HW9_4

7/30/2019 HW9_4

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Solutions to problems for Part 3

Sample Quiz Problems

Quiz Problem 1. Draw the phase diagram of the Ising Ferromagnet in an applied magnetic field. Indicate thecritical point. Plot the magnetization as a function of the applied field for three temperatures T < T c, T = T c, T > T c.

Solution. See lecture notes

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Quiz Problem 2. Plot the behavior of the magnetization of the Ising ferromagnet as a function of the temperature,for three applied field cases: h < 0, h = 0, h > 0. Indicate the critical point.

Solution. See lecture notes

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Quiz Problem 3. Write down the definition of the critical exponents α, β e, γ , δ , η and ν . What values do theseexponents take within mean field theory.

Solution.

C V ∼ t−α; m ∼ tβ; χ ∼ t−γ ; m(T c) ∼ h1/δ; c(r) ∼ e−r/ξ/rd−2+η (1)

where ξ = t−ν , and t = |T − T c|. Within mean field theory α = 0, β = 1/2, γ = 1, δ = 3, η = 0, ν = 1/2

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Quiz Problem 4. Write down the mean field equation for the Ising ferromagnet in an applied field, on a latticewith co-ordination number z and exchange constant J . From this equation find the critical exponent δ for the Isingferromagnet within mean field theory.

Solution.

m = Tanh(βJzm + βh) ∼ (βJzm + βh) −1

3(βJzm + βh)3 (2)

at the critical point βJ z = 1, so m ∼ h1/3 and hence δ = 3.

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Quiz Problem 5. Write down the van der Waals equation of state. Draw the P, v phase diagram of the van derWaals gas and indicate the critical point.

Solution.

P =kbT

v − b−

a

v2(3)

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Quiz Problem 6. Make plots of the van der Waals equation of state isotherms, for T > T c, T < T c and forT = T c. For the case T < T c explain why the non-convex part of the curve cannot occur at equilibrium and the

Maxwell construction to obtain a physical P, v isotherm.

Solution. The Maxwell construction requires that,

P ∗(vg − vl) =

vgvl

P dv (4)

The is required as the oscillation in the Van der Waals equations occur in a non-convex region of the Helmholtz freeenergy graph of this model. The non-convex region is not an equilibrium or thermodynamic state as it is possible tolower the free energy by constructing a tie line below the non-convex region which corresponds to a mixture of thetwo phases at the end points of the tie line.

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Quiz Problem 7. Write down the Landau free energy for the Ising and fluid-gas phase transitions. Explain thecorrespondences between the quantities in the magnetic and classical gas problems.

Solution.

F = a(T − T c)y2 + by4 + cy (5)

For the Ising model y = m, c = h, for the van der Waals gas, y = vg − vl, c = P .

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Quiz Problem 8. Explain the meaning of second quantization. Discuss the way that it can be used in positionspace and in the basis of single particle wavefunctions. Write down the commutation relations for Bose and Fermisecond quantized creation and annihilation operators.

Solution.Second quantization is a formulation of quantum mechanics and of quantum field theory that is expressed in

terms of creation and annihilation operators. In many body quantum physics creation and annihilation operatorscreate and destroy particles in many body basis sets constructed from single particle wave functions. In the caseof Fermions a many body basis function is a determinant, while for Bosons it is a permanent. The commutation

relations for Fermions and Bosons are similar, except that for fermions we have anticommutators and for Bosons wehave commutators. In many body quantum mechanics we have,

[ai, a†j ] = δ ij ; for bosons; and {ai, a†j} = δ ij ; for fermions; (6)

while for quantum fields, we have,

[ψ(x), ψ†(x)] = δ (x − x); for bosons; and {ψ(x), ψ†(x)} = δ (x − x); for fermions; (7)

These days, new theories are often formulated using creation and annihilation operators rather than the Heisenbergor Schrodinger formulations of quantum theory.

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Quiz Problem 9. Write down the Hamiltonian for BCS theory, and the decoupling scheme used to reduce it to asolvable form. Explain the physical reasoning for the decoupling scheme that is chosen.

Solution.In the s-wave BCS theory a singlet state is assumed, so that,

H pair − µN = kσ

( kσ − µ) a† kσa kσ + k l

V k l a† k↑a†− k,↓

a− l↓a l↑, (8)

where N =

kσ n kσ is the number of electrons in the Fermi sea. We carry out an expansion in the fluctuations,

a− l↓a l↑ = bl + (a− l↓a l↑ − bl); a† k↑a†− k↓

= b∗k + (a† k↑a†− k↓

− b∗k) (9)

The mean field Hamiltonian keeps only the leading order term in the fluctuations so that,

H MF − µN = kσ


V k l (a† k↑a†− k,↓

b l + b∗ ka− l↓a l↑ − b∗ kb l) (10)

This is the Hamiltonian that leads to the BSC solution.

