Instructor Solutions Manual for Solid State Physics: Essential Concepts 032156636X, 9780321566362

Solutions to all the in-text exercises are available electronically, with selected exercises worked out in Mathematica.

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Instructor Solutions Manual for Solid State Physics: Essential Concepts
 032156636X,  9780321566362

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Contents 1 Solutions to Chapter 1 Exercises

3

2 Solutions to Chapter 2 Exercises

43

3 Solutions to Chapter 3 Exercises

57

4 Solutions to Chapter 4 Exercises

77

5 Solutions to Chapter 5 Exercises

89

6 Solutions to Chapter 6 Exercises

107

7 Solutions to Chapter 7 Exercises

125

8 Solutions to Chapter 8 Exercises

143

9 Solutions to Chapter 9 Exercises

165

10 Solutions to Chapter 10 Exercises

183

11 Solutions to Chapter 11 Exercises

205

c Copyright 2009 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

2

Chapter 0

Problems marked with * are more difficult. Problems marked with † are substantially affected by typos which are included in the Errata.

c Copyright 2009 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

Chapter 1

Solutions to Chapter 1 Exercises Exercise 1.1. According to the prescription of the text, we assume that the solutions in the two regions take the form Ψ1 (x) = A1 sin(kx) + B1 cos(kx) , (1) and Ψ2 (x) = A2 eκx + B2 e−κx .

(2)

There is no need to write down the solution for Ψ3 (x), since the complete solution is either symmetric, in which case the derivative of Ψ2 (x) vanishes at x = 0, or antisymmetric, therefore, Ψ2 (x) itself vanishes at x = 0. The boundary conditions of the problem are         b b b = 0 −→ A1 sin k −a − + B1 cos k −a − =0. (3) Ψ1 x = −a − 2 2 2           b b b b b b Ψ1 x = − = Ψ2 x = − −→ A1 sin k − + B1 cos k − = A2 e−κ 2 + B2 eκ 2 2 2 2 2 (4)       b b Ψ2 b Ψ1 b = −→ A1 k cos k − − B1 k sin k − = A2 κeκ 2 − B2 κe−κ 2 , (5) dx x=− b dx x=− b 2 2 2

2

and finally, Ψ2 (x = 0) = 0 −→ A2 + B2 = 0 i.e., A2 = −B2 for the antisymmetric case, or Ψ2 = 0 −→ A2 κ − B2 κ = 0 , dx x=0

(6)

(7)

i.e., A2 = B2 for the symmetric case. Regarding the amplitudes A1 , B1 , A2 and B2 as the unknowns, we get a homogeneous set of four linear equations, which means that in order to have a solution,

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4

Chapter 1

the determinant of the corresponding matrix must vanish. From this condition, we arrive at the equation κb κb κ sin ak cosh + k cos ak sinh =0 (8) 2 2 for the antisymmetric case, and κ sin ak sinh

κb κb + k cos ak cosh =0 2 2

for the symmetric one. Dividing by cos ak and cosh

bκ bκ , or sinh , respectively, we get 2 2 tanh

tan ak k

= −

= −

bκ 2

(10)

bκ 2

(11)

κ coth

tan ak k

(9)

κ

for the antisymmetric and symmetric case, respectively. We have an additional equation, which links k and κ. Namely, ~2 k 2 2m ~2 κ2 2m

= E

(12)

= V0 − E ,

(13)

2mV0 − k2 . ~2

(14)

i.e., r κ=

Then the two equations above, Eqs. (10-11), lead to two equations for k, where a, b and 2mV0 /~2 play the role of parameters. Let us note that in Eq. (14), κ becomes imaginary, when ~2 k 2 /2m > V0 . This means that in that case, we have real sine and cosine solutions in the barrier, which is a simple consequence of the fact that the particle’s energy is larger than the “confining” potential, i.e., the particle is not bound in that region. The attached Mathematica code contains the derivation and the graphical solutions of the two equations above. A typical case is shown in Fig. 1, where a = 1, 2mV0 /~2 = 100, and b = 0.1, or b = 0.02. We plotted only k > 0, since the equations are invariant under the transformation k ↔ −k. By trying various values for b, we notice that as we increase b, the energy of the symmetric solution drops rapidly, while that of the antisymmetric is more or less constant. In particular, when b → 0, the right hand side of Eq. (11) tends to −∞, which means that the first solution will be at k = π/2. At the same time, the right hand side of Eq. (10) tends to 0, i.e., all solutions of that equation will be at integar multiples of π. This immediately answers the second question of the problem, because the wave number of the symmetric solution is exactly half of that of the asymmetric solution. Also, if we keep the thickness of the barrier constant, and increase the potential, the two solutions separate more and more.

c Copyright 2009 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

D.W. Snoke, Solid State Physics: Essential Concepts

1

1

0

0

-1

-1

-2

5

-2 0

2

4

6 k [a.u.]

8

10

0

2

4

6

8

10

k [a.u.]

Figure 1.1: The functions in Eqs. (10)-(11) of Exercise 1.1 for a = 1, b = 0.1, and 2mV0 /~ = 100 on the left hand side, and for a = 1, b = 0.02, and 2mV0 /~2 = 100 on the right hand side. The common left hand side of the equations is shown in solid red, long-dashed green is the right hand side of Eq. (10), the asymmetric solution, while the short-dashed blue line is the right hand side of Eq. (11), the symmetric solution. Also shown are the corresponding solutions for b = 0.1 on the left hand side.

Next, we calculate the first roots of Eqs.(10)-(11) as a function of the barrier width, b. We see from Fig. 1.1 that both of these roots are in the interval [π/2, π], and no other roots are to be found there. This means that we can use this interval for bracketing the solutions. The results are shown in Fig. 1.2. We can notice that for b → 0, we indeed have a factor of 2 in the values of the wave number, while for b → ∞, the two energies will virtually be the same. This behavior can be understood, if we notice that as b → ∞, the overlap of the wavefunctions in the central region goes to zero, so the solutions become decoupled. Once we have the value of k, we can solve for A1 , B1 , A2 and B2 , which give the wavefunctions. Two typical solutions are shown in Fig. 1.3, for a = 1, 2mV0 /~ = 100, and b = 0.1, or b = 0.3.

* Exercise 1.2.

a) The eigenvectors belonging to the matrix of the Hamiltonian,   E1 −V12 H= ∗ −V12 E2

(1)

are v1

=

v2

=

! p 4|V12 |2 + (E1 − E2 )2 + E1 − E2 1, 2V12 ! p 4|V12 |2 + (E1 − E2 )2 + E2 − E1 1, − 2V12

(2)

(3)

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6

Chapter 1

k [a.u.]

π

3π/4

π/2

0

0.2

0.4

0.6

0.8

1

b

Figure 1.2: The first roots of Eqs.(10)-(11) of Exercise 1.1 as a function of b, for a = 1, and 2mV0 /~2 = 100. The solid red line belongs to the antisymmetric solution, while the dashed green line to the symmetric case.

Figure 1.3: The symmetric (solid red line) and antisymmetric (dashed green line) wave functions for Exercise 1.1 for a = 1, b = 0.1, and 2mV0 /~ = 100 on the left hand side, while a = 1, b = 0.3, and 2mV0 /~ = 100 on the right hand side.

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D.W. Snoke, Solid State Physics: Essential Concepts

7

with the eigenvalues ˜1 E ˜2 E

=

E1 + E2 − 2

=

E1 + E2 + 2

s s

E1 − E2 2

2

E1 − E2 2

2

+ |V12 |2

(4)

+ |V12 |2 .

(5)

In the case of two identical atoms, i.e., E1 = E2 , the two eigenvectors reduce to   |V12 | v1 = 1, V12   |V12 | , v2 = 1, − V12

(6) (7)

while the eigenvalues become ˜1 E ˜2 E

= E1 − |V12 |

(8)

= E1 + |V12 | .

(9)

Therefore, the magnitudes of the coefficients of the second basis vector, |Ψ2 i, are the same, but the signs are opposite, while the coefficients of the first basis vector are equal. Since the new wavefunction should still be normalized, i.e., 1 = hΨ|Ψi = (c∗1 hφ1 | ± c∗1 hφ2 |) (c1 |φ1 i ± c1 |φ2 i) = |c1 |2 hφ1 |φ1 i +|c1 |2 hφ2 |φ2 i = 2|c1 |2 , | {z } | {z } =1

(10)

=1

√ and the phases of the wavefunctions can be chosen at will, we conclude that c1 = 1/ 2 and √ c2 = ±1/ 2. Note that in the last step we made use of the fact that the overlap between different wavefunctions is small, i.e., |φ1 i and |φ2 i are orthogonal. Since for c1 = c2 the energy is lowered, c1 = c2 corresponds to the bonding state, while c1 = −c2 , raising the energy, gives the antibonding configuration. b) When there is substantial overlap between the atomic wavefunctions, we have to calculate the energy from ˆ hΨ|H|Ψi E= . (11) hΨ|Ψi For the case of identical atoms, the eigenstates are still of the form 1 |Ψi = √ (|φ1 i ± |φ2 i) , 2

(12)

i.e., hΨ|Ψi = =

1 1 1 √ (hφ1 | ± hφ2 |) √ (|φ1 i ± |φ2 i) = (2 + hφ1 |φ2 i + hφ2 |φ1 i) 2 2 2

(13)

1 + Re I12

(14)

c Copyright 2009 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

8

Chapter 1

where I12 = hφ1 |φ2 i

(15)

is the overlap integral of the two states. The expectation value of the Hamiltonian operator for the case of two identical atoms is given above. Therefore, the energy of the system is E=

ˆ hΨ|H|Ψi E1 + E2 ∓ V12 = . hΨ|Ψi 1 + Re I12

(16)

c) In the third part of the problem, we first have to determine the two basis functions. Taking one quantum well, the potential is infinite on one side, while it has height V0 on the other side. That is, in the notation of the figure, for the lowest-lying state, we can search for the solution as ΨI (x)

= A sin(kx)

(17)

ΨII (x)

= Be−κx ,

(18)

=

0

(19)

ΨII (x = a) dΨII . dx x=a

(20)

with the boundary conditions ΨI (x = 0)

ΨI (x = a) = dΨI = dx x=a

(21)

With the specific choice in Eq.(17), the boundary condition in Eq.(19) is automatically satisfied, while the second and third boundary condition leads to the two equations A sin(ka) Ak cos(ka)

= =

Be−κa −Bκe

(22)

−κa

,

(23)

or, in matrix form,  

sin(ka)

−e−κa

k cos(ka) −κe−κa



A

 =0.



(24)

B

This homogeneous equation has solutions for A and B if and only if the determinant of the matrix is zero, i.e., if κ sin(ka) + k cos(ka) = 0 , (25) or

tan ka 1 =− . (26) k κ (This latter result can also easily be obtained by dividing Eq.(22) by Eq.(23).) As in Exercise 1, we still have r 2mV0 κ= − k2 . (27) ~2 From the same numerical approach as used in Exercise 1.1, we find for the case a = 1, b = 0.1 and 2mV0 /~2 = 100, that the lowest k that satisfies Eq.(26) is k = 2.85. The reason for choosing this value of V0 is that this is the one that we thoroughly studied in Exercise 1.1. c Copyright 2009 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

D.W. Snoke, Solid State Physics: Essential Concepts

9

The value of A and B still has an ambiguity, because we can multiply them by any number, and still get a valid solution. The value of A and B are fixed by the condition that the wavefunction is normalized. We could simply pick A = 1, and leave the wavefunction un-normalized, since it will not affect the energyies. Alternatvely, we can impose the normalization condition 1 = A2

Z

a

dx sin2 kx + B 2

Z



dx e−2κx =

a

0

A2 (2ka − sin 2ka) B 2 e−2aκ + . 4k 2κ

(28)

Using the continuity condition in Eq. (22), we can rewrite the normalization as A2 (2ka − sin 2ka) A2 sin2 ka + , 4k 2κ

1=

(29)

i.e.,  A=

(2ka − sin 2ka) sin2 ka + 4k 2κ

−1/2 ,

(30)

eκa sin ka .

