[] Solutions Manual for Equilibrium and Non-Equili(BookFi.org)

7/28/2019 [] Solutions Manual for Equilibrium and Non-Equili(BookFi.org)

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ERRATApage 205, equation(4.82)

G(r, r ) =M(r)

B (r )or G1(r, r ) =

2L[M]M(r)M(r )

page 209, equation (4.92)

G1ij (r, r) =

L[M]Mi(r)Mj(r )

M1=v,M2=0

page 244, equation following (4.179)

D 4

D

2 1

= 4 D

page 245, first equation of the page

Z1 =d

dq2G1(q)

q2=0

page 359, equation following (6.81)T(u = 0) = T

= P

page 366, first equation

(x, t) = 0 + 0Dt

0

dx exp

(x x)2

4Dt

page 512, read: An example of the increase of the H-function is given by Jaynes [60].ignore: but the argument is incomplete.


AB(t) =1

2[A(t),B(0)]


t n +1

m gL = 0

equation following (9.196)

(

V) =

(

V)

2V

page 584, first line of question 11. By examining the moments X, P, (P)2, (X)2 and PX...


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Solutions for selected exercises and problems

Solutions for chapter 1

Exercise 1.6.2 Internal variable at equilibrium

1. Considering the energy as a function of S and y, one obtains from (1.36)

E

y

S

y

S

E

S

E

y

= 1

At equilibrium (E/S)y = T, and

E

y

(eq)

S

= T Sy

(eq)

E

= 0

because one must have by definition (S/y)E = 0 at equilibrium.

Let us also consider a function1 f(S(E, y), y) :

f

y

E

=S

y

E

f

S

y

+f

y

S

=f

y

S

the last equality being valid at equilibrium, and let us use this result in (1.64)

2S

y2

(eq)E

= y

S

E

y

S

S

E

y

=E

y S2S

Ey

1

T

2E

y2 S

The first term vanishes at equilibrium, which demonstrates the second equation of (1.63), taking (1.62)into account.

2. The condition of maximum entropy reads

dS =

1

T1dE1 +

1

T2dE2

+

P1T1

dV1 +P2T2

dV2

Since dE1 = P1dV1, the condition dS = 0 is automatic.3. The condition of minimum energy at fixed entropy, volume and particle number, is

dE = (T1

T2)dS1

(P1

P2)dV1 + (1

2)dN1 = 0

Because the superfluid does not transport entropy, there cannot be entropy transfer between the twocompartments: dS1 = 0. Furthermore dV1 = 0 because the volumes are fixed. The only condition isthat of equality of the chemical potentials: 1 = 2. Assume that the temperature of compartment 1is increased by T. As the chemical potential remains unchanged, the condition d1 = 0 added to theGibbs-Duhem relation (1.41)

d = SN

dT +V

NdP

1The following equality is a particular case of a general identity which we shall encounter many times. Let f(xi) be afunction of several variables, which are themselves functions of another set of variables: xi(yj). One easily shows that

f

yj

yj=k=X

i

f

xi

xl=i

xi

yj

yj=k


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shows that the pressure must increase by P = (S1/V1)T.

Problem 1.7.1 Reversible and irreversible free expansion of an ideal gas

1. One starts from the expression of (E/V)T

E

V

T

= TS

V

T

P = T PT

V

P

where we have used the Maxwell relation (1.27). By assumption, this expression must vanish, which givesanother proof of (1.61). taking into account

P

T

V

=1

V(T)

One gets T (T) = (T) and by integration (T) = aT.

2. In the case of an ideal gas

SVT

= PTV

= RV

The expression giving S is

S(T, V) = S(T0, V0) +

TT0

dTCVT

+

VV0

dVR

V

that is

S(T, V) = S(T0, V0) + CV lnT

T0+ R ln

V

V0

In order to express S as a function of the pressure, and not of the volume, one uses

VV0

= TT0

P0P

and, with CP = CV + R

S(T, P) = S(P0, T0) + CP lnT

T0+ R ln

P0P

3.

S(T, V) = S(T0, V0) + R ln

Tl/2V

Tl/20 V

A transformation at constant entropy thus corresponds to

Tl/2V = const ou T1/(1)V = const

with = (l + 2)/l. If we use the pressure instead of the volume

T P(1)/ = const Tf = Ti

PfPi

(1)/One can compute explicitly the work provided to the gas by integrating the formula for infinitesimal work

Wif = fi

PdV

for a reversible adiabatic transformation: P V =const, but it is simpler to remark that the increase inwork is equal to that in internal energy, as there is no heat exchange

Wif = Eif = CV(Tf Ti)


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the last equality being valid only for an ideal gas, and CV independent of T. It is a good exerciseto check that this result coincides with that obtained from calculating the work; one will notice thatCV = R/( 1)

4. The work supplied to the gas is determined by the external pressure, which is equal to the finalpressure Pf: Wif = Pf(Vf Vi). As the transformation is adiabatic and as the gas is ideal, we have

once moreWif = Eif = CV(T

f Ti)

Using the ideal gas law, Vf = RTf/Pf, one determines T

f

Tf =Ti

1 + ( 1) Pf

Pi

The final temperature in the irreversible expansion is higher than that of the reversible expansion: thiscan be seen by using the concavity of the curve giving Tf as a function of x = Pf/Pi.

5. In the case of the reversible transformation, the entropy variation S vanishes, while in the irreversible

case S > 0. Since the final pressures are identical, one may use in the comparison the result of question 2

S(Tf, Pf) = S(Ti, Pi) + Cp lnTfTi

+ R lnPiPf

whence

CP lnTfTi

> CP lnTfTi

and thus Tf > Tf. The variation of internal energy is minus the work supplied to the external mediumW = CV(Ti Tf). As Tf > Tf, Wirr < Wrev. In order to convince oneself that the external gas does notplay any role in the entropy balance, one remarks that it can be replaced by a mass M on the piston, withPfA = M g. The external medium is then entirely mechanical, and does not contribute to the entropybalance.

If one puts again the mass on the piston, Pf = Pi, and the expression of the entropy shows that Tf > Tibecause one must have S > 0 after completion of both operations. An explicit computation gives

Tf =Ti2

1 + ( 1) Pi

Pf

1 + ( 1) Pf

Pi

It is easy to check in this expression that Tf > Ti by setting x = Pf/Pi and by remarking that (x+1/x) 2.

6. Since the expansion is adiabatic and does not supply any work to the system, and since the gas is ideal,the internal energy and therefore the temperature remains unchanged. From the results of question 2,dS = R ln[(V + dV)/V] RdV /V > 0. Since dQ = 0, the transformation cannot be quasi-static:dQ = TdS does not hold.

7. Since the internal energy does not vary (adiabatic wall and no work supply), Tf = Ti, and S =R ln(Vf/Vi), although no heat was supplied to the gas.

Problem 1.7.3 Equation of state for a solid

1. The knowledge of the equation of state allows us to express the temperature

T =E

S

V

= f(V)g(S) = S1/3

f(V)eS/3R

3R(S+ 4R)

(1)

The quantity between square brackets staying finite when S 0, one sees in the preceding expressionthat S and T vanish simultaneously. This behaviour is in agreement with the third law.

2. Expression (1) allows one to display the dominant behaviours of S, and thus ofCV , in the two limiting

cases


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1. low temperature: T S1/3 S T3 CV T3

2. high temperature T eS/3R S 3R ln T CV 3R

3. From (1.68), one gets

P = EV

S

= f(V)g(S) = 2Ab(V V0)eb(VV0)2

g(S) (2)

The physical interpretation of V0 is clearly displayed in (2): V0 is the volume at zero pressure, namelythe maximum volume of the solid. For V > V0, the atoms which form the solid are organized in anotherphase of matter described by another equation of state. One can point out an inconsistency of the model:it predicts a finite pressure for a vanishing volume! At zero pressure, the dilatation coefficient vanishes,since V0 is temperature independent.

4. Let us apply (1.36) to the variables (S,V,T)

S

V T = TV S TSV1

= f(V)g(S)f(V)g(S)

There are many possible proofs of relation (1.69), we propose below the proof which seems to us theshortest. Examination of the relevant variables for the calculation of the dilatation coefficient lead us towork with the Gibbs potential G(T, P) = E T S+ P V

dG = SdT + VdPOne may then use the Maxwell relation

V

T

P

=

P

S

T

1Relation (2) provides us with the expression for P in terms of the variables (S, V). Using a procedurewhich should be familiar by now, we display explicitly the partial derivatives of P with respect to thesevariables

P

S

T

=P

S

V

+P

V

S

V

S

T

= (f(V))2(g(S))2 f(V)f(V)g(S)g(S)

f(V)g(S)

= 1CV

(f(V))2(g(S))2 f(V)f(V)g(S)g(S)f(V)g(S)

The last line is obtained by noticing that

CV = TS

T

V

= T T

S

V

1=

g(S)g(S)

We invite the reader to work out another demonstration starting from

V

T

P

T

P

V

P

V

T

= 1

Problem 1.7.5 Surface tension of a soap film

1. The TdS equation reads in this case

dE = TdS+ dx


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Let us also write the differential of the free energy

dF = SdT + dx

and the Maxwell relation which follows from it2

S

x

T

= T

x

2. Using the differentiation rule for implicit functions and the preceding Maxwell relation, one showsthat

E

T

x

=E

S

x

S

T

x

= TS

T

x

= Cx

E

x

T

=E

x

S

+E

S

x

S

x

T

=

T

T

x

3. At a constant temperature

dE = T T

x

dx + dx

The second term is identified with the work: dW = dx, and thus the energy which is exchanged inthe from of heat is

dQ = T T

x

dx = a0T dx > 0

One must supply heat to the film in order to stretch it.