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Quiz Problem 10. Consider the inverse Bogoliubov-Valatin transformation,

γ kσ = u∗ ka kσ − σv ka†− k−σ

. (11)

Show that if the operators a, a† obey standard fermion anti-commutator relations, then the operators γ, γ † also obeythese relations, provided,

|uk|2 + |vk|2 = 1 (12)

Solution The anticommutator,{γ kσ, γ lα} = {u∗ ka kσ − σv ka†

− k−σ, u∗ l a lα − αv la

†

− l−α}. (13)

Expanding the anticommutator gives,

{γ kσ, γ lα} = (u∗ ka kσ − σv ka†− k−σ

)(u∗ l a lα − αv la†

− l−α) + (u∗ l a lα − αv la

†

− l−α)(u∗ ka kσ − σv ka†

− k−σ) (14)

which reduces to,

{γ kσ, γ lα} = u∗ ku∗ l {a kσ, a lα} + σαv k

v l{a† −k−σ, a†

− l−α} − αu∗ kv l{a kσ

, a†− l−α

} − σv ku∗ l {a −k−σ

, a† lα} (15)

The first two anticommutators are zero. The second two anticommutators are finite when the conditions δ (k, −l)δ (σ, −α)hold. However when this condition holds, the two commutators are equal and opposite, so they sum to zero. Takingthe adjoint of Eq. (15) shows that {γ kσ, γ lα} = 0. Now consider,

{γ kσ, γ † lα} = (u∗ ka kσ − σv ka†− k−σ

)(u la† lα

− αv∗ l a− l−α) + (u la† lα

− αv∗ l a− l−α)(u∗ ka kσ − σv ka†− k−σ

) (16)

{γ kσ, γ † lα} = u∗ ku l{a kσ, a† lα} + σαv kv∗ l {a† −k−σ, a− l−α} − αu∗ kv∗ l {a kσ, a− l−α} − σv ku l{a† −k−σ

, a† lα} = 0 (17)

which reduces to

{γ kσ, γ † lα} = δ kl|u k|2{a kσ, a† lα} + σα|v k|2{a† −k−σ

, a− l−α} = δ klδ σα(|u k|2 + |v k|2). (18)

This anticommutator thus requires Eq. (12) in order for the γ operators to obey Fermion anti commutator relations.

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Quiz Problem 11. Given that the energy of quasiparticle excitations from the BCS ground state have the

spectrum,E = [( − F )

2 + |∆|2]1/2, (19)

where ∆ is the superconducting gap and E F is the Fermi energy, show that the quasiparticle density of states if givenby,

D(E ) =N (F )E

(E 2 − ∆2)1/2(20)

Solution We use the relation,

N (F )d = D(E )dE ; and dE = − F

[( − F )2 + ∆2]1/2d (21)

to find,

D(E ) =N (F )[( − F )

2 + ∆2]1/2

− F (22)

and using,

( − F )2 = E 2 − |∆|2 (23)

yields,

D(E ) =N (F )E

[E 2 − |∆|2]1/2(24)

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Quiz Problem 12. Describe the physical meaning of the superconducting gap, and the way in which BCS theorydescribes it.

SolutionThe superconducting gap is the energy required to generate a quasiparticle excitation from the superconducting

ground state. In BCS theory, the quasiparticles behave like non-interacting fermions and the energy required togenerate a quasiparticle is at least 2∆(T ).

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Quiz Problem 13. (Omit this question) BCS theory works very well even quite near the superconductingtransition, despite the fact that it is a mean field theory. Use the Ginzburg criterion to rationalize this result.

Solution

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Quiz Problem 14. Given the general solutions to the BCS mean field theory

∆ k = − l

V k l∆ l

2E l, E k = (( k − µ)2 + |∆ k|2)1/2 (25)

Describe the assumptions that are made in deducing that,

1 =N (F )V

2

F +hωc

F −hωc

d

(( − F )2 + |∆|2)1/2=

N (F )V

hωc/∆0

dx

(1 + x2)1/2= N (F )V Sinh−1(

hωc

∆) (26)

and hence,

∆ = 2hωcExp[−1

N (F )V ] (27)

Solution

We assume an isotropic gap, and that the attractive coupling between electrons is constant −V , over the rangeF − hωc < < F + hωc. The density of states is assumed constant with value N (F ).