(31)

and  B=

(2ka − sin 2ka) sin2 ka + 4k 2κ

−1/2

This gives us one of the basis functions, which we denote by φ1 (x). We get the other one by “centering” it on the second well, i.e., after reflecting it with respect to x = 0, and shifting it to the right by 2a + b, we get φ2 (x) = φ1 (−x + 2a + b). Having obtained the two basis functions, we can calculate the energy of the system using Eq. (16). When we do not assume orthogonality, we need the overlap integral I12 , which we can calculate noting that while the wavefunctions extend to infinity or minus infinity, their product is non-zero over a finite interval [0, 2a + b] only, because outside this interval one of the wavefunctions is zero. Therefore, Z



Z dx φ1 (x)φ2 (x)

2a+b

=

−∞

dx φ1 (x)φ2 (x) 0

Z =

2AB

a κx −κ(2a+b)

dx sin kx e e 0

+B

2

Z

a+b

dx e−κx eκx e−κ(2a+b)

a

 k (κ sin ak − k cos ka)eκa + 2 − + B 2 be−κ(2a+b) = 2ABe κ2 + k 2 κ + k2  2  ~ ((κ sin ak − k cos ka)eκa + k) + B 2 be−κ(2a+b) , (32) = 2ABe−κ(2a+b) 2mV0 −κ(2a+b)



where we used the result of Eq.(27). We will also need the integral for V12 , i.e., Z V12 =

2a+b

 dx φ2 (x)

0

 ~2 d2 − V (x) φ∗1 (x) . 2m dx2

(33)

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10

Chapter 1

Since the Hamiltonian is Hermitian, V21 = V12 . This also wavefunctions. Now, since V (x) = V0 for a < x < a + b and 0  2 2 ~ k      2 2  2m φ1 (x) , if ~ d − V (x) φ1 (x) =  2m dx2 2 2    ~ κ φ1 (x) , if 2m

follows from the symmetry of the otherwise, 0 1, b -> 0.1}, DetMat2 /. {kappa -> Sqrt[20 - k^2], a -> 1, b -> 0.1}}, {k, 0, 4}, PlotStyle -> {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}] (* ==== Red is the symmetric, green is the antisymmetric solution. Also, this is the case of strong coupling, for b=0.01 ==== *) (* ==== Now, we try to find the solutions Note that we need to use eps, in order to avoid a run-away solution ===== *) eps = .1 Ksym = k /. FindRoot[DetMat3 /. { kappa -> Sqrt[20 - k^2], a -> 1, b -> 0.1} , {k, eps, 4}] Kanti = k /. FindRoot[DetMat2 /. {kappa -> Sqrt[ 20 - k^2], a -> 1, b -> 0.1} , {k, eps, 4}] (* === We fix a=1, b=0.1, 2mV/hbar^2=20 here === *) a = 1; b = .1; kappa1 = Sqrt[20 - Ksym^2] ; Bs = (-Sin[Ksym*b/2] + Tan[Ksym*(a + b/2)]*Cos[Ksym*b/ 2])/(Exp[-kappa1*(b/2)] + Exp[kappa1*(b/2)]) ; ps1 = Plot[Sin[Ksym*x] + Tan[Ksym*(a + b/2)]*Cos[Ksym*x], { x, -a - b/2, -b/2}, PlotStyle -> RGBColor[1, 0, 0]] ; ps2 = Plot[Bs*(Exp[kappa1*(x)] + Exp[kappa1*(x)]), {x, -b/2, b/2}, PlotStyle -> RGBColor[1, 0, 0]] ps3 = Plot[-Sin[Ksym*x] + Tan[Ksym*(a + b/2)]* Cos[Ksym*x], {x, b/2, a + b/2}, PlotStyle -> RGBColor[1, 0, 0]] ; Show[ps1, ps2, ps3]

(* ===== The antisymmetric solutions can be obtained in a similar way ===== *) a = 1; b = .1; kappa2 = Sqrt[20 - Kanti^2] ; Ba = (-Sin[Kanti*b/2] + Tan[Kanti*(a + b/2)]* Cos[Kanti*b/2])/(Exp[-kappa2*(b/2)] - Exp[kappa2*(b/2)]) ;

c Copyright 2009 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

D.W. Snoke, Solid State Physics: Essential Concepts

33

pa1 = Plot[Sin[Kanti*x] + Tan[Kanti*(a + b/2)]*Cos[Kanti*x], { x, -a - b/2, -b/2}, PlotStyle -> RGBColor[0, 1, 0]] ; pa2 = Plot[Ba*(Exp[kappa2*(x)] Exp[-kappa2*(x)]), {x, -b/2, b/2}, PlotStyle -> RGBColor[0, 1, 0]] pa3 = Plot[Sin[Kanti*x] - Tan[Kanti*(a + b/2)]*Cos[Kanti*x], {x, b/2, a + b/2}, PlotStyle -> RGBColor[0, 1, 0]] ; Show[{pa1, pa2, pa3}]

ChI-2.nb Clear[eps,A,B,a,b,k,kappa] (* ======== We define all necessary functions here ============= *) k[a_,b_,V_]:=FindRoot[Tan[x*a]/x+1/Sqrt[V-x*x]==0,{x,PI/2+eps,PI-eps}] kappa[a_,b_,V_]:=Sqrt[V-k[a,b,V]*k[a,b,V]] A[a_,b_,V_]:=((2*k[a,b,V]*a-Sin[2*k[a,b,V]*a])/(4*k[a,b,V])+ Sin[k[a,b,V]*a]*Sin[k[a,b,V]*a]/(2*kappa[a,b,V]))^(-1/2) B[a_,b_,V_]:=A[a,b,V]*Exp[kappa[a,b,V]*a]*Sin[k[a,b,V]*a] (* ======== Numerical integration takes place here ============= *) OverLap1[a_,b_,V_]:=2*A[a,b,V]*B[a,b,V]* NIntegrate[Sin[k[a,b,V]*x]*Exp[kappa[a,b,V]*(x-2*a-b)]],{x,0,a}] OverLap1[a_,b_,V_]:=B[a,b,V]*B[a,b,V]*b*Exp[-kappa[a,b,V]*(2*a+b)] OverLap[a_,b_,V_]:= OverLap1[a,b,V] + OverLap2[a,b,V] (* ============ Potential terms are defined here =============== *) PotV1[a_,b_,V_]:=NIntegrate[Exp[kappa[a,b,V]*(x-2*a-b)]*Sin[k[a,b,V]*x], {x,0,a}] PotV2[a_,b_,V_]:=b*Exp[-kappa[a,b,V]*(2*a+b)] PotV3[a_,b_,V_]:=NIntegrate[Exp[-kappa[a,b,V]*x]*Sin[(2*a+b-x)*k[a,b,V]], {x,a+b,2*a+b] PotV[a_,b_,V_]:=A[a,b,V]*B[a,b,V]*k[a,b,V]*k[a,b,V]*Pot1[a,b,V] + B[a,b,V]*B[a,b,V]*k[a,b,V]*k[a,b,V]*Pot2[a,b,V] + A[a,b,V]*B[a,b,V]*kappa[a,b,V]*kappa[a,b,V]*Pot3[a,b,V] Energy1[a_,b_,V_]:=k[a,b,V]*k[a,b,V] + PotV[a,b,V] Energy2[a_,b_,V_]:=k[a,b,V]*k[a,b,V] - PotV[a,b,V] (* Note that c1=c2=1/sqrt(2) *) EnergyOverlap1[a_,b_,V_]:=(k[a,b,V]*k[a,b,V] + PotV[a,b,V])/(1+2*OverLap[a,b,V]) EnergyOverlap1[a_,b_,V_]:=(k[a,b,V]*k[a,b,V] - PotV[a,b,V])/(1+2*OverLap[a,b,V]) (* We plot the results here. Only variable is b *)

c Copyright 2009 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

34

Chapter 1

Plot[{EnergyOverlap1[1, x, 100], EnergyOverlap2[1, x, 100]}, {x,0,1}]

ChI-3.nb Clear[Mat] Mat := {{1, 1, -1, -1}, {I*K, -I*K, -kappa, kappa}, {Exp[I*K*a], Exp[-I*K*a], -Exp[-kappa*b]*Exp[I*k*(a + b)], -Exp[kappa*b]* Exp[I*k*(a + b)]}, {(I*K + I*k)*Exp[I*K*a], -(I*K - I*k)* Exp[-I*K*a], -(kappa + I*k)*Exp[-kappa*b]* Exp[I*k*(a + b)], (kappa - I*k)*Exp[kappa*b]*Exp[I*k*(a + b)]}} MatrixForm[Mat] DetMat = Det[Mat] Detmat2 = Simplify[DetMat] Detmat3 = Detmat2/Exp[-I*a*K - b*kappa] FullSimplify[Re[Detmat3], Assumptions -> {Im[kappa] == 0, Im[K] == 0, Im[a] == 0, Im[b] == 0, Im[k] == 0}]

ChI-6.nb k[EE_, V_, b_] := ArcCos[(V - EE)*b/(2*Sqrt[EE])* Sin[Sqrt[EE]] + Cos[Sqrt[EE]]] Plot[{k[EE, 0.5, 0.1]}, {EE, 1, 140}, PlotStyle -> {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, AxesLabel -> {"E", "k"}]

ChI-8.nb False, ViewPoint -> {4, -.8, 1.2}] (*Perovskite*) Show[ Graphics3D[ Sphere[.6 ]] , TranslateShape[Graphics3D[{SurfaceColor[GrayLevel[ 1]], Sphere[.6 ]}] , {6, 6, 6}] , TranslateShape[

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D.W. Snoke, Solid State Physics: Essential Concepts

37

Graphics3D[ {SurfaceColor[GrayLevel[.1]], Sphere[.6 ]}] , {0, 6, 6}] , TranslateShape[ Graphics3D[ {SurfaceColor[GrayLevel[.1]], Sphere[.6 ]}] , {6, 0, 6}] , TranslateShape[Graphics3D[ {SurfaceColor[GrayLevel[.1]], Sphere[.6 ]}] , { 6, 6, 0}] , TranslateShape[ Graphics3D[ {SurfaceColor[GrayLevel[.1]], Sphere[.6 ]}] , {12, 6, 6}] , TranslateShape[Graphics3D[ {SurfaceColor[ GrayLevel[.1]], Sphere[.6 ]}] , {6, 12, 6}] , TranslateShape[Graphics3D[ {SurfaceColor[GrayLevel[.1]], Sphere[.6 ]}] , {6, 6, 12}] , TranslateShape[Graphics3D[ Sphere[.6 ]] , {0, 12, 0}] , TranslateShape[Graphics3D[ Sphere[.6]] , {0, 0, 12}], TranslateShape[Graphics3D[ Sphere[.6 ]] , {12, 12, 0}], TranslateShape[Graphics3D[ Sphere[.6]] , {0, 12, 12}], TranslateShape[Graphics3D[ Sphere[.6 ]] , {12, 12, 12}], TranslateShape[Graphics3D[ Sphere[.6 ]] , {12, 0, 12}], TranslateShape[Graphics3D[ Sphere[.6 ]] , {12, 0, 0}], Graphics3D[Line[{{0, 0, 0}, {12, 0, 0}}]], Graphics3D[Line[{{0, 0, 0}, {0, 12, 0}}]], Graphics3D[Line[{{0, 0, 0}, {0, 0, 12}}]], Graphics3D[Line[{{12, 12, 12}, {12, 12, 0}}]], Graphics3D[Line[{{12, 12, 12}, {12, 0, 12}}]], Graphics3D[Line[{{12, 12, 12}, {0, 12, 12}}]], Graphics3D[Line[{{0, 0, 12}, {12, 0, 12}}]], Graphics3D[Line[{{0, 0, 12}, {0, 12, 12}}]], Graphics3D[Line[{{0, 12, 0}, {12, 12, 0}}]], Graphics3D[ Line[{{12, 0, 0}, {12, 12, 0}}]], Graphics3D[Line[{{12, 0, 0}, {12, 0, 12}}]], Graphics3D[Line[{{ 0, 12, 0}, {0, 12, 12}}]], Graphics3D[{Line[{{0, 0, 0}, {0, 12, 0}}]} ], Graphics3D[{Line[{{0, 0, 0}, {12, 0, 0}}]} ], Graphics3D[{Dashing[{.01, .01}], Line[{{6, 6, 0}, {6, 6, 12}}]} ], Graphics3D[{Dashing[{.01, .01}], Line[{{6, 0, 6}, {6, 12, 6}}]} ], Graphics3D[{Dashing[{.01, .01}], Line[{{0, 6, 6}, {12, 6, 6}}]} ], Boxed -> False, ViewPoint -> {2.6, -1.2, .7}] (*Fluorite*) Show[ Graphics3D[ Sphere[.6 ]] , TranslateShape[Graphics3D[ {SurfaceColor[GrayLevel[.1]], Sphere[.6 ]}] , \ {3, 3, 3}] , Graphics3D[Line[{{0, 0, 0}, {3, 3, 3}}]], Graphics3D[Line[{{6, 0, 6}, {3, 3, 3}}]], Graphics3D[Line[{{0, 6, 6}, {3, 3, 3}}]], Graphics3D[Line[{{6, 6, 0}, {3, 3, 3}}]],