4. Let us show that (Cx/x)T = 0

Cxx T = T T Sx Tx = T

2

T2 x = 05. The surface tension, and thus f, are independent of the film area, and thus of x, whence

f

x

T

= 0 T =

One recognizes a well-known, and easily observable, property: any fluctuation destroys the soap film.

6. Let us express the TdS equation in terms of the variables (T, x)

TdS = Cx dT + a0T dx

When a transformation is done at constant entropy

dT

T= a0

Cxdx

Numerical application: dT = 4.3 102 dx (dx in m).

2One can derive directly this relation from (1.27) with the substitutions V x et P .


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Exercise 2.7.1 Density operator for spin-1/2

1. The second degree equation giving the eigenvalues of D is

2 + a(1 a) |c|2 = 0

and the product of the roots is given by

+ = a(1 a) |c|2

The positivity condition on D implies + 0 and the condition Tr D = 1, or + + = 1, so that0 + 1/4, whence (2.105). In the case of a pure state one of the eigenvalues vanishes (the otherbeing equal to one) and a(1 a) = |c|2. The density matrix of the state | is

D = ||2

||2and one can indeed check that ||2||2 = ()().2. One writes a = (1 + bz)/2, c = (bx iby)/2, whence (2.107) for D. As Tr i = 0, one deduces from(2.107)

Tr ij = 2 ij

and (2.108). One observes that a(1 a) = (1 b2z)/4 and that |c|2 = (b2x + b2y)/4. The inequalities (2.105)are equivalent to 0 b 2 1, the pure case corresponding to b 2 = 1.3. With B parallel to Oz, the Hamiltonian reads

H =

1

2

z

and the equation of evolution of the density matrix is

dD

dt=

1

i[H, D] = 1

2B(bxy byx)

which is equivalent todbxdt

= B bydbydt

= Bbx dbzdt

= 0

that isdb

dt= B b

This is nothing other than the evolution equation of a classical magnetic moment in a constant magneticfield.

4. The non-zero matrix elements of D are

D+;+ = D+;+ =1

2D+;+ = D+;+ = 1

2

On taking the partial trace, we obtain the density matrix of the first spin D(1)

D(1)++ = D

(1) =

1

2D(1)+ = D

(1)+ = 0

D(1) then describes a state with zero polarization. One has of course D(1) = D(2), and D(1)D(2), whichis a diagonal matrix with all elements equal to 1/4, is not equal to D, which would be the case were D a

tensor product.


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Exercise 2.7.7 Galilean transformation

The occupation probability Pr of a level r is unchanged under a Galilean transformation, otherwiseone could detect a translational motion without looking outside the system.This implies D = D andinvariance of the entropy. Under a Galilean transformation, the Hamiltonian becomes

H = H uP + 12

M u2

and the equality of the operator parts of D and D implies

H P = (H uP)that is = and u = . One also gets

Z = Z exp

1

2M u2

= Z exp

M

22

From (2.66), this gives PP = ln Z

=

M

= M u

which is the expected value. Concerning the energy one gets

E = ln Z

= ln Z

+

M2

22= E +

1

2M u2

The total energy is the sum of the rest energy E and the kinetic energy M u2/2 of the moving mass.

Exercise 2.7.8 Fluctuation-response theorem

1. Let us compute dK/dx

dK(x)

dx= ex(A+B)(A + B)exA ex(A+B)A exA = ex(A+B)B exA

On integrating this equation from x = 0 to x = 1 one gets

K(1) K(0) = e(A+B)eA 1 =1

0

dx ex(A+B)BexA

which leads to (2.118) if one multiplies the RHS by exp A. Applying this equation to exp(A + B) leadsto

eA+B = eA +

1

0 dx ex(A+B) (B) e(1x)A

One may replace exp(x(A + B)) by exp(xA) with an error on the order of 2. Using the invariance ofthe trace under circular permutation leads to (2.119)

Tr e(A+B) Tr eA + Tr (eAB)By definition

eA(+d) = eA()+dA() = eA() + d

d eA()

d

The substitution d in (2.119) leads to (2.120). Using once more the invariance of the trace undercircular permutation allows one to recover (2.57) in a particular case

d

d Tr eA()

= Tr dA()d eA() (3)


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2. From (2.66), or (2.124)

Ai =1

Z

iTr

exp

kkAk

=

1

ZTr

Ai exp

kkAk

= Tr (DAi)

In order to differentiate Ai with respect to j , one must use (2.120)

jTr

Ai exp

k

kAk

= Tr

Ai

10

dx exp

xk

kAk

Aj exp

(1 x)

k

kAk

= ZTr 10

dxAiDxAjD

(1x)

from which one deduces

Ai

j

=1

Z

j

Tr Ai expk kAk AiAj=

10

dx Tr

(Ai Ai)Dx(Aj Aj)D(1x)

Let us set B =

i ai(Ai Ai)

ij

aiajCij =

10

dx Tr (DxBD(1x)B) 0

because, in a basis where D is diagonal

Tr (DxBD(1x)B) =n,mDxnnBnmD(1x)mm Bmn =n,mDxnnD(1x)mm |Bnm|2 0using the hermiticity ofB and the fact that the diagonal elements of D are positive.

Exercise 2.7.9 Phase space volume for N free particles

The condition for the total energy to be E isp212m

+ . . . +p2N2m

E

and thus

N(E) =VN

N!hN p 21+...+p 2N2mEd3p1 . . . d

3pN

One expects the average energy per particle to be E/N, which suggest to introduce the dimensionlessvariable u defined as

p =

2mE

Nu u21 + . . . + u

2N N

since the average value of u is on the order of 1. This change of variables leads immediately to (2.122),with

C(N) =

u21+...+u

2NN

d3u1 . . . d3uN

In order to compute C(N) one calculates in two different ways the integral In

In =

+

dx1 . . .+

dxn e(x21+...+x2N)


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This integral is the product of n independent Gaussian integrals and its value is n/2. One can alsocompute this integral in polar coordinates in a n-dimensional space

In = An

0

dr er2

rn1

where An is the area of the unit sphere in a n-dimensional space. The change of variables v = r2 castsIn in the form

In =1

2An

0

dv evvn21 =

1

2Ann

2

which leads to

An =2 n/2

(n/2)

One computes C(N) in polar coordinates in a 3N-dimensional space (r = (u 21 + . . . + u2N)

1/2)

C(N) = A3N

N

0

dr r3N1 =23N/2(N)3N/2

3N(3N2 )=

3N/2(N)3N/2

( 3N2 + 1)

The final form of C(N) is obtained by using Stirling approximation

3N

2+ 1

3N

2e

3N/2One deduces N(E)

N(E) =VN

N!

4emE

3N h2

3N/2(4)

One remarks that the result is well approximated by

1N! ENN

where 1 is the number of levels calculated in (2.22) for a single particle. The density of states (E)is the derivative of N(E) with respect to E, (E) =

N(E), and (E) is given by

N(E)E.

Exercise 2.7.10 Entropy of mixing and osmotic pressure

1. The number of possible configurations available to place molecule B is proportional to the number ofA molecules, NA, as each molecule defines a possible localization for B. The entropy is then k ln NA+ afunction of (T, P). If one adds NB molecules, this can be done in

CNBNA (NA)NB

NB!

ways, whence the expression of S. This entropy is partly an entropy of mixing (see (2.93)).

2. The terms E and P V do not depend on NA because the energy and volume variations depend onlyon the interaction of molecule B with its neighbours A, and not on the number of A molecules: E andP V are local terms. If one adds NB molecules of type B, because of the condition NB NA, each Bmolecule interacts only with its neighbours of type A, and one can add the terms E and P V of each ofthe B molecules. To the Gibbs potential NA0(T, P) of the solvent one must finally add NBf(T, P) andthe term T S, whence (2.123). In a scale transformation, ln(eNA/NB) is unchanged, and G G.The chemical potentials follow by differentiation

A =G

NA= 0(T, P) NBkT

NA

B =G

NB = f(T, P) + kT lnNBNA


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One checks that G = ANA + BNB.

3. The chemical potentials of the type A molecules must be the same in both compartments

0(T, P0) = 0(T, P1) NBkT

NA

By performing a Taylor expansion

0(T, P0) = 0(T, P0) + (P1 P0) 0P

T

NBkTNA

= 0(T, P0) + (P1 P0) VNA

NBkTNA

namely

P1 P0 NBkTV

The osmotic pressure is identical to that of an ideal gas with NB

molecules in a volume V3. However thevolume V is typical of that of a liquid, on the order of 103 of that of a gas, and the osmotic pressurecan be very large.

3This remark must be taken as a mnemonic rule: as emphasized in our derivation, the osmotic pressure is of entropicorigin .


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Exercise 3.7.5 Solid and liquid vapour pressures

1. The chemical potentials will coincide when the two phases are in mutual equilibrium

s = g

only if the zeroes of energy have been chosen in an identical way in the two phases. One possibleconvention is to decide that an atom in the gaseous phase, located at infinity with a vanishing kineticenergy, has zero energy. On must then give the solid an internal energy at zero temperature whichcompensates exactly the ionization energy, so that if the solid is destroyed with all atoms sent to infinity,the total energy vanishes.

