = m+φℓ (x) êℓ + φ ~ t (x) êt , and m̄ the minimum of βH, • With m ~ (x) = (m̄ + φℓ (x)) êℓ + φ

(

2 t K t 2 2 d 4 2 ~t (∇φℓ ) + ∇φ + m̄ + um̄ + d x + 6um̄ φ2ℓ βH =V 2 2 2 )

2 t 2 ~ 2 3 2 4 2 2 2 ~ t + u φ + 2φ φ ~ ~ + 2um̄ φt + 4um̄ φℓ + φℓ φ . + ℓ ℓ t + φt 2

Z

Since m̄2 = −t/4u in the ordered phase (t < 0), this expression can be simplified, upon dropping the constant term, as βH =

Z

(

2

K 2 ~ ~t 2 d x (∇φℓ ) + ∇φt − tφ2ℓ + 4um̄ φ3ℓ + φℓ φ 2 )

2 ~t 2 + φ ~t 2 +u φ4ℓ + 2φ2ℓ φ . d

~ t as an unperturbed Hamiltonian βH0 , and the (b) Regard the quadratic terms in φℓ and φ ~ t as a perturbation U ; i.e. lowest order term coupling φℓ and φ U = 4um

Z

~ t (x)2 . dd xφℓ (x)φ

~ t (q).

Write U in Fourier space in terms of φℓ (q) and φ 69

• We shall focus on the cubic term as a perturbation U = 4um̄

Z

~ t (x)2 , dd xφℓ (x) φ

which can be written in Fourier space as U = 4um̄

Z

dd q ′

dd q d

d

(2π) (2π)

~ t (q) · φ ~ t (q′ ) . φℓ (−q − q′ ) φ

(c) Calculate the Gaussian (bare) expectation values hφℓ (q)φℓ (q′ )i0 and hφt,α (q)φt,β (q′ )i0 , and the corresponding momentum dependent susceptibilities χℓ (q)0 and χt (q)0 .

• From the quadratic part of the Hamiltonian, βH0 =

Z

1 d x 2 d

2 2 ~t K (∇φℓ ) + ∇φ − 2tφℓ , 2

we read off the expectation values     

d

(2π) δ d (q + q′ ) Kq 2 − 2t

hφℓ (q) φℓ (q′ )i0 =

d  (2π) δ d (q + q′ ) δαβ    hφt,α (q) φt,β (q′ )i0 = Kq 2

,

and the corresponding susceptibilities

    χℓ (q)0 =    χt (q)0 =

1 Kq 2 − 2t 1 Kq 2

.

~ t (q1 ) · φ ~ t (q2 ) φ ~ t (q′ ) · φ ~ t (q′ )i0 using Wick’s theorem. (Don’t forget that (d) Calculate hφ 1 2 ~ t is an (n − 1) component vector.) φ • Using Wick’s theorem, D E ~ t (q1 ) · φ ~ t (q2 ) φ ~ t (q′ ) · φ ~ t (q′ ) ≡ hφt,α (q1 ) φt,α (q2 ) φt,β (q′ ) φt,β (q′ )i φ 1 2 1 2 0

0 ′ = hφt,α (q1 ) φt,α (q2 )i0 hφt,β (q1 ) φt,β (q′2 )i0 + hφt,α (q1 ) φt,β (q′1 )i0 hφt,α (q2 ) φt,β (q′2 )i0

+ hφt,α (q1 ) φt,β (q′2 )i0 hφt,α (q2 ) φt,β (q′1 )i0 .

70

Then, from part (c), D

2d

~ t (q1 ) · φ ~ t (q2 ) φ ~ t (q′ ) · φ ~ t (q′ ) = (2π) φ 1 2 K2 0 E

(

2 δ

(n − 1)

d

(q1 + q2 ) δ d (q′1 + q′2 ) q12 q1′2

) δ d (q1 + q′2 ) δ d (q′1 + q2 ) δ d (q1 + q′1 ) δ d (q2 + q′2 ) + (n − 1) , + (n − 1) q12 q22 q12 q22 2

since δαα δββ = (n − 1) , and δαβ δαβ = (n − 1). (e) Write down the expression for hφℓ (q)φℓ (q′ )i to second-order in the perturbation U .

Note that since U is odd in φℓ , only two terms at the second order are non–zero.

• Including the perturbation U in the calculation of the correlation function, we have

′ −U ′ 2 φ (q) φ (q ) e φ (q) φ (q ) 1 − U + U /2 + · · · ℓ ℓ ℓ ℓ 0 0 hφℓ (q) φℓ (q′ )i = = . he−U i0 h(1 − U + U 2 /2 + · · ·)i0 Since U is odd in φℓ , hU i0 = hφℓ (q) φℓ (q′ ) U i0 = 0. Thus, after expanding the denominator to second order, 1 =1− 2 1 + hU /2i0 + · · ·

U2 2

0

+ O U3 ,

we obtain hφℓ (q) φℓ (q′ )i = hφℓ (q) φℓ (q′ )i0 +

1 2

φℓ (q) φℓ (q′ ) U 2 0 − hφℓ (q) φℓ (q′ )i0 U 2 0 .

(f) Using the form of U in Fourier space, write the correction term as a product of two 4–point expectation values similar to those of part (d). Note that only connected terms for the longitudinal 4–point function should be included.

• Substituting for U its expression in terms of Fourier transforms from part (b), the fluctuation correction to the correlation function reads

GF (q, q′ ) ≡ hφℓ (q) φℓ (q′ )i − hφℓ (q) φℓ (q′ )i0 Z d 1 d q1 dd q2 dd q1′ dd q2′ D 2 ~ t (q1 ) · φ ~ t (q2 ) = (4um̄) φℓ (q) φℓ (q′ ) φℓ (−q1 − q2 ) φ d d d d 2 (2π) (2π) (2π) (2π) E ~ t (q′ ) · φ ~ t (q′ ) − 1 hφℓ (q) φℓ (q′ )i U 2 , × φℓ (−q′1 − q′2 ) φ 1 2 0 0 2 0 71

i.e. GF (q, q′ ) is calculated as the connected part of 1 2 (4um̄) 2

Z

dd q1 dd q2 dd q1′ dd q2′ d

d

d

d

(2π) (2π) (2π) (2π)

hφℓ (q) φℓ (q′ ) φℓ (−q1 − q2 ) φℓ (−q′1 − q′2 )i0

D E ~ t (q1 ) · φ ~ t (q2 ) φ ~ t (q′ ) · φ ~ t (q′ ) , × φ 1 2 0

where we have used the fact that the unperturbed averages of products of longitudinal and transverse fields factorize. Hence Z d E d q1 dd q2 dd q1′ dd q2′ D ~ 1 2 ′ ~ t (q2 ) φ ~ t (q′ ) · φ ~ t (q′ ) φ (q ) · φ GF (q, q ) = (4um̄) t 1 1 2 d d d d 2 0 (2π) (2π) (2π) (2π) n × hφℓ (q) φℓ (−q1 − q2 )i0 hφℓ (q′ ) φℓ (−q′1 − q′2 )i0 o + hφℓ (q) φℓ (−q′1 − q′2 )i0 hφℓ (q′ ) φℓ (−q1 − q2 )i0 Z d E 1 d q1 dd q2 dd q1′ dd q2′ D ~ 2 ′ ′ ~ ~ ~ = 2 × (4um̄) φ (q ) · φ (q ) φ (q ) · φ (q ) t 1 t 2 t t 1 2 d d d d 2 0 (2π) (2π) (2π) (2π) × hφℓ (q) φℓ (−q1 − q2 )i0 hφℓ (q′ ) φℓ (−q′1 − q′2 )i0 .

Using the results of parts (c) and (d) for the two and four points correlation functions, and since u2 m̄2 = −ut/4, we obtain GF (q, q′ ) = 4u (−t)

Z

2d

dd q1 dd q2 dd q1′ dd q2′ (2π) d d d d (2π) (2π) (2π) (2π) K 2

(

(q1 + q2 ) δ d (q′1 + q′2 ) q12 q1′2 ) δ d (q1 + q′1 ) δ d (q2 + q′2 ) + δ d (q1 + q′2 ) δ d (q′1 + q2 ) + (n − 1) q12 q22 d

×

2 δ

d

(n − 1)

d

(2π) δ d (q − q1 − q2 ) (2π) δ d (q′ − q′1 − q′2 ) , Kq 2 − 2t Kq 2 − 2t

which, after doing some of the integrals, reduces to Z d 2 d q1 (q) δ d (q′ ) (n − 1) 2 4t q12 ) Z δ d (q + q′ ) dd q1 + 2 (n − 1) 2 2 .

(Kq 2 − 2t) q12 (q + q1 )

4u (−t) GF (q, q′ ) = K2

(

2 δ

d

(g) Ignore the disconnected term obtained in (d) (i.e. the part proportional to (n − 1)2 ),

and write down the expression for χℓ (q) in second order perturbation theory.

72

• From the dependence of the first term (proportional to δ d (q) δ d (q′ )), we deduce that

this term is actually a correction to the unperturbed value of the magnetization, i.e.

2(n − 1)u m→m 1− Kt

dd q1 q12

Z

,

and does not contribute to the correlation function at non-zero separation. The spatially varying part of the connected correlation function is thus d

(2π) δ d (q + q′ ) 8u (−t) δ d (q + q′ ) hφℓ (q) φℓ (q )i = + (n − 1) 2 Kq 2 − 2t K2 (Kq 2 − 2t)

Z

′

dd q1 q12 (q + q1 )

2,

leading to 8u (−t) (n − 1) 1 + χℓ (q) = 2 Kq − 2t K 2 (Kq 2 − 2t)2

Z

dd q1

1

d

2.

(2π) q12 (q + q1 )

(h) Show that for d < 4, the correction term diverges as q d−4 for q → 0, implying an infinite longitudinal susceptibility.

• In d > 4, the above integral converges and is dominated by the large q cutoff Λ. In

d < 4, on the other hand, the integral clearly diverges as q → 0, and is thus dominated by small q1 values. Changing the variable of integration to q′1 = q1 /q, the fluctuation correction to the susceptibility reads χℓ (q)F ∼ q

d−4

Z Λ/q 0

dd q1′

1

d

(2π) q1′2 (q̂ + q′1 )

2 = q

d−4

Z ∞ 0

dd q1′ d

(2π)

1

2 +O q1′2 (q̂ + q′1 )

which diverges as q d−4 for q → 0.

NOTE: For a translationally invariant system, hφ (x) φ (x′ )i = ϕ (x − x′ ) , which implies ′

hφ (q) φ (q )i = =

Z

Z

′

′

dd xdd x′ eiq·x+iq ·x hφ (x) φ (x′ )i dd (x − x′ ) dd x′ eiq·(x−x )+i(q+q )·x ϕ (x − x′ ) ′

d

= (2π) δ d (q + q′ ) ψ (q) .

73

′

′

q0 ,

Consider the Hamiltonian Z Z ′ d −βH = −βH + d xh (x) φ (x) = −βH +

dd q d

(2π)

h (q) φ (−q) ,

where −βH is a translationally invariant functional of φ (a one-component field for sim-

plicity), independent of h (x). We have

m (x = 0) = hφ (0)i =

Z

dd q d

(2π)

hφ (q)i ,

and, taking a derivative, ∂m = ∂h (q)

dd q ′

Z

d

(2π)

hφ (q′ ) φ (q)i .

At h = 0, the system is translationally invariant, and ∂m = ψ (q) . ∂h (q) h=0 Also, for a uniform external magnetic field, the system is translationally invariant, and Z ′ −βH = −βH + h dd xφ (x) = −βH + hφ (q = 0) , yielding ∂m = χ= ∂h

Z

dd q ′ d

(2π)

hφ (q′ ) φ (q = 0)i = ψ (0) .

******** 2. Crystal anisotropy:

Consider a ferromagnet with a tetragonal crystal structure. Cou-

pling of the spins to the underlying lattice may destroy their full rotational symmetry. The resulting anisotropies can be described by modifying the Landau–Ginzburg Hamiltonian to

K t 2 r 2 2 2 2 2 2 βH = d x (∇m) ~ + m ~ +u m ~ + m1 + v m1 m ~ , 2 2 2 Pn where m ~ ≡ (m1 , · · · , mn ), and m ~ 2 = i=1 m2i (d = n = 3 for magnets in three dimensions).

Z

d

Here u > 0, and to simplify calculations we shall set v = 0 throughout.

(a) For a fixed magnitude |m|; ~ what directions in the n component magnetization space are selected for r > 0, and for r < 0?

74

• r > 0 discourages ordering along direction 1, and leads to order along the remaining

(n − 1) directions.

r < 0 encourages ordering along direction 1. (b) Using the saddle point approximation, calculate the free energies (ln Z) for phases uniformly magnetized parallel and perpendicular to direction 1.

• In the saddle point approximation for m(x) ~ = mê1 , we have

t+r 2 4 ln Zsp = −V min , m + um 2 m where V =

R

dd x, is the system volume. The minimum is obtained for 3

(t + r)m + 4um = 0,

=⇒

m=

For t + r < 0, the free energy is given by fsp = −

0p

for t + r > 0 . −(t + r)/4u for t + r < 0

ln Zsp (t + r)2 =− .

V 16u

When the magnetization is perpendicular to direction 1, i.e. for m(x) ~ = mêi for i 6= 1, the

corresponding expressions are

t 2 4 ln Zsp = −V min m + um , 2 m

3

tm + 4um = 0,

and the free energy for t < 0 is fsp = −

m=

0p −t/4u

for t > 0 , for t < 0

t2 .

16u

(c) Sketch the phase diagram in the (t, r) plane, and indicate the phases (type of order), and the nature of the phase transitions (continuous or discontinuous).

• The saddle point phase diagram is sketched in the figure. (d) Are there Goldstone modes in the ordered phases?

• There are no Goldstone modes in the phase with magnetization aligned along direction 1, as the broken symmetry in this case is discrete. However, there are (n − 2) Goldstone

modes in the phase where magnetization is perpendicular to direction 1.

(e) For u = 0, and positive t and r, calculate the unperturbed averages hm1 (q)m1 (q′ )i0 and hm2 (q)m2 (q′ )i0 , where mi (q) indicates the Fourier transform of mi (x).

75

r disordered phase

1

discontinuous phase transition

continuous phase transitions

t 1

• The Gaussian part of the Hamiltonian can be decomposed into Fourier modes as βH0 =

Z

d

d q (2π)d

"

n X t

#

t+r K 2 2 q |m(q)| ~ + |m1 (q)|2 + |mi (q)|2 .

2 2 2 i=2

From this form we can easily read off the covariances  (2π)d δ d (q + q′ )  ′   hm1 (q)m1 (q )i0 = t + r + Kq 2 . d d ′    hm2 (q)m2 (q′ )i = (2π) δ (q + q ) 0 t + Kq 2 R (f) Write the fourth order term U ≡ u dd x(m ~ 2 )2 , in terms of the Fourier modes mi (q).

R dd q • Substituting mi (x) = (2π)d exp(iq · x)mi (q) in the quartic term, and integrating over

x yields U =u

Z

d

2 2

d x m ~

=u

Z

n

dd q1 dd q2 dd q3 X mi (q1 )mi (q2 )mj (q3 )mj (−q1 − q2 − q3 ).

(2π)3d i,j=1

(g) Treating U as a perturbation, calculate the first order correction to hm1 (q)m1 (q′ )i.

(You can leave your answers in the form of some integrals.)

76

• In first order perturbation theory hOi = hOi0 − (hOUi0 − hOi0 hUi0 ), and hence ′

′

hm1 (q)m1 (q )i = hm1 (q)m1 (q )i0 − u

Z

n

dd q1 dd q2 dd q3 X (2π)3d

i,j=1

c hm1 (q)m1 (q )mi (q1 )mi (q2 )mj (q3 )mj (−q1 − q2 − q3 )i0

Z (2π)d δ d (q + q′ ) u 4 8 dd k 4(n − 1) = 1− + + t + r + Kq 2 t + r + Kq 2 (2π)d t + Kk 2 t + r + Kk 2 t + r + Kk 2 ′

The last result is obtained by listing all possible contractions, and keeping track of how many involve m1 versus mi6=1 . The final result can be simplified to (2π)d δ d (q + q′ ) hm1 (q)m1 (q )i = t + r + Kq 2 ′

u 1− t + r + Kq 2

Z

dd k n−1 3 + (2π)d t + Kk 2 t + r + Kk 2

(h) Treating U as a perturbation, calculate the first order correction to hm2 (q)m2 (q′ )i.

• Similar analysis yields ′

′

hm2 (q)m2 (q )i = hm2 (q)m2 (q )i0 − u

Z

n

dd q1 dd q2 dd q3 X (2π)3d i,j=1

c hm2 (q)m2 (q′ )mi (q1 )mi (q2 )mj (q3 )mj (−q1 − q2 − q3 )i0

Z (2π)d δ d (q + q′ ) u dd k 4(n − 1) 4 8 = 1− + + t + Kq 2 t + Kq 2 (2π)d t + Kk 2 t + r + Kk 2 t + Kk 2

Z (2π)d δ d (q + q′ ) u 1 dd k n+1 = 1 − . + 2 2 d 2 t + Kq t + Kq (2π) t + Kk t + r + Kk 2

(i) Using the above answer, identify the inverse susceptibility χ−1 22 , and then find the transition point, tc , from its vanishing to first order in u.

• Using the fluctuation–response relation, the susceptibility is given by dd q hm2 (q)m2 (q = 0)i d x hm2 (x)m2 (0)i = (2π)d

Z u dd k 1 1 n+1 1− = + . t t (2π)d t + Kk 2 t + r + Kk 2

χ22 =

Z

Z

d

Inverting the correction term gives χ−1 22 = t + 4u

Z

dd k n+1 1 + + O(u2 ). d 2 2 (2π) t + Kk t + r + Kk 77

The susceptibility diverges at tc = −4u

Z

dd k n + 1 1 + O(u2 ). + d 2 2 (2π) Kk r + Kk

(j) Is the critical behavior different from the isotropic O(n) model in d < 4? In RG language, is the parameter r relevant at the O(n) fixed point? In either case indicate the

universality classes expected for the transitions.