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38

Chapter 1

TranslateShape[Graphics3D[ Sphere[.6 ]] , {0, 6, 6}] , TranslateShape[Graphics3D[ {SurfaceColor[GrayLevel[.1]], Sphere[.6 ]}] , {3, 9, 9}] , TranslateShape[Graphics3D[ Sphere[.6 ]] , {6, 6, 0}] , TranslateShape[Graphics3D[ {SurfaceColor[GrayLevel[.1]], Sphere[.6 ]}] , 9, 9, 3}] , Graphics3D[Line[{{6, 6, 0}, {9, 9, 3}}]], Graphics3D[Line[{{12, 12, 0}, {9, 9, 3}}]], Graphics3D[Line[{{6, 12, 6}, {9, 9, 3}}]], Graphics3D[Line[{{6, 6, 0}, {3, 9, 3}}]], Graphics3D[Line[{{6, 6, 0}, {9, 3, 3}}]], TranslateShape[Graphics3D[ {SurfaceColor[GrayLevel[.1]], Sphere[.6 ]}] , {3, 9, 3}] , Graphics3D[Line[{{12, 6, 6}, {9, 9, 3}}]], TranslateShape[Graphics3D[ Sphere[.6 ]] , {6, 0, 6}] , TranslateShape[Graphics3D[ {SurfaceColor[GrayLevel[.1]], Sphere[.6 ]}] , {9, 3, 9}] , TranslateShape[Graphics3D[ Sphere[.6 ]] , {0, 12, 0}] , TranslateShape[Graphics3D[ Sphere[.6]] , {0, 0, 12}], TranslateShape[Graphics3D[ Sphere[.6 ]] , {12, 12, 0}], TranslateShape[Graphics3D[ Sphere[.6]] , {0, 12, 12}], \ TranslateShape[Graphics3D[ Sphere[.6 ]] , {12, 12, 12}], TranslateShape[Graphics3D[ {SurfaceColor[GrayLevel[.1]], Sphere[.6 ]}] , {9, 9, 9}] , TranslateShape[Graphics3D[ Sphere[.6 ]] , {12, 0, 12}], TranslateShape[Graphics3D[ Sphere[.6 ]] , {12, 0, 0}], TranslateShape[Graphics3D[ Sphere[.6 ]] , {6, 6, 12}], TranslateShape[Graphics3D[ {SurfaceColor[ GrayLevel[.1]], Sphere[.6 ]}] , {3, 3, 9}] , TranslateShape[Graphics3D[ Sphere[.6 ]] , {12, 6, 6}], TranslateShape[Graphics3D[ {SurfaceColor[GrayLevel[.1]], Sphere[.6 ]}] , 9, 3, 3}] , TranslateShape[Graphics3D[ Sphere[.6 ]] , {6, 12, 6}], Graphics3D[Line[{{12, 12, 12}, {12, 12, 0}}]], Graphics3D[Line[{{12, 12, 12}, {12, 0, 12}}]], Graphics3D[Line[{{12, 12, 12}, {0, 12, 12}}]], Graphics3D[Line[{{0, 0, 12}, {12, 0, 12}}]], Graphics3D[Line[{{ 0, 0, 12}, {0, 12, 12}}]], Graphics3D[Line[{{0, 12, 0}, {12, 12, 0}}]], Graphics3D[Line[{{12, 0, 0}, {12, 12, 0}}]], Graphics3D[Line[{{12, 0, 0}, {12, 0, 12}}]], Graphics3D[Line[{{0, 12, 0}, {0, 12, 12}}]], Graphics3D[{Line[{{0, 0, 0}, {0, 12, 0}}]} ], Graphics3D[{Line[{{0, 0, 0}, {12, 0, 0}}]} ], Graphics3D[{Line[{{0, 0, 0}, {0, 0, 12}}]} ],

\

{

\

\

{

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D.W. Snoke, Solid State Physics: Essential Concepts

39

Boxed -> False, ViewPoint -> {2.6, -1.4, .6}] (*Cuprite*) Show[ Graphics3D[ Sphere[.6 ]] , TranslateShape[Graphics3D[ Sphere[.6 ]] , {6, 6, 6}] , TranslateShape[Graphics3D[ {SurfaceColor[GrayLevel[.1]], Sphere[.6 ]}] 3, 3, 3}] , TranslateShape[Graphics3D[ {SurfaceColor[GrayLevel[.1]], Sphere[.6 ]}] {9, 3, 9}] , TranslateShape[Graphics3D[ {SurfaceColor[GrayLevel[.1]], Sphere[.6 ]}] 3, 9, 9}] , TranslateShape[Graphics3D[ {SurfaceColor[GrayLevel[.1]], Sphere[.6 ]}] {9, 9, 3}] , TranslateShape[Graphics3D[ Sphere[.6 ]] , {0, 12, 0}] , TranslateShape[Graphics3D[ Sphere[.6]] , {0, 0, 12}], TranslateShape[Graphics3D[ Sphere[.6 ]] , {12, 12, 0}], TranslateShape[Graphics3D[ Sphere[.6]] , {0, 12, 12}], TranslateShape[Graphics3D[ Sphere[.6 ]] , {12, 12, 12}], TranslateShape[Graphics3D[ Sphere[.6 ]] , {12, 0, 12}], TranslateShape[Graphics3D[ Sphere[.6 ]] , {12, 0, 0}], Graphics3D[Line[{{0, 0, 0}, {6, 6, 6}}]], Graphics3D[Line[{{12, 0, 12}, {6, 6, 6}}]], Graphics3D[Line[{{12, 12, 0}, {6, 6, 6}}]], Graphics3D[Line[{{0, 12, 12}, {6, 6, 6}}]], Graphics3D[Line[{{0, 0, 0}, {12, 0, 0}}]], Graphics3D[Line[{{0, 0, 0}, {0, 12, 0}}]], Graphics3D[Line[{{0, 0, 0}, {0, 0, 12}}]], Graphics3D[Line[{{12, 12, 12}, {12, 12, 0}}]], Graphics3D[Line[{{12, 12, 12}, {12, 0, 12}}]], Graphics3D[Line[{{12, 12, 12}, {0, 12, 12}}]], Graphics3D[Line[{{0, 0, 12}, {12, 0, 12}}]], Graphics3D[ Line[{{0, 0, 12}, {0, 12, 12}}]], Graphics3D[Line[{{0, 12, 0}, {12, 12, 0}}]], Graphics3D[Line[{{12, 0, 0}, {12, 12, 0}}]], Graphics3D[Line[{{12, 0, 0}, {12, 0, 12}}]], Graphics3D[Line[{{0, 12, 0}, {0, 12, 12}}]], Graphics3D[{Line[{{0, 0, 0}, {0, 12, 0}}]} ], Graphics3D[{Line[{{0, 0, 0}, {12, 0, 0}}]} ], Boxed -> False, ViewPoint -> {2.6, -1.2, .7}]

, { , \ , { , \

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40

Chapter 1

ChI-15.nb Clear[MRing] MRing := {{E0, V, 0, 0, 0, V}, {V, E0, V, 0, 0, 0}, {0, V, E0, V, 0, 0}, {0, 0, V, E0, V, 0}, {0, 0, 0, V, E0, V}, {V, 0, 0, 0, V, E0}} MatrixForm[MRing] nu = 1; c[1] = 1.0; For[i = 1, i < 7, i++; c[i_] := 1*Exp[(i - 1)*Pi/3*I*nu]] Clear[MMR] MMR := MRing /. {E0 -> 1, V -> 0.1} vec = Table[c[k], {k, 1, 6}] res = MRing.vec Simplify[Extract[res, 6]/c[6]] Eigenvectors[MRing] Eigenvalues[MRing] vec[3] MatrixForm[MMR] Eigenvalues[MMR] Clear[MChain] MChain := {{E0, V, 0, 0, 0, 0}, {V, E0, V, 0, 0, 0}, {0, V, E0, V, 0, 0}, {0, 0, V, E0, V, 0}, {0, 0, 0, V, E0, V}, {0, 0, 0, 0, V, E0}} MatrixForm[MChain] MMC := MChain /. {E0 -> 1, V -> 0.1} MatrixForm[MMC] Eigenvalues[MMC] Eigenvalues[MMR]

ChI-18.nb k[EE_] = ArcCos[1/(2*Sqrt[EE])*Sin[Sqrt[EE]] + Cos[Sqrt[EE]]] Plot[{k[EE]}, {EE, 0, 92}, PlotStyle -> {RGBColor[1, 0, 0]},

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D.W. Snoke, Solid State Physics: Essential Concepts

41

AxesLabel -> {"K", "k"}] i = 0; For[x = 0, x < 100, x = x + 0.1; (*This is a simple forward scheme*) vec[i] = Abs[(k[x + 0.1] - k[x])/0.1]; i++; ] (list1 = Table[{l/10, vec[l]}, {l, 0, 1000}]); ListPlot[list1, PlotStyle -> RGBColor[1, 0, 0], AxesLabel -> {"E", "D(E)"}]

ChI-22.nb disp1[k_, p_, A_] = k^2/2 + Sqrt[A^2 + (k p)^2] disp2[k_, p_, A_] = k^2/2 - Sqrt[A^2 + (k p )^2] Plot[{disp1[k, 1, 0], disp2[k, 1, 0]}, {k, -5, 5}, PlotStyle -> {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, AxesLabel -> {"k", "E(k)"}] slope1[k_, p_, A_] = D[disp1[k, p, A], k] slope2[k_, p_, A_] = D[disp2[k, p, A], k] Plot[{slope1[k, 1, 0], slope2[k, 1, 0]}, {k, -5, 5}, PlotStyle -> {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, AxesLabel -> {"k", "dE(k)/dk"}] Plot[{disp1[k, 1, 0.1], disp2[k, 1, 0.1]}, {k, -4, 4}, PlotStyle -> {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, AxesLabel -> {"k", "E(k)"}] Plot[{slope1[k, 1, 0.1], slope2[k, 1, 0.1]}, {k, -4, 4}, PlotStyle -> {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, AxesLabel -> {"k", "dE(k)/dk"}]

ChI-23.nb Clear[HAMILTON] HAMILTON := {{1, 0, V, V}, {0, 1.3, V, 3*V}, {V, V, 1, 0},

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42

Chapter 1 {V, 3*V, 0, 1.3}}

MatrixForm[HAMILTON] n = 0; For[x = 0, x < 1, x = x + 0.01; Hamsubs := HAMILTON /. {V -> x}; eigenvalues = Eigenvalues[Hamsubs]; For[i = 0, i < 4, i++; vec[n] = {x, Extract[eigenvalues, i]}; n++; ] ] (list1 = Table[vec[n], {n, 0, 399}]); ListPlot[list1, PlotStyle -> RGBColor[1, 0, 0], AxesLabel -> {"V", "E"}] Eigenvectors[HAMILTON /. {V -> 0.6}] Eigenvalues[HAMILTON /. {V -> 0.6}]

ChI-27.nb Gap = 1.5 P[Eg_, Delta_] = 1/(Sqrt[2*Pi]*Delta)*Exp[-(Gap - Eg)^2/(2*Delta^2)] Dg[Energy_, Delta_] = NIntegrate[P[Eg, Delta]*Sqrt[Energy - Eg], {Eg, -Infinity, Energy}] Plot[{Dg[x, 0.1], Dg[x, 0.3], Dg[x, 0.5]}, {x, 0, 4}, PlotRange -> All, PlotStyle -> {RGBColor[1, 0, 0], RGBColor[0, 1, 0], RGBColor[0, 0, 1]}, AxesLabel -> {"E", "D(E)"}]

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Chapter 2

Solutions to Chapter 2 Exercises Exercise 2.1. In the scope of the Kronig-Penney model, we have derived the relation (c.f. Eq. (1.10) in the text) κ2 − K 2 sinh κb sin Ka + cosh κb cos Ka = cos k(a + b) , 2κK

(1)

where

2m 2m (V0 − E) and K 2 = 2 E . ~2 ~ Taking the limit b → 0 and V0 → ∞, but keeping the product V0 b finite yields Eq. (1.11), κ2 =

κ2 b sin Ka + cos Ka = cos ka . 2K

(2)

(3)

In the limit V0 → ∞, E is negligible by comparison, so we can use κ2 = 2mV0 /~2 . At zone center, i.e., when k → 0, we can assume that Ka is also small, and expand sin Ka and cos Ka up to order K 3 . This gives us   2m V0 b 1 1 1 3 Ka − (Ka) + 1 − (Ka)2 = 1 − (ka)2 , (4) ~2 2K 6 2 2 i.e.,

hence

1 ~2 ~2 V0 ba − V0 ba(Ka)2 − (Ka)2 = − (ka)2 , 6 2m 2m   ~2 K 2 mV0 ba ~2 k 2 V0 b 1+ = + . 2m 3~2 2m a

Solving for E = ~2 K 2 /2m then yields  −1  −1 ~2 k 2 mV0 ba V0 b mV0 ba E = 1+ + 1+ 2m 3~2 a 3~2 2 2 ~ k = + E0 . 2meff

(5)

(6)

(7)

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44

Chapter 2

This implies   mV0 ba meff = m 1 + , 3~2 which, as indicated in Fig. 1.12 of the text, is higher than the free electron mass. Also,   mV0 ba V0 b . E0 ' 1− a 3~2

(8)

(9)

This is the answer to Exercise 1.4 to second order in V0 b.

Exercise 2.2.

According to the equipartition theorem, 3 1 mv 2 = kB T , 2 2

(1)

from where for the velocity, we get r v=

3kB T . m

In the case of electrons, this is equal to s r 3kB T 3 × 8.6 · 10−5 eV/K × 300 K ve = = 3.69 · 107 cm/s , = mc 0.1 × 5.1 · 105 eV/c2

(2)

(3)

while for holes, r ve =

Exercise 2.3.

3kB T = mc

s

3 × 8.6 · 10−5 eV/K × 300 K = 9.54 · 106 cm/s , 1.5 × 5.1 · 105 eV/c2

(4)

With the units specified in the problem, 2

ωc =

Exercise 2.4.

