Solutions to Problems Fundamentals of Condensed Matter Physics Marvin L. Cohen University of California, Berkeley
Steven G. Louie University of California, Berkeley
c Cambridge University Press 2016
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Acknowledgement The authors thank Mr. Meng Wu, together with Mr. Felipe H. da Jornada, Mr. Fangzhou Zhao and Mr. Ting Cao, for their invaluable help in preparing this problem solutions manual.
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Sec. I I.1. Crystal structure of MgB2 . (a) The unit cell and Wigner-Seitz cell are displayed in Fig. 1. Top view
a2
B2
B1
Side view
Unit cell
a3
Mg B2
B1
Mg
Unit cell
Mg
Wigner-Seitz cell B1
B2
y
B
a1
B
B
a1
z
Wigner-Seitz cell
x
Mg
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 for the lattice vectors, ! √ 3 3 a1 = a, − a, 0 2 2 ! √ (1) 3 3 a2 = a, a, 0 2 2 a3 = (0, 0, c) , in Cartesian coordinates.
√
(b) The unit cell volume is Ωprim = |(a1 × a2 ) · a3 | = 3 2 3 a2 c, and the Brillouin zone volume is 2π ΩBZ = (2π)3 /Ωprim . The reciprocal lattice vectors are given by bi = Ωprim aj × ak ijk , where ijk is the Levi-Civita symbol, or ! √ 2π 3 3 b1 = ac, − ac, 0 Ωprim 2 2 ! √ 3 2π 3 b2 = ac, ac, 0 (2) Ωprim 2 2 ! √ 2π 3 3 2 b3 = 0, 0, a , Ωprim 2 in Cartesian coordinates. Note that other reciprocal lattice vectors can be obtained depending on 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. 3
Top view
Side view
b2
kz b3
Reciprocal cell
Brilluoin zone
b1
Reciprocal cell
Brilluoin zone
ky
b2
kx
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 operation: C6 , 2C6 , · · · , 6C6 = I. There are also 6 in-plane mirror symmetries that go along the B atoms or between the B atoms, including, for instance, the mirror symmetry along the x = 0 and y = 0 planes, denoted by σx and σy , respectively. One can also compose any of these 12 in-plane symmetry elements with a mirror operation along the z = 0 plane, σz . This gives the total of 24 symmetry elements. Note that this system has inversion symmetry, which is equivalent to σz C2 . (d) The irreducible part of the Brillouin zone is sketched below. It can be easily verified that, if we apply all the 24 symmetry elements, we will recover the full Brillouin zone. Top view
Side view
ky Irr. BZ
kz Irr BZ
kx
Figure 3: Irreducible BZ I.2. The GaN crystal. This problem is similar to Problem I.1, except that a here is the length of 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 = a2 =
! √ 1 3 a, − a, 0 2 2 ! √ 1 3 a, a, 0 2 2
(3)
a3 = (0, 0, c) , in Cartesian coordinates. Note that, in real space, the Wigner-Seitz cell is a hexagon rotated 90◦ relative to that in problem I.1. √ The unit cell volume is Ω = |(a1 × a2 ) · a3 | = 23 a2 c, and the reciprocal lattice vectors are, 2π a 2π b2 = a 2π b3 = c b1 =
1 1, − √ , 0 3 1 √ 1, ,0 3
(4)
(0, 0, 1) ,
in Cartesian coordinates. The Brillouin zone volume is ΩBZ = sketched in Fig. 4. Top view
16π 3 √ . 3a2 c
The Brillouin zone is
Side view b3
kz
b2 ky
Brilluoin zone
kx
Brilluoin zone
b1
b2
Figure 4: BZ (b) If we inspect the unit cell of GaN. which is enclosed by solid lines, we see that there are 2 N atoms (1 spread throughout the corners, and 1 inside), and there are 2 Ga atoms inside (1 spread along the edges of the unit cell, and 1 completely inside the cell). So, there are 4 atoms in 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 planes along 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, or 8 filled bands. There are no half-filled bands. so GaN is an insulator. So, it’s not surprising that this material has a bandgap. I.3. Born-Oppenheimer approximation for a molecule. Think about the simplest crude model that captures the physics of this problem. Imagine that we have two rigid ions with mass M 5
and effective charge +Q each, separated by a distance x. Consider that you put a single effective point charge of mass me and charge −2Q between the two ions, which corresponds to the effective charge 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 potential energy, and we assume that the kinetic energy comes essentially from quantum confinement via the uncertainty principle ∆p∆x ∼ ~, E(x) = T (x) + U (x) p2 ~2 T (x) = ∼ 2me 2me x2 2 Q Q(−2Q) Q2 U (x) = +2 =− x x/2 x 2 2 ~ Q E(x) = − 2 me x x Our model has a free parameter, Q. But we have a constraint: E 0 (x = a) = 0 =⇒ Q2 = E(x) =
~2 me
h
1 2x2
−
1 xa
i
(5) (6) (7) (8) ~2 ame .
