Solutions to Problems Fundamentals of Condensed Matter Physics · PDF fileSolutions to...

Solutions to ProblemsFundamentals of Condensed Matter Physics

Marvin L. CohenUniversity of California, Berkeley

Steven G. LouieUniversity of California, Berkeley

c© Cambridge University Press 2016

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Full file at http://TestBankSolutionManual.eu/Solution-Manual-for-Fundamentals-of-Condensed-Matter-Physics-by-Cohen

AcknowledgementThe authors thank Mr. Meng Wu, together with Mr. Felipe H. da Jornada,

Mr. Fangzhou Zhao and Mr. Ting Cao, for their invaluable help in preparingthis problem solutions manual.

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http://TestBankSolutionManual.eu/Solution-Manual-for-Fundamentals-of-Condensed-Matter-Physics-by-Cohen

Sec. II.1. Crystal structure of MgB2.

(a) The unit cell and Wigner-Seitz cell are displayed in Fig. 1.

a1 a1

x

y z

a3

a2

Mg

B1 B2

B1

B1

B2

B2

Top view

Wigner-Seitz cell

Unit cell

Unitcell

Wigner-Seitzcell

Side view

Mg Mg

Mg

B B B

Figure 1: Unit cell and Wigner-Seitz cell of MgB2

If a is the B-B distance, then |a1| = |a2| = a√

3. We will choose the following convention forthe lattice vectors,

a1 =(

32a, −

√3

2 a, 0)

a2 =(

32a,√

32 a, 0

)a3 = (0, 0, c) ,

(1)

in Cartesian coordinates.(b) The unit cell volume is Ωprim = |(a1 × a2) · a3| = 3

√3

2 a2c, and the Brillouin zone volume isΩBZ = (2π)3 /Ωprim. The reciprocal lattice vectors are given by bi = 2π

Ωprimaj × ak εijk, where

εijk is the Levi-Civita symbol, or

b1 = 2πΩprim

(√3

2 ac, −32ac, 0

)

b2 = 2πΩprim

(√3

2 ac,32ac, 0

)

b3 = 2πΩprim

(0, 0, 3

√3

2 a2),

(2)

in Cartesian coordinates. Note that other reciprocal lattice vectors can be obtained dependingon the orientation of the real-space lattice vectors.The reciprocal lattice vectors are display in Fig. 2: Using the convention from our unit cell,the atoms are at τMg = a3/2, τB1 = a1/3 + a2/3, τB2 = 2a1/3 + 2a2/3.

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kx

ky

kz

b1

b3

b2

b2

Brilluoinzone

Brilluoin zone

Reciprocalcell

Reciprocalcell

Top view Side view

Figure 2: Brillouin zone

(c) There are 24 point symmetry elements in MgB2 that leave the crystal structure invariant.There are 12 in-plane symmetry elements: 6 rotation operations, including the identity oper-ation: C6, 2C6, · · · , 6C6 = I. There are also 6 in-plane mirror symmetries that go along the Batoms or between the B atoms, including, for instance, the mirror symmetry along the x = 0and y = 0 planes, denoted by σx and σy, respectively. One can also compose any of these 12in-plane symmetry elements with a mirror operation along the z = 0 plane, σz. This givesthe total of 24 symmetry elements. Note that this system has inversion symmetry, which isequivalent to σzC2.

(d) The irreducible part of the Brillouin zone is sketched below. It can be easily verified that, ifwe apply all the 24 symmetry elements, we will recover the full Brillouin zone.

Irr. BZ Irr BZ

Top view Side view

kx

ky kz

Figure 3: Irreducible BZ

I.2. The GaN crystal. This problem is similar to Problem I.1, except that a here is the lengthof the in-plane lattice vectors. The details of the computation will be omitted.

(a) We will choose a different orientation for the unit cell vectors with respect to problem I.1. The

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lattice vectors are given by

a1 =(

12a, −

√3

2 a, 0)

a2 =(

12a,√

32 a, 0

)a3 = (0, 0, c) ,

(3)

in Cartesian coordinates. Note that, in real space, the Wigner-Seitz cell is a hexagon rotated90 relative to that in problem I.1.The unit cell volume is Ω = |(a1 × a2) · a3| =

√3

2 a2c, and the reciprocal lattice vectors are,

b1 = 2πa

(1, − 1√

3, 0)

b2 = 2πa

(1, 1√

3, 0)

b3 = 2πc

(0, 0, 1) ,

(4)

in Cartesian coordinates. The Brillouin zone volume is ΩBZ = 16π3√

3a2c. The Brillouin zone is

sketched in Fig. 4.

kx

ky

kz

b1

b3

b2

b2

Brilluoinzone

Brilluoin zone

Top view Side view

Figure 4: BZ

(b) If we inspect the unit cell of GaN. which is enclosed by solid lines, we see that there are 2 Natoms (1 spread throughout the corners, and 1 inside), and there are 2 Ga atoms inside (1spread along the edges of the unit cell, and 1 completely inside the cell). So, there are 4 atomsin the unit cell.