E(T = 0) = N u0

2. The solid being incompressible, the specific heats at constant volume and at constant pressure coincideCP = CV = C = N c; the energy of the solid becomes

E(T) = N u0 + NT0

dT c(T)

and its entropy

S(T) = N

T0

dTc(T)

T

This allows one to deduce the expression for the Gibbs potential

G = E(T) T S(T) + P N vs = N s(T)

= N

P vs u0 + T0

dT

1 TT

c(T)

whence (3.149). The equality of the Gibbs potentials s = g leads to an implicit equation for P(T) =kT/v. In the limit where vs v one gets

P(T) =

2 m

h2

3/2(kT)5/2 exp

s(T)

kT

3. The partition function for the N molecules of the liquid is

ZN =1

N!Nl l =

N v0h3

d3p ep

2/2meu0 =N v03

eu0

from which one derives the liquid Gibbs potential Gl

Gl = FN + P V = N

P v0 u0 kT lnev0

3

and l = G/N. Since V = N v0, N and V are not independent variables, one should not differentiatewith respect to N at fixed V. The equality of the chemical potentials g et l and the ideal gas lawP v = kT give the relation

u0kT + ln ev03 = ln kTP 3+ v0v


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and taking the exponential of this relation

eu0/kTev0 =kT

PePv0/kT

Since P v0/kT = v0/v, this quantity 1 if the specific volume of the liquid is very small with respectto that of the gas, an excellent approximation in the case of boiling water under standard conditionsv0/v 103. Within this approximation

P kTev0

eu0/kT

4. The entropies per molecules in the gas and in the liquid phases are deduced from (3.17) or from

g = gT

P

= k lnv

3+

5

2k

l =

l

T P = k ln v03 + 52 kso that

= T(g l) = kT ln vv0

On the other handdP

dT=

k

ev0

1 +

u0kT

eu0/kT =

P

T

1 +

u0kT

which allows one to check that the Clapeyron formula holds true, as must be the case because our startingpoint was the equality of chemical potentials. In the case ( i) one finds u0/kT 12.1 and in the case (ii)u0/kT 6.4. This difference by a factor of two illustrates the rough character of the model.

Problem 3.8.4 Models of a boundary surface

A. 1. The length of the interface is obtained from Pythagoras theorem

L =L0

dx

1 +

dy

dx

21/2 L + 1

2

L0

dx

dy

dx

2(5)

2. On differentiating (3.158) with respect to x one gets

dy

dx=

n=1

nL

An cos

nxL

whence the expression of the Hamiltonian

H =

2

L0

n,m=1

nL

mL

AnAm cos

nxL

cosmx

L

=

2

4L2

n,m=1

nmAnAm

L0

dx

cos

(n + m)x

L

+ cos

(n m)x

L

The result (3.159) for H is obtained thanks toL0

dx

cos

(n + m)x

L

+ cos

(n m)x

L

= Lnm

This result is nothing other than the decomposition of H into independent normal modes An.


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3. The probability law for the Ans is given by the Boltzmann weight exp[H]. Since the normal modesAn are uncoupled

P(An)

exp2

4LkTn2A2n P(An, Am) exp

2

4LkT n2A2n + m2A2mand average values follow from the equipartition theorem (3.53)

A2n =2LkT

2n2AnAm = 2LkT

2n2mn

This expression allows one to derive y(x)y(x)

y(x)y(x) = 2LkT2

n=1

1

n2sinnx

L

sin

nx

L

=2LkT

2

n=1

1

n2

cos

n(x x)

L

cos

n(x + x)L

Using the identity (3.160)

n=1

1

n2

cos

n(x x

L

cos

n(x + xL

=

2

L2x(L x)

we compute the average value

y(x)y(x) = kTL

x(L x)as well as

(y)2 = (y(x) y(x))2 = kTL

|x x|(L |x x|)When |x x| L, the fluctuation (y)2 is

(y)2 kT

|x x| (6)

One can also derive these last two results without appealing to normal modes, by discretizing the integralwhich gives L and by using the identity (A.45) on multiple Gaussian integrals. Indeed, the problem isequivalent to that of a random walk on a straight line x(t) (exercise 6.2), with the following correspon-dence: x y, t x et D kT /2, where D is the diffusion coefficient. Relation (6) is the analogueof: (x)2 = 2D|t t|. The fluctuation is largest for |x x| = L/2, and in that case (y)21/2 L.The interface width grows as

L, and is not a constant.

B. 1. The energy associated with a plane interface, corresponding to the ground state, is 2 JN. Let usobserve that in the preceding situation N is the length of the interface, in units of the lattice spacing. Ina general configuration where the interface is not a straight line, the interface length is

L =Np=1

(1 + |yp|)

and the energy

H = 2JL = 2JN + 2JN

p=1

|yp|

2. Let us define hp = 2J(1 + |yp|) and H =Np=1

hp: the steps are independent variables and the partition

function factorizes

Z =N

p=1Tr ehp = N


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15

where represents the partition function of a single step

=+

y=e2J(1+|y|) = e2J coth J

From the knowledge of (or of Z), one gets and f using (3.2) and (3.6)

= 2J

1 +

1

sinh 2J

f = 2J 1

lncoth J

To each of the extreme regimes corresponds a very different behaviour of the interface

1. J 1: the steps behave as independent random variables and the interface is very irregular 2J + kT f 2J kT ln(kT/J)

One then finds a kind of equipartition theorem with = 2. In this classical limit

=

+

dy e2J(1+|y|) = e2J1

J

2. J 1: the interface is in its ground state, meaning that it is a straight line. 2J f 2J

3. The constraint is obeyed in two situations y = p, so that

P(|y| = |p|) = 2e2J(1+|p|)

(7)

Setting = e2J = coth J and K = J in (7), one easily shows the two following results

|y| = 12

1

K=

1

sinh 2J

y2 = 14

1

2

K2=

1

2(sinh J)2

The first result allows one to check = 2J(1 + |y|.4. The central limit theorem tells us that 2y = q

2y . Moreover

2y = y2 because y = 0, whence

(y)2

= q

y2

=

q

2(sinh 2J)2

As in the continuum model (part A), one finds that (y)2 q |x x|.

Problem 3.8.5 Debye-Huckel approximation

1. The densities n(r) are proportional to the probabilities of finding a charge q located at positionr as measured from the ion at the origin of coordinates. They are thus proportional to the Boltzmannweight

n(r) expq(r)

kT

since the potential energy of an ion with charge q is q(r). The normalisation is fixed thanks to thecondition that, if r

, (r)

0 and n(r)

n. The condition of weak deformations, in other words

the fact that there are only small deviations with respect to the ideal gas case, corresponds to a regime


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16

where the modulus of the potential energy, |q| is very small with respect to the kinetic energy, kT.Therefore4

n(r) = neq(r) n (1 q(r)) (8)

Let us note that in the general definition (3.75) of the density, the origin of coordinates can be takenarbitrarily, in contrast to the present situation where the position r which appears in the densities n(r)is measured from a specific ion. In question 4, where we shall give the expression for the pair density, weshall see very clearly that n(r) must be interpreted in terms of conditional probabilities.

2. The second form of the Poisson equation illustrates very clearly the meaning of the ionic cloud: acentral charge, q(r), which modifies locally the charge distribution around it q(n+(r) n(r)). Using(8), the Poisson equation becomes

2(r) q0

2nq

kT(r) + (r)

and its Fourier transform reads

p2 = 2nq20kT + q

0= 1

b2 + q

0

so that = q0

1

p2 + b2

Using (3.161), the inverse Fourier transform of the preceding equation gives

(r) =q

40 rer/b

whence

n+(r)

n(r) =

1

4b2

r

er/b

3. The charge Q(R) contained in a sphere with radius R is

Q(R) = q

1 0rR

d3r (n+(r) n(r))

= q

1 1b2

R0

dr rer/b

= qeR/b(1 + R/b)

When R one expects that this charge tends to zero because the solution is electrically neutral. Onechecks this to hold true for R b, which allows us to interpret the Debye length b as measuring theextension of the ionic cloud: far beyond the Debye length, the central charge is completely screened.

4. The expression (3.162) of the average potential energy is obtained directly from the general relation(3.84), by identifying separately the contributions of the densities associated with the two kinds of ion,n+ and n. For the sake of simplicity, let us consider one of the two pair densities: n+2 (r

, r

)d3rd3r

is interpreted as the mean number of pairs of identical charges in d3rd3r around r and r

. However,assuming that an ion is located at r = 0, then n+(r)d3r represents the mean number of pairs of ions withidentical charge which one may build around the position r with the central ion. Hence

n2 (r, r

) = nn(r = r r ) = n2 eq(r)

The last equality gives the pair correlation function (3.77) of the problem.

4The expansion of the exponential is clearly not valid for r 0. However we only need it to be valid on the scale ofthe average distance between ions: q(r) kT, with r n1/3. As we shall see in question 5, the approximation is validprovided the density of potential energy is small with respect to the density of kinetic energy. If that were not the casen(r) would not be given simply by a local Boltzmann weight!


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17

The expression (3.162) of the potential energy is clearly of the form (3.84), once the result is multipliedby a factor of two in order to take into account the fact that two signs are possible for the central charge

Epot =q2

40 V d3rd3r 1|r r |n+2 (r , r ) n2 (r , r )= nV

q2

40

V

d3r1

r

n+(r) n(r)

= nV q

2

40b2

+0

dr er/b = nV q2

40b

5. The average potential energy between two ions (taken with the same sign to simplify the discussion)is

q2

40 r q

2

40 n1/3

The condition of weak deformations also readsq2

40 n1/3 kT

so that1

n2/3 40 kT

nq2= 8b2

which leads correctly to the condition r b; this last condition implies that the present model is onlyvalid if the volume b3 contains a large number of charges. In an equivalent way, one may observe thatthe density of potential energy nq2/(40b) is very small with respect to the density of potential energy(3/2)nkT provided b r.

Problem 3.8.7 Beyond ideal gas: first term of virial expansion

1. One deduces (3.166) from the expression (3.86) of the pressure

P =1

ln Z

V=

1

ln ZKV

+1

ln ZUV

and from /V = (1/N)(/v).