1 e × 1 V · s/m 1 s × 9 · 1016 m2 /s2 qB = = = 1.8 · 1012 s−1 . 2 m 5 · 104 m2 0.1 × 5.1 · 105 eV/c

(1)

The reduced mass of the exciton is 1 1 mr = = = 0.188 m0 , 1 1 1 1 + + me mh 0.275 m0 0.59 m0

(1)

where m0 is the free electron mass; i.e., the exciton Bohr radius is equal to aex

= =

2 8.3 × 6.6 × 1016 eV · s 4π~2  4π0 r ~2 1 = 2 = e2 mr e mr 1.44 × 10−7 eV · cm 0.188 × 5.1 · 105 eV/c2 2.32 · 10−7 cm = 23.2 ˚ A

(2)

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D.W. Snoke, Solid State Physics: Essential Concepts

Exercise 2.5.

45

On one hand, the total energy is √

 5/3 2 m3/2 N (3π 2 ) ~3 √ , U = 2V 2 3 5π ~ V 2 2 m3/2

(1)

while on the other hand, we can express it in terms of the kinetic energy of the particles, i.e.,   mv 2 U =N . (2) 2 Equating these two expressions yields √ √ 2  2 5/3 5/3  3π 4V 2 m1/2 N (3π 2 ) ~3 2/3 4 2 ~ √ √ =n , hv i = 2 3 2 2 3/2 N 5π ~ V 2 2 m 5π m 2 2 2

hence, p

hv 2 i

=n

√ 5/6  4 2 ~ 3π 2 √ √ . 5π m 2 2

1/3 2

(3)

(4)

By definition, at T = 0, Z

1 N

hki =

kF

k4πk 2 dk =

0

3 π 4 k = kF . N F 4

(5)

We will use the expression for the Fermi energy in order to calculate the value of kF . The Fermi energy is given by   2  3 2/3 3π ~2 kF2 ~ √ = n EF = . (6) 2m 2 2 m3/2 By re-arranging this, for kF we get r kF =

√ 2mEF = n1/3 2 2 ~



3π 2 √ 2 2

1/3 .

(7)

Hence, √ 3 hki = n1/3 2 4



3π 2 √ 2 2

1/3 .

(8)

With the values given in the problem, this becomes 3√ hki = 2 4



3π 2 √ 2 2

1/3

3

(1022 1/cm )1/3 = 4.99 · 107 cm−1 .

(9)

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46 Exercise 2.6. by

Chapter 2 In 3D, in an isotropic band with effective mass m∗ , the density of states is given 1 D(E) = 2π 2



2m∗ ~2

3/2 p

E − E0 ,

(1)

where E0 is the reference point of the energy (the bottom of the band). From this, for the conduction band we get  3/2 p 2me 1 E − E0 ; (2) Dc (E) = 2π 2 ~2 thus, the number of electrons is given by Ne

=

e

µ/kB T −Ec /kB T

Z



e−E/kB T Dc (E + Ec ) dE

e

(3)

0

Z



=

eµ/kB T e−Ec /kB T

=

0 µ/kB T −Ec /kB T 3/2 e e me I(T )

1 2π 2

e−E/kB T



2me ~2

3/2

√ E dE

,

(4) (5)

where  3/2 √ 2 1 E dE 2 2 2π ~ 0  3/2 Z ∞ √ 1 2kB T e−x x dx , 2 2 2π ~ 0

Z I(T )

= =



e−E/kB T

(6)

which depends on T only. Similarly, for the holes, we get 3/2

Nh = eµ/kB T e−Ev /kB T mh I(T ) .

(7)

Multiplication of Eq. (5) and Eq. (7) gives Ne Nh = e(Ev −Ec )/kB T (me mh )3/2 I 2 (T ) .

(8)

However, having Ne = Nh means that Ne = e(Ev −Ec )/2kB T (me mh )3/4 I(T ) .

(9)

Equations Eq. (5) to Eq. (9) then lead to eµ/kB T e−Ec /kB T m3/2 = e(Ev −Ec )/2kB T (me mh )3/4 I(T ) , e or eµ/kB T = e(Ev +Ec )/2kB T



mh me

(10)

3/4 I(T ) .

(11)

Hence, µ=

Ev + Ec 3 + kB T ln 2 4



mh me

 .

(12)

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D.W. Snoke, Solid State Physics: Essential Concepts In the particular case of

mh = 4, the expression above gives me µ=

(e)

Ev + Ec + 1.04 kB T . 2

(13)

(e)

Exercise 2.7. NQ

47

Eq. (2.24) involves NQ , which is obtained from the relation (2.19), Z



e−E/kB T Dc (E + Ec ) dE

= 0

Z

e

= 0

=

∞ −E/kB T

(1)

√ Z V m√ V m3/2 2 ∞ −E/kB T 1/2 2 2 3 2mE dE = e E dE 2π ~ π 2 ~3 0

√ √ V m3/2 2 π (kB T )3/2 . π 2 ~3 2

(2) (3)

The donors density is equal to Nd = 1016 . Eq. (2.25) gives the free electron density as ! s 1 −Ry/kB T (e) N D Ne (T ) = e NQ 1 + 4e−Ry/kB T (e) − 1 . 2 N

(4)

Q

The attached Mathematica code contains the function (the ratio of free electrons to impurities) as a function of temperature, which is shown in Fig. 2.1. A close-up between T = 11000 and T = 12000 K shows that this ratio is equal to 99% at around T = 11550 K. This is unrealistically large; other considerations will come in at high temperature (see Section 8.10.2). What this means is that at lower temperatures there will typically always be a substantial fraction of neutral donors.

Exercise 2.8.

From the problem we have Z P (r) = 1 −

r

Q(r0 ) dr0 ,

(1)

0

hence, dP = −Q(r) . dr However, from the definition of P (r) and Q(r), it follows that Q(r) = 4πnr2 P (r) .

(2)

(3)

By substituting Eq. (2) into Eq. (3), we get a differential equation for P (r), namely −

dP = 4πnr2 P (r) , dr

(4)

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48

Chapter 2

1 0.9 0.9904

0.8 0.7

Ne/Nd

Ne/Nd

0.9902

0.6

0.99 0.9898 0.9896

0.5

11000

0.4

11500

12000

Temperature [T]

0.3 0

2000

4000

6000

8000

10000

12000

Temperature [T] Figure 2.1: The ratio of free electrons to impurities as a function of temperature. The inset is a close-up in the vicinity of Ne /Nd = 0.99.

whose solution is straightforward. −

dP = 4πnr2 dr , P

(5)

which, after an integration, becomes ln P =

4πn 3 r +C , 3

(6)

where C is an integration constant. Taking the exponential of both sides gives us 4πn 3 r +C P (r) = e 3 . −

(7)

At this point we can drop the integration constant, because it only affects the way in which the distribution function is normalized. Then, differentiation of Eq. (7) with respect to r yields 4πn 3 − r dP 2 Q(r) = − = 4πnr e 3 . dr

(8)

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D.W. Snoke, Solid State Physics: Essential Concepts

49

Upon the definition of Q(r), for hri we get Z



hri =

Z Q(r) dr =



4πnr2 e



4πn 3 r 3 dr .

(9)

0

0

Changing the integration variable r to x := r3 , i.e., r = x1/3 , and dr = 31 x−2/3 dx, Z hri =



4πn 0

4πn   1/3 Z ∞ 1/3    1/3 x x1/3 − 3 3 4 1 dx = y 1/3 e−y dy = Γ e 3 ≈ 0.544 3 4πn 4πn 3 n 0 (10)

Exercise 2.9.

The band bending is given by Eq. (2.30) as r ∆ d= , en

(1)

3

with ∆ ≈ 1 V, n = 1016 1/cm and  = 14 0 . Hence, s 1 V × 14 × 8.85 · 10−14 C/V · cm d= = 3.58 · 10−5 cm = 358 nm . 3 1.6 · 10−19 C × 1016 1/cm

(2)

In order to answer the second question, we go back to the derivation of Eq. (2.29). The only thing that we have to keep in mind is that one side has five times higher doping, which can be re-phrased as saying that charge carriers from the other side need to go only one fifth of the distance to find a pair. Thus, in the derivation of Eq. (2.29), we will have 5d on one side, and d on the other, while the densities are n and 5n, respectively. Then the continuity condition at the boundary would give the equation V0 + hence

en 2 en 1 d = V0 + ∆ − (5d)2 , 2 2 5 en 2 d (1 + 5) = ∆ , 2

(3)

(4)

or solving for d, r d=

∆ . 3en

(5)

Exercise 2.10. When drawing the bands, we keep in mind that the chemical potential of the isolator is situated half-way between its two bands, and that the acceptor states in the p-doped material and the donor states in the n-doped material line up to this common chemical potential. With these considerations, the bending can be obtained as in the following figure.

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50

Chapter 2

p

n

i

µ

Figure 2.2: Band bending in a p-i-n structure.

Exercise 2.11.

By the definition of the density of states, in 1D we have D(E)dE =

L dk , 2π

i.e., D(E) =

L L dk = 2π dE 2π



dE dk

−1 =

L 2π



~2 k m

−1 =

(1) √ L m~ L m √ √ √ . = 2 2π ~ 2mE 2π ~ 2 E

(2)

Exercise 2.12. In order to calculate the number of particles for the quantum well, we first have to determine how many quantum levels are occupied, for which we would need the confinement energies. This is given by 2 2 · 10−5 eV · s × 3.142 2 ~2 π 2 2 2 E` = ` = (1) 2 ` = ` · 7.8 meV , mL2 5.1 · 105 eV × (10−6 cm) from which we can conclude that only the first, second and third confined levels will be occupied. The energies of these states are equal to 7.8, 31 and 70 meV, respectively. This means that the Fermi energies will be 100 meV, 69 meV, and 30 meV, respectively. Since for a 2D system the density of states is given by D2d (E) =

A m , 2π ~2

(2)

the number of particles is equal to Z N=

EF

D(E) dE = 0

A m EF , 2π ~2

(3)

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D.W. Snoke, Solid State Physics: Essential Concepts

51

i.e., after dividing by A, we get N m EF . = A 2π~2 With EF = 100 meV and m = m0 , this yields n=

n=

5.1 · 105 eV 2 × 3.14 × (2 · 10−5 eV · cm)

2

(4)

× 0.1 eV = 2.1 · 1013 cm−2 .

(5)

By the same approach the density in the second and third levels will be 1.4 · 1013 cm−2 and 0.6 · 1013 cm−2 , respectively. Thus, the number of particles can be expressed as N = (n1 + n2 + n3 )A = (2.1 + 1.4 + 0.6) · 1013 cm−2 × 1 mm2 = 4.1 · 1011 . In the case of a three dimensional electron gas, the density of states is √ 3/2 √ 2m E V D3d (E) = , 2π 2 ~3 i.e., the number of particles is obtained as √ 3/2 Z EF √ Z EF √ 2m V 2m3/2 2 3/2 V E dE = 2 N= E . D(E) dE = 2 2 2π ~3 π ~3 3 F 0 0

(6)

(7)

(8)

Hence, the particle density required is √ 3/2 2m N 3/2 = E n = V 3π 2 ~3 F √ = =

2 5.1 · 105 eV

3/2

3 × (3.14)2 × (2 · 10−5 eV · s)

3/2

3

× (0.1 eV)

7.1 · 1019 cm−3 .

(9)

Hence, in the same volume, i.e., in 10−5 mm3 = 10−8 cm3 , we would have N = 7.1 · 1011 electrons.

Exercise 2.13. The transition from Type I to Type II superlattice happens when the confinement energy of electrons located in the GaAs exceeds the difference between the energy of the Γ-valley of GaAs and the X-valley of AlAs. According to the text, this difference is 200 meV. The confinement energy is also known, and its value is ~2 π 2 2 N , (1) mL2 where we can set N = 1, since only those levels will be thermally occupied. Solving for the width, and setting E1 = 200 meV yields EN =

L= √

~π 6.6 · 10−16 eV · s × 3.14 =p = 7.91 · 10−7 cm = 7.9 nm . mE1 0.06 × 5.69 · 10−16 eV · s2 /cm2 × 0.2 eV

(2)

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52

Chapter 2

Exercise 2.14. Again, we have to make the various chemical potentials align. For the barriers, being n-doped, the chemical potential is equal to the donor levels, which are almost equal to the bottom of the conduction band. In the quantum well, the Fermi energy will be determined by the confinement energy, and if the density of electrons is not too high, then it will be equal to the energy of the first confined state. Thus the band bending is similar to that given in the following figure.

n

QW

n

µ

Figure 2.3: Band-bending in a quantum well structure, for Exercise 2.14.

Exercise 2.15. Let B point in the z direction. Then the electrons will orbit in the xy plane. Their motion in the z direction is determined by the quantization along z. This sets the lowest possible wavenumber, for kz = nπ/d, where n is an integer and d is the size along z. To all quantum numbers kz , there will be N = eBA/h Landau orbits. Now, the cyclotron energy is given by Ec = ~ωc =

~eB 6.6 · 10−16 eV · s × 1 e × 10 T = = 1.16 · 10−2 eV , m 0.1 × 5.1 · 105 eV/c2

(1)

while the thermal energy is obtained from ET = kB T = 8.6 · 10−5 eV/K × 1 K = 8.6 · 10−5 eV .