(c) A top view of the structure is displayed in Fig. 5.There are 6 point symmetry elements that do not involve glide or screw axes: 3 mirror planesalong the atoms that form the hexagons, and 3 C3 rotation operations (including identity),performed around the center of the hexagons.

(d) Ga has 3 valence electrons, and N has 5 valence electrons. So, one cell has 16 electrons, or8 filled bands. There are no half-filled bands. so GaN is an insulator. So, it’s not surprisingthat this material has a bandgap.

I.3. Born-Oppenheimer approximation for a molecule. Think about the simplest crudemodel that captures the physics of this problem. Imagine that we have two rigid ions with massM

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and effective charge +Q each, separated by a distance x. Consider that you put a single effectivepoint charge of mass me and charge −2Q between the two ions, which corresponds to the effectivecharge created by the bond. At rest, we know that x = a.The electronic contribution to the energy can be written as a sum over kinetic energy and potentialenergy, and we assume that the kinetic energy comes essentially from quantum confinement viathe uncertainty principle ∆p∆x ∼ ~,

E(x) = T (x) + U(x) (5)

T (x) = p2

2me∼ ~2

2mex2 (6)

U(x) = Q2

x+ 2Q(−2Q)

x/2 = −Q2

x(7)

E(x) = ~2

mex2 −Q2

x(8)

Our model has a free parameter, Q. But we have a constraint: E′(x = a) = 0 =⇒ Q2 = ~2

ame. So,

E(x) = ~2

me

[1

2x2 − 1xa

].

The electronic energy is the total energy at x = a, so Eel ∼ − ~2

2mea2 .The vibrational energy can be obtained if we realize that the total energy landscape as a functionof x can be approximated by a quadratic curve near x = a. This gives rise to a harmonic potential,where the vibrational modes are quantized, Evib = ~ω. But since we have a harmonic oscillator,ω =

√k/M with k = E′′(x = a). Thus, Evib ∼ ~2

a2√Mme.

Finally, the rotational energy can be calculated if we assume that the rotational motion is quan-tized. We get that Erot = ~2`(`+1)

2I with I = M(a/2)2. This gives Erot ∼ 2~2

Ma2 .So, the ratio of the electronic to vibrational to rotational energy is given by Eel : Evib : Erot ≈1 :√

meM : meM . The consequence of this finding is that, since M me, we can separate the three

degrees of freedom, and calculate the contribution to the vibrational and rotational degrees offreedom as a perturbation to the electronic one. This validates the use of the Born-Oppenheimerapproximation, which treats the ions as fixed particles when solving for the electronic contribution.

Figure 5: Crystal structure of GaN

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I.4. Hartree-Fock approximation. Let us first derive some useful identities related to Slaterdeterminants. Let’s define the antisymmetrization operator A as

A = 1N !

∑P

(−1)P P , (9)

where N is the number of particles and occupied orbitals, P is a permutation, and P is theoperator associated to the permutation P . If the overall permutation contains an odd numberof pairwise permutation, (−1)P = −1, otherwise the permutation is even and (−1)P = 1. Theidentity permutation is even. For instance, consider the following permutations and their signs:

P = (1, 2, 3, 4, . . . , N)→ (−1)P = 1P = (2, 1, 3, 4, . . . , N)→ (−1)P = −1P = (2, 3, 1, 4, . . . , N)→ (−1)P = 1

(10)

We denote the trial Hartree-Fock wavefunction in a shorthand notation,

ΨHF(1, 2, . . . , N) =√N ! A φ1(1)φ2(2) . . . φN (N), (11)

where φ is a single-particle orbitals, the subscript i in φi(j) denotes the orbital index, and theargument j denotes the space-time coordinate. We take a spinless wavefunction for the sake ofsimplicity, although the proof here can be trivially extended to the case with spin. We define thepermutation operator P acting on Ψ by permuting the orbitals indices,