2. Low values of the temperature enhance the effect of the attractive part of the potential, while highvalues enhance the repulsive part. The average number of molecules in a sphere of radius r0 is N1 nr30 1 if one chooses n to be arbitrarily small. Since the molecules are assumed to be independent,the probability P(N1) of finding N1 molecules in a sphere with radius r0 is a Poisson distribution. Theprobability of finding two molecules in the sphere is then N12/2 N1. If one draws around eachmolecule a sphere with radius r0, one sees that the average number of molecules whose distance is lessthan r0 is N nr

30/2. This number is negligible ifN nr

30 = n

2V r30 1, namely if one considers a sufficientlysmall volume of gas at a fixed density. This condition being satisfied, let us assume that the molecules 1and 2 are located at a distance less than r0.The probability of finding in the gas another pair of moleculeswhose distance is less than r0 is then (N nr30)2 and thus negligible. The integrand in (3.167) is differentfrom zero only if |r1 r2| < r0, while for all the other pairs one can take this distance to be larger thanr0, and the integral can be evaluated as

I V(N2)

d3r1 d3r2

eU(|r1r2|) 1

V(N1)

d3r

eU(|r|) 1

Since the number of pairs is N(N 1)/2 N2/2, one deduces (3.168) and (3.169) with

B(T) = 2 0 dr eU(|r|) 1 r2 (9)


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18

using polar coordinates. As B(T) r30, one has (N2/V)B(T) 1 and

ln ZU N2

VB(T) = N f(v, T)

One checks the extensivity of the free energy and obtains Pv/kT from ln ZU by using (3.166).

3. If the gas is sufficiently diluted, v b and

P =kT

v b a

v2 kT

v

1 +

b

v

a

v2

so that by identification with (3.170)

B(T) = b akT

4. In the case of a dilute gas, the probability of finding a molecule at a distance r from another molecule

is proportional to the Boltzmann weight exp[U(r)], because the potential energy of the set of thetwo molecules is U(r). This is not true for a dense gas, because this potential energy is modified bythe presence of neighbouring molecules, but this effect is negligible for a dilute gas. The normalisationlimr g(r) = 1 demands that g(r) exp[U(r)]. Putting this value ofg(r) in (3.172) and using polarcoordinates, one finds

P = nkT 2n2

3

0

r3dr U(r)eU(r)

One then integrates by parts, taking as a primitive of U(r)exp[U(r)] the function

1

eU(r) 1

so that the pressure is given by a convergent integral. The final result is

P = nkT 2n20

r2dr

eU(r) 1

which coincides with (3.170).

5.

1. Hard sphere gas

B(T) = 20

r2dr =2

33

The first virial coefficient is positive and independent of temperature. A naive reasoning relyingon excluded volume considerations would lead to a coefficient 4/3 in the van der Waals equation,instead of the correct value 2/3; actually it is the number of pairs of colliding molecules which isthe relevant quantity.

2. In this case the virial coefficient is given y

B(T) =2

3

3 (eu0 1)(r30 3)

When T 0, exp(u0) is large and the attractive part of the potential dominates: B(T) < 0. Onthe contrary, when T , the repulsive part of the potential dominates and B(T) (23)/3;the potential energy is then negligible with respect to the the kinetic energy, except if the repulsive

part is concerned. One can then predict a change of sign of B(T).


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19

3. Lennard-Jones potential

B(T) = 2

0

dr r2

exp

4u0

r 12

r 6

1

= 23

0

dx x2

exp4 u0

kT

1x

12 1

x

6 1

= 3f u0

kT

= 3f

1

If one draws B(T)/3 as a function of , one should obtain a universal curve, independent of the typeof molecule. However, at low temperatures, and for the lightest gases, hydrogen and helium, quantumeffects may not be neglected and one observes deviations with respect to the universal curve which relieson the classical theory.

Problem 3.8.8 Theory of nucleation

1. Using the results of subsection 3.5.4, one easily shows that

(E T0S 0N) 0 (10)

2. If the system exchanges work with the external medium, the expression of energy conservation becomes

E = Q + W + 0N

so that using once more Q T0S

W (E T0S 0N)

In the case of a reversible transformation we have

Stot = (E T0S 0N) W = 0

and one checksWmin = (E T0S 0N)

2. The entropy is an increasing function of the energy, as is clearly illustrated by the curve Stot(Etot).

3. in the vicinity of point D (or C), one may use a linear approximation

Stot dStotdEtot

D

Wmin = 1T0

Wmin

When going from D to B, one goes from a situation where the system is in equilibrium with the reservoir,and where the number of configurations is eq, to an off-equilibrium situation where the number ofconfigurations is . Therefore the probability of such a fluctuation is

P q

= e(StotSqtot)/k = eStot/k eWmin/kT0

4. One works in a container of volume V. Before the droplet is formed, the gas occupies the wholevolume, and thus (see (3.133))

Jin = P2VAfter formation of a droplet of volume V1,

Jfin = P2(V V1) P1V1


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20

The transformation gas-droplet is made at constant temperature and chemical potential. Indeed thesupersaturated vapour plays the role of the reservoir (figure 3.24), and the droplet that of the system A.The minimum work which is necessary in order to form the droplet is thus given by (3.173), when thesurface energy is taken into account

Wmin = J+ A = V1(P2 P1) + A

5. In the case of a spherical droplet, V = (4/3)R3 et A = 4R2, one has

dWmindR

= 4R2(P2 P1) + 8R

and R is given by

R =2

P1 P2 (11)

One observes on figure 3.36 that a droplet disappears if it is formed with a radius less than the criticalradius R. One must overcome a potential barrier of value Wmin(R) in order to obtain the dropletnucleation. This is the meaning of the expression activation energy which is given to the critical valueWmin = Wmin(R

)

Wmin =16

3

3

(P1 P2)2 (12)

R3/(P1 P2)

R

Wmin

16

3

3

(P1 P2)2

Figure 1: Curve giving Wmin(R).

7. Clearly the probability of nucleation is given by

P eWmin/kT

Taking (12) into account, one understands that droplet formation is more and more likely as the pressuredifference between the liquid and gaseous phases increases.

8. One must necessarily have 1(P1, T) = 2(P2, T) otherwise one would observe a flux of molecules:at the transition 1(P , T) = 2(P , T). Therefore, at fixed temperature

1(P1) 1(P) = 2(P2) 2(P)1(P + P1)

1(P) = 2(P + P2)

2(P)

v1P1 = v2P2


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21

where we have used (/P)T = v which is deduced from the Gibbs-Duhem relation (3.105). We cannow rewrite the denominator of (11) as

P1

P2

= P1

P2

=P2

v1(v2

v1

)

=P2v1

2P

T

1P

T

=

1

v1(2(P2, T) + 1(P2, T))

and one then obtains (3.174), namely an expression for the critical radius which does not involve theknowledge of the pressure inside the droplet.


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22


Exercise 4.6.1 High temperature expansion and the Kramers-Wannier duality

1. One first proves the identity

exp(KSiSj) = cosh K+ SiSj sinh K

by noticing that there are only two possibilities, SiSj = 1 and SiSj = 1, and that the identity is validin both cases. Then, expanding the expression of the partition function, one readily finds

Z(K) = (cosh K)LC

i,j

(1 + SiSj tanh K) (13)

Let us represent graphically the product of Ising spins

S

n1

1 Sni

i SnN

N

by drawing a line on the lattice between two Sis if they are nearest neighbours. The contribution of thespin product to the partition function will be non zero only if there is an even number of lines (0, 2 or 4)leaving from all sites. If a term in the expansion corresponds to an odd number of lines leaving any site,then it gives a vanishing contribution. Thus the contribution of the preceding spin product will be nonzero only if the lines form a closed polygon on the lattice.

2. Let us count the number of polygons for the first values of b, without paying attention to the boundaryconditions.

1. b = 4. The polygon is a square with perimeter 4 (in units of the lattice spacing). Since this squarecan be placed in N differents ways on the lattice, (4) = N.

2. b = 6. The polygons are rectangles with perimeter 6, whose largest side can be either horizontal(horizontal rectangle) or vertical (vertical rectangle). Then (6) = 2N.

3. b = 8. The leading term in N is obtained by placing two squares with b = 4 on the lattice. Thiscan be done in N2/2 ways. Other possibilities (i.e. rectangles with perimeter b = 8) give only afactor proportional to N. Thus (8) = N2/2. It is left to the reader to show that the exact result,using periodic boundary conditions, is (8) = N(N 5)/2.

4. b = 10. The leading term is given by a square with perimeter 4 and a rectangle with perimeter 6.This rectangle can be either vertical or horizontal, so that (10) = 2N2.

5. b = 12. The leading term corresponds to three squares with perimeter 4, giving (12) = N3/3!.

The first terms in the high temperature expansion of the partition function are then, setting tanh K = x

ZN(K) 2N(cosh K)L

1 + N x4 + 2N x6 +1

2N2x8

+ 2N2x10 +1

6N3x12 +

(14)The free energy FN is proportional to the logarithm of ZN; we have, to order x12

FN ln ZN N x4 + 2N x6 + 12

N2x8 + 2N2x10 +1

6N3x12

12

N x4 + 2N x6 +

1

2N2x82

+1

3N3x12 +

= N(x4 + 2x6 + ) + N2x12 +


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23

At least to order x10, we have thus checked the extensivity of the free energy, but our estimates of (b)are not precise enough to give the coefficient of N in the x8 term and to eliminate the N2 factor in thex12 term.

3. Let us flip all the spins of the dual lattice inside a given polygon of the original lattice. It is clear thatthere is a one-to-one correspondence between the links on the polygon and the number of broken linkson the dual lattice: each broken link on the dual lattice crosses a side of the polygon once and only once.Starting from a configuration with all spins up, the energy of a configuration with n broken links will be

En = KL + 2nK

because each broken link costs an energy 2K, hence the partition function is, in the low temperatureregion

Z(K) = 2 exp(KL)

1 +

n=4,6,8(n)exp(2nK)

(15)

where the factor 2 comes from the fact that we could have chosen to start from a configuation with all

spins down. Note that (13) converges for temperatures T > Tc1, while (15) is convergent for T < Tc2.with Tc2 Tc1: The temperatures Tc1 and Tc2 are defined as those temperatures which limit the radiusof convergence of the expansions (13) and (15) respectively.