• The parameter r changes the symmetry of the ordered state, and hence the universality

class of the disordering transition. As indicated in the figure, the transition belongs to the O(n − 1) universality class for r > 0, and to the Ising class for r < 0. Any RG

transformation must thus find r to be a relevant perturbation to the O(n) fixed point. ******** 3. Cubic anisotropy– Mean-field treatment:

Consider the modified Landau–Ginzburg

Hamiltonian βH =

Z

d

d x

"

n X t 2 K 2 2 2 (∇m) ~ + m ~ + u(m ~ ) +v m4i 2 2 i=1

#

,

for an n–component vector m(x) ~ = (m1 , m2 , · · · , mn ). The “cubic anisotropy” term Pn 4 i=1 mi , breaks the full rotational symmetry and selects specific directions. (a) For a fixed magnitude |m|; ~ what directions in the n component magnetization space

are selected for v > 0 and for v < 0? What is the degeneracy of easy magnetization axes in each case?

• In the figures below, we indicate the possible directions of the magnetization which are

selected depending upon the sign of the coefficient v, for the simple case of n = 2:

m

e^2

e^2 m

m m

e^1

m

m

e^1

m

m v > 0 diagonal order

v < 0 cubic axis order 78

This qualitative behavior can be generalized for an n-component vector: For v > 0, m lies along the diagonals of a n-dimensional hypercube, which can be labelled as m m = √ (±1, ±1, . . . , ±1), n and are consequently 2n -fold degenerate. Conversely, for v < 0, m can point along any of the cubic axes êi , yielding m = ±mêi , which is 2n-fold degenerate. (b) What are the restrictions on u and v for βH to have finite minima? Sketch these regions of stability in the (u, v) plane.

• The Landau-Ginzburg Hamiltonian for each of these cases evaluates to  t v   βH = m 2 + um 4 + m 4 , 2 n t   βH = m 2 + um 4 + vm 4 , 2

if

v>0 ,

if

v<0

implying that there are finite minima provided that   u + v > 0, n  u + v > 0,

for

v > 0,

for

v < 0.

Above, we represent schematically the distinct regions in the (u, v) plane.

u+v =0

u

allowed v

unphysical 79

u+

v =0 n

(c) In general, higher order terms (e.g. u6 (m ~ 2 )3 with u6 > 0) are present and ensure stability in the regions not allowed in part (b); (as in case of the tricritical point discussed in earlier problems). With such terms in mind, sketch the saddle point phase diagram in the (t, v) plane for u > 0; clearly identifying the phases, and order of the transition lines.

• We need to take into account higher order terms to ensure stability in the regions not allowed in part b). There is a tricritical point which can be obtained after simultaneously solving the equations t + 4(u + v)m2 + 6u6 m4 = 0 t + 2(u + v)m2 + 2u6 m4 = 0

)

,

=⇒

t∗ =

(u + v)2 , 2u6

m2 = −

(u + v)2 .

2u6

The saddle point phase diagram in the (t, v) plane is then as follows:

(d) Are there any Goldstone modes in the ordered phases?

• There are no Goldstone modes in the ordered phases because the broken symmetry

is discrete rather than continuous. We can easily calculate the estimated value of the transverse fluctuations in Fourier space as hφt (q)φt (−q)i =

(2π)d vt , Kq 2 + u+v

from which we can see that indeed these modes become massless only when v = 0, i.e., when we retrieve the O(n) symmetry.

80

******** 4. Cubic anisotropy– ε–expansion: (a) By looking at diagrams in a second order perturbation expansion in both u and v show that the recursion relations for these couplings are 

du   =εu − 4C (n + 8)u2 + 6uv dℓ

 dv  =εv − 4C 12uv + 9v 2 dℓ

,

where C = Kd Λd /(t + KΛ2 )2 ≈ K4 /K 2 , is approximately a constant.

• Let us write the Hamiltonian in terms of Fourier modes Z dd q t + Kq 2 m(q) ~ · m(−q) ~ βH = (2π)d 2 Z d d q1 dd q2 dd q3 mi (q1 )mi (q2 ) mj (q3 )mj (−q1 − q2 − q3 ) +u (2π)3d Z d d q1 dd q2 dd q3 mi (q1 )mi (q2 ) mi (q3 )mi (−q1 − q2 − q3 ), +v (2π)3d where, as usual, we assume summation over repeated indices. In analogy to problem set 6, after the three steps of the RG transformation, we obtain the renormalized parameters:  ′ t =b−d z 2 t̃       K ′ =b−d−2 z 2 K̃   u′ =b−3d z 4 ũ     ′ v =b−3d z 4 ṽ

,

with t̃, K̃, ũ, and ṽ, are the parameters in the coarse grained Hamiltonian. The dependence of ũ and ṽ, on the original parameters can be obtained by looking at diagrams in a second order perturbation expansion in both u and v. Let us introduce diagrammatic representations of u and v, as i

j

i

i

i

j

i

i

u

v

81

Contributions to u q1 q2

i i

q1 q2 q1

q

q3

j j

q4 q1 + q2 q q j q3 i

i l q1 + q2 q q4 q q3 j i

q2

m l q1 + q2 q

q1

i i

q2

Contributions to v

q

q4 j j

q1 + q2 q

q3 q4

q1

d 2 2 2n u2 (t +KK 2 )2 l 2

2 2 4

d

q2 q1

d d 2 2 (t +KK 2 )2 l

u2

q2

q j q3 i i j q1 + q2 q q4 q q3 i j i l q1 + q2 q q4

d 2 d 6 6 2 v2 (t +KK 2 )2 l

d d 6 4 2uv (t +KK 2 )2 l

2 d d 4 4 2 u2 (t +KK 2 )2 l

d d 2 6 2uv (t +KK 2 )2 l

where, again we have set b = eδℓ . The new coarse grained parameters are ( ũ = u − 4C[(n + 8)u2 + 6uv]δℓ , ṽ = v − 4C[9v 2 + 12uv]δℓ which after introducing the parameter ǫ = 4 − d, rescaling, and renormalizing, yield the

recursion relations

 du   =ǫu − 4C[(n + 8)u2 + 6uv] dℓ .   dv =ǫv − 4C[9v 2 + 12uv] dℓ

(b) Find all fixed points in the (u, v) plane, and draw the flow patterns for n < 4 and n > 4. Discuss the relevance of the cubic anisotropy term near the stable fixed point in each case.

• From the recursion relations, we can obtain the fixed points (u∗ , v ∗ ). For the sake of

simplicity, from now on, we will refer to the rescaled quantities u = 4Cu, and v = 4Cv, in terms of which there are four fixed points located at  u∗ = v ∗ = 0 Gaussian fixed point     ǫ    Ising fixed point u∗ = 0 v ∗ =   9 ǫ . ∗ v∗ = 0 O(n) fixed point u =   (n + 8)        u∗ = ǫ v ∗ = ǫ(n − 4) Cubic fixed point 3n 9n 82

Linearizing the recursion relations in the vicinity of the fixed point gives d A= dℓ

δu δv

!

=

ǫ − 2(n + 8)u∗ − 6v ∗

− 6u∗

−12v ∗

u∗ ,v ∗

ǫ − 12u∗ − 18v ∗

!

δu δv

!

.

As usual, a positive eigenvalue corresponds to an unstable direction, whereas negative ones correspond to stable directions. For each of the four fixed points, we obtain: 1. Gaussian fixed point: λ1 = λ2 = ǫ, i.e., this fixed point is doubly unstable for ǫ > 0, as A=

ǫ

0

0

ǫ

!

.

2. Ising fixed point: This fixed point has one stable and one unstable direction, as ǫ 3 A= ǫ −4 3 

0 −ǫ



,

corresponding to λ1 = ǫ/3 and λ2 = −ǫ. Note that for u = 0, the system decouples into n

noninteracting 1-component Ising spins.

3. O(n) fixed point: The matrix 

 A= 

−ǫ 0

 ǫ (n + 8)  ,  (n − 4) ǫ (n + 8) −6

has eigenvalues λ1 = −ǫ, and λ2 = ǫ(n − 4)/(n + 8). Hence for n > 4 this fixed point has one stable an one unstable direction, while for n < 4 both eigendirections are stable. This

fixed point thus controls the critical behavior of the system for n < 4.

4. Cubic fixed point: The eigenvalues of (n + 8)  3 A= (n − 4) −4 3 

−

−2 4−n



ǫ  , n

are λ1 = ǫ(4 − n)/3n, λ2 = −ǫ. Thus for n < 4, this fixed point has one stable and one

unstable direction, and for n > 4 both eigendirections are stable. This fixed point controls critical behavior of the system for n > 4.

83

v

v

n<4

n> 4

u

u

i

q1

q q2

i

Kd d l 2nu (t + K 2 ) t~ = t+4

q i

j

q2

q1 i

q1

q i

4u

q2

Kd d

(t + K 2) l

Kd d

(t + K 2 )

[(n+2)u+3v]

Kd d = 2 t+4 dl (t + K 2) [(n+2)u+3v]

dt

Kd d l 6v (t + K 2 )

In the (u,v) plane, v ∗ = 0 for n < 4, and the cubic term is irrelevant, i.e., fluctuations restore full rotational symmetry. For n > 4, v is relevant, resulting in the following flows: (c) Find the recursion relation for the reduced temperature, t, and calculate the exponent ν at the stable fixed points for n < 4 and n > 4.

• Up to linear order in ǫ, the following diagrams contribute to the determination of t̃:

After linearizing in the vicinity of the stable fixed points, the exponent yt is given by

 (n + 2) 1   ǫ + O(ǫ2 )  + 1 2 4(n + 8) = yt = 2 − 4C[(n + 2)u∗ + 3v ∗ ], =⇒ ν =  yt   1 + (n − 1) ǫ + O(ǫ2 ) 2 6n

for

n<4 .

for

n>4

(d) Is the region of stability in the (u, v) plane calculated in part (b) of the previous problem enhanced or diminished by inclusion of fluctuations? Since in reality higher order terms will be present, what does this imply about the nature of the phase transition for a small negative v and n > 4?

84

• All fixed points are located within the allowed region calculated in 1b). However, not

all flows starting in classically stable regions are attracted to stable fixed point. If the RG

flows take a point outside the region of stability, then fluctuations decrease the region of stability. The domains of attraction of the fixed points for n < 4 and n > 4 are indicated in the following figures:

u+v =0 u domain of attraction unphysical

v u + nv = 0

n>4 Flows which are not originally located within these domains of attraction flow into the unphysical regions. The coupling constants u and v become more negative. This is the signal of a fluctuation induced first order phase transition, by what is known as the Coleman–Weinberg mechanism. Fluctuations are responsible for the change of order of the transition in the regions of the (u,v) plane outside the domain of attraction of the stable fixed points. (e) Draw schematic phase diagrams in the (t, v) plane (u > 0) for n > 4 and n < 4, identifying the ordered phases. Are there Goldstone modes in any of these phases close to the phase transition?

• From the recursion relation obtained in 2c) for the parameter t, we obtain the following

non-trivial t∗

1 t∗ = − [(n + 2)u∗ + 3v ∗ ] ∝ −ǫ 2

Therefore, the phase diagrams in the (t,v) plane is schematically represented as As mentioned above, only when n < 4 fluctuations restore the full rotational symmetry. The parameter v is renormalized to zero, and there are Goldstone modes at the (u,v) plane, but only near the second order phase transition, where Kξt−2 = tv/(u + v) → 0. In

the ordered phases, the renormalized value of v is finite, albeit small, indicating that there are no Goldstone modes.

85

v

v disordered phase isotropic order

diagonal order

t

disordered phase

second order boundary

second order boundary

cubic axis order

cubic axis order

rst order boundary

t

rst order boundary

n>4

n<4 ******** 5. Exponents:

Two critical exponents at second order are,  (n + 2) (n + 2)(n2 + 23n + 60) 2 1   ǫ ,  ν = 2 + 4(n + 8) ǫ + 8(n + 8)3  (n + 2) 2  η = ǫ .

2(n + 8)2

Use scaling relations to obtain ǫ–expansions for two or more of the remaining exponents α, β, γ, δ and ∆. Make a table of the results obtained by setting ǫ = 1, 2 for n = 1, 2 and 3; and compare to the best estimates of these exponents that you can find by other sources (series, experiments, etc.). ******** 6. Anisotropic criticality:

A number of materials, such as liquid crystals, are anisotropic

and behave differently along distinct directions, which shall be denoted parallel and perpendicular, respectively. Let us assume that the d spatial dimensions are grouped into n parallel directions xk , and d − n perpendicular directions x⊥ . Consider a one–component

field m(xk , x⊥ ) subject to a Landau–Ginzburg Hamiltonian, βH = βH0 + U , with βH0 = and

U =u

Z

Z

n

d xk d

d−n

x⊥

L 2 t 2 K 2 2 (∇k m) + (∇⊥ m) + m − hm , 2 2 2

dn xk dd−n x⊥ m4 .

86

(Note that βH depends on the first gradient in the xk directions, and on the second

gradient in the x⊥ directions.)

(a) Write βH0 in terms of the Fourier transforms m(qk , q⊥ ).

R • In terms of the Fourier modes m(q) = dd xeiq·x m(x), βH0 =

Z

dn qk dd−n q⊥ (2π)d

4 Kqk2 + Lq⊥ +t

2

!

|m(q)|2 − hm(q = 0).

(b) Construct a renormalization group transformation for βH0 , by rescaling coordinates such that q′k = b qk and q′⊥ = c q⊥ and the field as m′ (q′ ) = m(q)/z. Note that parallel

and perpendicular directions are scaled differently. Write down the recursion relations for K, L, t, and h in terms of b, c, and z. (The exact shape of the Brillouin zone is immaterial

at this stage, and you do not need to evaluate the integral that contributes an additive constant.)

• The unperturbed Hamiltonian describes a set of independent modes labelled by a wave-

vector q. We can integrate out the modes that correspond to short distance fluctuations, and then rescale the parameters as suggested above to get (βH0 )′ = δfb0 + Z dn q′ dd−n q′ k

(2π)d

⊥ −n −(d−n)

b

c

2

4

Kb−2 q ′ k + Lc−4 q ′ ⊥ + t 2

!

z 2 |m′ (q′ )|2 − zhm′ (q′ = 0).

The renormalized bare Hamiltonian has the same form as βH0 , with the renormalized parameters  ′ K = b−(n+2) c−(d−n) z 2 K       L′ = b−n c−(d−n+4) z 2 L   t′ = b−n c−(d−n) z 2 t     ′ h = zh

.

(c) Choose c(b) and z(b) such that K ′ = K and L′ = L. At the resulting fixed point calculate the eigenvalues yt and yh for the rescalings of t and h.

• The requirements of K ′ = K and L′ = L lead to z 2 = bn+2 cd−n and z 2 = bn cd−n+4 , respectively. Consistency of these equations requires c = b1/2 ,

and 87

d+n

z = b1+ 4 .

Inserting these results for c and z in the recursion relations for t and h, we find  ( ′ d+n −n− d−n +2+ 2  yt = 2 2 2 t = b t t =b , =⇒ . d+n  hh = 1 + d + n h′ = zh = b1+ 4 h 4 (d) Write the relationship between the (singular parts of) free energies f (t, h) and f ′ (t′ , h′ ) in the original and rescaled problems. Hence write the unperturbed free energy in the homogeneous form f (t, h) = t2−α gf (h/t∆ ), and identify the exponents α and ∆.

• The RG transformation preserves the partition function Z, and the free energy density is

f ∼ ln Z/V . Under the rescaling transformation the volume changes to V ′ = b−n c−(d−n) V ,

and hence

d+n d+n d−n f (t, h) = b−n− 2 f (t′ , h′ ) = b− 2 f b2 t, b1+ 4 h .

We can now rescale to a point at which b2 t ∼ 1, at which point

d+n h f (t, h) = t 4 g , d+n 1 t2+ 8 corresponding to α=2−

d+n , 4

and

∆=

1 d+n + .

2 8

(e) How does the unperturbed zero–field susceptibility χ(t, h = 0), diverge as t → 0?

• Taking two derivatives of the above free energy with respect to h, we obtain 1 d+n 2 d+n ∂ 2f 4 t2+ 8 gχ ∼t χ∼ 2 ∂h

h t

d+n 1 2+ 8

∼ t−1

(for h = 0) .

The zero-field susceptibility diverges with an exponent γ = 1 at t = 0. (The same result can also be obtained by examining the two point correlation function.)

In the remainder of this problem set h = 0, and treat U as a perturbation. (f) In the unperturbed Hamiltonian calculate the expectation value hm(q)m(q ′ )i0 , and the

corresponding susceptibility χ0 (q) = h|mq |2 i0 , where q stands for (qk , q⊥ ).

• By examining the variance of the unperturbed Gaussian weight, we observe that hm(q)m(q ′ )i0 = (2π)d δ d (q + q ′ )χ0 (q),

88

with

χ0 (q) =

1 4 t + Kqk2 + Lq⊥

.

(g) Write the perturbation U , in terms of the normal modes m(q).

• After writing each of the four factors of m(x) in U in terms of m(q), and integrating

over all x, we obtain

U =u

Z

dq1 dq2 dq3 m(q1 )m(q2 )m(q3 )m(−q1 − q2 − q3 ),

where dq ≡ dn qk dd−n q⊥ /(2π)d . (h) Using RG, or any other method, find the upper critical dimension du , for validity of the Gaussian exponents.

• Under the rescaling steps outlined in the earlier parts of the problem, the coefficient of

the quartic perturbation changes to lowest order as 3

u′ = u(bc)−3 z 4 = ub−3n b− 2 (d−n) b4+(d+n) ≡ ubyu ,

with yu = 4 −

d+n .

2

The upper critical dimension is identified as the point at which yu = 0, and thus du = 8 − n.