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Assigned problems

Assigned Problem 1. Consider the Ising ferromagnet in zero field, in the case where the spin can take three valuesS i = 0, ±1. a) Find equations for the mean field free energy and magnetization. b) Find the critical temperature andthe behavior near the critical point. Are the critical exponents (β e, γ , α , δ ) the same as for the case S = ±1? Is thecritical point at higher or lower temperature than the spin ±1 case? c) Is the free energy for the the spin 0 , ±1 casehigher or lower than the free energy of the ±1 case? Why?

Solution. The partition function and Helmholtz free energy are,Z = [1 + 2Cosh(βJzm)]N ; F = −kBTNln[1 + 2Cosh(βJzm)] (28)

The mean field equation is,

m =2Sinh(βJzm)

1 + 2Cosh(βJzm)≈

2

3βJzm −

1

3(βJzm)3 + ... (29)

The critical point is at βJ z = 3/2, so the critical temperature is at kBT c = 2Jz/3 which is lower than that for spin1/2 due to the additional entropy of the spin one system. The critical exponents β and δ are clearly the same as forthe spin 1/2 case. The free energy is lower for the spin 1 case due to the higher entropy.

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Assigned Problem 2. (i) Starting from Eq. (36) of the lecture notes, prove Eq. (39) of the lecture notes. (ii)From Eq. (44) or (45) of the lecture notes prove Eq. (46) in three dimensions.

Solution. To fill in the details from Eq. (36) to (38), substitute the definitions of the Fourier transform and usetranslational invariance, as illustrated in the notes. To prove equation (39), note that the matrix J il is diagonalizedby the Fourier transform, and assuming only nearest neighbors, we then have,

J ( k) = eikxx + e−ikxx + eikyy + e−ikyy + .... (30)

which reduces to Eq. (39). The approximate expression is found by expanding for small k.To show Eq. (46) from Eq. (45), we use the integral,

∞0

e−ax−b/x

x3/2dx = e−2(ab)

1/2

(π

b)1/2. (31)

with a = 1/ξ 2, b = r2/4, d = 3, we find,

C (r) ≈1

re−r/ξ (32)

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Assigned Problem 3. The Dieterici equation of state for a gas is,

P =kBT

v − be−a/(kBTv) (33)

where v = V /N . Find the critical point and the values of the exponents β,δ,γ .

Solution. The critical point is found by solving,

P c =kBT cvc − b

e−a/(kBTvc); (34)

∂P

∂v = 0 = −kBT c

(vc − b)2 e−a/(kBTvc) +kBT cvc − b

a

kBT v2c e−a/(kBTvc) (35)

Which simplify to,

−1

vc − b+

a

kBT cv2c= 0; so

a

kBT c=

v2cvc − b

(36)

The second derivative yields,

∂ 2P

∂v2= 0 = 2

kBT c(vc − b)3

e−a/(kBTvc) − 2kBT

(vc − b)2a

kBT cv2ce−a/(kBT cvc)

−2

kBT cvc − b

a

kBT cv3c e−a/(kBT cvc)

+

kBT cvc − b (

a

kBT cv2c )2

e−a/(kBT cvc)

(37)

so that,

21

(vc − b)2− 2

1

vc − b

a

kBT cv2c− 2

a

kBT cv3c+ (

a

kBT cv2c)2 = 0 (38)

and using Eq. (),

−21

vc

1

vc − b+ (

1

vc − b)2 = 0; so vc = 2b (39)

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so we find that,

vc = 2b; kBT c =a

4b; P c =

a

4e2b2(40)

To find the critical exponents we write v = vc + δv,T = T c + δT,P = P c + δP , so that,

P c + δP =kB(T c + δT )

vc + δv − be−a/[kB(T c+δT )(vc+δv)] =

kBT cb

(1 + δT T c

)

1 + δvb

e−a/[kBT cvc(1+δT/T c)(1+δv/vc)] (41)

This reduces to,

1 + p = e21 + t

1 + 2xExp[−

2

(1 + t)(1 + x)] (42)

where p = δP/P c, t = δT/T c, x = δv/vc. To third order this expansion gives,

p = 3t + 2t2 −2

3t3 + (−2t − 4t2)x + 2tx2 −

2

3x3 (43)

Taking a derivative with respect to v at setting x = 0 leads to,

κT = (−V (∂P

∂V )T )

−1 ≈ |T − T c|−1 (44)

so γ = 1. The exponent δ is found by setting t to zero so that p ≈ x3, so that δ = 3. To find β we assume that pl = pg, xl = −xg, so that,

pl = 3t + 2t2 −2

3t3 + (−2t − 4t2)xl + 2tx2

l −2

3x3l (45)

p − g = 3t + 2t2 −2

3t3 − (−2t − 4t2)xl + 2tx2l +

2

3x3l (46)

Setting these equations to be equal yields,

2(−2t − 4t2)xl =2

3x3l , so that xl ≈ |T − T c|1/2 (47)

where we dropped the t2 term as it is higher order. In the above analysis the signs of the t and x are consistent buthave to be checked each time.