4. From the high temperature expansion we have

Z(K)

2N(cosh K)L=

1 + b=4,6,8,...

(b)(tanh K)b

while the low temperature expansion gives for a temperature T

Z(K)2 exp(KL)

=

1 +

n=4,6,8(n)exp(2nK)

The two ratios are equal if one chooses to relate K and K by

exp(2K) = tanh K

This gives a relation between the partition function at two different temperatures. If there is a uniquecritical temperature where both the high and low temperature expansions diverge, Tc1 = Tc2 Tc, weobtain this critical temperature Kc as

exp(2Kc) = tanh Kc or sinh2 Kc = 1

Note that the assumption that there is a unique critical temperature is a strong one: one easily findsmodels where this assumption is not satisfied, see for example problem 7.9.3.

Exercise 4.6.5 Shape and energy of an Ising wall

1. The equation of motion for Md2M

dz2= V(M)

is immediately deduced from the identities (A.59) and (A.60). In the mechanical analogy, the conservationof kinetic energy is

1

2

dx

dt

2+ U(x) = const

The particle starts at t =

with zero velocity from a point of abscissa x =

M0 and arrives at

t = + with zero velocity at the point x = M0. At the points x = M0, the kinetic energy vanishes and


24/98

24

the potential energy is U(M0), so that the constant in the preceding equation is U(M0). Transposingthis result to the original problem leads to

1

2 dMdz 2

= V(M) V(M0) (16)Of course, one could prove (16) directly from the equation of motion exactly as one proves energyconservation in classical mechanics. One checks that differentiating (16) with respect to z gives back theequations of motion, while the boundary conditions are limz M = M0 and lim|z|(dM/dz) = 0.2. Dividing equation (16) by |r0|M20 , one gets

1

2|r0|M20

dM

dz

2= 1

2M20(M2 M20 ) +

u04!|r0|M20

M4 M40

Now the zero field magnetization is

M20 =

6

|r0

|u0 whenceu0

4!|r0|M20 =1

4M40

This yields the equation for f

2

df

dz

2= 1

2(f2 1) + 1

4(f4 1) = 1

4

1 f2

Taking the square root of the preceding equation and choosing the positive sign (the negative sign wouldalso lead to an equally acceptable solution, corresponding to a decreasing function of z, one gets

df

1 f2 =dz

2

from which one deducesf(z) = tanh

z z02

(17)

Taking z0 = 0 simply fixes the position of the wall at z = 0.

3. The surface tension is the difference per unit area of the energy in the presence of a wall (first bracketin the following equation) and the energy without any wall (second bracket)

=

+

dz

12

dM

dz

2+ V(M)

V(M0)

Using (16) allows us to cast the expression for in the form

=

+

dzdM

dz

2= M20

+

dzdf

dz

2

= M20

+11

df

dzdf =

M202

+11

(1 f2)df = 23

M20

Exercise 4.6.6 The Ginzburg-Landau theory of superconductivity

1. The magnetic field B vanishes with the vector potential. Furthermore, decomposing into real andimaginary parts

= 1 + i2


25/98

25

we get||2 = |1 + i 2|2 = (1)2 + (2)2

so that ( N) becomes

N =

d3r

a||2 + b||4 + 2

2m

(1)2 + (2)2

which is exactly (4.88) with a suitable redefinition of the parameters a, b and .

2. The covariant derivativeD = i qA

transforms in the following way under a local gauge transformation

exp(iq(r ))(i qA)exp(iq(r )) = i qA(r) + q(r) = i qA(r)

which shows the invariance of ( N) under local gauge transformations.3. Let us give some details on the computation of functional derivatives. We begin with

1 =1

2

d3r B2 =

1

2

d3r ( A)2

We use the identity, easily proved from the techniques given at the end of 10.4.1 (V W) = W V V W (18)

to write 1 as

1 =

d3r( A) ( A) =

d3r B ( A)

= d3r A ( B) + (A B)=

d3r A ( B)

A (d2S B)

where we have used the divergence (Greens) theorem in going from the second to the third line. Thesecond term in the last line is a surface term, and neglecting this term for the moment yields

1

A(r)= B(r) (19)

Let us now compute the functional derivative with respect to A of 2

2 = 12md3r i qA2

=1

2m

d3r2||2 + iq( A ()2 A + q2||2 A2

from which one easily deduces

2

A=

iq

2m

() ()

+

q2

2m||2 A2 (20)

One recognizes the familiar expression for the probability current of the wave function in the presence ofa magnetic field. It remains to deal with

3 = d3r B H


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26

The variation of 3 is given by

3 =

d3r H A

= d3r (A H) + d3r A ( H)=

d3 A (d2S H)

We have used Greens theorem and the fact that H = 0 inside the superconductor. The surface termderived a few lines above combines with that obtained previously to give the final boundary condition

n ( B H) = 0 (21)

where n is a unit vector normal to the boundary of the superconductor: thus the tangential componentof ( H B) is continuous at the boundary between the normal and superconductor phases. The variation

with respect to leads to much easier calculations and we only quote the final results1. The equations of motion

0 =1

2m

i qA

+ a + 2b||2 (22)

B = iq02m

()

0q

2

mA ||2 (23)

2. The boundary conditions

n ( B H) = 0 n (i qA) = 0 (24)

4. In a uniform situation and zero magnetic field

||2 = a2b

a||2 + b||4 = a2

4b(25)

The difference of Gibbs potentials between the normal phase in the presence of an external field and inthe absence of such a field is

N(H) N(H = 0) = H0

dH B(H) = H2

20

while S(H) = S(H = 0) since B = 0 in the superconductor phase. The phase transition between the

normal and superconductor phases occurs when the Gibbs potentials are equal. This defines the criticalfield Hc(T)

S(H = 0) = N(H = 0) 120

H2c (T)

Now one uses (25)

S(H = 0) N(H = 0) = a2

4b

so that

H2c (T) =0a2

2b

The critical field vanishes linearly at T = Tc since

a(T) = a(T Tc)


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27

5. Let us assume a semi-infinite geometry where the normal-superconductor boundary is the plane z = 0.Then the order parameter obeys

|(z = 0)| = 0 limz |(z)| =|a|2bWe may choose to be real without loss of generality. As in exercise 4.6.5, we define a dimensionlessfunction f(z)

f(z) =

b

2|a| (z) f(z = 0) = 0 limz f(z) = 1

The equation of motion for (z) is

2

2m

d2

dz2+ a + 2b3 = 0

from which one deduces the corresponding equation for f

22d2f

dz2 f + f3 = 0

which is integrated to

2

df

dz

2= 1

2f2 +

1

4f4 + const

For z , df /dz 0 and f 1, which means that the constant is 1/4. One finally gets

2

df

dz

2=

1

4(1 f2)2 (26)

which is easily integrated as in the preceding exercise and allows one to obtain the z-dependence of theorder parameter

(z) =

|a|2b

tanh

z

2

(27)

Starting at z = 0 from a vanishing order parameter, the order parameter grows from zero toward its bulkvalue

|a|/(2b) over a length , the coherence length.6. We now assume = cst and A = 0. In other words, we consider the situation deep inside thesuperconductor, at distances from the boundaries. The electromagnetic currrent in (25) becomes

(r) = q2nSm

A (28)

so that the equation of motion (23) becomes in the Coulomb gauge A = 0

2 A 0q2nS

mA = 0 (29)

This equation defines a second characteristic length of the superconductor, the London penetrationlength

=

m

0q2nS

1/2=

2mb

0q2|a|1/2

(30)

In order to make explicit the fact that is indeed the penetration length, let us first express (29) in termsof the magnetic field by taking the curl of both sides

2 B 1

2 B = 0


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28

In a one-dimensional situation where B is perpendicular to the z-axis, B = Bu, u z = 0, one findsd2B

dz2 1

2B = 0

whose solution isB(z) = B0 exp

z

Thus the magnetic field vanishes in the bulk of the superconductor.

Exercise 4.6.11 Critical exponents to order for n 11. We write

4 =

(2 2) + 22 = 22 + 22(2 2) + (2 2)2We may omit 22, which adds a constant to H. Furthermore the term

u0n (

2

2

)2

1and may be neglected. With these simplifications, the Hamiltonian reads

H =

dDr

1

2((r))2 + 1

2r0

2(r) +u02n

22(r)

(31)

and is now of Gaussian type. The coefficient of 2 is

1

2

r0 +

u0n

2

and one may write immediately the correlation function from (4.171)

Gij(q) =ij

q2 + r0 +u0n

2 (32)

On the other handni=1

2i =ni=1

i(0)i(0) = n

dDq

(2)DGii(q)

which leads to the consistency equation

2 = nKD0

qD1dq

q2 + r0 +u0n

2(33)

where we have useddDq

(2)D=

SDqD1dq(2)D

= KD qD1dq

We may check in (33) the consistency of our assumption 2 n. The first two terms in H are oforder n. As 4 is of order n2, one must choose the combination u0/n so that all terms in H are of thesame order in n when n .2. From the fluctuation response theorem, the inverse susceptibility (T) is proportional to 1/G(q = 0),so that

(T) = r0(T) +u0n

2

= r0(T) + u0KD

0qD1dDq

q2 + (T)(34)


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29

The critical temperature Tc corresponds to an infinite susceptibility, and thus to (Tc) = 0. This leads to

0 = (Tc) = r0(Tc) + u0KD

0qD1dq

q2

(35)

Subtracting (35) from (34) we obtain

(T) = r0(T Tc) u0(T)KD0

qD1 dqq2(q2 + (T))

(36)

Let us make the change of variables q = x

. The equation for (T) becomes

(T) = r0(T Tc) + [(T)](D2)/2u0KD/

0xD1 dx

x2(x2 + 1)(37)

where is the ultraviolet cut-off.