(i) Write down the expansion for hm(q)m(q ′ )i, to first order in U , and reduce the correction term to a product of two point expectation values.

• Averages in perturbation theory are calculated as hOi = hOi0 −(hOU i0 − hOi0 hU i0 )+· · ·.

Using the expression for U in terms of Fourier modes, we find

hm(q)m(q ′ )i =hm(q)m(q ′ )i0 Z − u dq1 dq2 dq3 hm(q)m(q ′ )m(q1 )m(q2 )m(q3 )m(−q1 − q2 − q3 )ic0 + O(u2 ), where the superscript ‘c’ in the first order correction stands for the cumulant. Using Wick’s theorem the average of the six factors of m can be reduced to 12 equivalent pairwise contractions, such that hm(q)m(q ′ )i =hm(q)m(q ′ )i0 Z −12u dq1 dq2 dq3 hm(q)m(q1 )i0 hm(q ′ )m(q2 )i0 hm(q3 )m(−q1 − q2 − q3 )i0 + O(u2 ). (j) Write down the expression for χ(q), in first order perturbation theory, and identify the transition point tc at order of u. (Do not evaluate any integrals explicitly.)

89

• The two point expectation value has the form hm(q)m(q ′ )i = (2π)d δ d (q + q ′ )χ(q), with Z 1 12u 1 χ(q) = − + O(u2 ) dk 2 4 2 4 2 4 2 t + Kqk + Lq⊥ (t + Kqk + Lq⊥ ) t + Kkk + Lk⊥ #−1 " Z 1 2 4 . = t + Kqk2 + Lq⊥ + 12u dk 4 + O(u ) t + Kkk2 + Lk⊥ The inverse susceptibility χ(q = 0) should diverge at the critical point, t = tc , which is thus identified as tc = −12u

Z

dn kk dd−n k⊥ 1 2 2 4 + O(u ). d (2π) Kkk + Lk⊥ ********

7. Long–range interactions between spins can be described by adding a term Z Z d d x dd yJ(|x − y|)m(x) ~ · m(y), ~ to the usual Landau–Ginzburg Hamiltonian.

(a) Show that for J(r) ∝ 1/r d+σ , the Hamiltonian can be written as Z dd q t + K 2 q 2 + K σ q σ + · · · βH = m(q) ~ · m(−q) ~ (2π)d 2 Z d d q1 dd q2 dd q3 +u m(q ~ 1 ) · m(q ~ 2 ) m(q ~ 3 ) · m(−q ~ 1 − q2 − q3 ) .

(2π)3d • After changing variables to R = (x + y)/2, r = (x − y)/2, and writing the coarse-grained spin as a sum of Fourier modes, the long-range interaction term becomes Z Λ d Z Z 1 d q1 dd q2 ~ · m(y) ~ d d d d m(x) = d Rd r m(q ~ 1 ) · m(q ~ 2 )eiq1 ·(R+r) eiq2 ·(R−r) d xd y |x − y|d+σ r d+σ (2π)2d Z Λ d Z Z Λ d d q 1 d q σ d 2iq·r = m(q) ~ · m(−q) ~ d r d+σ e =I q m(q) ~ · m(−q). ~ d (2π) |r| (2π)d R The final expression is obtained by a rescaling of r = x/|q|, and I = dd xe2ix1 /|x|d+σ is independent of q = |q|. Including the ever-present short-range parts, we can write the full Hamiltonian as

dd q t + K 2 q 2 + K σ q σ + . . . m(q) ~ · m(−q) ~ (2π)d 2 .

Z d d q1 dd q2 dd q3 +u m(q ~ 1 ) · m(q ~ 2 ) m(q ~ 3 ) · m(−q ~ 1 − q2 − q3 ) (2π)3d

βH =

Z

90

(b) For u = 0, construct the recursion relations for (t, K2 , Kσ ) and show that Kσ is irrelevant for σ > 2. What is the fixed Hamiltonian in this case?

• For u = 0, the Gaussian integral over the outer-most shell will give us an additive

constant. Subsequent length q ′ = bq, and spin m′ = m/z, rescalings lead to the following recursion relations

 ′ t = b−d z 2 t    K2′ = b−d−2 z 2 K2 .    ′ Kσ = b−d−σ z 2 Kσ

If σ > 2, the fixed point (t∗ , K2∗ , Kσ∗ ) = (0, 0, Kσ ) (obtained by choosing Kσ′ = Kσ ) has no basin of attraction, as (

t′ = b σ t K2′ = bσ−2 K2

indicate two relevant directions; consequently, this fixed point does not describe the usual critical point of the system. On the other hand, the fixed point (t∗ , K2∗ , Kσ∗ ) = (0, K2 , 0), for which z 2 = bd+2 , leads to (

t′ = b 2 t Kσ′ = b2−σ Kσ

,

i.e., it has one relevant and one irrelevant direction, and is therefore, the fixed point controlling standard critical behavior of the system. In conclusion, if σ > 2, Kσ is irrelevant, and the fixed Hamiltonian in this case is K2 βH = 2 ∗

Z

dd q 2 q m(q) ~ · m(−q). ~ (2π)d

(c) For σ < 2 and u = 0, show that the spin rescaling factor must be chosen such that Kσ′ = Kσ , in which case K2 is irrelevant. What is the fixed Hamiltonian now?

• Conversely, for σ < 2, Kσ is the only relevant parameter at the fixed point (t∗ , K2∗ , Kσ∗ ) =

(0, 0, Kσ ), which is the one controlling the critical behavior in this case. At this point, the spin scaling factor z, is chosen such that z 2 = bd+σ , and the fixed Hamiltonian is Kσ βH = 2 ∗

Z

dd q σ q m(q) ~ · m(−q). ~ (2π)d

(d) For σ < 2, calculate the generalized Gaussian exponents ν, η, and γ from the recursion relations. Show that u is irrelevant, and hence the Gaussian results are valid, for d > 2σ.

91

• The divergence of the correlation length ξ, is controlled by the exponent ν = 1/yt . Since

ξ ′ = ξo t′ −ν = b−yt ν ξ, and ξ ′ = ξ/b, the exponent ν is then ν = 1/σ.

For σ < 2, the Gaussian correlation function in Fourier space is χ(q) = (t + Kσ q σ )−1 , and consequently, χ = lim χ(q) = t−1 ,

with

q→0

γ = 1.

In addition, at the critical point χ(q) ∼

1 q 2−η

,

so

η = 2 − σ.

If u 6= 0, at first order it will be renormalized to u′ = b−3d z 4 u = b2σ−d u, indicating that u is irrelevant and the Gaussian results are valid provided that d > 2σ. (e) For σ < 2, use a perturbation expansion in u to construct the recursion relations for (t, Kσ , u) as in the text.

• Let us obtain the recursion relations for (t, Kσ , u). As shown in the text, the first order diagrams

and

are zero; and

q1

q

q2

u

contributes an additive constant.

2n =b (2d q)d (t + K1 q ) = 2nu (t +KKd ) l Z

d

d

q u

q1

q2

4 =b (2d q)d (t + K1 q ) = 4u (t +KKd ) l Z

d

d

92

The above two diagrams contribute to the coarse-grained parameter t. Their respective contributions are obtained after considering b = eδl , and expanding the exponential to linear order in the infinitesimal δl. The coarse-grained parameter t̃ is then t̃ = t + 2

(2n + 4)uKd Λd δl + O(u2 ). σ t + Kσ Λ

At this order, the other parameters in the coarse grained Hamiltonian remain unchanged (ũ = u, K̃2 = K2 ).

Taking into account the results of section 1.(b)-(c), after the usual steps 2 and 3 of the RG transformation, we obtain the following differential recursion relations for the infinitesimal rescaling factor b ∼ 1 + δl:  ′ t = bσ t̃    K2′ = bσ−2 K2 ,    ′ u = b2σ−d u

(n + 2)Kd Λd δl dt = σt + 4u + O(u2 ) dl t + Kσ Λσ dK2 . = (σ − 2)K2 + O(u2 )  dl       du = (2σ − d)u + O(u2 ) dl        

=⇒

For σ < 2 and d > 2σ, the Gaussian fixed point, which is the only solution of these equations, has only one relevant direction, describing correctly the phase transition. However, if d < 2σ, there are two relevant directions, and we have to calculate the recursion relations at the next order to find other non-trivial fixed points.

At second order in u, there are several new interactions proportional to m2 , m4 , m2 q 2 , which will appear in the coarse grained Hamiltonian. But of all these, we are essentially interested in the second order diagrams which contribute to the renormalization of m ~ 4 . As shown in the text, these are: q1

q

q2

q1 + q2

q1 q2

q3

q1 + q2 q

q1 + q2

q4

0

q

q

q1 q2

q3

0

q

q q4 q3 q4

u2

2 2 2 2n

u2

Z

dd q 1 d (t + K q )(t + K q1 + q2 (2

) =b

2 2 2 4 2 u2

2 4 4 2

j

Kd d 2 = 4 nu q j ) (t + K )2 l

Z

dd q 1 d

(2 =b ) (t + K q )(t + K jq1 + q2

Z

dd q 1 d

=b (2 ) (t + K q )(t + K q1 + q2

j

93

Kd 2 j = 16u (t + K )2 l

d

q )

Kd d 2 = 16 u

q ) (t + K )2 l

j

so that the new coarse grained parameters are  (n + 2)Kd Λd   t̃ = t + 4u δl − u2 Ct  σ  t + K Λ  σ     K̃2 = K2 − u2 CK2  K̃σ = Kσ       (n + 8)Kd Λd   ũ = u − 4u2 δl (t + Kσ Λσ )2

.

Let us introduce the parameter ǫ = 2σ − d. After rescaling and renormalizing, we obtain

the following recursion relations:

 (n + 2)Kd Λd dt   =σt + 4u − u2 Ct  σ dl t + Kσ Λ . d  du  2 (n + 8)Kd Λ  =ǫu − 4u dl (t + Kσ Λσ )2

(f) For d < 2σ, calculate the critical exponents ν and η to first order in ǫ = d − 2σ.

[See M.E. Fisher, S.-K. Ma and B.G. Nickel, Phys. Rev. Lett. 29, 917 (1972).]

• From the recursion relations, we can now obtain the non-trivial fixed point (t∗ , u∗ ), which satisfies dt/dl = du/dl = 0, as

 Kσ2 Kσ2 Λǫ ∗    u = 4(n + 8)K ǫ ≈ 4(n + 8)K d

d=2σ

 K Λσ (n + 2)   t∗ = − σ ǫ σ(n + 8)

ǫ ,

where, up to linear order in ǫ, we have set t = 0 in the denominators of both dt/dl and du/dl, and performed other similar simplifications.

Linearizing the recursion relations in the vicinity of the non-trivial fixed point gives d dl

δt δu

!



=

σ−

(n + 2) ǫ (n + 8) 2

O(ǫ )

... −ǫ

 

δt δu

!

.

Hence the eigenvalues can be read directly from the diagonal elements of the matrix. The first one is positive and controls the critical behavior, whereas the second is negative and consequently irrelevant for d < 2σ.

94

The new exponent ν is then 1 ν= = yt

−1 (n + 2) 1 (n + 2) σ− ǫ = + ǫ + O(ǫ2 ), 2 (n + 8) σ (n + 8)σ

and the exponent η remains the same as before, η = 2 − σ + O(ǫ2 ), since the parameter

Kσ has not been renormalized at linear order in ǫ. In fact, it can be shown that the RG

procedure only generates analytic corrections, and hence the coefficient of the non-analytic term Kσ is unrenormalized to all orders, resulting in η = 2 − σ, exactly. (g) What is the critical behavior if J(r) ∝ exp(−r/a)? Explain!

• A short-range interaction term of the form J(r) ∝ exp(−r/a) will modify the actual

value of the critical temperature Tc , and other non-universal quantities, but it will not change the value of the critical exponents, since its Fourier transform is analytic: Z =

d

−

d

d xd ym(x) ~ · m(y)e ~ Z Λ

|x−y| a

=

Z Λ

dd q m(q) ~ · m(−q) ~ (2π)d

dd q C(q)m(q) ~ · m(−q), ~ (2π)d

where C(q) = 2πa

d

Z ∞

dxx

d−1 −x

e

Jo (aqx) = a

0

d

∞ X

n=0

********

95

Z

r

dd r e− a e2iq·r

cn (qa)2n .

Solutions to problems from chapter 6- Lattice Systems 1. Cumulant method: Apply the Niemeijer–van Leeuwen first order cumulant expansion P to the Ising model on a square lattice with −βH = K <ij> σi σj , by following these steps:

(a) For an RG with b = 2, divide the bonds into intra–cell components βH0 ; and inter–cell

components U .

• The N sites of the square lattice are partitioned into N/4 cells as indicated in the figure below (the intra-cell and inter-cell bonds are represented by solid and dashed lines respectively).

intra-cell bonds inter-cell bonds boundary of new unit blocks

The renormalized Hamiltonian βH′ [σα′ ] is calculated from βH′ [σα′ ] = − ln Z0 [σα′ ] + hUi0 −

1 2 2 U 0 − hUi0 + O U 3 , 2

where hi0 indicates expectation values calculated with the weight exp(−βH0 ) at fixed [σα′ ]. (b) For each cell α, define a renormalized spin σα′ = sign(σα1 + σα2 + σα3 + σα4 ). This choice P4 becomes ambiguous for configurations such that i=1 σαi = 0. Distribute the weight of

these configurations equally between σα′ = +1 and −1 (i.e. put a factor of 1/2 in addition

to the Boltzmann weight). Make a table for all possible configurations of a cell, the internal probability exp(−βH0 ), and the weights contributing to σα′ = ±1.

96

0 = 1

Weight

Weight

+

+

+

+

+

+

-

-

e4K

4

0 = 1

1 12

-

+

+

+

+

-

-

+

4

1

2

e 4K 21 Weight

Weight

-

-

-

-

+

+

-

-

e4K

4

1 12

+

-

-

-

+

-

-

+

4

1

2

e 4K 21

• The possible intracell configurations compatible with a renormalized spin σα′ = ±1, and

their corresponding contributions to the intra-cell probability exp(−βH0 ), are given below, resulting in Z0 [σα′ ] =

Y α

N/4 e4K + 6 + e−4K = e4K + 6 + e−4K .

(c) Express hUi0 in terms of the cell spins σα′ ; and hence obtain the recursion relation K ′ (K).

• The first cumulant of the interaction term is − hUi0 = K

X

hα,βi

hσα2 σβ1 + σα3 σβ4 i0 = 2K

1

2

1

α 4

2

β 3

4

97

3

X

hα,βi

hσα2 i0 hσβ1 i0 ,

where, for σα′ = 1, hσαi i0 =

e4K + (3 − 1) + 0 + 0 e4K + 2 = . (e4K + 6 + e−4K ) (e4K + 6 + e−4K )

Clearly, for σα′ = −1 we obtain the same result with a global negative sign, and thus hσαi i0 = σα′

e4K + 2 . (e4K + 6 + e−4K )

As a result, N ln −βH [σα′ ] = ′

4

4K

e

−4K

+6+e

+ 2K

e4K + 2 e4K + 6 + e−4K

2 X

σα′ σβ′ ,

hα,βi

corresponding to the recursion relation K ′ (K), ′

K = 2K

e4K + 2 e4K + 6 + e−4K

2

.

(d) Find the fixed point K ∗ , and the thermal eigenvalue yt . ∗

• To find the fixed point with K ′ = K = K ∗ , we introduce the variable x = e4K . Hence,

we have to solve the equation

1 x+2 √ = , x + 6 + x−1 2

√

or

√ 2 − 1 x2 − 6 − 2 2 x − 1 = 0,

whose only meaningful solution is x ≃ 7.96, resulting in K ∗ ≃ 0.52.

To obtain the thermal eigenvalue, let us linearize the recursion relation around this non-trivial fixed point, ∂K ′ = byt , ∂K K ∗

⇒

2

yt

= 1 + 8K

∗

∗ ∗ ∗ e4K e4K − e−4K − , e4K ∗ + 2 e4K ∗ + 6 + e−4K ∗

(e) In the presence of a small magnetic field h

⇒

yt ≃ 1.006.

P

i σi , find the recursion relation for h; and

calculate the magnetic eigenvalue yh at the fixed point.

• In the presence of a small magnetic field, we will have an extra contribution to the Hamiltonian

h

X α,i

hσα,i i0 = 4h

X e4K + 2 σ′ . (e4K + 6 + e−4K ) α α 98

Therefore,

e4K + 2 . h = 4h 4K (e + 6 + e−4K ) ′

(f) Compare K ∗ , yt , and yh to their exact values.

• The cumulant method gives a value of K ∗ = 0.52, while the critical point of the Ising model on a square lattice is located at Kc ≈ 0.44. The exact values of yt and yh for the two dimensional Ising model are respectively 1 and 1.875, while the cumulant method yields yt ≈ 1.006 and yh ≈ 1.5. As in the case of a triangular lattice, yh is lower than the exact result. Nevertheless, the thermal exponent yt is fortuitously close to its exact value. ******** 2. Migdal–Kadanoff method: Consider Potts spins si = (1, 2, · · · , q), on sites i of a hypercubic lattice, interacting with their nearest neighbors via a Hamiltonian X δsi ,sj . −βH = K <ij>

(a) In d = 1 find the exact recursion relations by a b = 2 renormalization/decimation process. Indentify all fixed points and note their stability.

• In d = 1, if we average over the q possible values of s1 , we obtain ( q X q − 1 + e2K if σ1 = σ2 ′ ′ K(δσ1 s1 +δs1 σ2 ) e = = eg +K δσ1 σ2 , q − 2 + 2eK if σ1 6= σ2 s1 =1

from which we arrive at the exact recursion relations: ′ ′ q − 1 + e2K eK = , eg = q − 2 + 2eK .