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Assigned Problem 4. Consider the Landau free energy,

F =1

2am2 +

1

4bm4 +

1

6cm6 (48)

where c > 0 as required for stability. Sketch the possible behaviors for a, b positive and negative, and show that thesystem undergoes a first order transition at some value a,b,c. Find the discontinuity in m at the transition.

Solution.The mean field equation is,

δF

δm= am + bm3 + cm4 = 0; (49)

Note that in general we do a variation with respect to m, so when we add fluctuations later, we need to use theEuler-Lagrange equation. Here the variation is the same as a partial derivative with respect to m. Solving the meanfield equation, we find five solutions.

m = 0, m = ±m±; m2± =

−b ± (b2 − 4ac)1/2

2c(50)

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Though there are always five solutions, only the real solutions are physical. Analysis of the behavior of the modelreduces to identifying the real solutions, and finding which real solution has the lowest free energy. We can understandthe nature of the solutions by looking at the second derivative,

∂ 2F

∂m2= a + 3bm2 + 5cm4, (51)

which enables us to distinguish between maxima and minima. We also use the fact that F is symmetric in m andthat at large m, because c is positive, F is large and positive for large |m|. Finally, without loss of generality, we candivide through by c, or equivalently set c = 1. Now consider the four cases for a, b.

(i) a > 0, b > 0. In this case b2 − 4ac < b, so m2± is always negative. Therefore the solutions m± are always

imaginary. The only real solution is m = 0, which is a minimum having F (0) = 0.(ii) a < 0, b > 0. In this regime, b2 + 4|a|c > b2, so m2

+ > 0, so that m+ is real. However −b − (b2 + 4|a|c)1/2

remains negative, so m− remains imaginary. The real solutions are thus a maximum at m = 0 and two symmetricminima at m2

+ = ±12([b2 + 4|a|]1/2 − b).

There is a phase transition between states (i) and (ii) that occurs at a = 0 where two new solutions emerge and theextremum at m = 0 changes from a maximum for a < 0 to a minimum for a > 0. The nature of the transition isfound by making a small |a| expansion of the solutions m+, which leads to m ≈ |a|1/2. This is the Ising/Van derWaals univesality class we have studied using mean field theory, where we found |a| ≈ |T − T c|

(iii) a < 0, b < 0. In this regime, b2 + 4|a|c > b2, so m+ is real but m− remains imaginary. Therefore, as in case(ii), there is a maximum at m = 0, and minima at ±m+.

(iv) a > 0, b < 0. In this regime there are several things going on. First, the discriminant b2 − 4a is negative for

b2 < 2a, so in this regime there is only one real solution, a minimum at m = 0. For b2 > 2a, there are five realsolutions because |b| ± (b2 − 4a)1/2 > 0. Moreover, we know that the solution at m = 0 is a minimum, so we knowthat ±m− are maxima, while ±m+ are minima.

The final issue we have to resolve is the behavior of the minima m+ as a function of a, b, in particular we need toknow if F (m+) is greater than or less than F (0). If the lowest free energy state changes it corresponds to a phasetransition. We can solve this problem by evaluating F (m+), or we can find solutions where F (m0) = 0 and then solvem0 = m+. The later leads to,

m20 =

3

2[|b|

2± (

b2

4−

4a

3)1/2] (52)

and setting m20 = m2

+, we find,

3

2 [|b|

2 ± (b2

4 −4a

3 )1/2] =1

2 [|b| + (b2 − 4a)1/2] (53)

which has solution

|b|∗ = 4(a

3)1/2; with m2

∗ = m20(|b|∗, a) = 2(

a

3)1/2 (54)

m∗ is the magnetization on the phase boundary defined by |b|∗. There is then a first order phase transition frommagnetization m∗ for |b| < |b|∗, to magnetization m = 0 for |b| > |b|∗. The behavior of the magnetization on thephase boundary is m∗ ≈ a1/4 ≈ |T − T c|1/4, which is the mean field result for the order parameter near a tricriticalpoint where a line of second order phase transitions meets a line of first order phase transitions.