3. The integral in (35) is divergent for D 2, and the theory can only be valid for D > 2. For 2 < D < 4,the integral in (37) is convergent at infinity, and one may write

(T) = r0(T Tc) + C(D2)/2 (38)

whence(T) (T Tc)2/(D2)

Indeed, for 2 < D < 4, 2/(D 2) < 1 and one may neglect (T) on the LHS (38). The critical exponent is then

=2

D 2 2 < D < 4For T = Tc, G(q) 1/q2, so that = 0. Finally

(T) ((T))1/2 (T Tc) 1D2 = = 1D 2

To summarize, for 2 < D < 4 and with an error O(1/n)

= 0 =1

D 2 =2

D 2which obeys the scaling laws (4.166). Note also that the results agree with the n limit of (4.203).For D > 4, one recovers, as one should, the mean-field exponents. This can be seen as follows. Sincethe integral over q is ultraviolet divergent and infrared convergent even when = 0, we may write theintegral in (36)

u0KD

0

qD1 dqq2(q2 + (T))

u0KD0

qD5dq =u0KDD 4

D4 = C()

The equation for (T) becomes(T) = r0(T Tc) C()(T)

so that (T) (T Tc), and = 1. The only effect of the interaction is to renormalize the value of r0,leaving the critical exponents unchanged.

Exercise 4.6.12 Irrelevant exponents


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30

1. From the definition u0 = g0 and (4.185) we deduce

u0 = g exp

g

0 dg

(g, )

+1

g (39)We now write the integral in (39) as

g0

dg

(g, )+

1

g

=

g0

dg

(g g) +g0

dg

(g, )+

1

g

(g g)

The second integral is convergent at g = g, so that it may be expanded in powers of g

A(g) = exp

g0

dg

(g, )+

1

g

(g g)

= 1 + c1g + c2g2 +

Performing the integrals in (39) yields

u0 gA(g) g

g g/ (40)

from which one deduces (g g)

g g = u/0 [gA(g)]/ g

or

g = g

1

u0

gA(g)

/(41)

namelyg g 1 + O()

The difference between the value g ofg at the fixed point and g is controlled by definition by the highestirrelevant exponent 3.2. We want to keep finite and are not allowed to take the limit 0 of the various expressions. Westart from the expression of (g, ) given in 4.5.3

(g, ) = g + 32

(/2)

(4)D/2g2 + O(g3)

= g + 32

B()g2 + O(g3)

withB() =

(/2)

(4)D/2

This gives the value of g

g =2

3B()

On the other hand, from the results of 4.5.3 and0

udu

(u + 1)=

( + 1)( 1)()

we get for (r0 r0c)

r0 r0c = 21 + u0 (/2)2(4)D/2(1 /2


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This leads to the following expression of Z(g)

Z(g) = 12

2d

d(r0 r0c)

u0= 1 +

gB()

2

from which one deduces

(g) = (g, )dZ

dg=

g + 3

2B()g2

B()

2= 1

2gB()

Subsituting the value of g in this expression leads to

(g) = 13

= 13

if D = 3

For D = 3 and using (4.196), the critical exponent becomes

=

3

5 = 0.6

instead of = 7/12 = 0.583 from the -expansion.

Exercise 4.6.13 Energy-energy correlations

1. The only temperature dependence in the Landau-Ginzburg Hamiltonian (4.96) is hidden in r0, asr0 = r0(T T0). We may write the partition function (4.95) in zero magnetic field as

Z =

dP[]exp

1

2r0

dDr2(r)

dP[]exp

1

2

dDr 2(r)

(42)

Indeed, since we are only interested in the singularity structure of the physical quantities, we do not pay

attention to the proportionality factors, so that we may make the replacement

r0 = r0(T T0)

by noticing that

r0(T T0) = r0k

1

1

0

r0

0 k20

It is important to notice that the factor dP[] is temperature independent

dP[] =

D(r)exp

1

2[]2 + 1

4!u0

4 12

r0k0

2

(43)

The average values are by definition

= 1Z

dP[]exp

1

2

dDr 2(r)

() (44)

Thus the average energy E is

E = ln Z

= 12

dDr 2(r)

We obtain the specific heat by differentiation

C =dE

dT= 1

kT2dE

d=

1

kT2d2 ln Z

d2=

1

4kT2

dDr dDr (r)(r )c (45)

This is of course an example of the fluctuation-response theorem.


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32

2. Using translation invariance, with u = r r and the scaling properties of 22c for u


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Exercise 5.6.4 Non-degenerate Fermi gas

1. We have shown in exercise 2.7.2 that the density of states not only depends on the dispersion law(p), but also on the dimension of space. For example in space-dimension two

() =mS

2= D

2. With a uniform density of states, one immediately obtains

F =N

DE =

1

2N F (48)

The classical approximation is valid when z = e 1 (section 5.1.2); in such a case

N = z

0

d ()e = zDkT

Then the classical limit corresponds to

N DkT (49)

The order of magnitude of the maximal energy is kT. The quantity DkT gives the number of levelsbetween = 0 and = kT and provides an estimate of the number of accessible levels. The condition(49) is nothing other than the validity condition of the Maxwell-Boltzmann statistics in the particular

case of a uniform density of states, namely that giving the small probability of occupation of the levels.

3. Let us make the change of variables y = x. Then, using 1, we determine immediately thedominant terms which we were looking for

+

dx1

ex + 1=

1

+

dy1

ey + 1 1

e (50)

+

dxx

ex + 1=

1

2

+

dyy

ey + 1 1

2e (51)

In order to demonstrate (5.86), we are going to use the symmetries of the integral. We call f(x) theFermi function written in terms of the variable x = and we set

(x) = S(x) + A(x) S,A(x) =1

2

(x) (x)

f(x) = fS(x) + fA(x) fS,A(x) =

1

2

f(x) f(x)

From f(x) = 1 f(x), we deduce

fS =1

2fA(x) = f(x) 1

2

As the integration domain is symmetrical with respect to x = 0 we may eliminate the integrands which


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are antisymmetric in x x in the calculation of (5.86)

dx (x)f(x) = 2

0 dx S(x)fS(x) + 2

0 dx A(x)fA(x)=

0

dx

S A

+ 2

0

dx A(x)f(x)

=

0

dx (x) +

0

dx(x) (x)

ex + 1

4. The chemical potential is given by

N = D

dx

1

ex + 1 = D

dx

1

ex + 1 + D

dx

1

ex + 1

Using (4.86) with (x) = 1, one obtains

N = D + O(e)

while, from (48) one gets

= F + O(e) (52)The average energy reads

E = D

dx

1

ex + 1+ D

dx

x

ex + 1

The first integral is identical to that which appears in the calculation of the chemical potential; the secondone features once more (5.86) with (x) = x, and one uses (51) in order to be able to use (A.51). Onefinally gets for the energy

E =1

2D2 +

2

6D(kT)2 + O(e) (53)

5. As a general rule, when one applies the Sommerfeld expansion of the Fermi distribution to a generictest function g() (g() = () for N and g() = () for E), one obtains its Taylor expansion in termsof the (small) parameter (kT/)2. The peculiarity of the two dimensional situation, in which the densityof states is uniform, leads to a Taylor expansion which is necessarily finite. Indeed

N = D

0

d + all other terms vanish

E = D

0

d +2

6D(kT)2

0

d ( ) + all other terms vanishing

The calculation of the preceding integrals leads indeed to the results (52) and (53). Clearly the correctionsof order e cannot be taken into account in an expansion in powers of (kT/)2. Exercise 5.6.5 gives,in the case of a Bose gas, a still more spectacular illustration of the influence of the space dimension onthe behaviour of a quantum gas. Let us also remark that the techniques developed in this exercise allowone to give an alternative proof of Sommerfeldss formula (5.29).

Problem 5.7.4 Quark-gluon plasma


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A. 1. Because the -meson has three different charge states, the density of states of the -meson gasreads

() d = 3V

(2)34p2 dp =

3V

222 d

For a boson gas with a vanishing chemical potential, the grand partition function is given by (see (5.7))

ln Q = 0

d () ln

1 e

= V22

0

d3

e 1 =2V

30T3

(54)

One uses the integral (A.53) to obtain the final result. From the equation of sate of an ultra-relativisticideal gas

P V = T ln

Q=

1

3

E

and using (54), one easily finds the pressure and the energy density

P(T) =2

30T4 (55)

(T) =2

10T4 (56)

2. cv(T) et s(T) are obtained from (56)

cv(T) =d

dT=

22

5T3

s(T) =

T0

dT cv(T)T = 2

2

15T3

A direct calculation gives the energy density

n(T) =3

22

0

d2

e 1 =3(3)

2T3 (57)

One notes that the proportionality of cv, s and n to T3 is the consequence of a simple dimensionalargument.

3. From (56) et (57) one can express the mean particle energy as

(T) = (T)n(T)

= 4

30(3)T

Lets use once more a reasoning which was already developed in the discussion of (5.67). From n and, one may build two length scales. One of them, d n1/3 T1, is a measure of the mean distancebetween particles, the other one, p1 1 T1, provides an order of magnitude of quantumeffects. If one compares these two typical distances, one observes that, whatever the temperature, /d 1.The -meson gas can never be treated as a classical gas.