K q − 2 + 2e

To find the fixed points we set K ′ = K = K ∗ . As in the previous problem, let us ∗ introduce the variable x = eK . Hence, we have to solve the equation q − 1 + x2 x= , or x2 + (q − 2)x − (q − 1) = 0, q − 2 + 2x whose only meaningful solution is x = 1, resulting in K ∗ = 0. To check its stability, we consider K ≪ 1, so that

q + 2K + 2K 2 K2 ′ K ≃ ln ≃ ≪ K, q + 2K + K 2 q which indicates that this fixed point is stable.

In addition, K ∗ → ∞ is also a fixed point. If we consider K ≫ 1, ′ 1 eK ≃ eK , =⇒ K ′ = K − ln 2 < K, 2 which implies that this fixed point is unstable.

*

* 99

(b) Write down the recursion relation K ′ (K) in d–dimensions for b = 2, using the Migdal– Kadanoff bond moving scheme.

• In the Migdal-Kadanoff approximation, moving bonds strengthens the remaining bonds by a factor 2d−1 . Therefore, in the decimated lattice we have d−1

K′

e

q − 1 + e2×2 K . = q − 2 + 2e2d−1 K

(c) By considering the stability of the fixed points at zero and infinite coupling, prove the existence of a non–trivial fixed point at finite K ∗ for d > 1.

• In the vicinity of the fixed point K ∗ = 0, i.e. for K ≪ 1, K′ ≃

22d−2 K 2 ≪ K, q

and consequently, this point is again stable. However, for K ∗ → ∞, we have ′

eK ≃

1 exp 2d − 2d−1 K , 2

=⇒

K ′ = 2d−1 K − ln 2 ≫ K,

which implies that this fixed point is now stable provided that d > 1.

• As a result, there must be a finite K ∗ fixed point, which separates the flows to the other

fixed points.

*

*

*

(d) For d = 2, obtain K ∗ and yt , for q = 3, 1, and 0.

• Let us now discuss a few particular cases in d = 2. For instance, if we consider q = 3,

the non-trivial fixed point is a solution of the equation x=

2 + x4 , 1 + 2x2

or

x4 − 2x3 − x + 2 = (x − 2)(x3 − 1) = 0,

which clearly yields a non-trivial fixed point at K ∗ = ln 2 ≃ 0.69. The thermal exponent

for this point

4K ∗ ∗ e2K 16 ∂K ′ e yt − = , = 2 = 4 4K ∗ ∗ 2K ∂K K ∗ e + 2 1 + 2e 9

=⇒

yt ≃ 0.83,

which can be compared to the exact values, K ∗ = 1.005, and yt = 1.2.

100

By analytic continuation for q → 1, we obtain ′

eK =

e4K . −1 + 2e2K

The non-trivial fixed point is a solution of the equation x=

x4 , −1 + 2x2

or

(x3 − 2x2 + 1) = (x − 1)(x2 − x − 1) = 0,

whose only non-trivial solution is x = (1 +

√

5)/2 = 1.62, resulting in K ∗ = 0.48. The

thermal exponent for this point h i ∗ ∂K ′ = 2yt = 4 1 − e−K , ∂K K ∗

=⇒

yt ≃ 0.61.

As discussed in the next problem set, the Potts model for q → 1 can be mapped onto the problem of bond percolation, which despite being a purely geometrical phenomenon, shows many features completely analogous to those of a continuous thermal phase transition.

And finally for q → 0, relevant to lattice animals (see PS#9), we obtain ′

eK =

−1 + e4K , −2 + 2e2K

for which we have to solve the equation x=

−1 + x4 , −2 + 2x2

or

x4 − 2x3 + 2x − 1 = (x − 1)3 (x + 1) = 0,

whose only finite solution is the trivial one, x = 1. For q → 0, if K ≪ 1, we obtain K′ ≃ K +

K2 > K, 2

indicating that this fixed point is now unstable. Note that the first correction only indicates marginal stability (yt = 0). Nevertheless, for K ∗ → ∞, we have ′

eK ≃

1 exp[2K], 2

=⇒

K ′ = 2K − ln 2 ≫ K,

which implies that this fixed point is stable.

*

* 101

******** 3. The Potts model:

The transfer matrix procedure can be extended to Potts model,

where the spin si on each site takes q values si = (1, 2, · · · , q); and the Hamiltonian is PN −βH = K i=1 δsi ,si+1 + KδsN ,s1 .

(a) Write down the transfer matrix and diagonalize it. Note that you do not have to solve a q th order secular equation as it is easy to guess the eigenvectors from the symmetry of the matrix.

• The partition function is Z=

X

{si }

< s1 |T |s2 >< s2 |T |s3 > · · · < sN−1 |T |sN >< sN |T |s1 >= tr(T N ),

where < si |T |sj >= exp Kδsi ,sj is a q × q transfer matrix. The diagonal elements of the

matrix are eK , while the off-diagonal elements are unity. The eigenvectors of the matrix are easily found by inspection. There is one eigenvectors with all elements equal; the corresponding eigenvalue is λ1 = eK + q − 1. There are also (q − 1) eigenvectors orthogonal

to the first, i.e. the sum of whose elements is zero. This corresponding eigenvalues are degenerate and equal to eK − 1. Thus Z=

X α

K λN α = e +q−1

N

+ (q − 1) eK − 1

N

.

(b) Calculate the free energy per site.

• Since the largest eigenvalue dominates for N ≫ 1,

ln Z = ln eK + q − 1 .

N (c) Give the expression for the correlation length ξ (you don’t need to provide a detailed derivation), and discuss its behavior as T = 1/K → 0.

• Correlations decay as the ratio of the eigenvalues to the power of the separation. Hence

the correlation length is

−1 K −1 λ1 e +q−1 ξ = ln . = ln λ2 eK − 1 102

In the limit of K → ∞, expanding the above result gives 1 eK = exp ξ≃ q q

1 .

T

******** 4. The spin–1 model:

Consider a linear chain where the spin si at each site takes on

three values si = −1, 0, +1. The spins interact via a Hamiltonian −βH =

X

Ksi si+1 .

i

′

(a) Write down the transfer matrix hs|T |s′ i = eKss explicitly.

• The explicit form of the 3x3 transfer matrix is   K e 1 e−K 1 1 .

T = 1 −K e 1 eK

(b) Use symmetry properties to find the largest eigenvalue of T and hence obtain the expression for the free energy per site (ln Z/N ).   1  • Because of the symmetry of T , we guess an eigenvector of the form v1 = r . For an 1 eigenvalue λ, we then obtain the pair of equations

( K e + e−K + r = λ 2 = λ − 2 cosh K. , =⇒ r = λ−1 2 + r = rλ This leads to the quadratic equation λ2 − λ (1 + 2 cosh K) − 2 (1 − cosh K) = 0, which has solutions 2λ± = 1 + 2 cosh K ±

p 1 + 4 cosh K + 4 cosh2 K + 8 − 8 cosh K.

The partition function is related to the largest eigenvalue in the limit of large N , and hence i h p ln Z = ln λ+ = ln 1 + 2 cosh K + 1 + 4 cosh K + 4 cosh2 K + 8 − 8 cosh K − ln 2.

N 103

(c) Obtain the expression for the correlation length ξ, and note its behavior   as K → ∞.

1 • The second largest eigenvalue corresponds to the eigenvector v2 =  0 , and is equal −1 to λ2 = eK − e−K = 2 sinh K. The ratio of eigenvalues is related to the correlation length by # " p 2 1 1 + 4 cosh K + 4 cosh K + 8 − 8 cosh K 1 + 2 cosh K + ξ= . = ln−1 ln(λ1 /λ2 ) 4 sinh K In the limit of large K,

λ1 ≃ eK 1 + 4e−2K + · · · ,

and

such that the correlation length diverges as

λ2 ≃ eK 1 − 2e−2K + · · · ,

e2K . lim ξ(K) ≃ K→∞ 5

(d) If we try to perform a renormalization group by decimation on the above chain we find that additional interactions are generated. Write down the simplest generalization of βH whose parameter space is closed under such RG. • The only symmetry present in the Hamiltonian is inversion, {si } → {−si }, and the

most general spin-1 Hamiltonian consistent with this symmetry is B(s, s′ ) = g + Kss′ +

µ 2 s + s′2 + ∆(ss′ )2 .

2

******** ectionClock model: Each site of the lattice is occupied by a q-valued spin si ≡

1, 2, · · · , q, with an underlying translational symmetry modulus q, i.e. the system is invariant under si → (si + n)modq . The most general Hamiltonian subject to this symmetry

with nearest–neighbor interactions is

βHC = −

X

<i,j>

J(|si − sj |modq ),

104

where J(n) is any function, e.g. J(n) = J cos(2πn/q). Potts models are a special case of Clock models with full permutation symmetry, and Ising model is obtained in the limit of q = 2. (e) For a closed linear chain of N clock spins subject to the above Hamiltonian show that the partition function Z = tr [exp(−βH)] can be written as Z = tr [hs1 |T |s2 ihs2 |T |s3 i · · · hsN |T |s1 i] ; where T ≡ hsi |T |sj i = exp [J(si − sj )] is a q × q transfer matrix.

• This is a simple generalization of the previous examples, and again the partition function

for N sites with period boundary conditions is Z = tr T N , with the q × q matrix whose

elements are T ≡ hsi |T |sj i = exp [J(si − sj )].

(f) Write down the transfer matrix explicitly and diagonalize it. Note that you do not have to solve a q th order secular equation; because of the translational symmetry, the eigenvalues are easily obtained by discrete Fourier transformation as q X

2πink λ(k) = exp J(n) + . q n=1 • Because of the symmetry T ≡ hsi |T |sj i = T (|s −s′ |modq ), we anticipate that eigenvectors

of the matrix have the simple form 1 ωk ω 2k .. .



 1  vk = √  q 



ω (q−1)k

  ,  

ω = e2πi/q

where

is a q th root of unity.

It can then be explicitly verified that T vk = λk vk , with q X

2πink . λk = exp J(n) + q n=1

(g) Show that Z =

Pq

k=1 λ(k)

N

≈ λ(0)N for N → ∞. Write down the expression for the

free energy per site βf = − ln Z/N .

105

• For periodic boundary conditions Z = tr T

N

N N = λN 0 + λ1 + · · · = λ0

"

1+

N q−1 X λk k=1

λ0

#

.

For N → ∞, since λ0 is the largest eigenvalue, we obtain # "q−1 X ln Z βf = − = − ln λ0 = − ln eJ(k) .

N k=0

(h) Show that the correlation function can be calculated from δsi ,si+ℓ

q

1 X tr Πα T ℓ Πα T N−ℓ , = Z α=1

where Πα is a projection matrix. Hence show that δsi ,si+ℓ c ∼ [λ(1)/λ(0)]ℓ. (You do not

have to explicitly calculate the constant of proportionality.)

• Consider the matrix Πα that has zero elements everywhere, except for a 1 for its (α, α) element. Then

δsi ,si+ℓ

q

1 X tr Πα T ℓ Πα T N−ℓ . = Z α=1

Evaluating the trace in the basis of (Fourier) eigenvectors gives q

δsi ,si+ℓ

1 XX = < k ′ |Πα |k > λℓk < k|Πα |k ′ > λN−ℓ . k′ Z α=1 ′ k,k

The main contributions to the sum come from the the two largest eigenvalues, λ0 and λ1 ; hence δsi ,si+ℓ

" ℓ #

q ∗ α 1 1 λ1 (ωω ) 1 X N 1 λ0 2 + λN−ℓ 1+ ≃ e−ℓ/ξ . = λℓ1 = N 0 2 q q q λ0 q λ0 α=1

(Note that if the second eigenvalue is degenerate, there will also be a degeneracy factor.)

The correlation length is given as −1

ξ = ln

λ1 λ0

******** 106

.

5. XY model:

Consider two component unit spins s~i = (cos θi , sin θi ) in one dimension, PN with the nearest neighbor interactions described by −βH = K i=1 s~i · si+1 ~ .

(a) Write down the transfer matrix hθ|T |θ ′ i, and show that it can be diagonalized with eigenvectors fm (θ) ∝ eimθ for integer m.

• In the basis of the angles {θi },

hθ|T |θ ′ i = exp [K cos (θ − θ ′ )] .

The above transfer ‘matrix’ is diagonalized in the Fourier basis with hθ|mi = eimθ (not

normalized), as it is easily checked that ′

′

hθ|T |θ i hθ |mi =

Z 2π 0

dθ ′ K cos(θ−θ′ )+imθ′ e = eimθ 2π

with λm =

Z 2π 0

Z 2π 0

dφ K cos(φ)+imφ e = λm hθ|mi, 2π

dφ K cos(φ)+imφ e .

2π

(b) Calulate the free energy per site, and comment on the behavior of the heat capacity as T ∝ K −1 → 0.

• For large K the eigenvalues can be approximated by the saddle point method as eK λm ≃ 2π

Z ∞

2 eK − m2 e−Kθ /2+imθ = √ e 2K .

2πK −∞

They are ordered with λ0 being the largest, and thus 1 ln Z ≃ ln λ0 ≈ K − ln(2πK) + · · · .

N 2 The dimensionless parameter K can be related to a bond energy J by K = βJ, where β = 1/(kB T ). Thus the energy of the chain in the limit of low temperatures is given by

∂ ln Z ∂ ln Z 1 kB T E=− =J = −N J 1 − + · = −N J + N + ···. ∂β ∂K 2K 2 The first term is just the energy of the perfectly ordered chain. The second is the energy of excitations into the harmonic modes. Since there are N such quadratic modes, the potential energy of excitations is N × (kB T )/2. Note that since quantum effects are ignored, there

is no quenching of the oscillator energies, and a finite heat capacity of N kB /2 at zero temperature.

107

(c) Find the correlation length ξ, and note its behavior as K → ∞.

• As usual, the correlation length is related to the ratio of the two largest eigenvalues as −1

ξ = − ln

λ1 λ0

≈(K→∞) 2K ∝

1 .

T

Note that the divergence of the correlation length as T → 0 is very different for cases where there is a gap in the excitation energy (above the ground state), and when there is none. ******** 6. One dimensional gas:

The transfer matrix method can also be applied to a one

dimensional gas of particles with short-range interactions, as described in this problem. (a) Show that for a potential with a hard core that screens the interactions from further neighbors, the Hamiltonian for N particles can be written as H=

N X p2 i

2m i=1

+

N X i=2

V(xi − xi−1 ).

The (indistinguishable) particles are labeled with coordinates {xi } such that 0 ≤ x1 ≤ x2 ≤ · · · ≤ xN ≤ L, where L is the length of the box confining the particles.

• Each particle i interacts only with adjacent particles i − 1 and i + 1, as the hard cores from these nearest neighbors screen the interactions with any other particle. Thus we need only consider nearest neighbor interactions, and, included the kinetic energies, the Hamiltonian is given by H=

N X p2 i

2m i=1

+

N X i=2

V(xi − xi−1 ),

for

0 ≤ x1 ≤ x2 ≤ · · · xN ≤ L.

(b) Write the expression for the partition function Z(T, N, L). Change variables to δ1 = x1 , δ2 = x2 − x1 , · · · , δN = xN − xN−1 , and carefully indicate the allowed ranges of integration and the constraints.

108

• The partition function is 1 Z(T, N, L) = N h

Z L

dx1

0

Z L x1

dx2 · · ·

Z L

xN −1

"

dxN exp −β "

N X i=2

V(xi − xi−1 )

#

# p2i · dp1 · · · dpN exp −β N 2m −∞ −∞ i=1 # " Z L Z L Z L N X 1 V(xi − xi−1 ) , dx1 dx2 · · · dxN exp −β = N λ 0 x1 xN −1 Z ∞

Z ∞

X

i=2

√ where λ = h/ 2πmkB T . (Note that there is no N ! factor, as the ordering of the particles is specified.) Introducing a new set of variables δ2 = x2 − x1 ,

δ1 = x1 ,

···

δn = xN − xN−1 ,

or equivalently x1 = δ1 ,

x2 = δ1 + δ2 ,

···

xN =

1 Z(T, N, L) = N λ

dδ1

0

Z L−δ1

dδ2

Z L−(δ1 +δ2 ) 0

0

δi ,

i=1

the integration becomes Z L

N X

dδ3 · · ·

Z L−PN

′

"

i=1

δi

−β

dδN e

0

PN

i=2

V (δi ) .

This integration can also be expressed as 1 Z(T, N, L) = N λ

Z

dδ1 dδ2 · · · dδN

with the constraint 0≤

N X i=1

exp −β

N X i=2

#

V(δi ) ,

δi ≤ L.

This constraint can be put into the equation explicitly with the use of the step function ( 0 for x<0 Θ(x) = , 1 for x≥0 as 1 Z(T, N, L) = N λ

Z ∞ 0

dδ1

Z ∞ 0

dδ2 · · ·

Z ∞ 0

"

dδN exp −β

109

N X i=2

#

V(δi ) Θ L −

N X i=1

δi

!

.

(c) Consider the Gibbs partition function obtained from the Laplace transformation Z(T, N, P ) =

Z ∞

dL exp(−βP L)Z(T, N, L),

0

and by extremizing the integrand find the standard formula for P in the canonical ensemble.

• The Gibbs partition function is Z(T, N, P ) =

Z ∞

dL exp(−βP L)Z(T, N, L).

0

The saddle point is obtained by extremizing the integrand with respect to L, ∂ = 0, exp(−βP L)Z(T, N, L) ∂L T,N which implies that βP =

∂ , ln Z(T, N, L) ∂L T,N

=⇒

P = kB T

∂ ln Z . ∂L T,N

From thermodynamics, for a one–dimensional gas we have dF = −SdT − P dL,

=⇒

P =−

∂F . ∂L T,N

Further noting that F = −kB T ln Z, again results in Pcanonical = kB T

(d) Change variables from L to δN+1 = L−

∂ ln Z . ∂L T,N

PN

i=1 δi , and find the expression for Z(T, N, P )

as a product over one-dimensional integrals over each δi .

• The expression for the partition function given above is 1 Z(T, N, L) = N λ

Z ∞ 0

dδ1

Z ∞ 0

dδ2 · · ·

Z ∞ 0

"

dδN exp −β

110

N X i=2

#

V(δi ) Θ L −

N X i=1

δi

!