(2)

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D.W. Snoke, Solid State Physics: Essential Concepts

53

Since ET  Ec , only the first Landau level will be occupied. The number of states in this level is N=

eBA 1 e × 10 T × 1 cm2 = = 2.41 · 1011 . h 2 × 3.14 × 6.6 · 10−16 eV · s

(3)

Then we still have to count the number of states along z. kz will be quantized according to kz = nπ/d = nπ/(1 cm) = 3.14 n cm−1 , and the cut-off wavenumber is determined by the thermal 2 energy. If we set ET = ~2 kmax /(2m), then s r 2mET 2 × 0.1 × 5.1 · 105 eV/c2 × 8.6 · 10−5 eV = 1.48 · 105 cm−1 . (4) = kmax = 2 ~2 (6.6 · 10−16 eV · s) Thus the number of states along z is nz = 2 × 1.48 · 105 cm−1 /3.14 cm−1 = 48068 Therefore the total number of states is given by 48068 · N = 1.16 · 1016 .

Exercise 2.16.

By definition, µ=

∂U , ∂B

(1)

where the energy can be calculated from     1 eB 1 U = N Eν = N ~ωc ν − = ν− . 2 m 2

(2)

Thus the derivative at ν = 1 is ∂ eBN µ= ∂B m



1 ν− 2

 =

eN . 2m

(3)

The condition that the Landau level is half-filled used only when taking the derivative with respect to B: an infinitesimal variation in B does not force electrons to the next Landau level, i.e., the variation in the energy of the occupied Landau level is continuous.

Exercise 2.17.

On one hand, the electric field in the y direction is given by   h Jx Ey = ν , e2 ∆y

(1)

while on the other hand, it can be expressed as the gradient of the potential, namely, Ey = −

∆V . ∆y

By equating the two forms of the electric field, we arrive at the relation   h ν Jx = −∆V e2

(2)

(3)

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54

Chapter 2

i.e.,  Jx = −∆V For the first Landau level ν = 1, and Jx = −

e2 h

 .

(4)

h = 25812 Ω. Hence, e2

10−5 V = 3.87 · 10−10 A = 387 pA. 25812 Ω

(5)

Exercise 2.18. We will use induction to prove the statement of the problem. First, for N = 2, the statement is trivially true, since the wavefunction is of the form Φ(0, z) ∼ rl eilΘ .

(1)

Now, let us suppose that the statement holds true for N . Then with Ψ(z1 , . . . zN ) =

N Y

(zn − zm )lnm

Ψ0 (z1 , . . . zN +1 ) =

n All, AxesLabel -> "Energy"] (*===============================================================*) (*2D case*) SPINS = 30; ITERATIONS = 300; T = 1.6; J = 1; Clear[s, Energy, Magnetisation]; Array[s, {SPINS, SPINS}];

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D.W. Snoke, Solid State Physics: Essential Concepts

203

For[i = 0, i "Energy"] ChX-2.nb (*For the sake of simplicity, T_c = 1. Also, alpha/k_B = 1*) f[m_, T_, H_] := m*(T - 1)/T + 1/3*m^3*(1/T)^3 + (m^2(1/T)^2 - 1)*H/T; Clear[n, vec]; n = 0; For[H = -10, H < 10, H = H + 0.01; s = Table[m /. N[Solve[f[m, 1.9, H] == 0, m]]]; i = 0; Do[ (*We walk through the 3 solutions to see if they are real. *) If[ Im[Extract[s, i]] == 0, {vec[n] = {H, Extract[s, i]}; n++;} ],

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204

Chapter 10 {i, 3} ]

] Print[n] (list1 = Table[vec[k], {k, 1, n - 1}]); ListPlot[list1] ChX-14.nb

Hamiltonian:={{y^2+z^2,0,-x*y+I*z,z*(y+I*x),-z*(x+I*y),y*(y+I*x)}, {0,y^2+z^2,z*(I*x-y),-I*z^2-x*y {z*(y-I*x),z^2*I-x*y,0,x^2+z^2,x*(I*x-y),-z*(y+I*x)}, {z*(I*y-x),y*(I*x+y),-z*(I*x+y),-x*I*x+y),x^ MatrixForm[Hamiltonian] n = 0; (* First, we calculate the k_x = k_y = k_z case, i.e., the [111] direction *) For[X = 0, X < 1, X = X + 0.01; Hamsubs := Hamiltonian /. {x -> X, y -> X, z -> X}; eigenvalues = Eigenvalues[Hamsubs]; For[i = 0, i < 6, i++; vec[n] = {X, Extract[eigenvalues, i]}; n++; ] ] (list1 = Table[vec[n], {n, 0, 599}]); ListPlot[list1, PlotStyle -> RGBColor[1, 0, 0], AxesLabel -> {"k", "E"}] n = 0; (* Then, the k_x = k_y, k_z = 0 case, i.e., the [110] direction *) For[X = 0, X < 1, X = X + 0.01; Hamsubs := Hamiltonian /. {x -> X, y -> X, z -> 0}; eigenvalues = Eigenvalues[Hamsubs]; For[i = 0, i < 6, i++; vec[n] = {X, Extract[eigenvalues, i]}; n++; ] ] (list2 = Table[vec[n], {n, 0, 599}]); ListPlot[list2, PlotStyle -> RGBColor[1, 0, 0], AxesLabel -> {"k", "E"}]

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Chapter 11

Solutions to Chapter 11 Exercises Exercise 11.1.

In general, the number of particles is given by N=

∞ X

gn , exp(β( n − µ)) − 1 n=0

(1)

where gn is the degeneracy of the nth energy level. In the case of an isotropic 3D harmonic trap, n = n~ω (we can drop the zero-point energy, since the zero of the energy is arbitrary). We also have to take into account the fact that the energy levels are gn = 21 (n + 1)(n + 2)-times degenerate, and that in a condensate µ = 0. With these in mind, Eq. (1) takes the form N=

∞ 1 X 2 (n + 1)(n + 2) . exp(nβ~ω) − 1 n=0

(2)

Applying the E = n~ω, i.e., (n + 1)(n + 2) ≈ (E/~ω)2 substitution, we are led to N

 3 Z ∞ Z ∞ dE 1 E 2 ∞ 1 X 1 1 E 2 dE ~ω 2 ~ω 2 (n + 1)(n + 2) ≈ N0 + = N0 + = exp(nβ~ω) − 1 exp(nβ~ω) − 1 2 ~ω exp(βE) − 1 0 0 n=0 3 Z ∞  3 1 y 2 dy 1 kB T ζ(3)Γ(3) = N0 + ~ωβ exp(βy) − 1 2 ~ω 2 0  3 1 1 ≈ N0 + 1.2 · . 2 ~ωβ = N0 +

1 2



(3)

Therefore, the critical temperature and the critical number is linked by the relation Ncr = 1.2 ·

1 2



1 ~ωβ

3 .

(4)

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206

Chapter 11

Exercise 11.2.

For the sake of brevity, we will use u~k = q

e−iθ

A~ eiθ v~k = q k . 1 − A~2

and

1 − A~2

k

(1)

k

Then for the four operators appearing in the Hamiltonian in Eq. (11.14), the transformations take on the form a~k

= u~k ξ~k − v~k∗ ξ † ~

−k ∗ † v~k ξ~ k

a−~k

= u~k ξ−~k −

a~† k † a ~ −k

= u~∗k ξ~† − v~k ξ−~k =

k ∗ † u~k ξ ~ −k

− v~k ξ~k

Using these definitions, the particular terms in the Hamiltonian can be expressed as    a~† a~k = u~∗k ξ~† − v~k ξ−~k u~k ξ~k − v~k∗ ξ † ~ k

=

k 2 † |u~k | ξ~ ξ~k k

−k



u~∗k v~k∗ ξ~† ξ † ~ k −k

− v~k u~k ξ−~k ξ~k + |v~k |2 ξ−~k ξ † ~ −k | {z }

1+ξ † ~ ξ−~k −k

=

|u~k |2 ξ~† ξ~k k

+

|v~k |2 ξ † ~ ξ−~k −k



u~∗k v~k∗ ξ~† ξ † ~ k −k

− v~k u~k ξ−~k ξ~k + |v~k |2

(2)

Similarly, a~† a† ~ e−2iθ k −k

   = e−2iθ u~∗k ξ~† − v~k ξ−~k u~∗k ξ † ~ − v~k ξ~k −k k   † −2iθ ∗2 † † ∗ = e u ~k ξ~ ξ ~ − u~k v~k ξ~ ξ~k − v~k u~∗k ξ−~k ξ † ~ + v~k2 ξ−~k ξ~k −k k k −k   2 † † † † = e−2iθ u∗~k ξ~ ξ ~ − u~∗k v~k ξ~ ξ~k − v~k v~k∗ ξ ~ ξ−~k − v~k u~∗k + v~k2 ξ−~k ξ~k .

(3)

   = e2iθ u~k ξ~k − v~k∗ ξ † ~ u~k ξ−~k − v~k∗ ξ~† −k k   † 2iθ 2 ∗ ∗ = e u~k ξ~k ξ−~k − u~k v~k ξ~k ξ~ − v~k u~k ξ † ~ ξ−~k + v ∗~2k ξ † ~ ξ~† k −k −k k   † † = e2iθ u~2k ξ~k ξ−~k − u~k v~k∗ ξ~ ξ~k − u~k v~k∗ − v~k∗ u~k ξ ~ ξ−~k + v ∗~2k ξ † ~ ξ~† .

(4)

k −k

k

−k

And finally, a~k a−~k e2iθ

−k

k

−k k

The Hamiltonian in Eq. (11.14) reads as H=

 X X L~  † †  1 k L0 N + E~k + L~k a~† a~k + a~ a ~ e−2iθ + a~k a−~k e2iθ . k k −k 2 2 ~ k

(5)

~ k

After collecting the terms, we have  X  1 H = L0 N + E~k + L~k |u~k |2 ξ~† ξ~k + |v~k |2 ξ † ~ ξ−~k − u~∗k v~k∗ ξ~† ξ † ~ − u~k v~k ξ−~k ξ~k + |v~k |2 k −k k −k 2 ~ k

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D.W. Snoke, Solid State Physics: Essential Concepts

+

X L~ h k

2

~ k

207

  e−2iθ u∗~2k ξ~† ξ † ~ − u~∗k v~k ξ~† ξ~k − v~k u~∗k ξ † ~ ξ−~k − v~k u~∗k + v~k2 ξ−~k ξ~k k −k

−k

k

 i +e2iθ u~2k ξ~k ξ−~k − u~k v~k∗ ξ~† ξ~k − u~k v~k∗ − v~k∗ u~k ξ † ~ ξ−~k + v ∗~2k ξ † ~ ξ † ~ =

−k −k

−k

k

X   1 L0 N + E~k + L~k |u~k |2 + |v~k |2 ξ~† ξ~k k 2 ~ k    X   2iθ ∗ 2 ∗ ∗ −2iθ ∗ 2 L~ k ξ~† ξ † ~ + −u~k v~k E~k + L~k + e v ~k + e u ~k k −k 2 ~ k L  X   ~ k + −u~k v~k E~k + L~k + v~k2 e−2iθ + u~2k e2iθ ξ~k ξ−~k 2 ~ k

oL Xn ~ k † ξ ξ~ + −u~∗k v~k e−2iθ − v~k u~∗k e−2iθ − u~k v~k∗ e2iθ − v~k∗ u~k e2iθ 2 ~k k ~ k

+

X

E~k + L~k



|v~k |2

2

+

~ k

 L   L~k −2iθ  ~ e −v~k u~∗k + k e2iθ −u~k v~k∗ . 2 2

(6)

Now, the last sum does not contain operators, nor does the term 12 L0 N ,so they obviously are al  q ready diagonal. Then, upon the definition of u~k and v~k , and A~k = E~k + L~k − E~2 + 2E~k L~k /L~k , k

L   ~ k −u~∗k v~k∗ E~k + L~k + e2iθ v ∗~2k + e2iθ u∗~2k =0, 2

(7)

and similarly, its complex conjugate, L   ~ k −u~k v~k E~k + L~k + e−2iθ v~k2 + e−2iθ u~2k =0. 2

(8)

This means that in the Hamiltonian, all operators of the form ξ~† ξ † ~ and ξ~k ξ−~k will have a vanishing k −k

coefficient, i.e., the Hamiltonian will contain only ξ~† ξ~k and constants. k

Exercise 11.3. definition,

First we prove that the operators ξ~k obey bosonic commutation relations. By ξ k − Ak ξ † ~ −k −iθ e = u~k ξ~k − v~k∗ ξ † ~ , a~k = q −k 1 − A~2

(1)

k

with u~k = q

e−iθ 1 − A~2

k

A~ e−iθ and v~k = q k 1 − A~2

(2)

k

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208

Chapter 11

(Here we have made use of the fact that from the definitions, u~k = u−~k and u~k = u−~k , since A~k = A−~k .) The transformation then reads as !

a~k a† ~

 =

−k

u~k −v~k

−v~k∗ u~∗k

ξ~k ξ† ~



! (3)

−k

which can be inverted to yield ξ~k ξ† ~

!