P φ1(1)φ2(2) . . . φN (N) = φP1(1)φP2(2) . . . φPN (N) (12)

The following properties can be easily proved for A,

A2 = A, A† = A (13)

In addition, A commutes with any operator that preserves particle indistinguishability, such asthe crystal Hamiltonian,

[A, Hxtal

]= 0. In particular, we can show that:A, ∑

i1,...,iM

f(i1, . . . , iM )

= 0 (14)

We can proof a couple of useful properties involving A:

(a) 〈ΨHF|ΨHF〉 = 1(b) 〈ΨHF|

∑i h(i)|ΨHF〉 =

∑i〈φi|h(1)|φi〉

(c) 〈ΨHF|∑ij g(i, j)|ΨHF〉 =

∑ij [〈φiφj |g|φiφj〉 − 〈φiφj |g|φjφi〉]

We will prove relation 3 in the following:

〈ΨHF|∑ij

g(i, j)|ΨHF〉 = N ! 〈Aφ1(1) . . . φN (N)|∑ij

g(i, j)|Aφ1(1) . . . φN (N)〉

= N !∑ij

〈φ1(1) . . . φN (N)|g(i, j)|Aφ1(1) . . . φN (N)〉

=∑ij

∑P

(−1)P 〈φ1(1) . . . φN (N)|g(i, j)|φP1(1) . . . φPN (N)〉

(15)

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We will now particularize to the case where i = 1 and j = 2, Since g(1, 2) doesn’t depend on anycoordinate larger than 2 and the orbitals are orthogonal to each other, we can separate the orbitalsinto two sets, one with indices 1 or 2, where g(1, 2) will act upon, and another set with indiceslarger than 2. There can be no cross-terms between orbitals because they are orthogonal. Thisway, we get:

〈ΨHF|g(1, 2)|ΨHF〉 =∑P

(−1)P 〈φ1(1) . . . φN (N)|g(1, 2)|φP1(1) . . . φPN (N)〉

=∑P

(−1)P 〈φ1(1)φ2(2)|g(1, 2)|φP1(1)φP2(2)〉

=∑P

(−1)P 〈φ1(1)φ2(2)|g(1, 2)|φP1(1)φP2(2)〉

× 〈φ3(3)|φP3(3)〉 · · · 〈φN (N)|φPN (N)〉=∑P

(−1)P 〈φ1(1)φ2(2)|g(1, 2)|φP1(1)φP2(2)〉 δ3P3 · · · δN PN

(16)

Now, there are only two permutations such that Pi = i for i > 2, namely:

P = (1, 2, 3, 4, . . . , N)→ (−1)P = 1P = (2, 1, 3, 4, . . . , N)→ (−1)P = −1

(17)

So,〈ΨHF|g(1, 2)|ΨHF〉 = 〈φ1(1)φ2(2)|g(1, 2)|φ1(1)φ2(2)〉

− 〈φ1(1)φ2(2)|g(1, 2)|φ2(1)φ1(2)〉.= 〈φ1φ2|g|φ1φ2〉 − 〈φ1φ2|g|φ2φ1〉,

(18)

where we assume that the coordinate indices (e.g., (1) and (2)) follow the same order in the bras,kets, and in the operators. A generalization for other orbital indices give relation 3, as desired.Relations 1 and 2 are easier cases that can be proved in the same fashion, but in those cases onlythe identity permutation yields a non-zero contribution to the matrix element.Now, we can write the crystal Hamiltonian as

Hxtal =N∑i

h(i) +N∑i 6=j

g(i, j), (19)

where, in Rydberg atomic units,

h(i) = − 12m∇

2i + Vext(ri)

g(i, j) = 12−e2

|ri − rj |.= 1

2v(i, j)(20)

In order to find the Hartree-Fock equations, we wish to minimize the expectation value of the totalenergy with respect to the single-particle orbitals that make up the Hartree-Fock wavefunction,with the constraint that the orbitals are orthonormal, which can be introduced via a Lagrangemultiplier,

L = 〈ΨHF|Hxtal|ΨHF〉 − λ (〈ΨHF|ΨHF〉 − 1) (21)

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We set δL = 0 and obtain,

δ

N∑i

〈φi|h|φi〉+ 12

N∑ij

[〈φiφj |v|φiφj〉 − 〈φiφj |v|φjφi〉]