.

The Laplace transform of this equation is 1 Z(T, N, P ) = N λ

Z ∞

Z ∞

Z ∞

dL exp(−βP L) dδ1 dδ2 · · · 0 0 ! # " N N X X δi V(δi ) Θ L − · exp −β 0

1 = N λ βP

dδ1

0

Z ∞ 0

1 = N λ (βP )2

dδ2 · · ·

Z ∞ 0

dδN

#

"

0

i=1

i=2

Z ∞

Z ∞

Z ∞ 0

dδ2 · · ·

"

dδN exp −β

Z ∞ 0

(

N X i=2

dδN exp −

V(δi ) exp −βP

N X

N X

δ1

i=1

)

!#

[βV(δi ) + βP δi ] .

i=2

Since the integrals for different δi′ s are equivalent, we obtain 1 Z(T, N, P ) = N λ (βP )2

Z ∞

dδ exp [−β (V(δ) + P δ)]

0

N−1

.

This expression can also be obtained directly, without use of the step function as follows.

1 Z(T, N, P ) = N λ

Z L

dδ1

dδ2

·

Z ∞

1 = N λ

Z L

Z L−(δ1 +δ2 ) 0

0

0

0

Z L−δ1 "

N X

dL exp −βP L − β

dδ1

0

Z L−δ1 0

(

· exp −βP

Change variables to δN+1 ≡ L −

i=1

δi

dδN

N X i=1

Z ∞ −

0

L−

δi

dδN

V(δ)

i=1

δi +

i=1

dδ3 · · · 0 !#

Z L−PN

dδ2 · · ·

"N X

i=2

Z L−PN

δi

!#

−β

PN

i=1

N X i=2

δi

d L−

V(δi )

!)

N X i=1

δi

!

.

PN

′ i=1 δi , and note that each of the δ s indicates the

distance between neighboring particles. The size of the gas L, has been extended to any value, hence each δ can be varied independently from 0 to ∞. Thus the Gibbs partition 111

function is 1 Z(T, N, P ) = N λ

1 = N λ · 1 = N λ (βP )2

Z ∞

Z ∞

dδ2 · · ·

· exp −βP

δi

0

dδ1 "

Z ∞

0 Z ∞

0

N+1 X i=1

Z ∞

dδN

0

!

−β

Z ∞

0 N X i=2

dδ · exp [−βV(δ) − βP δ]

dδN+1 !#

V(δi )

N−1 Z ∞

dδ1 exp(−βP δ1 )

0

dδN+1 exp(−βP δN+1 )

0

Z ∞

dδ exp [−β (V(δ) + P δ)]

0

N−1

.

(e) At a fixed pressure P , find expressions for the mean length L(T, N, P ), and the density n = N/L(T, N, P ) (involving ratios of integrals which should be easy to interpret).

• The mean length is ∂ ln Z(T, N, P ) ∂(βP ) T,N R∞ dδ · δ · exp [−βV(δ) − βP δ] 2 , = + (N − 1) 0R ∞ βP dδ · exp [−βV(δ) − βP δ] 0

L(T, N, P ) = −kB T

and the density n is given by N n= =N L(T, N, P )

(

)−1 R∞ dδ · δ · exp [−β (V(δ) − P δ)] 2kB T + (N − 1) 0R ∞ .

P dδ · exp [−β (V(δ) − P δ)] 0

Note that for an ideal gas V i.g. (δ) = 0, and

Li.g. (T, N, P ) = leading to n(p)i.g. =

(N + 1)kB T , P

N P .

N + 1 kB T

Since the expression for n(T, P ) in part (e) is continuous and non-singular for any choice of potential, there is in fact no condensation transition for the one-dimensional gas. By contrast, the approximate van der Waals equation (or the mean-field treatment) incorrectly predicts such a transition.

112

(f) For a hard sphere gas, with minimum separation a between particles, calculate the equation of state P (T, n).

• For a hard sphere gas δi ≥ a,

i = 2, 3, · · · , N,

for

the Gibbs partition function is 1 Z(T, N, P ) = N λ (βP )2

Z ∞ a

dδ exp (−βV(δ) − βP δ)

N−1

Z ∞ N−1 1 = N dδ exp (−βP δ) λ (βP )2 a N+1

1 1 N−1 exp (−βP a) . = N λ βP From the partition function, we can calculate the mean length ∂ ln Z (N + 1)kB T L = −kB T = + (N − 1)a, ∂P T,N P

which after rearrangement yields (N + 1) n + 1/L βP = = ≈ (n + 1/l)(1 + (n − 1/L)a + (n − 1/L)2 a2 + · · ·).

L − (N − 1)a 1 − (n − 1/L)a For N ≫ 1, n ≫ 1/L, and βP ≈ n(1 + na + n2 a2 + · · ·) = n + an2 + a2 n3 + · · · ,

which gives the virial coefficients Bℓ (T ) = aℓ−1 .

The value of B3 = a2 agrees with the result obtained in an earlier problem. Also note that the exact ‘excluded volume’ is (N − 1)a, as opposed to the estimate of N a/2 used in deriving the van der Waals equation. ******** 7. Potts chain (RG): Consider a one-dimensional array of N Potts spins si = 1, 2, · · · , q, P subject to the Hamiltonian −βH = J i δsi ,si+1 . (a) Using the transfer matrix method (or otherwise) calculate the partition function Z, and the correlation length ξ.

•

(b) Is the system critical at zero temperature for antiferromagnetic couplings J < 0?

•

113

(c) Construct a renormalization group (RG) treatment by eliminating every other spin.

Write down the recursion relations for the coupling J, and the additive constant g.

•

(d) Discuss the fixed points, and their stability.

•

(e) Write the expression for ln Z in terms of the additive constants of successive rescalings.

• (f) Show that the recursion relations are simplified when written in terms of t(J) ≡

e−1/ξ(J) .

•

114

(g) Use the result in (f) to express the series in (e) in terms of t. Show that the answer can be reduced to that obtained in part (a), upon using the result ∞ X

n=0

1 2n+1

ln

n

1 + t2 1 − t2 n

= − ln(1 − t).

•

P (h) Repeat the RG calculation of part (c), when a small symmetry breaking term h i δsi ,1 P is added to −βH. You will find that an additional coupling term K i δsi ,1 δsi+1 ,1 is gen-

erated under RG. Calculate the recursion relations in the three parameter space (J, K, h).

115

•

(i) Find the magnetic eigenvalues at the zero temperature fixed point where J → ∞, and obtain the form of the correlation length close to zero temperature.

• ******** 8. Cluster RG: Consider Ising spins on a hexagonal lattice with nearest neighbor interactions J. (a) Group the sites into clusters of four in preparation for a position space renormalization group with b = 2.

• (b) How can the majority rule be modified to define the renormalized spin of each cluster.

•

116

(c) For a scheme in which the central site is chosen as the tie–breaker, make a table of all possible configurations of site–spins for a given value of the cluster–spin.

• (d) Focus on a pair of neighboring clusters. Indicate the contributions of intra–cluster and inter–cluster bonds to the total energy.

• (e) Show that in zero magnetic field, the Boltzmann weights of parallel and anti-parallel clusters are given by R(+, +) = x8 + 2x6 + 7x4 + 14x2 + 17 + 14x−2 + 7x−4 + 2x−6 , and R(+, −) = 9x4 + 16x2 + 13 + 16x−2 + 9x−4 + x−8 , 117

where x = eJ .

• Summing over the contributions in the previous tables gives the answer quoted above. (f) Find the expression for the resulting recursion relation J ′ (J).

• 118

(g) Estimate the critical ferromagnetic coupling Jc , and the exponent ν obtained from this RG scheme, and compare with the exact values.

• The non-trivial fixed point occurs at J ∗ ≈ 1.05, while the exact value of the hexagonal lattice critical coupling is Jc ≈ 0.66. The value of the eigenvalue yt and hence the exponent ν = 1/yt is close to the exact value of 1. (h) What are the values of the magnetic and thermal exponents (yh , yt ) at the zero temperature ferromagnetic fixed point?

•

(i) Is the above scheme also applicable for anti-ferromagnetic interactions? What symmetry of the original problem is not respected by this transformation?

• The distinction between the two sublattices of the hexagonal lattice is not respected by this RG scheme, and hence it does not reproduce the anti-ferromagnetic phase transition. ******** 9. Transition probability matrix: Consider a system of two Ising spins with a coupling K, which can thus be in one of four states. (a) Explicitly write the 4 × 4 transition matrix corresponding to single spin flips for a Metropolis algorithm. Verify that the equilibrium weights are indeed a left eigenvector of this matrix. (b) Repeat the above exercise if both single spin and double spin flips are allowed. The two types of moves are chosen randomly with probabilities p and q = 1 − p. ******** 119

Solutions to problems from chapter 7 - Series Expansions 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 J~si · ~sj .

In fact, if we are only interested in universal properties, any generalized interaction f (~si ·~sj ) leads to the same critical behavior. By analogy with the Ising model, a suitable choice is exp [f (~si · ~sj )] = 1 + (nt)~si · ~sj , 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 Y Z = {Dsi } [1 + (nt)si · sj ] , hiji

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 Z Z 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 2

6 5

3 4

120

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.

We shall rescale all integrals over spin by the n-dimensional solid angle, such that R

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 Z

ds′ (sα s′α )(s′β s′′β ) =

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 {Dsi } (s1α s2α )(s2β s3β )(s3γ s4γ )(s4δ s5δ )(s5η s6η )(s6ν s1ν ) = = 6.

6 n n 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 X Z= nNℓ tNb , 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.

m

n

121

The correlation function can also be calculated graphically from Z Y 1 {Dsi }snα smβ [1 + (nt)si · sj ] .

Gαβ (n − m) = hsnα smβ i = Z hiji

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 1 X 2 R Wℓ (R), R2 = Wℓ R

where Wℓ =

P

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|−(d−2+η) , up to the correlation length ξ, which diverges as ξ ∝ (tc − t)−ν . Hence, X R2 G(R) ∝ ξ d+2−(d−2+η) = (tc − t)−ν(4−η) = (tc − t)−γ−2ν .

R

We noted above that G(t, R) is the generating function of Wℓ (R), in the sense that P P ℓ ℓ W (R)t = G(t, R).

Similarly ℓ ℓ ℓ Wℓ t is the generating function of Wℓ , and is related to the susceptibility χ, by X X Wℓ tℓ = G(R) = χ ∝ (tc − t)−γ . ℓ

R

122

To obtain the singular behavior of Wℓ from its generating function, we perform a Taylor expansion of (tc − t)−γ , as X

Wℓ t

ℓ

l

= t−γ c

−γ ℓ

X t Γ(1 − γ) t −γ = tc , 1− tc Γ(1 + ℓ)Γ(1 − γ − ℓ) tc ℓ

which results in Wℓ =

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 expression of the gamma function, we obtain

Γ(γ + ℓ) −ℓ t ∝ ℓγ−1 t−ℓ c , Γ(1 + ℓ) c P P and, similarly one can estimate R R2 Wℓ (R) from R R2 G(R), yielding Wℓ ∝

R2 ∝

ℓ2ν+γ−1 t−ℓ c = ℓ2ν . −ℓ γ−1 ℓ tc

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 P −βH = K <ij> δsi ,sj . (a) To treat this problem graphically at high temperatures, the Boltzmann weight for each bond is written as

exp Kδsi ,sj = C(K) [1 + T (K)g(si, sj )] ,

with g(s, s′) = qδs,s′ − 1. Find C(K) and T (K).

• To determine the two unknowns C(K) and T (K), we can use the expressions ( K e = C [1 + T (q − 1)] if si = sj , 1 = C [1 − T ] if si 6= sj 123

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

g(s1 , s)g(s, s2) = q δs1 s δs2 s − q(δs1 s + δs2 s ) + 1 = q (qδs1 s2 − 1) = qg(s1 , 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 hg(sm , sn )i for a one–dimensional chain.

• The factor T (K) will be our high temperature expansion parameter. Each bond conP 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=

XY

{si } hiji

C(K) [1 + T (K)g(si, sj )] = C(K)N−1 q N = q eK + q − 1

N−1

.

Ignoring the boundary effects, i.e., that there are N − 1 bonds in the chain, the free energy per site is obtained as −

βF = ln eK + q − 1 .

N 124

With the same method, we can also calculate the correlation function hg(snsm )i. To get a nonzero contribution, we have to consider a graph that directly connects these two sites.

Assuming that n > m, this gives hg(sn sm )i = =

{si }

hiji

X C(K)N T (K)n−m g(sn sm )g(sm , sm+1 ) · · · g(sn−1 , sn ) Z N

=

Y 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=

XY

{si } hiji

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

eK̃ − 1 eK − 1 = q, 125

and consequently, if there is a single singular point Kc , it must be self-dual point, Kc = K̃c ,

√ 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.

P (a) Including a magnetic field h i δsi ,1 , show that the partition function takes the form Z(q, K, h) =

X

all graphs

Y

clusters c in graph

i h c c v nb × q − 1 + ehns ,

where ncb and ncs are the numbers of bonds and sites in cluster c. This is known as the random cluster expansion.

126

• Including a symmetry breaking field along direction 1, the partition function Z=

XY

[1 + v(K)δ(si , sj )]

Y

ehδsi ,1 ,

i

{si } hiji

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 P 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 Z(q, v, h) =

X

all graphs

Y

clusters in graph

h

i

s c v(K)nb q − 1 + ehnc ,

where ncb 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.

127

(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 W (graph) = (1 − p)

Y

zN

clusters in graph

p 1−p

ncb

.

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 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 limit of q → 1 from ∂ ln Z(q, v) = ∂q q=1

X

[probablility of graph]

all graphs

X

s

e−hnc .

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 temperature is played by the percolation threshold pc , which we can calculate using the expression obtained in problem 2 as pc = 1/2 (after noting that v ∗ = 1).

An alternative way of obtaining this threshold is to find a duality rule for the percolation 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.

128

• The partition function Z(q, v, h) goes to zero at q = 0, but again information 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. Ising model in a field:

Consider the partition function for the Ising model (σi = ±1)

on a square lattice, in a magnetic field h; i.e.   X X X σi  . σi σj + h Z= exp K i

<ij>

{σi }

(a) Find the general behavior of the terms in a low–temperature expansion for Z. • At low temperatures, almost all the spins are oriented in the same direction; low energy

excitations correspond to islands of flipped spins. The general behavior of the terms in a low-temperature expansion of the partition function is then of the form e−2KL e−2hA , where L is the perimeter of the island, or the number of unsatisfied bonds, and A is the area of the island, or the number of flipped spins. + +

-

+

+

+ L=4, A=1

+

+

-

-

+

+

L=6, A=2

129

+

(b) Think of a model whose high–temperature series reproduces the generic behavior found in (a); and hence obtain the Hamiltonian, and interactions of the dual model.

• To reproduce terms that are proportional to the area, we may consider plaquette-like interactions of the form X

h̃

σ1 σ2 σ3 σ4 ,

plaquettes

where σ is a spin variable defined on each bond of the lattice. Thus, we can define a (lattice gauge) model whose Hamiltonian is given by

−βH = K̃

X

σ + h̃

bonds

X

σ1 σ2 σ3 σ4 ,

plaquettes

and, as usual, rewrite

= cosh K̃ 1 + tanh K̃ σ, e

eh̃σ1 σ2 σ3 σ4 = cosh h̃ 1 + tanh h̃ σ1 σ2 σ3 σ4 .

K̃σ

The high temperature series of the partition function will be then of the form,

Z∝

X

X

bonds

plaquettes

tanh K̃

Nb

Np tanh h̃ ,

where Nb and Np , are the number of bonds and plaquettes involved in each term of the expansion. Nevertheless, we will only have non-zero contributions when the bond variables σ, appear an even number of times, either in a plaquette or as a bond. For example, the lowest order contributions contain 1 plaquette (so that A = 1), and 4 extra bonds (L = 4), or 2 plaquettes (A = 2), and 6 bonds (L = 6), etc. σ σ

σ σ

L=4, A=1

L=6, A=2

130

In general, we obtain terms of the form

L A tanh K̃ tanh h̃ , with L and A, the perimeter and the area of the graph considered. Hence, as in the Ising model, we can write the duality relations e−2K = tanh K̃,

and

e−2h = tanh h̃.

******** 5. 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 −βH = K

X

δsi ,sj .

<ij>

(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 2NK

Z(K) = qe

2N 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).

• As discussed in problem 3 of this chapter, the low temperature series takes the form

Z = qe2NK 1 + N (q − 1)e−4K + · · · ≡ qe2NK Ξ e−K ,

while at high temperatures

" # K K 2N 4 e −1 e +q−1 N 1 + N (q − 1) K q +··· Z= q e +q−1 K

K 2N e −1 −N ≡q e +q−1 Ξ K . e +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 − 1 131

and a critical (self-dual) point of Kc = K̃c ,

√ Kc = ln ( q + 1) .

=⇒

(b) Starting from the duality expression for Z(K), derive a similar relation for the internal energy U (K) = hβHi = −∂ 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 K

e −1 qeK eK ′ ln Ξ . + = 2N K e + q − 1 (eK + q − 1)2 eK + q − 1

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 2N −

ln Ξ′c 2N ln Ξ′c √ = √ + √ , 1+ q q 1+ q

=⇒

q−1 ln Ξ′c = √ N, q

and, U (Kc ) =N − Kc

q−1 2− √ , q+q

=⇒

U (Kc ) = N Kc

√

q+1 √ . q

******** 6. 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.

132

• Defining h0, 0| W (ℓ) |x, yi 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 hx, y| T ℓ |qx , qy i = hx, y| T ℓ |x′ , y ′ i hx′ , y ′ | qx , qy i x′ ,y ′

ℓ

= (2tx cos qx + 2ty cos qy ) hx, y| qx , qy i ,

√ P where hx, y| qx , qy i = eiqx x+iqy y / N . Since W (x, y) = ℓ h0, 0| W (ℓ) |x, yi, its Fourier

transform is calculated as XX W (qx , qy ) = h0, 0| T ℓ |x, yi hx, y| qx , qy i 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 Z π 2 Z π d q d2 q e−iqx x−iqy y −iqx x−iqy y .