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Assigned Problem 5. The BCS pairing Hamiltonian is a simplified model in which only pairs with zero center of mass momentum are included in the analysis. We also assume that the fermion pairing that leads to superconductivityoccurs in the singlet channel. The BCS Hamiltonian is then,

H pair − µN = kσ


V k l a† k↑a†− k,↓

a− l↓a l↑, (55)

where N =

kσ n kσ is the number of electrons in the Fermi sea. Defining,

b k =< a− k↓a k↑ >, and b∗ k =< a† k↑a†− k↓

> . (56)

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carry out a leading order expansion in fluctuations, leading to,

H MF − µN = kσ

( kσ − µ) a† kσa kσ +

k l

V k l (a† k↑a†− k,↓

b l + b∗ ka− l↓

a l↑ − b∗ kb l) (57)

This is the Hamiltonian that we will solve to find the thermodynamic behavior of superconductors, using an atomisticmodel.

Solution. The mean-field Hamiltonian corresponding to Eq. (38) can be considered as a first order expansion in

the fluctuations i.e.a− k↓a k↑ =< a− k↓a k↑ > +[a− k↓a k↑− < a− k↓a k↑ >] (58)

Substition of this into (38), along with the definitions (39) lead to H MF .

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Assigned Problem 6. Using the Bogoliubov-Valatin transformation (Eq. 107), show that the mean field BCSHamiltonian (Eq. (106)) reduces to Eq. (110), provided Equations (111) and (112) are true.

Solution. With this transformation (ie. plug (6) into (3)), the mean field Hamiltonian looks messy,

H MF − µN =

k

( k − µ) (a† k↑a k↑ + a† k↓

a k↓) − k

(∆ ka† k↑

a†− k,↓

+ ∆∗ k

a− k↓a k↑ − b∗ k∆ k) = k

( k − µ) ([u∗ kγ † k↑ + v kγ − k↓][u kγ k↑ + v∗ kγ †− k↓

] + [u∗ kγ † k↓− v kγ − k↑][u kγ k↓ − v∗ kγ †

− k↑])

−∆ k[u∗ kγ † k↑ + v kγ − k↓][u∗ kγ †− k↓

− v kγ k↑] − ∆∗ k

[u kγ − k↓ − v∗ kγ † k↑][u k

γ k↑ + v∗ kγ †− k↓

] + b∗ k∆ k

Expanding this yields,

k

( k − µ)[u∗ ku kγ † k↑

γ k↑ + u∗ kv∗ kγ † k↑γ †− k↓

+ v ku kγ − k↓γ k↑ + v kv∗ kγ − k↓γ †

− k↓]

+( k − µ)[u∗ ku kγ † k↓γ k↓ − u∗ kv∗ kγ † k↓

γ †− k↑

− v ku kγ − k↑γ k↓ + v kv∗ kγ − k↑γ †− k↑

]

−∆ k[u∗ ku∗ kγ † k↑γ †− k↓

− u∗ kv kγ † k↑γ k↑ + v ku∗ kγ

− k↓γ †− k↓

− v kv kγ − k↓

γ k↑]

−∆∗ k

[u ku kγ − k↓

γ k↑ + u kv∗ kγ − k↓

γ †− k↓

− v∗ ku kγ † k↑γ k↑ − v∗ kv∗ kγ † k↑

γ †− k↓

] + b∗ k∆ k

We now collect the terms in this expression into three catagories: Those which have no operators in them; those whichcan be reduced to diagonal form ie. those which are of the form γ †kγ k and; those that are off diagonal (e.g. γ †γ †).The first stage is to collect together the terms which look to be in these three catagories. First the constant term,

H MF − µN = k

b∗ k∆ k

The following terms can be converted to diagonal form,

+( k − µ) (|u k|2γ † k↑γ k↑ + |v k|2γ − k↓γ †

− k↓+ |u k|2γ † k↓

γ k↓ + |v k|2γ − k↑γ †− k↑

)

−∆ k[−u∗ kv kγ † k↑γ k↑ + v ku∗ kγ

− k↓γ †− k↓

] − ∆∗ k

[u kv∗ kγ − k↓

γ †− k↓

− v∗ ku kγ † k↑γ k↑]

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Finally the off-diagonal terms are,

+( k − µ) (u∗ kv∗ kγ † k↑γ †− k↓

− u∗ kv∗ kγ † k↓γ †− k↑

+ v ku kγ − k↓γ k↑ − v ku kγ − k↑γ k↓)