 =

−k

u~∗k v~k

v~k∗ u~k

a~k a† ~



! (4)

−k

When taking the inverse, we made use of |u~k |2 − |v~k |2 = 1, which is a trivial consequence of the definitions of u~k and v~k . Thus, for ξ~k we have ξ~k = u~k∗ a~k + v~k∗ a† ~

(5)

ξ~† = u~k a~† + v~k a−~k .

(6)

−k

which can be transposed to give ξ~† as k

k

k

Hence, the commutation relation is obtained as i h h i h i h i ξ~k , ξ~†0 = u~∗k a~k + v~k∗ a† ~ , u~k0 a~† 0 + v~k0 a−~k0 = u~k u~∗k0 a~k , a~† 0 +v~k v~k∗0 a−~k , a−~k0 k −k k k | {z } | {z } =δ~k,k~0

=

2

=−δ~k,k~0

2

|u~k | δ~k,k~0 − |v~k | δ~k,k~0 = δ~k,k~0 ,

(7)

for |u~k |2 − |v~k |2 = 1. Similarly, i h i h i   h ξ~k , ξ~k0 = u~∗k a~k + v~k∗ a† ~ , u~∗k0 a~k0 + v~k∗0 a† ~ 0 = u~∗k v~k∗0 a~k , a† ~ 0 +v~k∗ u~∗k0 a† ~ , a~k0 = 0 . −k −k −k −k | {z } | {z } =δ~k,−~k0

(8)

=−δ~k,−~k0

And finally, h i  † ξ~† , ξ †~0 = ξ~k0 , ξ~k = 0 . k

Exercise 11.4. Hint

= =

k

By definition, the interaction Hamiltonian is given by Z 1 dr1 dr2 U (~r1 − ~r2 ) 2 X Z 1 1 ~ ~ ~ ~ dr1 dr2 2 δ(~r1 − ~r2 )ei(−k4 ·~r1 −k3 ·~r2 +k2 ·~r1 +k1 ·~r2 ) b~† b~† b~k2 b~k1 k4 k3 2 V

(9)

(1) (2)

~ k1 ,~ k2 ,~ k3 ,~ k4

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D.W. Snoke, Solid State Physics: Essential Concepts

209

~ = (~r1 + ~r2 )/2 and ~r = ~r1 − ~r2 substitutions, the integral can be re-written If we implement the R as Z 1 1 ~ ~ ~ ~ dr1 dr2 2 δ(~r1 − ~r2 )ei(−k4 ·~r1 −k3 ·~r2 +k2 ·~r1 +k1 ·~r2 ) b~† b~† b~k2 b~k1 k4 k3 2 V Z 1 ~ ~ ~ ~ ~ ~ ~ ~ ~ = d3 R d3 r ei(k1 +k2 −k3 −k4 )·R ei(k1 −k2 +k3 −k4 )·~r/2 2 Z (2π)3 (3) ~ ~ ~ ~ ~ ~ ~ ~ δ (k1 + k2 − k3 − k4 ) d3 r δ(~r)ei(k1 −k2 +k3 −k4 )·~r/2 = V2 =

(2π)6 (3) ~ δ (k1 + ~k2 − ~k3 − ~k4 ) . V2

(3)

By summing over the wave vectors, we arrive at Hint =

(2π)3 2

X

δ~k1 +~k2 −~k3 −~k4 ,0 b~† b~† b~k2 b~k1 = k4 k3

~ k1 ,~ k2 ,~ k3 ,~ k4

(2π)3 2

X

b~† b~† b~k2 b~k1 , k4 k3

(4)

~ k1 ,~ k2 ,~ k3

where the value of ~k4 is determined by momentum conservation. We also used the fact that we have to multiply by V 2 /(2π)3 , when we replace the Dirac delta by the Kronecker delta.

~g p~ ~ n0 ∇θ, the velocity of the superfluid is simply , for ~v = = m n0 m ~∇θ ~ × ~g = 0 implies ∇ ~ × ~v = 0. Let us take the . Therefore, assuming that n0 is constant, ∇ m vector identity ~ ~ × (∇ ~ × ~v ) , ∇2~v = ∇(∇ · ~v ) − ∇ (1) Exercise 11.5.

Since ~g =

and substitute it into the Navier-Stokes equation ∂~v ~ v = − 1 ∇P ~ + ν∇2~v , + ~v · ∇~ ∂t ρ

(2)

∂~v ~ v = − 1 ∇P ~ + ν(∇(∇ ~ ~ × (∇ ~ × ~v )) . + ~v · ∇~ · ~v ) − ∇ ∂t ρ

(3)

that is,

~ × ~v = 0, and in the incompressible case ∇ · ~v = 0, the last two terms drop out, and we are Since ∇ left with ∂~v ~ v = − 1 ∇P ~ , + ~v · ∇~ (4) ∂t ρ where the viscosity does not play any role.

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210

Chapter 11

* Exercise 11.6.

a) The original equation reads as 0 = −U0 n0 Ψ + U0 |Ψ|2 Ψ −

~2 2 ∇ Ψ. 2m

(1)

Using the assumption Ψ=



n0 eiθ f (r/r0 ) ,

(2)

and moreover, that n0 is constant and that the azimuthal angle φ is equal to the condensate phase θ, the second derivative becomes √ √ √ √ √ 1 ∂2 1 ∇2 Ψ = ∇2 { n0 eiθ f (r/r0 )} = n0 eiθ ∇2r f + n0 f 2 2 eiθ = n0 eiθ ∇2r f − n0 f 2 eiθ . r ∂θ r √ Upon substitution into Eq. (1), and division by n0 eiθ , we get 0 = −U0 n0 f + U0 n0 f 3 −

~2 2 ~2 ∇r f + f , 2m 2mr2

(3)

(4)

that is, 0 = f − f3 +

~2 ~2 f , ∇2r f − 2mU0 n0 2mU0 n0 r2 | {z } | {z } r02

(5)

r02

hence 0 = f − f 3 + ∇2y f −

f , y2

(6)

because y = r/r0 , therefore, r02 ∇2r f (r/r0 ) = ∇2y f (y). This was to be proved. b) The Appendix gives Mathematica code for a numerical solution of the equation. The numerical method can also be implemented by programming a straightforward difference method solution. We multiply by y, thus Eq. (6) takes the form   f ∂ ∂f f ∂f ∂2f 0 = yf − yf 3 + y∇2y f − = yf − yf 3 + y − = yf − yf 3 + +y 2 . (7) y ∂y ∂y y ∂y ∂y The last two terms in Eq. (7) can be turned into a difference equation. Note, that in this way, by making use of the boundary condition f (y = 0) = 0, we can avoid any singularities. So, the last two terms at y = yi are given as ∂f ∂2f +y 2 ∂y ∂y

fi+1 − fi−1 fi+1 − 2fi + fi−1 + yi 2∆ ∆2       1 yi 2yi yi 1 = fi+1 + 2 − fi + f − , i−1 2∆ ∆ ∆2 ∆2 2∆

=

(8)

where ∆ = yi − yi−1 . According to Eq. (7), this is equal to yi fi3 − yi fi , thus we can express fi+1 as c Copyright 2009 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

D.W. Snoke, Solid State Physics: Essential Concepts

 fi+1 =

1 yi + 2∆ ∆2

211

−1      2yi yi 1 yi fi3 − yi fi + fi − f − . i−1 ∆2 ∆2 2∆

(9)

In Eq. (9), no singularities are present, and we can easily solve for f (y), if we fix the derivative at y = ∆, i.e., the value of f (∆). (This practically turnes the original boundary value problem into an initial value problem.) The validity of our solution will depend on whether f (y = ∞) = 1 is fulfilled or not. So, we take various values for the derivative at y = ∆, and calculate f according to Eq. (9). In our calculations, we took 100 points on the interval [0, 10], thus ∆ = 0.1. As is seen in Fig. 11.1, depending on the initial value, f (10) is lower (green) or higher (blue) than 1. We can use these two cases as the two initial guesses for a successively approximating scheme: the interval of the initial derivatives is bisected, and then the function is evaluated at the middle point. If the value is lower than 1, this will become the lower end of the new interval, while if it is higher than 1, then this will be the new higher end. These steps are repeated until convergence is achieved (red curve).

3

True solution Undershooting solution Overshooting solution

2.5 2

f(y)

1.5 1 0.5 0 -0.5 -1 0

1

2

3

4

5

6

7

8

9

10

y Figure 11.1: Three solutions of the differential equation in Eq. (6), for Exercise 11.6. The green curve undershoots the desired solution, while the blue one overshoots it. The red curve corresponds to the solution with the boundary conditions f (y = 0) = 0 and f (y = 10) = 1, achieved by the successive approximation scheme described in the text. The green curve corresponds to f 0 (∆) = 0.58, the blue is f 0 (∆) = 0.58195, and the red is the true solution with f 0 (∆) = 0.581922.

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212

Chapter 11

Exercise 11.7.

According to the transformations in Eq. (11.16), ξ~k − A~k ξ † ~ −k −iθ e a~k = q 1 − A~2

ξ~† − A~k ξ−~k iθ a~† = kq e . k 1 − A~2

and

k

(1)

k

Hence, a~† a~k k

= = =

   1 ξ~† − A~k ξ−~k ξ~k − Ak ξ † ~ 2 −k 1 − A~ k k h i 1 † † † † 2 ξ ξ − A ξ ξ − A ξ ξ + A ξ ξ ~ ~ ~ ~ ~ ~ ~ k ~ k −k k k −k −~ k −~ k k 1 − A~2 ~k k k h  i 1 † † † 2 † ξ ξ + A ξ ξ − A ξ ξ + ξ ξ . ~ ~ ~ ~ k k −~ k ~ k k −~ k −~ k −~ k 1 − A~2 ~k k

(2)

k

Similarly, a† ~ a−~k = −k

h  i 1 † † † † 2 ξ ξ + A ξ ξ − A ξ ξ + ξ ξ , ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ −k k −k −k k k −k k 1 − A~2 −k −k

(3)

k

i.e., a~† a~k − a† ~ a−~k −k

k

=

h  i 1 ξ~† ξ~k − ξ † ~ ξ−~k + A~k2 ξ † ~ ξ~k − ξ~† ξ~k 2 −k −k k 1 − A~ k

=

ξ~† ξ~k − ξ † ~ ξ−~k .

k

(4)

−k

k

Therefore, X ~2~k · ~k0 m

~ k

a~† a~k k

=

~ k

=

Exercise 11.8.

 1 X ~2 ~ ~  † (k · k0 ) a~ a~k − a† ~ a−~k k −k 2 m 1 X ~2 2

~ k

 X ~2  (~k · ~k0 ) ξ~† ξ~k − ξ † ~ ξ−~k = (~k · ~k0 )ξ~† ξ~k k −k k m m

(5)

~ k

a) The scattering cross section for hard sphere scattering is σ = 4πa2 ,

(1)

where a is the radius of the scatterers. On the other hand, Z 2 2 dσ m2 3 i~ k·~ r = m = d re U (~ r ) 4 dΩ 4π~ 4π~4

Z 2 2 d3 rei~k·~r U0 δ(~r) = m U02 . 2 4π ~4

(2)

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D.W. Snoke, Solid State Physics: Essential Concepts Comparing the two equations, and keeping in mind that Z dσ dσ σ= dΩ = 4π , dΩ dΩ

213

(3)

results in

4π~2 a . m b) Now, from the discussion after Eq. (11.38), we know that the critical velocity is r r √ N U0 nU0 2~ nπa v = = = mV m m √ −16 2 × 6.6 · 10 eV · s × 1022 cm−3 × 3.1 × 1.4 · 10−8 cm ≈ = 6626 cm/s , 2 4 × 9.4 · 108 eV/c U0 =

(4)

(5)

˚ for the radius of the He atom. This is the so-called van der Waals where we used a = 1.4 A radius, which is used in most hard-sphere approximations of atomic physics. The mass was simply taken as m = 4mP , because there are two protons and two neutrons with nearly equal mass in a He nucleus. The value of the critical velocity is suprisingly close to the experimentally measured 50 m/s.

Exercise 11.9. Here we simply have to compile the specific meaning of the various terms. First, from Eq. (11.17) and Eq. (11.18),  q 1  (1) A~k = E~k + L~k − E~2 + 2E~k L~k , k L~k where

N ~2 k 2 and E~k = . V 2m By setting N/V = 1 and ~2 /2m = 1, we have  p 1  2 A~k = k + U0 − k 4 + 2k 2 U0 , U0 L~k = U (~k)

(2)

(3)

where the only parameter is U0 . The Mathematica code in the appendix contains a possible implementation of plotting the distribution of excited particles based on this last expression. The distribution for three values of the interaction potential is shown in Fig. 11.2.