W (q , q ) e = W (x, y) = x y 2 2 −π (2π) 1 − (2tx cos qx + 2ty cos qy ) −π (2π) 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?

• 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 1 − 2 (tx + ty ) + tx qx2 + ty qy2 .

Then, with qi′ ≡

√

ti qi , we have ′

Z ∞

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 = √t , √ . As the 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 y2 x2 + = constant. tx ty 133

(c) How does the average number of steps, hℓi = hℓx + ℓy i, diverge as t approaches tc ?

• The weight of all walks of length ℓ, irrespective of their end point location, is X x,y

ℓ

ℓ

h0, 0| W (ℓ) |x, yi = h0, 0| T ℓ |qx = 0, qy = 0i = (2tx + 2ty ) = (4t̄ ) .

Therefore, " # ℓ X ℓ (4 t̄ ) ∂ 1 4t̄ ∂ ℓ hℓi = Pℓ ln (4 t̄ ) = 4 t̄ ln = , = 4 t̄ ℓ ∂ (4t̄ ) ∂ (4t̄ ) 1 − 4t̄ 1 − 4t̄ (4 t̄ ) ℓ ℓ P

i.e.

hℓi =

t̄ , tc − t̄

diverges linearly close to the singular value of t̄.

******** 7. Anisotropic Ising model:

Consider the anisotropic Ising model on a square lattice

with a Hamiltonian −βH =

X

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 .

• 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 ) tℓxx tℓyy ,

where ti = tanh Ki , and the sum runs over all closed graphs. By extension of the isotropic case, ℓ

X tℓx tyy ln Z x f= h0| W ∗ (ℓx , ℓy ) |0i , = ln (2 cosh Kx cosh Ky ) + N ℓx + ℓy ℓx ,ℓy

134

where

′

1X number of crossings h0| W (ℓx , ℓy ) |0i = (−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 1 f = ln (2 cosh Kx cosh Ky ) + 2

Z

d2 q

∗ tr ln 1 − T (q) , 2 (2π)

where

∗ ∗ tr ln 1 − T (q) = ln det 1 − T (q)

= 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 ln (cosh 2Kx cosh 2Ky − sinh 2Kx cos qx − sinh 2Ky cos qy ) .

(2π)

(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 p 1 Kx p cosh 2Ky − 1 − e−Kx cosh 2Ky + 1 . = e 2

Therefore, the critical line is given by s 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 135

i.e. the critical boundary can be described as Kx = K̃y , where the dual interactions, K̃ 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   Z

2 2 X 2 p p 1 d q qi fS = sinh 2Ki  . ln  eKx cosh 2Ky − 1 − e−Kx cosh 2Ky + 1 + 2 2 2 (2π) 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 =

r

2 q′ , sinh 2Ki i

and δt = eKx

p p 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 = p sinh 2Kx sinh 2Ky

d2 q ′

Z

2 ln

(2π)

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 amplitude itself does vary along the critical line.) ******** 8. Müller–Hartmann and 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 labeled by heights hn for 1 ≤ n ≤ L. Show that for an anisotropic Ising model of interactions (Kx , KY ), the energy of an interface along the x-direction is

−βH = −2Ky L − 2Kx 136

X n

|hn+1 − hn | .

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+ + +

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

- - - - h1 - - - - - - - -

-

-

-

+ + + + + + + + - - - - - - - - - - - - -

-

1

2

.

.

.

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 P 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 hh|T |h′ i, and diagonalize it.

• We can define

or, in matrix form,  1 −2Ky  −2Kx T =e e ···

hh| T |h′ i ≡ exp (−2Ky − 2Kx |h′ − h|) ,

e−2Kx 1

e−4Kx e−2Kx

··· e−HKx H · · · e−2( 2 −1)Kx

e−HKx H e−2( 2 +1)Kx

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

i 2π k

e

i 2π k ·2

,e

i 2π k ·3

,e

i 2π k ·(H+1)

,···,e

137

,

is associated the eigenvalue −2Ky

λk = e

H+1 X

2π

T1n ei 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 e−2Kx n  T1n = e−2Ky 1 + 2 λ1 = e−2Ky n=1

n=1



−2Ky 

=e



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

exp −2Kx

X

d≥0

yielding

L X

n=1

|hn+1 − hn |

!

= e−2Ky L

L

e−2Kx d − 1 = e−2Ky coth Kx

" X d

L

exp (−2Kx |d|)

#L

,

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 as computed in the previous problem?

• The interfacial free energy vanishes for coth Kx = e2Ky , which coincides with the condition found in problem 2. This illustrates that long wavelength fluctuations, such as interfaces, are responsible for destroying order at criticality. ******** 138

9. Anisotropic Landau theory:

Consider an n–component magnetization field m(x) ~ in

d–dimensions. (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, " # Z d X t K ∂ m ~ ∂ m ~ i 2 4 −βH = dd x m ~ (x) + · + um ~ (x) .

2 2 ∂xi ∂xi i=1 (b) Are such anisotropies “relevant?”

• Clearly, the apparent anisotropy can be eliminated by the rescaling r K ′ xi . 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 inter-plane 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 inter-plane 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. ******** 10. Energy by duality: P −βH = K <ij> σi σj .

Consider the Ising model (σi = ±1) on a square lattice with 139

(a) Starting from the duality expression for the free energy, derive a similar relation for the internal energy U (K) = hHi = −∂ ln Z/∂ ln K.

• The low temperature partition function has the form X

Z(K) = e2KN

e−2KL ,

islands

whereas the high temperature expansion looks like Z(K) = 2 cosh2 K

N X

(tanh K)L .

loops

What is more, we found that we can write a duality relation of the form tanh K = e−2K̃ ,

or

sinh 2K sinh 2K̃ = 1,

which we can use to rewrite the partition function as Z(K) = 2 cosh2 K yielding

N

e−2K̃N Z(K̃) = (sinh 2K)N Z(K̃),

−βF (K) = N ln(2 sinh 2K) − βF (K̃).

The internal energy U (K) is then obtained from 2 cosh 2K ∂ ln Z(K) = −N K −K U (K) = −K ∂K sinh 2K

∂ ln Z(K̃) ∂ K̃

∂ K̃ . ∂K

After taking the logarithm of the relation sinh 2K sinh 2K̃ = 1, and differentiating it, we obtain ∂ K̃ tanh 2K̃ =− , ∂K tanh 2K and, as a result U (K) = −2N K coth 2K −

tanh 2K̃ K U (K̃) tanh 2K K̃

or, equivalently U (K)

tanh 2K̃ tanh 2K = −2N. + U (K̃) K K̃

(b) Using (a), calculate the exact value of U at the critical point Kc .

140

• Using the last result, we can calculate the exact value of U at the critical point, K ∗ =

K = K̃,

√ √ N 2 N K∗ = − ln( U (K ) = − 2 + 1), tanh 2K ∗ 2 √ where we have used K ∗ = ln( 2 + 1)/2, and sinh 2K ∗ = 1. ∗

******** 11. Clock model duality:

Consider spins si = (1, 2, · · · , q) placed on the sites of a square

lattice, interacting via the clock model Hamiltonian βHC = −

X

<i,j>

J (|(si − sj )modq |) ,

(a) Change from the N site variables to the 2N bond variables bij = si − sj . Show

that the difference in the number of variables can be accounted for by the constraint that around each plaquette (elementary square) the sum of the four bond variables must be zero modulus q.

• Around any closed loop, e.g. for the four bonds making up an elementary square, Sp ≡

X X p (si − sj ) = 0, bij =

as any si appears once with a positive and once with a negative sign. We may insist that all bond variables are positive, using bij = (si − sj )modq , in which case the above equality

is also valid modulus q.

(b) The constraints can be implemented by adding “delta–functions”

q 1 X 2πinp Sp δ [Sp ]modq = exp , q n =1 q p

for each plaquette. Show that after summing over the bond variables, the partition function can be written in terms of the dual variables, as Z = q −N

X Y

{np } hp,p′ i

λ (np − np′ ) ≡

X

{np }

exp 

where λ(k) is the discrete Fourier transform of eJ(n) .

141



X

hp,p′ i



J˜ (np − np′ ) ,

• In terms of the constrained bond variables, the partition function is Z=

XY

δ [Sp ]modq

{bij } p

Y

J(bij )

e

=

{bij } p

<ij>

2πinp X p bij exp q q

X YX 1 np

Y

eJ(bij ) .

<ij>

Now we can do the sum over each bond variable bij independently. Note, however, that a given bij appears twice in the constraints appropriate to its adjacent plaquettes, say p and p′ , and thus contributes X bij

2πibij ˜ exp (np − np′ ) + J (bij ) ≡ λ(np − np′ ) ≡ eJ (np −np′ ) . q

The difference in the exponent is again meaningful modulus q, and thus Z = q −N

X Y

˜

eJ (np −np′ ) =

{np } hp,p′ i

X

{np }



exp 

X

hp,p′ i



J˜ (np − np′ ) .

(c) Calculate the dual interaction parameter of a Potts model, and hence locate the critical point Jc (q).

• For the Potts model, the discrete Fourier transform has only two values. If np = np′ , then

λ(np = np′ ) = eJ + q − 1, and otherwise λ(np 6= np′ ) = eJ − 1, since the sum of all roots of unity (excluding one itself) is minus one. The ratio of these two cases gives the dual Potts interaction as ˜

eJ =

eJ + q − 1 . eJ − 1

Assuming that there is only one critical point (which is correct), it has to be at the self-dual point jc (q) = J˜c (q). This leads to a quadratic equation for x = eJc (q) , 2

x − 2x − (q − 1) = 0,

=⇒

x=1+

p

q − 1,

142

=⇒

h

Jc (q) = ln 1 +

p

i q−1 .

(d) Construct the dual of the anisotropic Potts model, with X

−βH = Jx δsx,y ,sx+1,y + Jy δsx,y ,sx,y+1 ; x,y

i.e. with bonds of different strengths along the x and y directions. Find the line of self–dual interactions in the plane (Jx , Jy ).

• The dual of the anisotropic square lattice is another anisotropic square lattice with Jy

e J˜x = ln

+q−1 , J e y −1

Jx

e and J˜y = ln

+q−1

eJx − 1

.

It is easy to check that the J˜˜ = J, and hence Jy = J˜x is a line that maps onto itself under duality. It turns out that this is also the critical line of the anisotropic Potts model on the square lattice. ******** 12. Triangular/hexagonal lattice Ising model: For any planar network of bonds, one can define a geometrical dual by connecting the centers of neighboring plaquettes. Each bond of the dual lattice crosses a bond of the original lattice, allowing for a local mapping.

Clearly, the dual of a triangular lattice is a hexagonal (or honeycomb) lattice, and vice versa. (a) Consider the Ising models on a hexagonal lattice with nearest neighbor interaction strength Kh . Note that the hexagonal lattice is bipartite, i.e. can be separated into two sublattices. In the partition function, do a partial sum over all spins in one sublattice.

Show that the remaining spins form a triangular lattice with nearest neighbor interaction Kt (Kh ). (This is called the star–triangle transformation.)

•

143

(b) Show that the dual of a triangular Ising model is a hexagonal Ising model with the usual duality relation K̃(K).

•

144

(c) By combining the previous results, obtain the critical couplings Kt∗ and Kh∗ of triangular and hexagonal lattices.

•

******** 13. Triangular/hexagonal lattice Potts model: The steps of the previous problem can be repeated for a general Potts model. (a) Consider Potts spins (si = 1, 2, · · · , q) on a hexagonal lattice with nearest neighbor

interaction Kh δsi ,sj . Perform the star-triangle decimation to show that the remaining spins form a triangular lattice with nearest neighbor interaction Kt (Kh ), and a three spin interaction L(Kh ). Why is L absent in the Ising model?

• (b) What is the dual of the Potts model on the triangular lattice?

• 145

(c) Clearly, the model is not self–dual due to the additional interaction. Nonetheless, obtain the critical value such that K̃t (Kc ) = Kc . Then check that L(Kc ) = 0, i.e. while the model in general is not self–dual, it is self–dual right at criticality, leading to the exact value of Kc (q)!

•

146

******** 14. Cubic lattice: The geometric concept of duality can be extended to general dimensions d. However, the dual of a geometric element of dimension D is an entity of dimension d−D. For example, the dual of a bond (D = 1) in d = 3 is a plaquette (D = 2), as demonstrated in this problem. (a) Consider a clock model on a cubic lattice of N points. Change to the 3N bond variables bij = si − sj . (Note that one must make a convention about the positive directions on the

three axes.) Show that there are now 2N constraints associated with the plaquettes of this

lattice.

147

(b) Implement the constraints through discrete delta-functions by associating an auxiliary variable np with each plaquette. It is useful to imagine np as defined on a bond of the dual lattice, perpendicular to the plaquette p. (c) By summing over the bond variables in Z, obtain the dual Hamiltonian ˜ = βH

X p

J˜ (np12 − np23 + np34 − np41 ) ,

where the sum is over all plaquettes p of the dual lattice, with npij

indicating the four

bonds around plaquette p.

˜ is left invariant if all the six bonds going out of any site are simultaneously (d) Note that βH increased by the same integer. Thus unlike the original model which only had a global translation symmetry, the dual model has a local, i.e. gauge symmetry. (e) Consider a Potts gauge theory defined on the plaquettes of a four dimensional hypercubic lattice. Find its critical coupling Jc (q).

148

******** Percolation Fluids do not pass through a solid with a small concentration of holes. However, beyond a threshold concentration, the holes overlap, and the fluid can percolate through 149

a connected channel in the material. Percolation is a classical geometric phase transition, and has been used as a model of many breakdown or failure processes. The loss of rigidity in an elastic network, conductivity in resistor nets, magnetization in diluted magnets are but a few examples.

In simple models of percolation, elements of a lattice (sites or bonds) are independently occupied with a probability p. A cluster is defined as a connected (by neighboring bonds) set of these occupied elements. At small p, only small clusters exist, and the probability that two sites, separated by a distance r, are connected to each other decays as exp (−r/ξ).

The correlation length ξ(p) grows with increasing p, diverging at the percolation threshold pc as ξ(p) ∼ |pc − p|−ν . An infinite cluster first appears at this threshold, and percolates through the (infinite) system for all p > pc . The analog of the order parameter is the probability P (p) that a site belongs to this infinite cluster. On approaching pc from above, it vanishes as P (p) ∼ |pc − p|β . While the value of pc depends on the details of the model, the exponents β and ν are universal, varying only with the spatial dimension d.

In the following problems we shall focus on bond percolation, i.e. p denotes the probability that a bond on the lattice is occupied.

15. Cayley trees: Consider a hierarchical lattice in which each site at one level is connected to z sites at the level below. Thus the n-th level of the tree has z n sites. (a) For z = 2, obtain a recursion relation for the probability Pn (p) that the top site of a tree of n levels is connected to some site at the bottom level.

•

150

(b) Find the limiting behavior of P (p) for infinitely many levels, and give the exponent β for this tree.

•

(c) In a mean-field approximation, P (p) for a lattice of coordination number z is calculated self-consistently, in a manner similar to the above. Give the mean-field estimates for pc and the exponent β.

•

151

(d) The one-dimensional chain corresponds to z = 1. Find the probability that end-points of an open chain of N + 1 sites are connected, and hence deduce the correlation length ξ.

•

********

16. Duality has a very natural interpretation in percolation: If a bond is occupied, its dual is empty, and vice versa. Thus the occupation probability for dual bonds is p̃ = 1 − p ≡ q.

Since, by construction, the original and dual elements do not intersect, one or the other percolates through the system. (a) The dual of a chain in which N bonds are connected in series, has N bonds connected

in parallel. What is the corresponding (non-) percolation probability?

• Consider N bonds forming a circle (1d chain with periodic boundary conditions). The

dual system has one point inside the circle and one point outside. While on the original chain neighboring points are connected in series with probability p, on the dual lattice, the inside and outside points are connected by N dual bonds in parallel with probability q = 1 − p, i.e. a connected path on the original chain blocks a possible dual path. The

probability that the original chain is connected is pN . The probability that the two dual points are not connected is also pN . (b) The bond percolation problem on a square lattice is self-dual. What is its threshold pc ?

• The dual to the square lattice is another square lattice whose sites are the centers of the

squares of the original lattice. If a bond is not connected on the original lattice, a path exists between the centers of neighboring squares. Thus p̃ = 1 −p = q. The self-dual point, which is the bond percolation threshold occurs at pc = 1/2.

(c) Bond percolation in three dimensions is dual to plaquette percolation. Is it possible to percolate while maintaining solid integrity in d = 3?

• Yes, there is an intermediate range of values between pc and p̃c in which both bonds and

plaquettes percolate.

152

(d) The triangular and hexagonal lattices are dual to each other. Combined with the star– triangle transformation that also maps the two lattices, this leads to exact values of pc for these lattices. However, this calculation is not trivial, and it is best to follow the steps for calculations of the critical couplings of q-state Potts models in an earlier problem. ******** 17. Position–space renormalization group (PSRG): In the Migdal–Kadanoff approximation, a decimation is performed after bonds are moved to connect the remaining sites in a one-dimensional geometry. For a PSRG rescaling factor b on a d–dimensional hypercubic lattice, the retained sites are connected by bd−1 strands in parallel, each strand having b bonds in series. This scheme is naturally exact for d = 1, and becomes progressively worse in higher dimensions. (a) Apply the above scheme to b = d = 2, and compare the resulting estimates of pc and yt = 1/ν to the exact values for the square lattice (pc = 1/2 and yt = 3/4).

•

153

(b) Find the limiting values of pc and yt for large d. (It can be shown that ν = 1/2 exactly, above an upper critical dimension du = 6.)