−∆ k[(u∗ k)2γ † k↑γ †− k↓

− v2 kγ − k↓γ k↑] − ∆∗ k

[u2 k

γ − k↓γ k↑ − (v∗ k)2γ † k↑γ †− k↓

]

We have to rearrange the terms catagorized as diagonal above, as we need them in the form γ †γ . We do this usingthe commutation relation, i.e. γγ † = 1 − γ †γ . This yields,

H MF − µN = k

b∗ k∆ k + 2( k − µ)|v k|2 − ∆ ku∗ kv k − ∆∗ k

u kv∗ k

+( k − µ) (|u k|2γ † k↑γ k↑ − |v k|2γ †

− k↓γ − k↓ + |u k|2γ † k↓

γ k↓ − |v k|2γ †− k↑

γ − k↑)

+∆ k[u∗ kv kγ † k↑γ k↑ + v ku∗ kγ †

− k↓γ − k↓] + ∆∗

k[u kv∗ kγ †

− k↓γ − k↓ + v∗ ku kγ † k↑

γ k↑]

Finally the off-diagonal terms are (as before),

+( k − µ) (u∗ kv∗ kγ † k↑γ †− k↓

− u∗ kv∗ kγ † k↓γ †− k↑

+ v ku kγ − k↓γ k↑ − v ku kγ − k↑γ k↓)

−∆ k

[(u∗

k

)2γ †

k↑

γ †

− k↓

− v2

k

γ − k↓

γ k↑

] − ∆∗

k

[u2

k

γ − k↓

γ k↑

− (v∗

k

)2γ †

k↑

γ †

− k↓

]

We note that n kσ = n− kσ, and that the expectation of other terms that transform into one another through the

transformation k → − k, are equivalent. Collecting terms then leads to Eqs. (110)-(112) of the lecture notes.

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Assigned Problem 7. We define

v k =g k

(1 + |g k|2)1/2(59)

show that Eq. (111) reduces to (117).

Solution Starting with,

2( k − µ)(1 − |v k|

2

)

1/2

v k + ∆ kv

2

k − ∆

∗

k(1 − |v k|

2

) = 0, (60)We have,

|u k|2 = 1 − |v k|2 =1

1 + |g k|2; uk =

1

(1 + |g k|2)1/2. (61)

Substitution into Eq. () leads to,

2( k − µ)g k + ∆ kg2 k − ∆∗ k

= 0. (62)

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Assigned Problem 8. Show that E k as defined in Eq. (118) is in agreement with Eq. (112).

Solution We need to show that the definitions,

E k = ( k − µ)(|u k|2 − |v k|2) + ∆ ku∗ kv k + ∆∗ kv∗ ku k. (63)

and

E k = [( k − µ)2 + |∆ k|2]1/2 (64)

are the same. To do this we use the results of Problem 9 in Eq. (53) to write,

E k = ( k − µ)([E k + ( k − µ)]2

4E 2 k−

[E k − ( k − µ)]2

4E 2 k) + ∆ k

∆∗ k

2E k+ ∆∗

k

∆ k

2E k. (65)

which reduces to Eq. (54) as required.

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—————–

Assigned Problem 9. Prove the relations Eq. (119-121).

Solution We have,

g k =E k − ( k − µ)

∆ k

; E k = [( k − µ)2 + |∆ k|2]1/2 (66)

so that,

|∆ k|2 = E 2 k − ( k − µ)2; (67)

so that,

|g k|2 =E k − ( k − µ)

∆ k

=[E k − ( k − µ)]2

E 2 k − ( k − µ)2=

E k − ( k − µ)

E k + ( k − µ)(68)

We then find,

|v k|2 =|g k|2

1 + |g k|2=

E k − ( k − µ)

2E k; |u k|2 =

E k + ( k − µ)

2E k(69)

and hence,

u kv∗ k =g k

1 + |g k|2= g k|u k|2 =

∆∗ k

2E k(70)

(70)

—————–

Assigned Problem 10. Prove the relation (135).

Solution We start with,

b k =< a− k↓a k↑ >=< (u kγ − k↓ − v∗ kγ † k↑)(u kγ k↑ + v∗ kγ †− k↓

) > (71)

To evaluate this expression at finite temperature, we must evaluate,

< O >=tr(Oe−βH )

tr(e−βH )(72)

The only terms that survive are the terms that are diagonal, such as n kσ. We thus find,

b k =< a− k↓a k↑ >=< u kv∗ kγ − k↓γ †− k↓

− u kv∗ kγ † k↑γ k↑ >= u kv∗ k[1− < n k↑ > − < n− k↓ >] (73)

Finally we use the fact that < n kσ >=< n− kσ > to find Eq. (135).