So,
. 2
The electronic energy is the total energy at x = a, so Eel ∼ − 2m~e a2 . The vibrational energy can be obtained if we realize that the total energy landscape as a function of x can be approximated by a quadratic curve near x = a. This gives rise to a harmonic potential, where are quantized, Evib = ~ω. But since we have a harmonic oscillator, pthe vibrational modes 2 00 ω = k/M with k = E (x = a). Thus, Evib ∼ a2 √~M m . e
Finally, the rotational energy can be calculated if we assume that the rotational motion is quan2 2~2 with I = M (a/2)2 . This gives Erot ∼ M tized. We get that Erot = ~ `(`+1) . 2I a2 So, q the ratio of the electronic to vibrational to rotational energy is given by Eel : Evib : Erot ≈ me e 1: m M : M . 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 of freedom as a perturbation to the electronic one. This validates the use of the Born-Oppenheimer approximation, 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 Slater determinants. Let’s define the antisymmetrization operator A as 1 X (−1)P Pˆ , Aˆ = N! P
(9)
where N is the number of particles and occupied orbitals, P is a permutation, and Pˆ is the operator associated to the permutation P . If the overall permutation contains an odd number of pairwise permutation, (−1)P = −1, otherwise the permutation is even and (−1)P = 1. The identity permutation is even. For instance, consider the following permutations and their signs: P = (1, 2, 3, 4, . . . , N ) → (−1)P = 1 P = (2, 1, 3, 4, . . . , N ) → (−1)P = −1
(10)
P
P = (2, 3, 1, 4, . . . , N ) → (−1) = 1 We denote the trial Hartree-Fock wavefunction in a shorthand notation, √ ΨHF (1, 2, . . . , N ) = N ! Aˆ 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 the argument j denotes the space-time coordinate. We take a spinless wavefunction for the sake of simplicity, although the proof here can be trivially extended to the case with spin. We define the permutation operator Pˆ acting on Ψ by permuting the orbitals indices, Pˆ φ1 (1)φ2 (2) . . . φN (N ) = φP1 (1)φP2 (2) . . . φPN (N )
(12)
ˆ The following properties can be easily proved for A, Aˆ† = Aˆ
ˆ Aˆ2 = A,
(13)
In addition, Aˆ commutesh with any i operator that preserves particle indistinguishability, such as ˆ Hxtal = 0. In particular, we can show that: the crystal Hamiltonian, A,
ˆ A,
X
fˆ(i1 , . . . , iM ) = 0
(14)
i1 ,...,iM
ˆ We can proof a couple of useful properties involving A: (a) hΨHF |ΨHF i = 1 P ˆ P ˆ (b) hΨHF | i h(i)|Ψ HF i = i hφi |h(1)|φi i (c) hΨHF |
P
ij
gˆ(i, j)|ΨHF i =
P
ij
[hφi φj |ˆ g |φi φj i − hφi φj |ˆ g |φj φi i]
We will prove relation 3 in the following: hΨHF |
X
ˆ 1 (1) . . . φN (N )| gˆ(i, j)|ΨHF i = N ! hAφ
ij
X
ˆ 1 (1) . . . φN (N )i gˆ(i, j)|Aφ
ij
= N!