− δ

N∑i

〈φi|φi〉

= 0 (22)

Note that we don’t have to impose orthogonality constraint because, for small perturbations inthe orbitals, the change in the orthogonality will be quadratic in |δφi〉. Then,

N∑i

〈δφi|h|φi〉 − λN∑i

〈δφi|φi〉+ 12

N∑ij

[〈δφiφj |v|φiφj〉 − 〈δφiφj |v|φjφi〉]

+ 12

N∑ij

[〈φiδφj |v|φiφj〉 − 〈φiδφj |v|φiφj〉] + h.c. = 0(23)

Note that the double sums can be combined into a single sum, since v(1, 2) = v(2, 1). We seekcritical solutions for any orbitals. Because the linear independence of the orbitals, we only needto solve the linear system for the bras. For a given orbital with index i, we obtain

h|φi〉+N∑j

[〈φj(2)|v|φi(1)φj(2)〉 − 〈φj(2)|v|φj(1)φi(2)〉]− λ|φi〉 = 0. (24)

This equation can be identified as the Hartree-Fock equation,

h|φi〉+ F|φi〉 = εi|φi〉, (25)

where we define the Fock operator as,

F |φi〉.=

N∑j

[〈φj(2)|v|φi(1)φj(2)〉 − 〈φj(2)|v|φj(1)φi(2)〉] . (26)

In real space, the Hartree-Fock equation reads,

− 12m∇

2φi(x) + Vext(x)φi(x)− e2N∑j=1

[∫d3x′|φj(x′)|2 φi(x)|x− x′|

−∫

d3x′φ∗j (x′)φi(x′)φj(x)

|x− x′|

]= εiφi(x).

(27)The total energy of the system is not the sum of the orbital eigenvalues εi, but rather

E = 〈ΨHF|Hxtal|ΨHF〉

=N∑i

〈φi|h|φi〉+ 12

N∑ij

[〈φj(2)|v|φi(1)φj(2)〉 − 〈φj(2)|v|φj(1)φi(2)〉]

=N∑i

〈φi|h|φi〉+ 12

N∑i

⟨φi∣∣∣F ∣∣∣φi⟩

=N∑i

εi −12

N∑i

⟨φi∣∣∣F ∣∣∣φi⟩

(28)

For the second part of the question, we assume that we promote one electron from an occupiedorbital a to an unoccupied orbital b, i.e., we assume that single-particle orbitals remain frozen, and

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that orbital a, which was previously occupied, becomes unoccupied, and that orbital b, which aspreviously unoccupied, is now occupied. We need to assume that single-particle orbitals are frozen:if we solve the HF equations self-consistently, we always arrive at the ground-state orbitals). Theexcitation energy ∆E is the energy difference from the ground-state energy, E0, to the excitedstate energy, Ea→b.The energy difference is composed of two parts: ∆E1, which depends on h(1), and ∆E2, whichdepends on g(1, 2). It is clear that ∆E1 = 〈φb|h|φb〉 − 〈φa|h|φa〉. The change ∆E2 is a little bitharder to compute as the whole Fock operator changes as an electron gets promoted,

∆E2 = 12

N∑j

[〈φbφj |v|φbφj〉 − 〈φbφj |v|φjφb〉]

− 12

N∑j

[〈φaφj |v|φaφj〉 − 〈φaφj |v|φjφa〉](29)

The excitation energy, within the Hartree-Fock approximation, is therefore

∆Ea→b = ∆E1 + ∆E2 = (εb − εa)−12

N∑j

[〈φbφj |v|φbφj〉 − 〈φbφj |v|φjφb〉]

+12

N∑j

[〈φaφj |v|φaφj〉 − 〈φaφj |v|φjφa〉] .(30)

Note that, within the HF approximation, the excitation energy is not just the difference between theeigenvalues, but it also contains a correction term, which can be understood as an approximationto the exciton binding energy.

I.5. Born-von Karman boundary condition.

(a) See Fig. 6.

(a) (b) (c)

E(k)

-π/a-2π/a-3π/a π/a 2π/a 3π/a

E(k)


E(k)


Figure 6: Simple bandstructure diagrams for a one dimensional periodic solid in the limit of V (r)→0 expressed in the extended zone (a), repeated zone (b), and reduced zone (c) scheme.

(b) ∑k

1 = Ω(2π)3

∫d3k = ΩΩBZ

(2π)3 = ΩΩc

= N (31)

I.6. Energy bands of elemental solids.

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