—————–

Assigned Problem 11. Starting from Eq. (137), prove the relation (145).

Solution In the case of the weak-coupling s-wave model Eq. (137) reduces to,

1

N (F )V =

hωc0

dTanh(β

2 (2 + ∆2)1/2)

(2 + ∆2)1/2(74)

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As shown in lectures, the critical point is found from the equation,

1

N (F )V = ln(β chωc/2)Tanh(∞) + a1; where, a1 = −

∞0

ln(x)Sech2(x)dx = 0.8188. (75)

The critical temperature is then,

kBT c = 1.13hωcExp[−1

N (F )V ] (76)

To find the behavior of the gap near T c, we carry out a first order Taylor expansion of Eq. (64) using ∆ 2 as the smallquantity. This leads to,

Tanh(β2 (2 + |∆|2)1/2)

(2 + |∆|2)1/2≈

Tanh(β2 (1 + ∆2

22 ))

(1 + ∆2

22 )≈

Tanh(β2 )

+

β |∆|2

4

Sech2(β2 )

2−

|∆|2

2

Tanh(β2 )

3(77)

Taking the gap to be real, we find,

1

N (F )V =

hωc

0

dTanh(β

2 )

+ ∆2

hωc

0

d [β

4

Sech2(β2 )

2−

1

2

Tanh(β2 )

3] (78)

This expression may be written in the form,

1N (F )V

= ln( 12

β hωc) + a1 + ∆2 β 28

a2 (79)

where,

a1 = .8188 ; and a2 =

∞0

dx[Sech2(x)

x2−

Tanh(x)

x3] = −0.853 (80)

We expand β about β c, but we only need to consider carry out this expansion in the first term on the right hand side,so that,

1

N (F )V = ln[

hωc

2kBT c(1 − t)] + a1 + ∆2 β 2

8a2; where t = (T c − T )/T c (81)

We then find that,

ln(1 − t) ≈ −t = −∆2 β 2

8|a2|; so that

∆(T )

kBT c≈ (

8

|a2|)1/2t1/2 (82)

so that,

∆(T )

kBT c≈ 3.06(1 −

T

T c)1/2; T → T c (83)

This is correct near T c. Also note that Eq. (127) and (141) immply that,

2∆(0)

kBT c≈

4

1.13≈ 3.52, (84)

—————–

Assigned Problem 12. Consider a ferromagnetic nearest neighbor, spin 1/2, square lattice Ising model where theinteractions along the x-axis, J x, are different than those along the y-axis, J y. Extend the low and high temperatureexpansions Eq. (150) and Eq. (152) to this case. Does duality still hold? From your expansions, find the internalenergy and the specific heat.

Solution

Z =

{S i=±1}

<ij>

eK ijS iS j = (Cosh(K x))N (Cosh(K y))N

{S i=±1}

<ij>

(1 + tijS iS j). (85)

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tij = tx for horizontal bonds and tij = ty for vertical bonds, where tx = tanh(K x), ty = tanh(K y). The diagramsused are similar, but now we have to treat subclasses with different numbers of hirizontal and vertical bonds, so that,

−βF = ln(Z ) =zN

2ln(Cosh(K )) + N ln(2) + ln[1 + N t4 + 2N t6 + N (N +

9

2)t8 + 0(t10)] (86)

becomes

−βF = N [ln(2) + ln(Cosh(K x)) + ln(Cosh(K x)) + t2xt2y + t2xt2y(t2x + t2y) +5

2

(t4xt4y) + t2xt6y + t6xt2y + ....] (87)

Similarly, the extension of the low temperature expansion,

Z =

{S i=±1}

<ij>

eKS iS j = eKzN [1 + N s4 + 2N s6 + N (N +9

2)s8 + O(s10)] (88)

to the anisotropic case leads to,

−βF = N [K x + K y + s2xs2y + s2xs2y(s2x + s2y) +5

2s4xs4y + s2xs6y + s6xs2y...] (89)

where sx = e−2K x , sy = e−2K y . Duality holds for both x and y directions.

—————–

Assigned Problem 13. Find the second virial coefficient for six cases: (i) the classical hard sphere gas; (ii)Non-interacting Fermions; (iii) Non-interacting Bosons; (iv) The van der Waals gas.