B. 1. The reader is referred to subsection 5.1.3.

2. Because massless particles are ultra-relativistic, the density of states of this Fermion gas reads

() d =V

2 2 d


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Expression (5.105) is obtained thanks to the change of variables x = in

ln

Q=

0 d () ln1 + e()+ ln 1 + e

(+)=

V

32

0

d 3

1

1 + e() + 1

1 + e(+) (58)

For = 0, one gets (see appendix A.5.2)

ln Q = 72V

180T3 P(T) =

72

180T4 n(T) =

3(3)

2T3

The entropy is deduced from the grand potential J= T ln Q

s(T) = 1V JT V = 72

45 T3

3. Let us call respectively I et I+ the two integrals between square brackets in the second line of (58).One can cast the quantity T ln Q into the form

T ln Q = V32

0

d 3 [I + I+]

The condition T = 0 implies that the two integrals are to be calculated in the limit . Let usassume > 0

lim

I = ( ) lim

I+ = 0 if > 0

lim

I = 0 lim

I+ = (( + )) if < 0

From these equation it follows that

T ln Q = V32

0

d 3 =V

1224

P() =T ln Q

V=

1

1224

() = 3P(T) = 142

4

n() =1

V

T ln Q

T,V

=1

32||3

The entropy of a fully degenerate Fermi gas of course vanishes.

4. Let us start from (5.105) and set

I1 =

+0

dxx3

e(x) + 1I2 =

+0

dxx3

e(x+) + 1

We do not seek to compute separately I1 and I2, but rather we seek to cast them into a form where their


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sum has a simple expression

I1 =

0 dy(y )3

ey

+ 1

+

+

0 dy(y + )3

e

y

+ 1

= 0

dy (y )3

1 1ey + 1

+

+0

dy(y + )3

ey + 1

I2 = 0

dy(y )3

ey + 1+

+0

dy(y )3

ey + 1

The sum I1 + I2 is then

I1 + I2 =

0 dy (y )3 + 2

+

0 dyy3

ey + 1+ 62

+

0 dyy

ey + 1(59)

The first integral in (59) is elementary integral, the other two are given in appendix A.5.2. Inserting (59)in (5.105), one finally gets the expected result (5.106).

C. 1. In order to estimate the contribution of the quark gas to the plasma pressure, one can simply usethe results for the Fermion gas; it is only necessary to take properly into account the degeneracy linkedto the three colour states

Pquark = 2 3 72

180T4

For the contribution of the gluon gas, one simply uses the expression derived previously for the -mesongas

Pgluon = (2

8)

1

3 2

30

T4The gluons have spin one, but since they are massless, their spin degeneracy gives only a factor two. Thefactor 1/3 comes from the fact that gluons have only one charge state. On gathering all contributions,one gets the expression (5.107) of the pressure of the quark-gluon plasma.

2. The additional term proportional to the volume in the expression of the free energy leads to a constantterm in the pressure

Pplasma = FV

T

=372

90T4 B

3. Figure 2 shows the behaviour of the pressure as a function of temperature for the -meson gas and forthe quark-gluon plasma. The most stable phase is that which minimizes the free energy, or equivalently,that which has the highest pressure. The transition temperature Tc is fixed by the equality of the pressures

372

90T4c B =

2

30T4c

a condition leading to

Tc =

45

172

1/4B1/4 144 Mev

This rough estimate of Tc is in qualitative agreement with the results from lattice quantum chromody-namics.

5. The entropy densities of the two phases are different

s(T) =

T T ln Q

V =P

T

=

742

45T3 for the plasma

22

15 T3 for the meson gas


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P

plasma

TTc

B

P(T)

Pplasma(T)

mesons

Figure 2: Curves giving P(T) for the -mesons and the quark-gluon plasma.

At the transition, the entropy variation does not vanish and is given by

s =682

45T3c

The phase transition is thus first order and is accompanied by a latent heat per unit volume = Tcs.

Problem 5.7.5 Bose-Einstein condensates of atomic gases

1. One uses the fact that the potential is slowly varying over distances . The total energy is =

1

2

p2x +p

2y +p

2z

+ U(r) px =

h

nx etc.

so that n2x + n

2y + n

2z

2mh2

( U(r))Counting the number of states in the box is done exactly as in the U = 0 case, making the substitution U(r). In order to get the total density, it suffices to integrate in the whole space; however oneshould limit the integration to the region where U(r), because (n2x + n2y + n2z) 0.In the second method, one starts from the single-particle Hamiltonian

H(p, r) =p 2

2m+ U(r)

The semi-classical approximation reads

() =

d3pd3r

h3

p

2

2m U(r)

=

2

h3

d3r

d(p2)p

p

2

2m U(r)

=

2 (2m)3/2

h3

d3r

U(r) ( U(r))

(60)

In the case U(r) = 0 in a volume V, U(r) = + outside V, (60) is in agreement with the familiar result

() =2 (2m)3/2

h3 V


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2. In the case of a harmonic potential, (60) becomes

() =82 (2m)3/2

h3

0 dr r2

1

2

m2r2 1

2

m2r2=

82 (2m)3/2

h3

0

dr r2

1 m

2r2

2

2

m2 r2

=82 (2m)3/2

h3

q2

m20

dr r2

1 m

2r2

2

On setting u = m2r2/2, one can feature the integral (5.108) and obtain

() =1

2

1

(

)3

2 (61)

3. As in (5.68), one can isolate the contribution of the ground state 0 = 3/2

N =z

e0 z +0

d()

z1e 1 = N0 + N1

with

N1 0

d()

e(0) 1

Indeed, at the transition, the chemical potential takes its maximum value = 0

: the ratio N0

/N cantend toward a finite value only if z = exp(0). It is more convenient to choose the ground state energyas a reference energy, and thus to redefine the chemical potential as follows

= 0 = 32 z = e

The critical temperature Tc (c = 1/kTc) is then fixed by the condition

N = N1 =

0

d()

ec 1

namely, taking into account (61),

N = kTc3 (3) (62)

In the case T Tc, z = 1 andN = N0 +

kT

3(3) (63)

On comparing (62) et (63), one obtains the expression (5.109) of the ratio N0/N.

4. With the data given in the text

N

(3)

1/3= 1.86 1029

in MKS units, which leads to a critical temperature Tc = 1.35 K.


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40

5. At T = Tc particles occupy excited states and

N =

0 d()

e

1The potential U(r) being slowly variable, one can define, by using (60), a local density of states as

n(, r) =2 (2m)3/2

h3

U(r)e 1 ( U(r))

(5.110) is obtained by integrating n(, r) over . To establish this formula, we have assumed, as inquestion 3, that the chemical potential is zero when one chooses the ground state energy to be zero. Onenotices that for U(r) = 0, the density in the center of the trap corresponds to the density (5.70) at thecritical temperature. With the data given in the text and Tc = 1.35 K, one finds for the density

n(r = 0) = 2mkTch2 3/2

(3/2) = 8.6 1019 atoms.m3

.

Problem 5.7.6 Solid-liquid equilibrium for helium-3

A. 1. From the expression (5.64) of the specific heat in the Debye model, one deduces that of the entropyper atom

=S

N=

4

54k

T

TD

3and that of the energy per atom (5.112) thanks to f/T =

. For a value of the temperature T = 1K,

kT 0.1 meV, while (T /TD)3 103. The last term of (5.112) is then on the order of 6 103 meV andit is negligible compared to the other two terms which are on the order of the meV.

2. The entropy of a system of N disordered spins 1/2 is kN ln 2. One must add to f the term T =kT ln 2, because, in the solid, interactions between nuclear moments are negligible with respect tothermal motion, except when T reaches values on order of a mK.

3. To go from the free energy F to the Gibbs potential G, one must add P V; as G = N, one must addvsP to f. This expression of does not obey the third law s k ln 2 for T 0, and it does not vanishwhen T = 0. In fact the spins begin to be ordered below 2 mK: solid helium-3 becomes anti-ferromagneticbelow this temperature and the entropy vanishes5

B The logarithm of the grand partition function is given by

ln Q = V(2m)3/2

223

0

d

ln (1 + exp(( )))

One integrates by parts

P =1

Vln Q = V(2m)

3/2

323

0

d3/2

exp(( )) + 1

5This nuclear anti-ferromagnetism occurs for an exceptionally high temperature in the case of solid helium-3, because ofan exchange interaction of quantum origin. In general nuclear anti-ferromagnetism, due to a magnetic interaction betweennuclear moments, occurs only at the K level.


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At low temperatures, the integral is computed to the order (kT/)2 thanks to the Sommerfeld formula(5.29) with the result

P =

2

15

(2m)3/2

23 5/21 + 528 kT

2

= B5/2

1 +

52

8

kT

F(P)

2B =

2

15

(2m)3/2

23

The replacement of by F(P) is correct to this order in T. This substitution allows one to obtain asa function of P

= AP2/5

1

2

4

kT

F(P)

2A = B2/5

At T = 0

F(P) = AP2/5 = F(P0)

P

P02/5

and one finally obtains (5.114). Let us compute the compressibility6 at T = 0

(T = 0) =1

n

n

P=

ln n

P=

3

5P

Using the expression (5.22) of the pressure yields

(T = 0) =3

5

5m

2

1

(32n)2/3n=

3m

2

(32n)1/3

32n2=

1

n2(F)

C. 1. The expressions of and v are derived from (5.111)

= T = 2kbkTA PP0

2kbkTA (64)v =

P

= A

P0

P

P0

11 bkT

A

2 A

P0

P

P0

1(65)

Differentiating the entropy with respect to T allows one to obtain the specific heat

cv cP = T T

2kb

kT

A

As n = 1/v P(1) one gets

(T = 0) =ln n

P=

1 P

2. The transition line is given by the equality of the chemical potentials of the liquid and solid phases

u0 + 98D kT ln2 + vsP = A

PP0

1 bkT

A

2(66)

Differentiating this equation with respect to T, or equivalently, using Clapeyrons formula (3.97), onegets

dP

dT=

s vs v =

k ln 2 2bk(kT/A)vs (4A/5P0) (67)

Writing (dP/dT = 0) leads tob

A=

ln 2

2kTm 12.5 (meV)1

6Using the Gibbs-Duhem relation gives a more direct proof of (5.117).