• (c) In an alternative PSRG scheme, first bd−1 bonds are connected in parallel, the resulting groups are then joined in series. Repeat the above calculations with this scheme.

•

********

154

Solutions to problems from chapter 8 - Beyond Spin Waves 1. Anisotropic nonlinear σ model: Consider unit n-component spins, ~s(x) = (s1 , · · · , sn ) P with α s2α = 1, subject to a Hamiltonian βH =

Z

1 2 2 d x (∇~s ) + gs1 .

2T d

For g = 0, renormalization group equations are obtained through rescaling distances by T , and lead to the flow a factor b = eℓ , and spins by a factor ζ = bys with ys = − (n−1) 4π

equation

(n − 2) 2 dT = −ǫT + T + O(T 3 ), dℓ 2π where ǫ = d − 2.

155

(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 ~s → ζ~s ′ , 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.

•

g 1 continuous phase transitions

discontinuous phase transition

disordered phase

T 1

156

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). ********

2. 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 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 K βH = 4

Z

dd x tr ∇M (x) · ∇M T (x)

,

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=

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 elements Mij (i, j = 1, · · · , N ). Describe the ground state of the system.

• In terms of the matrix elements, the Hamiltonian reads K βH = 4

Z

dd x

X i,j

∇Mij · ∇Mij ,

and the orthogonality condition becomes X

Mik Mjk = δij .

k

Since ∇Mij · ∇Mij ≥ 0, any constant (spatially uniform) orthogonal matrix realizes a

ground state.

157

(b) Define the symmetric and anti-symmetric matrices 

1  σ = M + M T = σT 2   π = 1 M − M T = −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 h i K K 2 2 d d d x tr [∇ (σ + π) · ∇ (σ − π)] = d 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−

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= where Dπ (x) =

Z

Q

Dπ (x) exp

j>i Dπij (x),

K 4

Z

2 1 2 T d x tr (∇π) − ∇ ππ , 4 d

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 158

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,

∇ ππ T

2

2 2 = ∇ π2 = [(∇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, i h

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 magnitude 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 hπij (q)πkl (q′ )i =

(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

KX βH0 = 2 i<j

Z

dd q 2 q |πij (q)|2 , d (2π)

leading to the bare expectation values d

(2π) δ d (q + q′ ) , hπij (q) πij (q )i0 = Kq 2 ′

159

and hπij (q) πkl (q′ )i0 = 0, if the pairs (ij) and (kl) are different.

Furthermore, since π is anti-symmetric, hπij (q) πji (q′ )i0 = − hπij (q) πij (q′ )i0 , and in particular hπii (q) πjj (q′ )i0 = 0. These results can be summarized by d

hπij (q) πkl (q′ )i0 =

(2π) δ d (q + q′ ) (δik δjl − δil δjk ) .

Kq 2

We shall now construct a renormalization group by removing Fourier modes M > (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 >

1 π2 > > tr π 2 0 +··· ≈N+ htr σi0 = tr I + 2 2 0 * * +> +> 2 > 1 1 X 2 1 X πij πji πij =N− N 2 − N πij =N− =N+ 0 2 2 2 i6=j i6 = j 0 0 !

Z Λ d N −1 d q 1 N −1 =N 1− =N 1− Id (b) . d 2 2 2K Λ/b (2π) Kq 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).

(g) Use perturbation theory to calculate the coarse grained coupling constant K̃. Evaluate only the two diagrams that directly renormalize the (∇πij )2 term in βH, and show that N K̃ = K + 2

Z Λ

dd q 1 d 2 Λ/b (2π) q

160

.

• Distinguishing between the greater and lesser modes, we write the partition function as Z Z > < > 0 < > −βH +U π ,π < > −βH< ] [ 0 0 = Dπ < e−δfb −βH0 eU 0 , Z = Dπ Dπ e where H0 denotes the quadratic part, and K X U =− 8

i,j,k,l

K X = 8

i,j,k,l

Z

dd x [(∇πij ) πjk · (∇πkl ) πli + πij (∇πjk ) · (∇πkl ) πli ]

Z

dd q1 dd q2 dd q3 3d

(2π)

[(q1 · q3 + q2 · q3 ) ·

·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)

E> dd q1 dd q2 dd q3 D > > < < π (q ) π (−q − q − q ) (q1 · q3 ) πij (q1 ) πkl (q3 ) 2 1 2 3 jk li 3d 0 (2π) i,j,k,l  ! Z Z Λ d ′ Λ/b d X d q d q 2 K 1 < <  = q πij (q) πji (−q) , d d ′2 8 (2π) Λ/b (2π) Kq 0 i,j K X 8

Z

and KX (ii) 8

j,k,l

Z

dd q1 dd q2 dd q3 3d

(2π)

*

X

i6=j,l

> πij (q2 ) πli> (−q1 − q2 − q3 )

+>

< < (q2 · q3 ) πjk (q2 ) πkl (q3 )

0

 # Z Z Λ d ′ Λ/b d X K d q 1 d q 2 < <  = q πjk (q) πkj (−q) .

(N − 1) d d ′2 8 (2π) Λ/b (2π) Kq 0 j,k "

Adding up the two contributions results in an effective coupling K K K̃ = + N 4 4 8

Z Λ

dd q

1 d Kq 2

Λ/b (2π)

,

i.e.

K̃ = K +

N Id (b) .

2

(h) Using the result from part (f), show that after matrix rescaling, the RG equation for K ′ is given by: K ′ = bd−2

"

N −2 K− 2

161

Z Λ

dd q 1 d 2 Λ/b (2π) q

#

.

• After coarse-graining, renormalizing the fields, and rescaling,

N N −1 Id (b) K 1 + Id (b) 1− K = b ζ K̃ = b K 2K

N −2 d−2 Id (b) + O (1/K) , K− =b 2 ′

d−2 2

d−2

i.e., to lowest non-trivial order, K ′ = bd−2

"

N −2 K− 2

# dd q 1 . d 2 Λ/b (2π) q

Z Λ

(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

N −2 dK = (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

N −2 dT = (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∗ =

2 (d − 2) 4πǫ = + O ǫ2 . d−2 (N − 2) Kd Λ N −2

In the vicinity of the fixed point, the flows are described by

d δT = 1 + dT ′

dT dℓ

= (1 + ǫδℓ) δT.

T∗

δℓ δT = 1 + (2 − d) + (N − 2) Kd Λd−2 T ∗ δℓ δT 162

Thus, from δT ′ = byT δT = (1 + yT δℓ) δT, we get yT = ǫ, and ν=

1 . ǫ

(j) Consider a small symmetry breaking term −h

R

dd x tr(M ), added to the Hamiltonian.

Find the renormalization of h, and identify the corresponding exponent yh .

• As usual, h renormalizes according to

N −1 d−2 h = b ζh = (1 + dδℓ) K − Kd Λ δℓ h 2K

N −1 2 d−2 δℓ + O δℓ h.

Kd Λ = 1+ d− 2K ′

d

From h′ = byh h = (1 + yh δℓ) h, we obtain yh = d −

N −1 N −1 ǫ d−2 2 K Λ = d − (d − 2) = 2 − + O ǫ . d 2K ∗ N −2 N −2

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). ******** 3. The roughening transition:

In an earlier problem we examined a continuum interface

model which in d = 3 is described by the Hamiltonian Z K d2 x (∇h)2 βH0 = − 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) , 163

to the Hamiltonian. Treat −βU as a perturbation, and proceed to construct a renormalization group as follows: (a) Show that *

"

exp i

X

qα h(xα )

α

#+

0



= exp 

1 X

K

α<β



qα qβ C(xα − xβ )

P for α qα = 0, and zero otherwise. (C(x) = ln |x|/2π is the Coulomb interaction in two dimensions.)

P • The translational invariance of the Hamiltonian constrains hexp [i α qα h (xα )]i0 to P vanish unless α qα = 0, as implied by the following relation !* " #+ * ( )+ X X X exp iδ qα exp i qα h (xα ) = exp i qα [h (xα ) + δ] α

α

α

0

=

*

"

exp i

X α

0

qα h (xα )

#+

.

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 X 1 exp i qα h (xα ) = exp − qα qβ hh (xα ) h (xβ )i0  2 α αβ 0   E D X 1 2 . qα qβ (h (xα ) − h (xβ )) = exp  4 0 αβ

Note that the quantity hh (xα ) h (xβ )i0 is ambiguous because of the symmetry h(x) → P 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  #+ * "

 X X qα qβ Z d2 q eiq·xα − eiq·xβ e−iq·xα − e−iq·xβ  exp i qα h (xα ) = exp  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   X 1 qα qβ C (xα − xβ ) , = exp  K α<β

164

where C (x) =

1 |x| d2 q 1 − cos (q · x) = ln , 2 2 q 2π a (2π)

Z

is the Coulomb interaction in two dimensions, with a short distance cutoff a. (b) Prove that |h(x) − h(y)|2 = −

d2 , Gk (x − y) dk 2 k=0

where Gk (x − y) = exp ik h(x) − h(y) .

• From the definition of Gk (x − y),

D E d2 2 G (x − y) = − [h (x) − h (y)] exp [ik (h (x) − h (y))] . k dk 2 Setting k to zero results in the identity D

2

[h (x) − h (y)]

E

=−

d2 .

Gk (x − y) dk 2 k=0

(c) Use the results in (a) to calculate Gk (x − y) in perturbation theory to order of y02 .

(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 −U = y0

Z

y0 d x cos (2πh) = 2 2

Z

d2 x e2iπh + e−2iπh .

The perturbation expansion for Gk (x − y) = hexp [ik (h (x) − h (y))]i ≡ hGk (x − y)i is calculated as

hGk i = hGk i0 − (hGk U i0 − hGk i0 hU i0 )

1 2 + Gk U 2 0 − 2 hGk U i0 hU i0 + 2 hGk i0 hU i0 − hGk i0 U 2 0 + O U 3 .

2

From part (a),

hU i0 = hGk U i0 = 0, and

k2 − 2πK

k2 |x − y| hGk i0 = exp − C (x − y) = .

K a

165

Furthermore, U

2

y02 = 0 2 y2 = 0 2

Z

d2 x′ d2 x′′ hexp [2iπ (h (x′ ) − h (x′′ ))]i0 y2 d2 x′ d2 x′′ hG2π (x′ − x′′ )i0 = 0 2

Z

"

# 2 (2π) d2 x′ d2 x′′ exp − C (x′ − x′′ ) , K

Z

and similarly, exp [ik (h (x) − h (y))] U 2 0 = ( Z 2 y02 k2 (2π) = d2 x′ d2 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 Thus, the second order part of Gk (x − y) is 2 Z

k 4π 2 y02 2 ′ 2 ′′ ′ ′′ exp − C (x − y) d x d x exp − C (x − x ) · 4 K K

2πk ′ ′′ ′′ ′ (C (x − x ) + C (y − x ) − C (x − x ) − C (y − x )) − 1 , · exp K and 2

− kK C(x−y)

Gk (x − y) = e

Z

2πk 2

′ ′′ y02 C x −x D 4 2 ′ 2 ′′ − 4π ( ) 1+ d xd x e K e K − 1 + O y0 , 4

where 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 − kK C(x−y)

Gk (x − y) = e

2

− kK C(x−y)

=e

(

y2 1 1+ 0 × 4 2

2πk K

2

× C (x − y) × 2π

π 3 k2 2 1+ y C (x − y) K2 0 166

Z

3 −

drr e

2π ln(r/a) K

Z

drr 3 e

.

2π ln(r/a) − K

)

The second order term can be exponentiated to contribute to an effective coupling constant Keff , according to

1 π3 1 = − 2 a2π/K y02 Keff K K

Z ∞

drr 3−2π/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 1 1 π3 = − 2 a2π/K y02 Keff K K

Z ab a

drr

3−2π/K

π3 − 2 a2π/K × y02 b4−2π/K × K

Z ∞

drr 3−2π/K .

a

(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 π 3 2π/K 2 1 = − a y0 K′ K K2

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



dK   = π 3 a4 y02 + O y04 dℓ .

  dy0 = 2 − π y + O y 3 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 167

irrelevant, and the flows terminate on a fixed line with y0 = 0 and K ≤ π/2, corresponding to a rough phase at high temperatures.

|x − y|, find the magnitude of the discontinuous jump in

(g) For large separations |h(x) − h(y)|

2

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 C (x − y) .

Gk (x − y) → hGk i0 = exp − π From part (b), D

2

[h (x) − h (y)]

E

=−

d2 4 2 = G (x − y) C (x − y) = ln |x − y| . k dk 2 π π2 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

[h (x) − h (y)]

2

E

→ 0.

D E 2 Thus, the magnitude of the jump in [h (x) − h (y)] at the transition is 2 ln |x − y| . π2 ******** 4. 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 βH =

K X |hi − hj |∞ 2 <i,j>

,

where the “∞” power means that there is no energy cost for ∆h = 0; an energy cost of K/2 for ∆h = ±1; and ∆h = ±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: 168

(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.

R 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 −βH = −

KX ∞ |nij | .

2 hiji

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 X

Z=

e−βH

Y α

{nij }

δP

hiji

nα ,0 , ij

where the index α labels the N plaquettes, and nα ij is non-zero and equal to nij only if the bond hiji belongs to plaquette α. Expressing the Kronecker delta in its exponential representation, we get Z=

X

−K 2

e

{nij }

P

hiji

|nij |∞ Y α

Z 2π 0

dθα iθα e 2π

P

hiji

nα ij

.

(iii) As each bond belongs to two neighboring plaquettes, we can label the bonds by αβ rather than ij, leading to

Z=

γ

=

2π

!

dθγ 2π

!

Y Z 2π dθγ 0

Y Z 2π γ

0

X

{nαβ }





X K ∞ exp  − |nαβ | + i (θα − θβ ) nαβ  2

YX

hαβi nαβ

hαβi

exp

K ∞ − |nαβ | + i (θα − θβ ) nαβ 2

169

.

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=

Y Z 2π dθγ γ

0

2π

!



exp 

X

hαβi

 i h K ln 1 + 2e− 2 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 considered in the problems of the previous chapter.)

• This is the loop gas model introduced earlier. For K large, h i K K ln 1 + 2e− 2 cos (θα − θβ ) ≈ 2e− 2 cos (θα − θβ ) , and

Z=

Y Z 2π dθγ γ

0

2π

! P e

K

hαβi

2e− 2 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 problem 1, 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

,

have a roughening transition?

• In one dimension, we can directly sum the partition function, as ! !

X X X KX K ∞ ∞ Z= exp − |hi − hi+1 | = exp − |ni | 2 i 2 i {hi } {ni }

N Y YX K ∞ 1 + 2e−K/2 = 1 + 2e−K/2 , = = exp − |ni | 2 i i n i

170

(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.

******** 5. Nonlinear σ model with long–range interactions: Consider unit n-component spins, P ~s(x) = (s1 , s2 , · · · , sn ) with |~s(x)|2 = i si (x)2 = 1, interacting via a Hamiltonian Z Z d βH = d x dd y K(|x − y|) ~s(x) · ~s(y) . (a) The long-range interaction, K(x), is the Fourier transform of Kq ω /2 with ω < 2.

What kind of asymptotic decay of interactions at long distances is consistent with such decay? (Dimensional analysis is sufficient for the answer, and no explicit integrations are required.)

• The dependence of the interaction strength on the separation x between spins in real space is obtained by Fourier transformation, as

Z K dd q iq·x Kq ω ∝ e K(x) = d (2π) 2 |x|d+ω

for |x| → ∞,

where the asymptotic dependence on |x| is obtained by dimensional analysis. (b) Close to zero temperature we can set ~s (x) = (~π (x), σ(x)) , where ~π (x) is an n − 1 com-

ponent vector representing small fluctuations around the ground state. Find the effective Hamiltonian for ~π (x) after integrating out {σ(x)}.

• Employing the constraint of unit vectors, each integration over ~s can be decomposed as Z Z

n 2 d ~s δ s − 1 → dn−1~π dσ δ π 2 + σ 2 − 1 → Z n−1 Z

i

h p p p d ~π dσ n−1 √ δ σ − 1 − π2 . d ~π dσ δ σ − 1 − π 2 σ + 1 − π 2 → 2 1 − π2 √ The δ-function can now be used to replace the field σ with 1 − π 2 . This leads to a final partition function

Z=

Z

D~π (x) exp [−βHeff (~π )] ,

with the effective Hamiltonian Z h i p p βHeff = dd x dd y K(|x − y|) ~π (x) · ~π (y) + 1 − π 2 (x) 1 − π 2 (y) + Z p

ρ dd x ln 1 − π 2 (x) , 2 171

where ρ = N/V is the density of degrees of freedom. (c) Fourier transform the quadratic part of the above Hamiltonian focusing only on terms proportional to K, and hence calculate the expectation value hπi (q)πj (q′ )i0 .

• To leading order in temperature (K −1 ), the unperturbed Hamiltonian (after Fourier

transformation) is

βH0 =

Z

dd q Kq ω ~π (q) · ~π (−q).

(2π)d 2

From this Gaussian form, we can read hπi (q)πj (q′ )i0 =

δij (2π)d δ d (q + q′ ) .

Kq ω

We shall now construct a renormalization group by removing Fourier modes, ~π > (q), with q in the shell Λ/b < |q| < Λ.

2 (d) Calculate the coarse grained expectation value for hσi> 0 to order of π after removing

these modes. Identify the scaling factor, ~s ′ (x′ ) = ~s < (x)/ζ, that restores ~s ′ to unit length.

• Due to thermal fluctuations of the modes ~π > (q), the length of the spin is reduced to 1 n−1 hσi = 1 − h~π · ~π i = 1 − 2 2

Z Λ

dd q n − 1 Kd Λd−ω 1 − bω−d 1 = 1 − . d ω 2 K d−ω Λ/b (2π) Kq

The renormalized spins must be rescaled by this factor to remain unit vectors. In particular, for an infinitesimal rescaling b = 1 + δℓ, n − 1 Kd Λd−ω δℓ. ζ =1− 2 K

(e) A simplifying feature of long–range interactions is that the coarse grained coupling constant is not modified by the perturbation, i.e. K̃ = K to all orders in a perturbative calculation. Use this information, along simple with dimensional analysis, to express the renormalized interaction, K ′ (b), in terms of K, b, and ζ.