Exercise 11.10.

We start with the definitions of the c operators, i.e.,  X 1 † ~ ~ cK = φ K − k b† ~ ~ b† ~ ~ ↓,K−k ↑,k 2 ~ k   X 1 ~ 0 ~0 cK = φ∗ K − k b↑,~k0 b↓,K ~0 ~ 0 −~ k0 , 2

(1)

~ k0

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0.9

U=1 U=3 U=10

0.8

Occupation [a.u]

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

2

4

6

8

10

k [a.u.] Figure 11.2: The distribution function of particles for three different values of the interaction potential, for Exercise 11.9. Plotted are U = 1 (solid red curve), U = 3 (dashed green line), and U = 10 (dotted blue line).

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215

where we have dropped the principal quantum number, n, for it does not appear in the commutation relations.    i h i X 1 1 ~ 0 ~0 h † † † † † † ~ ~ . (2) b K − k φ K − k b b , b cK , c = φ 0 ~ ~ ~ ~ ~ 0 −~ K k0 ↓,K− k ↑,~ k ↓,K k0 ↑,~ 2 2 ~ k,~ k0

Expanding the commutator yields h i b† ~ ~ b† ~ , b† ~ 0 ~ 0 b† ~ 0 = ↓,K−k ↑,k

↓,K −k

b†

− b†

= b†

+ b†

= b†

+ b†

= b†

− b†

↑,k

b† b† ~ 0 ~ 0 b† ~ 0 ~ ~ ↓,K− k ↑,~ k ↓,K −k ↑,k b† b† ~ 0 ~ 0 b† ~ 0 ~ ~ ↓,K− k ↑,~ k ↓,K −k ↑,k b† b† ~ 0 ~ 0 b† ~ 0 ~ ~ ↓,K− k ↑,~ k ↓,K −k ↑,k b† b† ~ 0 ~ 0 b† ~ 0 ~ ~ ↓,K− k ↑,~ k ↓,K −k ↑,k

b† 0 b† ~ ~ b† ~ ~ 0 −~ ↓,K k0 ↑,~ k ↓,K−k ↑,k b† ~ ~ b† ~ 0 b† ~ ~ 0 −~ ↓,K k0 ↓,K− k ↑,k ↑,k b† ~ 0 ~ 0 b† ~ b† ~ 0 ~ ~ ↓,K− k ↓,K −k ↑,k ↑,k b† b† ~ 0 ~ 0 b† ~ 0 ~ ~ ↓,K− k ↑,~ k ↓,K −k ↑,k

=0.

(3)

Hence, i † , c c†K ~0 = 0 , ~ K

(4)

which also implies that after a conjugation,   cK ~ , cK ~0 = 0 .

(5)

h

† We proceed with the commutation relation of cK ~ and cK ~ 0 by inserting the definition of the operators into the definition of the commutator, i.e.,     i X h i 1~ ~ 1 ~ 0 ~0 h † † † ∗ = φ , c cK K − k φ K − k b b , b b . (6) ~ ~ ~ ~0 ~ 0 −~ ↑,~ k ↓,K− k ↓,K K k0 ↑,~ k0 2 2 ~ ~0 k,K

Then the commutator can be expanded as h i † † † b↑,~k b↓,K− , b b = b↑,~k b↓,K− ~ ~ ~ ~ ~0 ~ 0 ~0 k kb ↓,K −k

b† 0 ~ 0 −~ ↓,K k0 ↑,~ k

↑,k

− b†

b† 0 b↑,~k b↓,K− ~ ~ ~ 0 −~ k ↓,K k0 ↑,~ k

b ~ ~k )b† ~ 0 ~ 0 −~ ↓,K k0 ↓,K− ↑,k

=

† b↑,~k (δK ~ 0 −~ ~ ~ k0 ,K− k −b

=

† † δK ~ 0 −~ ~ ~ k0 ,K− k b↑,~ kb ~0 − b ↑,k

− b†

b k b† ~ 0 b↓,K− ~ ~ ~ 0 −~ k ↓,K k0 ↑,~ ↑,k

† † † = δK ~ 0 −~ ~ ~ k0 ,K− k b↑,~ kb ~0 + b ~0 b ↑,k

b† 0 b↑,~k b↓,K− ~ ~ ~ 0 −~ k k ↓,K k0 ↑,~

↑,k

b k b↓,K− ~ ~ ~ 0 −~ k ↓,K k0 ↑,~

− b†

b† 0 b↑,~k b↓,K− ~ ~ ~ 0 −~ k ↓,K k0 ↑,~ k

− δ~k,~k0 b†

b ~ ~k ~ 0 −~ ↓,K k0 ↓,K−

−b†

b† 0 b↑,~k b↓,K− ~ ~ ~ 0 −~ k ↓,K k0 ↑,~ k

† † = δK ~ 0 −~ ~ ~ k0 ,K− k b↑,~ k b ~ 0 − δ~ k,~ k0 b ↑,k

b ~ ~k ~ 0 −~ ↓,K k0 ↓,K−

† † = δK ~ 0 −~ ~ ~ ~ 0 −~ ~ ~ k0 ,K− k δ~ k,~ k 0 − δK k0 ,K− k b ~ 0 b↑,~ k − δ~ k,~ k0 b ↑,k

b ~ ~k ~ 0 −~ ↓,K k0 ↓,K−

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(7)

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Chapter 11

After summing over ~k and ~k 0 , and making use of the fact that the wave function is normalized to unity, we arrive at the last commutation relation in Eq. (11.56).

† Exercise 11.11.

The vertex describing the Fr¨ohlich interaction is given in Section 8.5 as  1/2 r e ~ωLO 1 1 M~k = − ; (1) k (∞) (0) 2V

thus the second-order matrix element in Eq. (11.62), represented by the diagram of Fig. 11.6, is   1 1 ~ωLO 2~ωLO e2 (2) − . (2) Vif = 2 k (∞) (0) 2V (∆E)2 − (~ωLO )2 This implies that without taking screening into account, the second order matrix element (11.62) is   1 e2 / e2 1 1 ~ωLO 2~ωLO Mk = + 2 − 2 V k k (∞) (0) 2V (∆E)2 − (~ωLO )2   2 2 (~ωLO ) 1 e / 1+ (3) = V k2 (∆E)2 − (~ωLO )2 under the assumption (0)  (∞) given in the problem. In Section 8.10 we computed the screening by constructing an infinite series of diagrams as shown in Fig. 8.13. We can construct a series the same way here for the phonon-mediated interaction, but using this revised interaction instead of the bare Coulomb interaction in the series of Fig. 8.13. Therefore we can use the same result (8.71), but substituting the present vertex, i.e. S=

Ce2 / 1 , 2 V k − (1/V )(Ce2 /)Πk

(4)

where C = 1 + (~ωLO )2 /((∆E)2 − (~ωLO )2 ) and Πk is the electron polarizability loop calculated in the same way as in Section 8.10. We can rewrite this in the form   1 e2 / (~ωLO )2 Mk = 1 + , (5) V k 2 + κ2ph (∆E)2 − (~ωLO )2 where κ2ph = −(1/V )(e2 /)Πk C. Note that C appears in the screening constant κph , which was not mentioned in the statement of the problem. This implies that in the low ∆E limit, the effective screening is weaker than the Thomas-Fermi formula for electron interaction alone, and in the long-wavelength limit (low exchanged k), the overall interaction can become repulsive, since Πk is typically negative in ∆E → 0 limit. For an alternate derivation of the Bardeen-Pines interaction, which accounts for screening effects on the ion motion of the optical phonons, see Leggett 2006, Appendix 5B.

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Exercise 11.12. Since the multiplicative factors are the same in both equations, we only need to show that X X X X φ∗ (~k)b↑,~k b↓,−~k h0| φ∗ (~k 0 )b↑,~k0 b↓,−~k0 φ(~ p0 )b†↓,−~p0 b†↑,~p0 |0i φ(~ p)b†↓,−~p b†↑,~p ~ k

=

~ k0

X

p ~0

p ~



 φ∗ (~k)φ∗ (~k 0 )φ(~ p)φ(~ p0 ) δ~k,~p0 δ~k0 ,~p + δp~,~k δp~0 ,~k0 − δ~k,~p δ~k,~p0 δp~,~p0 δ~k0 ,~p − δ~k,~p0 δ~k,~p δ~k0 ,~p0 δ~k0 ,~p .

~ k,~ k0 ,~ p,~ p0

(1) Now, h0|

X

φ∗ (~k)b↑,~k b↓,−~k

X

~ k

X

=

φ∗ (~k 0 )b↑,~k0 b↓,−~k0

X

~ k0

φ(~ p)b†↓,−~p b†↑,~p

X

φ(~ p0 )b†↓,−~p0 b†↑,~p0 |0i

p ~0

p ~

φ∗ (~k)φ∗ (~k 0 )φ(~ p)φ(~ p0 )h0|b↑,~k b↓,−~k b↑,~k0 b↓,−~k0 b†↓,−~p b†↑,~p b†↓,−~p0 b†↑,~p0 |0i

,

(2)

~ k,~ k0 ,~ p,~ p0

i.e., we have destruction and creation operators sandwiched between two vacuum states. In principle, we could put the operators in normal order using the anticommutation relations, but we can make the problem simpler by realizing that the factor will only be non-zero if all creation operators are paired, in other words, if for each particle created there is a destruction operator. Since we have got to take the spins into account, the possible pairings are the following: ~k = p~,

~k 0 = p~0

or ~k = p~0 ,

~k 0 = p~,

(3)

which give two terms with factors δ~k,~p δ~k0 ,~p0 and δ~k,~p0 δ~k0 ,~p . However, if p~0 = p~ or ~k 0 = ~k, there will be double creation of a particle in the same state, which gives zero. We must therefore substract off the terms for these two cases, which correspond to the last two terms with four Kronecker deltas in (1). Now, picking two sums (the other two can be treated in a similar fashion), X

φ∗ (~k)φ∗ (~k 0 )φ(~ p)φ(~ p0 )δ~k,~p0 δ~k0 ,~p =

~ k,~ k0 ,~ p,~ p0

X

φ∗ (~k)φ∗ (~k 0 )φ(~k)φ(~k 0 ) =

~ k,~ k0

X

|φ(~k)|2 |φ(~k 0 )|2 ,

(4)

~ k,~ k0

while X

φ∗ (~k)φ∗ (~k 0 )φ(~ p)φ(~ p0 )δ~k,~p δ~k,~p0 δp~,~p0 δ~k0 ,~p =

~ k,~ k0 ,~ p,~ p0

=

X ~ k,~ k0

X

φ∗ (~k)φ∗ (~k 0 )φ(~ p)φ(~ p)δ~k,~p δ~k,~p δ~k0 ,~p

~ k,~ k0 ,~ p

φ (~k)φ (k )φ(~k)φ(~k)δ~k,~k0 = ∗



~0

X

|φ(~k)| . 4

(5)

~ k

Obviously, this last sum is simply one term in the sum above, namely, the one for which ~k = ~k 0 , and as such, is much smaller than the sum in Eq. (4).

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Exercise 11.13. We will prove only that the operators in Eq. (11.75) obey fermionic anticommutation relation, for the case of Eq. (11.76) can be shown in the same way. By definition, † † ξ↑,~ ~ b↑,~ ~ b↓,−~ p , p = up p + vp

(1)

i.e., o n † † = ξ↑,~ p , ξ↑,~ p0 =

o n   up~ b†↑,~p + vp~ b↓,−~p , up~0 b†↑,~p0 + vp~0 b↓,−~p0 o n o n up~ up~0 b†↑,~p , b†↑,~p0 + up~ vp~0 b†↑,~p , b↓,−~p0 o n +up~0 vp~ b↓,−~p , b†↑,~p0 + vp~ vp~0 {b↓,−~p , b↓,−~p0 } = 0 .

(2)

Consequently, {ξ↑,~p , ξ↑,~p0 } = 0 . Finally, n o † ξ↑,~ p0 p , ξ↑,~

(3)

n

  o up~ b†↑,~p + vp~ b↓,−~p , up~0 b↑,~p0 + vp~0 b†↓,−~p0 n o n o = up~ up~0 b†↑,~p b↑,~p0 + up~ vp~0 b†↑,~p b†↓,−~p0 n o  +vp~ up~0 {b↓,−~p b↑,~p0 } + vp~ vp~0 b↓,−~p b†↓,−~p0 = u2p~ + vp~2 δp~,~p0 = δp~,~p0 =

Exercise 11.14. First, we have

(4)

Let us deal with the terms in the Hamiltonian in Eq. (11.77) one by one. b†↑,~p

† = up~ ξ↑,~ ~ ξ↓,−~ p p − vp

b†↓,~p

† = up~ ξ↓,~ ~ ξ↑,−~ p p + vp

b↑,~p

† = up~ ξ↑,~p − vp~ ξ↓,−~ p

b↓,~p

† = up~ ξ↓,~p + vp~ ξ↑,−~ p .