ˆ 1 (1) . . . φN (N )i hφ1 (1) . . . φN (N )|ˆ g (i, j)|Aφ
X ij
=
XX ij
(−1)P hφ1 (1) . . . φN (N )|ˆ g (i, j)|φP1 (1) . . . φPN (N )i
P
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(15)
We will now particularize to the case where i = 1 and j = 2, Since gˆ(1, 2) doesn’t depend on any coordinate larger than 2 and the orbitals are orthogonal to each other, we can separate the orbitals into two sets, one with indices 1 or 2, where gˆ(1, 2) will act upon, and another set with indices larger than 2. There can be no cross-terms between orbitals because they are orthogonal. This way, we get: hΨHF |ˆ g (1, 2)|ΨHF i =
X
=
X
=
X
(−1)P hφ1 (1) . . . φN (N )|ˆ g (1, 2)|φP1 (1) . . . φPN (N )i
P
(−1)P hφ1 (1)φ2 (2)|ˆ g (1, 2)|φP1 (1)φP2 (2)i
P
(−1)P hφ1 (1)φ2 (2)|ˆ g (1, 2)|φP1 (1)φP2 (2)i
(16)
P
× hφ3 (3)|φP3 (3)i · · · hφN (N )|φPN (N )i =
X
(−1)P hφ1 (1)φ2 (2)|ˆ g (1, 2)|φP1 (1)φP2 (2)i δ3 P3 · · · δN PN
P
Now, there are only two permutations such that Pi = i for i > 2, namely: P = (1, 2, 3, 4, . . . , N ) → (−1)P = 1 P = (2, 1, 3, 4, . . . , N ) → (−1)P = −1
(17)
So, hΨHF |ˆ g (1, 2)|ΨHF i = hφ1 (1)φ2 (2)|ˆ g (1, 2)|φ1 (1)φ2 (2)i − hφ1 (1)φ2 (2)|ˆ g (1, 2)|φ2 (1)φ1 (2)i . = hφ1 φ2 |ˆ g |φ1 φ2 i − hφ1 φ2 |ˆ g |φ2 φ1 i,
(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 only the identity permutation yields a non-zero contribution to the matrix element. Now, we can write the crystal Hamiltonian as Hxtal =
N X
ˆ + h(i)
i
N X
gˆ(i, j),
(19)
i6=j
where, in Rydberg atomic units, ˆ = − 1 ∇2 + Vext (ri ) h(i) 2m i 1 −e2 . 1 gˆ(i, j) = = v(i, j) 2 |ri − rj | 2
(20)
In order to find the Hartree-Fock equations, we wish to minimize the expectation value of the total energy 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 Lagrange multiplier, L = hΨHF |Hxtal |ΨHF i − λ (hΨHF |ΨHF i − 1) (21)
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We set δL = 0 and obtain, δ
N X
ˆ ii + hφi |h|φ
i
N 1X
2
[hφi φj |ˆ v |φi φj i − hφi φj |ˆ v |φj φi i]
ij
−δ
(N X
)
hφi |φi i
=0
(22)
i
Note that we don’t have to impose orthogonality constraint because, for small perturbations in the orbitals, the change in the orthogonality will be quadratic in |δφi i. Then, N X
ˆ ii − λ hδφi |h|φ
i
N X
hδφi |φi i +
i
1 + 2
N X
N 1X [hδφi φj |ˆ v |φi φj i − hδφi φj |ˆ v |φj φi i] 2 ij
(23)
[hφi δφj |ˆ v |φi φj i − hφi δφj |ˆ v |φi φj i] + h.c. = 0
ij
Note that the double sums can be combined into a single sum, since vˆ(1, 2) = vˆ(2, 1). We seek critical solutions for any orbitals. Because the linear independence of the orbitals, we only need to solve the linear system for the bras. For a given orbital with index i, we obtain N X
ˆ ii + h|φ
[hφj (2)|ˆ v |φi (1)φj (2)i − hφj (2)|ˆ v |φj (1)φi (2)i] − λ|φi i = 0.