Solution

P v

kBT =

∞l=1

al(T )(λ3

v)l−1 (90)

where the first virial coefficient a1(T ) = 1, and the second virial coefficient is a2(T ). The virial expansion is mostoften carried out in the grand canonical ensemble, where we may write,

P

kBT =

1

λ3

∞l=1

blzl;

N

V =

1

λ3

l=1

lblzl (91)

so that,

P v

kBT =

∞l=1 blzl∞l=1 lblzl

=∞l=1

al(T )(λ3

v)l−1 (92)

which gives relations between the quantities al(T ) and bl(T ). For the second virial coefficient the relationship isa2(T ) = −b2(T ). We have already calculated the coefficients bl for the ideal Bose and Fermi gases, with the results,

bl =(−1)l+1

l5/2; (Ideal F ermi) bl =

1

l5/2; (Ideal Bose) (93)

so we have,

P vkBT = 1 − 125/2 ( λ

3

v ) + ....; (Ideal Bose) (94)

and

P v

kBT = 1 +

1

25/2(

λ3

v) + ....; (Ideal Fermi) (95)

The van der Waals equation of state may be expanded to find the second virial coefficient,

P =kBT

v − b−

a

v

2, so that

P v

kBT =

1

(1 − b/v)−

a

kBT v≈ 1 +

b − a/kBT

v+ ... (96)

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For the classical interacting gas, b2 is given by,

b2 =1

2V λ3

dr1d3r2[e−βu(r1− r2 − 1] =

1

2λ3

d3r[e−βu(r) − 1] (97)

For the hard sphere problem, with a hard core radius of R, we then find that b2 = −2πR3/(3λ3), so the virialexpansion for this case is,

P v

kBT = 1 +

2πR3

3v + .... (98)

The behavior of the second virial coefficient as a function of temperature can be used to deduce the interactionpotential, and the importance of quantum effects as they have different temperature dependences.

—————–

PHY831 - Quiz 5, Friday November 11, 2011Answer all questions. Time for quiz - 25 minutes

Name:

1. (i) Plot the behavior of the magnetization of the Ising ferromagnet as a function of the temperature, for threeapplied field cases: h < 0, h = 0, h > 0. Indicate the critical point.

(ii) Write down the definition of the critical exponents α, β , γ , δ , η and ν for the Ising ferromagnet critical point.What values do these exponents take within mean field theory?(iii) Write down (or derive) the mean field equation for the spin 1/2 Ising ferromagnet in an applied field, on a

lattice with co-ordination number z and exchange constant J . From this equation find the critical exponent δ for theIsing ferromagnet within mean field theory.

2. (i) Write down (or derive) the van der Waals equation of state. Make plots of the van der Waals equation of state isotherms, for T > T c, T < T c and for T = T c.

(ii) For the case T < T c explain why the non-convex part of the curve cannot occur at equilibrium and showthe Maxwell construction to obtain a physical P, v isotherm. Write the mathematical statement of the Maxwellconstruction.

(iii) Write down the definition of the critical exponents α, β , γ , δ , η and ν for the liquid-gas critical point. Whatvalues do these exponents take within Van der Waals theory.

3. Write down the Hamiltonian for BCS theory, and the decoupling scheme used to reduce it to a solvable form.Explain the physical reasoning for the decoupling scheme that is chosen.

—————–

PHY831 - Quiz 6, Friday November 11, 2011Answer all questions. Time for quiz - 25 minutes

Name:

1. (i) Describe the physical meaning of the superconducting gap, and the way in which BCS theory describes it.(ii) Given that the energy of quasiparticle excitations from the BCS ground state have the spectrum,

E = [( − F )2 + |∆|2]1/2, (99)

where ∆ is the superconducting gap and F is the Fermi energy, show that the quasiparticle density of states is givenby,

D(E ) =N (F )E

(E 2 − |∆|2)1/2(100)

Draw this density of states and compare it to the density of states of the system in the absence of the pairing termin the Hamilonian.

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2. (i) Explain the importance of “linked-cluster” theorems in the perturbation theory of many particle systems.(ii) Draw the low temperature series expansion diagrams to order s8 (where s = Exp[−2βJ ]) for the square lattice,

nearest neighbor, spin half Ising ferromagnet partition function. What is the degeneracy of each of these diagrams?Use these terms to write down an expansion for the Helmholtz free energy and give a physical reason why only theterms of order N are kept.

3. (i) Write down the mathematical form of the virial expansion for many particle systems and explain why it isimportant. What physical properties can be extracted from the second virial coefficient?

(ii) Given that,

bl =1

l!λ3l−3V (sum over all l − connected − cluster integrals) (101)

find the second virial coefficient for the classical gas with hard core repulsive interactions, where the hard core radiusis R.

HW9_4

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