• The parameters of the Hamiltonian are changed as x → bx′ and ~s → ζ~s′ , resulting in a

renormalization of the interaction constant to

1 K ′ (b) = Kb2d d+ω ζ 2 = ζ 2 bd−ω K. b

172

(f) Obtain the differential RG equation for T = 1/K by considering b = 1 + δℓ.

• Setting b = 1 + δℓ, and expanding to lowest order in δℓ gives

2 dK n − 1 Kd Λd−ω K + δℓ = [1 + (d − ω)δℓ] 1 − δℓ K. dℓ 2 K The differential recursion relations are now obtained as dK = (d − ω)K − (n − 1)Kd Λd−ω , dℓ

⇒

dT = −(d − ω)T + (n − 1)Kd Λd−ω T 2 . dℓ

(g) For d = ω + ǫ, compute Tc and the critical exponent ν to lowest order in ǫ.

• Setting dT /dℓ = 0 gives T ∗ = Tc =

d−ω ǫ = + O(ǫ2 ). d−ω (n − 1)Kd Λ (n − 1)Kω

Linearizing around this fixed point gives dδT = −(d − ω)δT + 2(n − 1)Kd Λd−ω T ∗ δT = +ǫδT. dℓ This corresponds to yt = ǫ to lowest order, and hence ν = 1/yt ≈ 1/ǫ.

R (h) Add a small symmetry breaking term, −~h · dd x ~s(x), to the Hamiltonian. Find the renormalization of h and identify the corresponding exponent yh .

• Under the rescalings x → bx′ and ~s → ζ~s′ , the magnetic field changes (in magnitude) to h′ (b) = bd ζ h, and b = 1 + δℓ, we get

n − 1 Kd Λd−ω dh h. = d− dℓ 2 K

Evaluated at the fixed point 1/K ∗ = T ∗ , this gives yh = d −

d−ω d+ω = .

2 2

******** 6. The XY model in 2 + ǫ dimensions: The recursion relations of the XY model in two dimensions can be generalized to d = 2 + ǫ dimensions, and take the form:  dT   = −ǫT + 4π 3 y 2 dℓ .

  dy = 2 − π y dℓ T 173

(a) Calculate the position of the fixed point for the finite temperature phase transition.

• There is a trivial fixed point at T ∗ = y ∗ = 0 which is the sink for flows inside the ordered

phase. Flows in the high temperature phase go towards y → ∞ and T → ∞. In between

there is a fixed point at finite T ∗ and y ∗ , given by π T = , 2 ∗

and

∗

y =

r

ǫ .

8π 2

[Note that the location of the fixed point is accurate only to order of

√

ǫ.

(b) Obtain the eigenvalues at this fixed point to lowest contributing order in ǫ.

• Linearizing the recursion relations in the vicinity of this fixed point gives √   !  −ǫ !  −ǫ !

2π 2 2ǫ 8π 3 y ∗ δT δT δT d  =  √2ǫ =  πy ∗ . π  dℓ δy 2 − δy δy 0 T ∗2 T∗ π2 The eigenvalues satisfy the equation y 2 + ǫy − 4ǫ = 0. To lowest order, the eigenvalues √ are ±2 ǫ. There is one irrelevant eigenvalue, the positive (relevant) eigevalue is identified

with yT .

(c) Estimate the exponents ν and α for the superfluid transition in d = 3 from these results.

[Be careful in keeping track of only the lowest nontrivial power of ǫ in your expressions.]

• The divergence of the correlation length is controlled by the exponent ν = 1/yT = √ 1/(2 ǫ), which evaluates to 1/2 for ǫ = 1 in d = 3. The heat capacity is governed by the √ exponent α = 2 − dν, whose correct expansion to order of ǫ gives α=2−

1 2+ǫ = − √ + O(1), yT ǫ

which evaluates to α = −1 for ǫ = 1 in d = 3.

********

7. Symmetry breaking fields: Let us investigate adding a term Z −βHp = hp d2 x cos (pθ(x)), to the XY model. There are a number of possible causes for such a symmetry breaking field: p = 1 is the usual ‘magnetic field,’ p = 2, 3, 4, and 6 could be due to couplings to an underlying lattice of rectangular, hexagonal, square, or triangular symmetry respectively.

174

As hp → ∞, the spin becomes discrete, taking one of p possible values, and the model

becomes equivalent to the previously discussed clock models.

(a) Assume that we are in the low temperature phase so that vortices are absent, i.e. the vortex fugacity y is zero (in the RG sense). In this case, we can ignore the angular nature of θ and replace it with a scalar field φ, leading to the partition function Z=

Z

Z

K 2 2 (∇φ) + hp cos(pφ) .

Dφ(x) exp − d x 2

This is known as the sine–Gordon model, and is equivalent to the roughening transition.

Obtain the recursion relations for hp and K. • The spin waves are represented by the angle φ(x), which ignoring its angular character

can be integrated over all values. Forgetting the angular character is equivalent to leaving out vortices by setting their fugacity to y = 0. This leads to the above Sine-Gordon model.

Following the steps of problem 3 of chapter 8, on the roughening transition, we then arrive to recursion relations

p2 hp , 2− 4πK −1  πp2 h2p −2   dK =− K . dℓ 2    

dhp = dℓ

(b) Show that once vortices are included, the recursion relations are

p2 hp , 2− 4πK πp2 h2p −2 dK −1 = − K + 4π 3 y 2 ,   dℓ 2     dy  = (2 − πK) y. dℓ        

dhp = dℓ

• The usual recursion relation for the XY model (without the symmetry breaking field) are

 dK −1   = 4π 3 y 2 , dℓ  dy  = (2 − πK) y. dℓ

When both hp and y are small, we can add the two sets of recursion relations to obtain the three equations above.

175

−1 < π2 , and thus only (c) Show that the above RG equations are only valid for 8π p2 < K

apply for p > 4. Sketch possible phase diagrams for p > 4 and p < 4. In fact p = 4 is rather special as there is a marginal operator h4 , and the transition to the 4-fold phase (cubic anisotropy) has continuously varying critical exponents!

• The above perturbative recursion relations remain valid only if both hp and y asymp-

totically go to zero under RG. Hence, we must require 8π π −1 < K < , p2 2

or

4 1 1 < η = < . p2 2πK 4

Depending on the value of p, different types of behavior are possible: (1) For p > 4 (e.g. p = 6), there is a finite range of K −1 over which the above inequalities hold. In this range, it is possible to have a phase which is controlled by spin-waves. The spin-correlations in this phase decay algebraically with an exponent that ranges from 4/p2 to 1/4. At higher temperatures vortices appear and the resulting disordered phase has algebraic decay of correlations. At low-enough temperatures, hp becomes relevant, fixing the spins along one of p discrete directions. This phase (like the Ising model) has no Goldstone modes.

(2) For p < 4 (e.g. the Ising model at p = 2) there can be no intermediate phase with algebraic order. We expect the usual transition from ordered to disordered states (in the Ising universality for p = 2, and the 3-state Potts universality for p = 4.

(3) The case p = 4 is rather special. The field h4 is a marginal operator and the exponents vary continuously at the transition between ordered and disordered phases. ******** 8. Inverse-square interactions: Consider a scalar field s(x) in one-dimension, subject to an energy K −βHs = 2

Z

s(x)s(y) dxdy + |x − y|2

Z

dxΦ[s(x)].

The local energy Φ[s] strongly favors s(x) = ±1 (e.g. Φ[s] = g(s2 − 1)2 , with g ≫ 1). (a) With K > 0, the ground state is ferromagnetic. Estimate the energy cost of a single domain wall in a chain of length L. You may assume that the transition from s = +1 to s = −1 occurs over a short distance cutoff a.

• Let us consider a domain wall at x = 0, separating s = +1 on the left, and s = −1 on the right. Spins at −x and +y are now connected by a dissatisfied bond at the energy cost

of 2K/|x + y|2 . Note that there are two factors of 2 involved: the first occurs because each 176

R pair of points is considered once, rather than twice in (K/2) dxdy, the second accounts

for the factor of 2 in the difference of energies of a satisfied and unsatisfied bond. Adding up all such energies leads to an energy cost of

Z L/2 L 1 , ≈ 2K ln dxdy βE1 = 2K 2 |x + y| a a/2

where we have ignored subleading corrections. (b) From the probability of the formation of a single kink, obtain a lower bound for the critical coupling Kc , separating ordered and disordered phases.

• Since the kink can be placed in anyone of approximately (L/a) locations along the chain,

its weight scales as

1−2K L L L W1 ∝ exp −2K ln ∝ . a a a Even if all other excitations are suppressed, a single kink configuration will spontaneously appear for K < 1/2 due to its large weight for (L/a) → ∞. Since other excitations are also possible, this is only a lower bound on the critical value of K.

(c) Show that the energy of a dilute set of domain walls located at positions {xi } is given

by

−βHQ = 4K

X i<j

qi qj ln

|xi − xj | a

+ ln y0

X i

|qi |,

where qi = ±1 depending on whether s(x) increases or decreases at the domain wall.

(Hints: Perform integrations by part, and coarse-grain to size a. The function Φ[s] only contributes to the core energy of the domain wall, which results in the fugacity y0 .)

• Performing two integrations by part on the nonlocal part of the Hamiltonian gives

Z Z K s(x)s(y) |x − y| K dxdy dxdy∂x s(x)∂y s(y) ln . = 2 |x − y|2 2 a

(The additional factor of ln a makes no difference if s(x) is the same at the edges.) The function s(x) is mostly constant at low temperatures, except at the domain walls where it rapidly switches between -1 and +1 over a distance a. Assuming that such domain walls are well-separated, we can integrate out fluctuations over scales shorter than a. In the R vicinity of a kink at xi , integration gives dx∂x s(x) = 2qi , where qi = ±1. The nonlocal part then results in an interaction

X |xi − xj | |xi − xj | KX (2qi )(2qj ) ln = 4K qi qj ln + core contributions.

2 i,j a a i<j 177

There are also contributions to the core energy from the local part Φ[s]. The core energies are simply proportional to the number of domain walls (irrespective of their sign), resulting in −βHQ = 4K

X

qi qj ln

i<j

|xi − xj | a

+ ln y0

X i

|qi |.

(d) The logarithmic interaction between two opposite domain walls at a large distance L, is reduced due to screening by other domain walls in between. This interaction can be calculated perturbatively in y0 , and to lowest order is described by an effective coupling (see later) K → Kef f = K − 2Ky02

Z ∞

drr

a

a 4K r

+ O(y04 ).

(1)

By changing the cutoff from a to ba = (1 + δℓ)a, construct differential recursion relations for the parameters K and y0 .

• Let us break the above integration into two parts, to get Kef f = K − 2Ky02

Z ba a

a 4K

drr r Z ∞

= K ′ − 2K ′ y02 b2−4K

− 2Ky02

dr ′ r ′

a

Z ∞

a 4K r′

ba

drr

a 4K r

+ O(y04 )

+ O(y04 ),

where r ′ = br, and we can identify the renormalized parameters  Z ba a 4K   K ′ = K − 2Ky 2 ≈ K − 2Ky02 a2 δℓ drr 0 r . a   ′ 1−2K yo = y0 b ≈ y0 + (1 − 2K)y0 δℓ

The last two equations are obtained by setting b = 1 + δℓ, and result in the differential recursion relations

 dK   = −2Ky02 a2 dℓ .   dy0 = (1 − 2K)y 0 dℓ

(e) Sketch the renormalization group flows as a function of T = K −1 and y0 , and discuss the phases of the model.

178

yo

K -1 •

The trajectories at low temperature with y0 → 0, correspond to the ferromagnetic

phase, where there are asymptotically no domain walls. (Note that because of the longrange interactions, it is possible to have an ordered phase even in d = 1.) The high temperature phase, where y0 becomes large, is disordered. (f) Derive the effective interaction given above as Eq.(1). (Hint: This is somewhat easier than the corresponding calculation for the two-dimensional Coulomb gas, as the charges along the chain must alternate.)

• Consider two domain walls separated by a distance L. The direct interaction between

them can be partly screened by another pair of domain walls. Indicating the distances of the screening pair of domain walls from the left-most one by R + r/2 and R − r/2, the effective interaction is calculated as

R

1 + y02 dRdr e−4K ln(r/a)+4KD(R,r) + O(y04 ) −βV (L) −4K ln(L/a)

R

e =e × 1 + y02 dRdr e−4K ln(r/a) + O(y04 ) # " Z L

dRdr e−4K ln(r/a) e4KD(R,r) − 1 + O(y04 ) , = e−4K ln(L/a) 1 + y02 a

with the dipole-dipole interaction

r r r r − ln R − − ln L − R + + ln L − R − D(R, r) = ln R + 2 2 2 2 d [ln (R) + ln (L − R)] + O(r 3 ). =r dR We next set exp(4KD) − 1 = 4KD + O(D2 ), and use the above form for D to obtain (" Z −βV (L)

e

−4K ln(L/a)

=e

1 + y02

L

a

drdR e−4K ln(r/a) ×

#)

d [ln (R) + ln (L − R)] + O(r 3 ) + O(y04 ) .

4Kr dR 179

The integration over R is now easily performed to give # Z L L drre−4K ln(r/a) + O(y04 ) e−βV (L) = e−4K ln(L/a) 1 + 8Ky02 ln a a ( " #) , Z L r a 4K ≈ exp −4K ln drr 1 − 2y02 + O(y04 ) a r a "

from which we can read off the expression for the effective coupling Kef f given earlier. ******** 9. Melting:

The elastic energy cost of a deformation ui (x), of an isotropic lattice is 1 −βH = 2

Z

dd x [2µuij uij + λuii ujj ] ,

where uij (x) = (∂i uj + ∂j ui ) /2, is the strain tensor. (a) Express the energy in terms of the Fourier transforms ui (q), and find the normal modes of vibrations.

• In terms of Fourier modes uij (q) = i[qi uj (q) + qj ui (q)]/2, and dd q [2µuij (q)uij (−q) + λuii (q)ujj (−q)] (2π)d Z i

1 dd q h µ ∗ ∗ ∗ = (q u (q) + q u (q)) q u (q) + q u (q) + λq u (q)q u (q) i j j i i j j i i i j j 2 (2π)d 2 Z Z

dd q 2 dd q 1 1 ∗ = Mij (q)ui (q)u∗j (q). µq δ + (µ + λ)q q u (q)u (q) ≡ ij i j i j 2 (2π)d 2 (2π)d

1 −βH = 2

Z

The normal modes are obtained by making the matrix Mij (q) diagonal. The longitudinal phonons have ~q k ~uℓ , while for transverse phonons have ~q ⊥ ~ut , resulting in 1 −βH = 2

Z

dd q 2 2 2 2 (2µ + λ)q |~ u (q)| + µq |~ u (q)| . ℓ t (2π)d

(b) Calculate the expectation value hui (q)uj (q′ )i.

• From the Gaussian form of the interaction we conclude −1 hui (q)uj (q′ )i = (2π)d δ d (q + q′ )Mij (q).

180

−1 By symmetry, the inverse matrix must also have the form Mij (q) = A(q)δij + B(q)qi qj ,

with the functions A(q) and B(q) obtained from δik =

X j

−1 (q)Mjk (q) = [A(q)δij + B(q)qi qj ] µq 2 δjk + (µ + λ)qj qk Mij

= µq 2 Aδik + (µ + λ)A + µq 2 B + (µ + λ)q 2 B qi qk .

Equating the two sides of the above equation gives A(q) =

1 , µq 2

and B(q) = −

(µ + λ)A µ+λ =− , 2 (2µ + λ)q µ(2µ + λ)q 4

resulting in −1 (q) = Mij

1 µ + λ qi qj δij − . µq 2 2µ + λ q 2

(c) Assuming a short-distance cutoff of the order of the lattice spacing a, calculate U 2 (x) ≡ E D 2 (~u(x) − ~u(0)) .

• Using the above expectation value of Fourier components, we obtain U 2 (x) =

X i

h(ui (x) − ui (0)) (ui (x) − ui (0))i

µ + λ qi qi dd q 2 − 2 cos(q · x) X δii − = (2π)d µq 2 2µ + λ q 2 i

2 µ+λ = d− Cd (x). µ 2µ + λ Z

The Coulomb potential Cd (x) satisfies ∇2 Cd (x) = δ d (x), and is given by Cd (x) = |x|2−d /[Sd (2 − d)] + c0 . Choosing the constant of integration such that U 2 (a) = 0, where

a is the short distance cutoff, leads to 2 U (x) = µ 2

µ+λ d− 2µ + λ

x2−d − a2−d .

Sd (2 − d)

(d) The (heuristic) Lindemann criterion states that the lattice melts when deformations grow to a fraction of the lattice spacing, i.e. for lim|x|→∞ U (x) = cL a. Assuming that µ = µ̂/(kB T ) and λ = λ̂/(kB T ), use the above criterion to calculate the melting temperature Tm . Comment on the behavior of Tm as a function of dimension d.

181

• In dimensions d > 2, the maximum extent of fluctuations is finite, and related to temperature T by

2kB T lim U 2 (x) = µ̂ |x|→∞

d−

µ̂ + λ̂ 2µ̂ + λ̂

!

a2−d .

Sd (d − 2)

According to the Lindemann criterion, the melting temperature is obtained by equating the above expression to (cL a)2 , leading to Tm =

µ̂ad (d − 2)Sd c2L h i.

2kB d − (µ̂ + λ̂)/(2µ̂ + λ̂)

Note that the melting temperature vanishes at d = 2. This is again a consequence of the divergence of fluctuations in d ≤ 2 due to proliferation of (phonon) Goldstone modes. The

Lindemann criterion is too crude to account for the quasi-long-range translational order in d = 2, and its destruction by dislocations. ********

182