(24)
j
This equation can be identified as the Hartree-Fock equation, ˆ i i + F|φi i = εi |φi i, h|φ
(25)
where we define the Fock operator as, N
. X ˆ ii = [hφj (2)|ˆ v |φi (1)φj (2)i − hφj (2)|ˆ v |φj (1)φi (2)i] . F|φ
(26)
j
In real space, the Hartree-Fock equation reads, N X 1 2 − ∇ φi (x) + Vext (x)φi (x) − e2 2m j=1
"Z
|φj (x0 )|2 φi (x) − d 3 x0 |x − x0 |
Z
φ∗j (x0 )φi (x0 )φj (x) = εi φi (x). d3 x0 |x − x0 | #
(27) The total energy of the system is not the sum of the orbital eigenvalues εi , but rather E = hΨHF |Hxtal |ΨHF i =
N X i
= =
N X ˆ ii + 1 hφi |h|φ [hφj (2)|ˆ v |φi (1)φj (2)i − hφj (2)|ˆ v |φj (1)φi (2)i] 2 ij
N X
N D E X ˆ ii + 1 hφi |h|φ φi Fˆ φi 2 i
i N X i
εi −
(28)
N D E 1X φi Fˆ φi 2 i
For the second part of the question, we assume that we promote one electron from an occupied orbital a to an unoccupied orbital b, i.e., we assume that single-particle orbitals remain frozen, and 9
that orbital a, which was previously occupied, becomes unoccupied, and that orbital b, which as previously 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). The excitation energy ∆E is the energy difference from the ground-state energy, E 0 , to the excited state energy, E a→b . ˆ The energy difference is composed of two parts: ∆E 1 , which depends on h(1), and ∆E 2 , which 1 ˆ ˆ depends on gˆ(1, 2). It is clear that ∆E = hφb |h|φb i − hφa |h|φa i. The change ∆E 2 is a little bit harder to compute as the whole Fock operator changes as an electron gets promoted, ∆E 2 =
N 1X [hφb φj |ˆ v |φb φj i − hφb φj |ˆ v |φj φb i] 2 j
(29)
N 1X − [hφa φj |ˆ v |φa φj i − hφa φj |ˆ v |φj φa i] 2 j
The excitation energy, within the Hartree-Fock approximation, is therefore ∆E a→b = ∆E 1 + ∆E 2 = (εb − εa ) −
N 1X [hφb φj |ˆ v |φb φj i − hφb φj |ˆ v |φj φb i] 2 j
N 1X + [hφa φj |ˆ v |φa φj i − hφa φj |ˆ v |φj φa i] . 2 j
(30)
Note that, within the HF approximation, the excitation energy is not just the difference between the eigenvalues, but it also contains a correction term, which can be understood as an approximation to the exciton binding energy. I.5. Born-von Karman boundary condition. (a) See Fig. 6. E(k)
E(k)
-3π/a -2π/a -π/a
π/a 2π/a 3π/a -3π/a -2π/a -π/a
(a)
E(k)
π/a 2π/a 3π/a -3π/a -2π/a -π/a
(b)
π/a 2π/a 3π/a
(c)
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) X k
1=
Ω (2π)3
Z
d3 k =
I.6. Energy bands of elemental solids. 10
ΩΩBZ Ω = =N 3 (2π) Ωc
(31)