(1)

Hence, the first term in the Hamiltonian can be re-written in the new variables as    † † b†↑,~p b↑,~p = up~ ξ↑,~ up~ ξ↑,~p − vp~ ξ↓,−~ ~ ξ↓,−~ p p − vp p † † † † 2 = u2p~ ξ↑,~ p − vp ~ up ~ ξ↓,−~ p ξ↑,~ p − up ~ vp ~ ξ↑,~ p ξ↓,−~ ~ ξ↓,−~ p + vp p p ξ↑,~ p ξ↓,−~ | {z }

† 1−ξ↓,−~ ξ p p ↓,−~



† † † 2 † = vp~2 + up2~ ξ↑,~ p ξ↓,−~ p + ξ↑,~ p − vp p − vp ~ up ~ ξ↑,~ ~ ξ↓,−~ p p ξ↓,−~ p ξ↓,−~ p ξ↑,~



.

(2)

Similarly, b†↓,~p b↓,~p

=

   † † up~ ξ↓,~ + v ξ u ξ + v ξ p ~ ↑,−~ p p ~ ↓,~ p p ~ p ↑,−~ p

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219

† † † † 2 = u2p~ ξ↓,~ p + up ~ vp ~ ξ↓,~ ~ up ~ ξ↑,−~ p ξ↓,~ p + vp p ξ↑,−~ ~ ξ↑,−~ p ξ↓,~ p ξ↑,−~ p + vp p | {z }

† 1−ξ↑,−~ ξ p p ↑,−~



† † † 2 † = vp~2 + up2~ ξ↓,~ p + vp p + up ~ vp ~ ξ↓,~ p ξ↓,~ p ~ ξ↑,−~ p ξ↓,~ p ξ↑,−~ p ξ↑,−~ p + ξ↑,−~



.

(3)

By collecting the terms in question, we arrive at the assertion of the problem.

Exercise 11.15. factors) as

From Eq. (11.85), the density of states is given (up to irrelevant multiplicative D(E) =

where E~k =

~2 k 2 −µ 2m

1 k 2 (E) , |∇~k E| and

E=

q

(1)

∆2 + E~2 .

(2)

k

These two relations also imply E~k =

p

E 2 − ∆2 ,

(3)

and

  2m p 2m 2 − ∆2 + µ E + µ = E . ~ k ~2 ~2 Since the dispersion relation is spherically symmetric, we can take the gradient as k2 =

|∇~k E|

=

=

dE~k2 dE~ 1 dq 2 1 ∆ + E~2 = q =q E~k k k dk dk 2 ∆2 + E~2 dk ∆2 + E~2 k k qp q 2p 2 2p 2 E − ∆2 E~k + µ = E − ∆2 E 2 − ∆2 + µ . E E

(4)

(5)

Putting these results together yields  E 1 1 mE 2m p 2 2+µ = p√ D(E) = √ E − ∆ 2 2 2 2 2 2 E −∆ ~2 E −∆ +µ ~

p√

E 2 − ∆2 + µ √ E 2 − ∆2

(6)

Since m and ~ are again multiplicative constants, we drop them. Obviously, the density of states is not defined for |E| < ∆. (Or, to be more precise, the density of states is zero for that interval.) Thus the integral we have to calculate is Z eV Z eV p√ 2 E E − ∆2 + µ √ I(V ) = D(E)dE = dE . (7) E 2 − ∆2 ∆ ∆ This is implemented in the Mathematica code at the end of the chapter. The resulting current profile is given in Fig. 11.3.

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Chapter 11

2

I [a.u.]

1

0

-1

-2 -4

-3

-2

-1

0

1

2

3

4

V [a.u.] Figure 11.3: The dependence of the current on the applied voltage for Exercise 11.15, as given by the density of states in Eq. (11.85)

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221

Exercise 11.16. Let us fix the units in such a way that the prefactor in Eq. (11.87), U0 D(EF )/V = 1. Let also kB = 1. Then the self-consistent equation for ∆(T ) is  q Z 1 1 Ec 2 + E 2 /2T dEk p 2∆ , (1) 1= tanh k 2 0 2∆2 + Ek2 where the value of Ec is determined by the condition that at T = 0, i.e., when the tanh function tends to 1, Z 1 Ec 1 1= dEk p . (2) 2 2 0 2∆ (0) + Ek2 In this particular case, the integral can analytically be calculated, yielding ! √ Z 1 1 2Ec 1 Ec −1 dEk p = sinh , 1= 2 0 2 2∆(0) 2∆2 (0) + Ek2 i.e., Ec =



2∆(0) sinh 2 .

(3)

(4)

(Note that with these units, the value of Tc is given by the condition ∆(Tc ) = 0, i.e., Z Ec dEk 2= tanh (Ek /2Tc ) , Ek 0

(5)

where Ec is already fixed, as we discussed above.) The complete integral itself can be brought to a simpler form by the change of variable Ek , x= √ 2∆

(6)

when the self-consistent equation becomes Z 2= 0

E √ c 2∆

dx √ tanh 1 + x2



2∆ p 1 + x2 2T

! ,

(7)

which is then calculated as a function of ∆ and T . Since at a particular temperature T is simply a parameter, we solve for ∆. In the implementation presented in the appendix, we use the FindRoot function. The gap as a function of the temperature is shown in Fig. 11.4.

Exercise 11.17. rendered as

~ is zero, thus Eq. (11.94) can be Inside the superconductor the current, J, ~. ~∇θ = q A

Integration along a closed loop then yields I I ~ ~ · d~l. ~∇θ · dl = q A

(1)

(2)

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1

∆(T)/∆(0)

0.8 0.6 0.4 0.2 0 0

0.2

0.4

0.6

0.8

1

Temperature [a.u.] Figure 11.4: The dependence of the gap on the temperature for Exercise 11.16, as dictated by Eq. (11.87)

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223

Now, by virtue of Stoke’s theorem, the integral on the right hand side can be transformed into a ~ i.e., surface integral over the curl of A, I Z ~ · d~l = q q (3) A d~a · ∇ × A = qΦ , | {z } S ~ B

while the integral on the left hand side is equal to an integer multiple of 2π, since I ~∇θ · d~l = ~(θ2 − θ1 ) ,

(4)

where θ1 and θ2 are the values of the phase at the two “end points” of any parametrization of the curve. However, by the very definition of the contour integral, those two end points describe the same physical place, thus the quantum mechanical phase can be off by an integer of 2π only. Therefore, I ~∇θ · d~l = 2π~m = hm , (5) where m is an integer. Putting everything together, 2πm = qΦ ,

(6)

or

~m h = m, q 2e where we accounted for the fact that the charge is 2e, given that we have pairs of electrons. Φ=

(7)

Exercise 11.18. Here we can make use of the fact that the density is equal to the square of the wave function, for the wave function is normalized in such a way that its absolute value integral over space yields the number of particles. The mean-field solution gives b=−

a , |Ψ|2

(1)

and from Eq. (11.100) s Hc =

a2 = µ0 b

s

a|Ψ|2 . µ0

(2)

Thus, Bc

p 4π · 10−9 V · s2 /(C · cm) × 5 · 10−2 eV × 1022 cm−3

=

µ0 Hc =

=

p 4π · 10−9 V · s2 /(C · cm) × 5 · 10−2 V × 1022 cm−3 (1.6 · 10−19 C)

=

0.001

=

10 T.

(3)

V· s cm2 (4)

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224

Chapter 11

Typical laboratory magnets can produce a magnetic field of about 10-15 T, while the best DC magnets are capable of producing a field of 50 T.

Exercise 11.19.

a) Using the relationship  q  ~ , ~n0 ∇θ − qn0 A J~ = m

(1)

and the fact that the current is zero, we infer that ∇θ =

q~ A. ~

(2)

When integrating the equation above around a closed loop in the superconductor, we have to watch out for the discontinuity in ∇θ at the two junctions. Thus, the integral can be written as I Z b1 Z a2 ∇θ · d~l = ∇θ d~l + (θb2 − θb1 ) + ∇θ d~l + (θa1 − θa2 ) a1 b1

Z =

b2

∇θ · d~l +

Z

a1

a2

b2

∇θ · d~l + (θb2 − θb1 ) + (θa1 − θa2 ) | {z } | {z } ∆θ1

I =

∇θ · d~l + ∆θ1 + ∆θ2 = ∆θ1 + ∆θ2 ,

∆θ2

(3)

regular

where a and b denote the two junctions, and 1 and 2 indexes the two sides of the junctions. The last contour integral vanishes, because we do not have vortices enclosed by the loop. On the other hand, integrating the right hand side of Eq. (2), we have I Z Z q ~ · d~l = q ~ d~a = q ~ d~a = q ΦA . A (∇ × A) B (4) ~ ~ A ~ A ~ Therefore,

q ΦA − ∆θ1 . (5) ~ b) The current through the junctions is related to the phase difference between the two sides, namely, Ia = I0 sin ∆θ1 (6) ∆θ2 =

and Ib = −I0 sin ∆θ2 .

(7)

(The negative sign is the consequence of the direction of the current flow.) Thus, the total current is given by     ∆θ1 + ∆θ2 ∆θ1 − ∆θ2 cos Itot = Ia + Ib = I0 (sin ∆θ1 − sin ∆θ2 ) = 2I0 sin 2 2         eΦA ∆θ1 − ∆θ2 eΦA eΦA = 2I0 cos sin = 2I0 cos sin ∆θ1 − . (8) ~ 2 ~ 2~ c Copyright 2009 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

D.W. Snoke, Solid State Physics: Essential Concepts

225

The maximum value of the total current is controlled by the cos eΦA /~ factor, which oscillates as e a function of the flux, with being the scaling factor. ~



Exercise 11.20.

From Eq. (11.119), A is defined as A=

e2 hxi2 N GT − 1 e2 hxi2 T (GT − 1) T =n . ~V GT + 1 GT + 1

(1)

(Note typo in book; there should be an ~ in the denominator.) Thus, the condition A/0  1 translates into GT + 1 1 ~0 . (2) n GT − 1 T e2 |hxi|2 hxi can be expressed as hxi = hc|x|vi = i

hc|p|vi , mω0

(3)

i.e., |hxi|2 =

|hpi|2 . m2 ω02

(4)

However, by definition, the oscillator strength (which we assume to be equal to unity) is f=

2|hpi|2 =1, mEg

(5)

which means that

mEg . 2 By plugging this result into the exression for |hxi|2 , we get |hpi|2 =

|hxi|2 =

~ mEg = . 2m2 ω02 2ω0 m

(6)

(7)

Thus

GT + 1 1 20 ~ω0 m . (8) GT − 1 T e2 ~ Assuming that the first fraction is of order unity, and the lifetime is a couple of ps, we arrive at n

n



= =

1 10−12 1 10−12

2 × 8.9 · 10−14 A · s/(V · cm) × 2 eV × 5.1 · 105 eV/c2 s e2 × 6.6 · 10−16 eV · s

(9)

2 × 8.9 · 10−14 × 6.2 · 1018 e/(V · cm) × 2 eV × 5.1 · 105 eV s e2 × 6.6 · 10−16 eV · s × (3 · 1010 cm/s)2

1.9 · 1018 cm−3 .

(10)

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Chapter 11

Mathematica source codes ChXI-5.nb solution1 = NDSolve[ {D[f[y],y]]== f1[y], D[f1[y],y]==f[y]^3 - f[y]*(1 - 1/y^2) - f1[y]/y, f[ .01] == 0.01, f1[ 0.01] == .166517}, {f, f1}, {y, 0.01, 10} ]; s1 = N[Table[{y, 1*(f[y] /. solution1)?1?}, {y, 0.01,10, .1}]]; ListPlot[s1,PlotRange -> {0, 2}] ChXI-9.nb Clear[A, Dist] A[k_, U_] = 1/U*(k^2 + U - Sqrt[k^4 + 2*k^2*U]) Dist[k_, U_] = k^2*A[k, U]*A[k, U]/(1 - A[k, U]*A[k, U]) Plot[Dist[k, 11.1], {k, 0, 10}, PlotLabel -> {"k", "Occupation"}] ChXI-15.nb (* Functions are defined here *) Clear[En,Delta,mu] Density[En_,mu_,Delta_] := En*Sqrt[Sqrt[En^2-Delta^2]+mu]/Sqrt[En^2-Delta^2] f[En_,mu_] := 1/(1+Exp[En-mu]) (* We fix the temperature T=1 *) Current[V_,mu_,Delta_] := NIntegrate[Density[En,mu,Delta]]*f[En,m],{En,Delta,V}] Plot[{Current[V,1,0.02]},{V,1,4}] ChXI-16.nb Clear[Delta0, Ec, Delta, n] n = 0 Delta0 = 1 Ec = Sqrt[2]*Delta0*Sinh[2] Int[T_, Delta_] := NIntegrate[ 1/Sqrt[1 + x^2]*Tanh[Sqrt[2]*Delta/(2*T)*Sqrt[1 + x^2]], {x, 0, Ec/Sqrt[2]/Delta}] For[t = 0.01, t < 1.30, t = t + 0.01; vec[n] = {t, y /. Table[FindRoot[2 == Int[t, y], {y, 2}]]}; n++; ] (list = Table[vec[k], {k, 0, n - 1}]); ListPlot[list]

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