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3. One may cast (67) into the form

dP

dT=

k ln 2(1 T /Tm)vs v

hence (vsv) 0.251029 m3, in good agreement with the value given in the introduction: (vsv) 0.22 1029 m3.4. A = 54 vP0 1.1 meV, A/k 13 K, while

b =ln 2

2Tm

A

k 14

For T = 1K, b(kT/A)2 0.08. In the case of an ideal Fermi gas with specific volume v, one findsF 0.55 meV, and

b

A=

2

4F 4.5 (meV)1

while

A = 215 (2m)3/223 2/5

P2/50 2.4meVusing the value = 2/5 of the ideal Fermi gas. The numerical result of the ideal gas and that of theactual liquid differ by a factor 2-3. The Debye frequency follows from the value of A

9

8D = u0 vsP0 + A

leading to D 0.75 meV, that is TD 9 K, of the same order of magnitude as the experimental valueof 16K, and a sound velocity

cs =D

(62ns)1/3 102 m.s1

5. As (s ) 0 for T Tm, the quantity of heat which is absorbed in in the solidliquid transfor-mation is Q = T(s ) 0.6. Because of the third principle, one must have s(T = 0) = (T = 0) = 0, and (dP/dT) in (67) mustvanish, since vs v = 0. The transformation being adiabatic and quasi-static, the entropy is constantand is represented by an horizontal line on figure 5.21. The method relying on the Pomeranchuk effectdoes not allow one to cool down below a temperature 1 mK. In the solid, the spins can be identified:they form a paramagnetic system for T > 2 mK, and the entropy per atom of the solid is dominatedby the term k ln 2. In the same range of temperatures, the entropy of the liquid tends to zero, becausePaulis principle implies that the ground state is unique.

D. 1. One finds numerically that HeB/k = 0.78102 K, and HeB/kT = 0.78 for T = 0.01 K, whence

2Sz

= tanhHeB

kT 0.65

2. As the term k ln 2 is absent in the entropy of the fully polarized solid, s 0, and from (67),the pressure is an increasing function of T and does go through a minimum as in the unpolarized state.The Pomeranchuk effect is no longer present. As expected, the polarized solid is more organized thanthe polarized solid.

Problem 5.7.7 Superfluidity for hardcore bosons

1. Expressing the ais and the ai s in terms of the spin operators

Si gives for the Hamiltonian

H = tij

S+i S

j + S

i S

+j

i

Szi +

1

2

=

N

2 tij S+i Sj + Si S+j i Szi (68)


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2. The occupation number operator at site i is

ni = aiai = S

zi +

1

2

(we have used a system of units where = 1). We have for the spin system

Szi | =

1

2|i Szi |+ =

1

2|+i

so that the action of the occupation number operator ni on the states |i is

ni|i = 0 ni|+i = |+iSince S+i |+i = 0, the only possible eigenvalues of ni are ni = 0 or ni = 1. If site i is in the state |i,then the site is not occupied, and it is occupied if the state is |+i. From these results, the total numberof bosons in the ground state |0 vanishes

Nb|0 = Nbi

|i = 0

3. Because of the relation between the occupation number ni and the z-component of the spin operatorSi

Szi = ni

1

2

the average value of

i Szi given by

i

Szi =i

ni N2

= Nb N2

Let us compute the average value ofi Szi in the state (5.123). We remark (see footnote 34, chap. 5)that for i = 0

|i = ei()Syi |i (69)

The operator exp(iSyi ) rotates spin i by an angle around Oy

eiSyi S

zi e

iSyi = Szi cos Sxi sin (70)

Thus, with =

Szi = |

eiSyi S

zi e

iSyi

|

=|

Szi cos

Sxi sin

|=

1

2cos =

1

2cos

where we have used |Sx| = 0. We thus obtain the average value iSzi in the state |i

Szi =N

2cos (71)

from which follows the relation between Nb and N

N

2cos = Nb N

2

In terms of the density = Nb/N this relation reads

sin2 = 4(1 ) (72)


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4. The number of bosons in the condensate is

N0 = |a(k = 0)a(k = 0)| = 1Nij

|ajai|

=1

N

ij

0|

l

eiSyl

S+j S

i

q

eiSyq

|0

This can be simplified by noting that operators referring to different sites commute, so that

0|

l

eiSyl

S+j S

i

q

eiSyq

|0 = 0|eiSyi eiSyj S+j Si ei

Syi ei

Syj |0

We use once more the fact that exp(iSy) is a rotation operator

eiSy

S ei

Syl = Sx cos + Sz sin iSy

and the property |Sx| = |Sy | = 0 to derive at once

|eiSy S eiSyl | = 12

sin =1

2sin

so that

N0 =N

4sin2 =

N

4[4(1 )] (73)

or in terms of the superfuid density0 = (1 )

In general 0 < , unlike the ideal gas case where 0 = (remember that we are working at zerotemperature). When 1, the bosons rarely collide, so that the system is approximately free and weregain 0 = .6. The grand potential is given from (5.132) and the expression (68) by

J= N2

i

|Szi | tij

| S+i Sj + Si S+j | (74)The second term in (74) is evaluated from the results of question 3 and the third from those of question 4

J= N2

N2

cos tNsin2 (75)

7. The minimization equation dJ/d = 0 gives at once the value of

cos = 4t

(76)

from which we derive the mean field values of , 0 and J

=1

2+

8t0 =

1

4

8t

2J= N

t +

2

16t

(77)

8. The state |T is obtained from the ground state |0 by a site dependent rotation of ( ) around Oyfollowed by a rotation of around Oz. Clearly, the rotation around Oz does not affect the calculationof the average value ofSzi , so that we may use the result of question 3

T|i Szi |T =N

2 cos


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On the other hand, from the transformation properties ofSx and Sy under a rotation around Oz (or froma simple matrix multiplication)

eiSz

S eiS

z

= ei S

and we get

T|S+i Sj |T =1

4sin2 ei(ij)

T|S+j Si |T =1

4sin2 ei(ij)

This leads to ij

T|S+i Sj + Si S+j |T =N

2sin2

1 + cos

2

L

where we have used the fact that there are N nearest neighbours in the x direction as well as in the ydirection. Gathering all the preceding results gives the grand potential

JT = N2

N2

cos tN2

(1 + cos )sin2 (78)

The minimization with respect to gives the following value of

cos =

2t(1 + cos )

9. In the limit 1 the preceding value of becomes

cos 4t

1 +

1

4()2

It is then straightforward to compute the difference (JT J

) per site

1

N(JT J) = J

N= t(1 )()2

But J can also be written in terms of the superfluid density s and the velocity v asJN

=1

2msv

2

Comparing the two expressions of J/N and using v = 2t leads to

s = (1 ) (79)

10. The condensate density 0 and the superfluid density s are thus identical within the mean fieldapproximation

0|MF = s|MFThis equality does not persist if one includes the effect of spin waves. If one takes spin waves into account,the analytical results are in excellent agreement with those of numerical simulations.


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Exercise 6.4.5 Lord Kelvins model of earth cooling

1. In the region x 0, the temperature (x, t) = 0, while in the region x < 0, the temperature (x, t)obeys the diffusion equation

t= D

2

t2(80)

The most general solution of (80) is (without taking into account boundary conditions)

(x, t) = A +B

4Dt

+

dx g(x) exp

(x x)2

4Dt

(81)

in terms of an arbitrary function g(x) and of two constants A and B. This result is easily proved byshowing that (81) obeys (80) thanks to

2

x2 =

2

(x x)2and the fact that

14Dt

exp

x

2

4Dt

obeys (80). For t = 0 we have

x 0 (x, t = 0) = 0x < 0 (x, t = 0) = A + B g(x) = 0

These conditions imply that one may choose g(x) = 1 and A + B = 0 for x < 0, and they allow us tocompute (x = 0, t)

(x = 0, t) = A +

B

2 = 0

whenceA = 0 B = 20

The explicit expression for (x, t) (for x 0) follows from (81)

(x, t) = 0 + 0Dt

0

dx exp

(x x)2

4Dt

(82)

2. The gradient is easily computed by making the change of variable u = x x

(x, t) = 0 +0Dt

x

du exp u24Dt (83)with the result for x = 0

x

x=0

= 0Dt

(84)

3. The numerical result for the age of the Earth is

t = 8.5 1015 s = 2.7 108 y

4. In three dimensions, we may use the Laplacian in spherical coordinates

2 f(r) =1

r

d2

dr2 [rf(r)]


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valid for a function f(r) which is regular at the origin. The function (r, t) = r(r, t) obeys the equation

t= D

2

r2(85)

As in Question 1, the general solution of (85) may be written in terms of three constants A, A and B as

(r, t) = Ar + A +B

4Dt

+

dr exp

(r r)2

4Dt

(86)

from which we derive A = 0 (because must be finite at r = 0) and g(r) = r so that

(r, t) = A +B

r

4Dt

+RR

rdr exp

(r r)2

4Dt

(87)

The symmetric integration from R to +R ensures that (r, t) be finite at r = 0. Furthermore we have

(r = 0, t) = A + B 1 + O(eR2/(Dt))so that A + B = 0. Let us now compute (R, t), with the change of variable u = r r

(R, t) = A +B

R

4Dt

2R0

du (R u)exp

u2

4Dt

= A +B

2

1 + O(eR2/(Dt))

so that A + B/2 = 0. As in question 1, A = 0 and B = 20. The final result is then

(r, t) = 0 + 0r

Dt

r+R

rRdu (r u)exp

u

2

4Dtfrom which we derive the expression of the gradient

r

r=R

=0Dt

1 + eR2/(Dt) + 2R0

duu

R2exp

u

2

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