Marder M.P. Condensed Matter Physics

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Brief Survey of the Periodic Table 283

ε?

(A)

il O.

T.t. T.

(B)

X W

Energy density of states D(£)

Figure 10.8. (A) Energy bands of

krypton,

calculated with the plane-wave pseudopotential

code VASP of Kresse and Hafner (1994) and Kresse and Furthmiiller (1996). All energies

are measured relative to the Fermi energy. The calculation uses around 1000 plane waves,

and it assumes an fee structure with lattice parameter a = 5.72 Â. The notation for the path

traversed in reciprocal space for this figure is given in Figure 7.8. The ionization energy

of an isolated krypton atom is

£j

the experimental band gap £™

is 10.5 eV, and the

calculated band gap is around 6 eV. (B) Density of states of krypton. The electrons are

restricted to three narrow bands, holding six electrons, lying just below the Fermi surface.

An additional band at —19 eV, not depicted, holds an additional two electrons. Peaks in

the density of states are van Hove singularities, as described in Section 7.2.5

10.4.3 Semiconductors

The fourteenth column of the periodic table separates insulators from conductors,

and it is largely populated with semiconductors. Carbon is a semimetal in the form

of graphite and is a very wide band semiconductor in the form of diamond; gray

tin is semiconducting while white tin is metallic.

The band structure of graphene appears in Figure 10.9 (see also Problem 8.5).

Graphene consists in a single atomic layer of carbon atoms in a honeycomb struc-

ture.

The most notable feature of the band structure is the occurrence of a two-fold

degeneracy right at the Fermi surface at symmetry point K. The uppermost degen-

erate band is unoccupied in the ground state, while the lower one is fully occupied.

The density of states at the Fermi surface is extremely small, making graphene, like

graphite, a semimetal. When graphene folds around an axis to create a nanotube,

the periodic boundary conditions imposed on the wave function make it possible

for the tube either to remain a semimetal, or to become a narrow or wide-gap semi-

conductor. The relation of electrical properties to the geometry of the nanotube is

the subject of Problem 3.

The single most important band structure of the elements appears in Figure

10.10:

the band structure of

silicon.

Understanding fine details of electronic excita-

tion and transport and silicon constitutes the foundation of the electronics industry,

and will occupy much of Chapter 19. The calculations represented in Figure 10.10

contain valuable information, but also reveal a characteristic weakness of density

functional calculations. As with krypton, it is tempting to regard the gap between

284 Chapter 10. Realistic Calculations in Solids

M T K Energy density of states D{E)

Figure 10.9. (A) Energy bands of graphene, calculated with the plane-wave pseudopo-

tential code VASP of Kresse and Hafner (1994) and Kresse and Furthmiiller (1996). All

energies are measured relative to the Fermi energy. Notation for the symmetry points in

k space in this figure is given by Figure 7.10. Note that at symmetry point K a filled and

empty band are degenerate and meet at the Fermi surface. (B) Energy density of states

of graphene. At the Fermi level, the density of states falls to zero, reflecting the fact that

graphene is a semi-metal. Peaks in the density of states are van Hove singularities, as

described in Section 7.2.5

the highest occupied and lowest unoccupied state as the energy that will be charac-

teristic of excitations. The existence of such an energy gap is correct, but the value

visible in Figure 10.10 is too small by about a factor of two when compared with

experiments, such as optical absorption to be discussed in Section 21. Theoretical

curves displayed in Section 23.6.2 cure this problem in two ways—some by de-

liberately adjusting parameters in the pseudopotentials to improve correspondence

with experiment, and others by combining additional information about electron

interactions in with ordinary density functional theory.

10.4.4 Transition Metals

The transition metals lying in columns 3-8 of the periodic table are not well de-

scribed by the nearly free electron model. The 3d, Ad and 5d shells fill when

passing left to right among these elements, but always strongly mixed with the As,

5s,

and 6s states, as shown by the calculated band structure of vanadium, in Figure

10.11.

The wave functions of silver shown in Figure 10.2 illustrate the fact that

the d orbitals are restricted to a radius of about 2 Â about the atom—in contrast

to the s and p orbitals, which extend out many times as far. Therefore, the most

appropriate simple starting point for considering the d electrons is a tight-binding

model. However, all simple calculations of this type are risky, and only partly be-

cause of the need to mix s and p states in with d. Several of the transition metals

have magnetic structures in their ground states, a sure sign that the electrons are

developing correlations of a sort not readily comprehensible in the one-electron

framework.

Brief Survey of the Periodic Table

285

X W L

Energy density of states D(£)

Figure 10.10. (A) Energy bands of silicon, calculated with the plane-wave pseudopotential

code VASP of Kresse and Hafner (1994) and Kresse and Furthmiiller (1996). All energies

are measured relative to the Fermi energy. The calculation assumes that silicon adopts

the diamond structure, with a lattice spacing of 5.43 Â, and uses several thousand plane

waves. The calculation correctly determines that silicon is a semiconductor, and that the

lowest lying spot in the conduction band lies between Γ and X. However the size of the

energy gap £

is computed to be around 0.5 eV, which is half the value obtained by more

sophisticated calculations and experimental measurements shown in Figure 23.16. (B)

Energy density of states of silicon. Note that the density of states falls to zero in the gap.

Peaks in the density of states are van Hove singularities, as described in Section 7.2.5

Energy density of states D(£)

Figure

10.11.

(A) Energy bands of vanadium, as calculated with the plane-wave pseu-

dopotential code VASP of Kresse and Hafner (1994) and Kresse and Furthmiiller (1996).

All energies are measured relative to the Fermi energy. The calculation uses the fact that

vanadium adopts the bcc structure, and it requires about 1000 plane waves. Notation for

the symmetry points in k space in this figure is given by Figure 7.9. The location of the

Fermi level in the midst of the complicated d bands is characteristic of the transition met-

als.

(B) Energy density of states of vanadium, showing the large density of states due to d

electrons near the Fermi surface. Peaks in the density of states are van Hove singularities,

as described in Section 7.2.5

286 Chapter 10. Realistic Calculations in Solids

10.4.5 Rare Earths

The difficulties of applying band theory to the transition metals become even more

acute in the case of the lanthanides. In this series of elements, the 4/ orbitals are in-

complete. These orbitals are even more tightly peaked about individual atoms than

d orbitals, but nevertheless the solids remain metallic. While the close association

of electrons to their respective atoms might seem to make a simple tight-binding

model accurate, the actual situation is that the one-electron picture breaks down.

For example, it is generally unfavorable for two electrons to occupy the same spa-

tial state with opposite spin, a fact that a one-electron point of view will not be

able to comprehend. The origin of these difficulties is Coulomb repulsion between

the / electrons. Roughly speaking, Coulomb repulsion for wave functions of ra-

dius R will contribute terms to the energy going as e

exp [—/?/£]//?, where ξ is

the screening length. Because screening lengths are typically on the order of 0.5

Â, screening cannot much reduce the interactions of the / electrons, and they are

driven to imaginative solutions involving local magnetic moments that band theory

does not capture.

Problems

Details of the augmented plane wave: When there is more than one type of

atom per unit cell, one surrounds each atom with a sphere of radius R

, places

the center of each sphere at r

, and proceeds to deduce that for augmented

plane waves,

(Φεΐ\ΰ-ε\Φεψ)

£ )

n-W

+ E^'^îW'

(

°)

with

iW =

4irR

'tfq·? _Λ M\q-?\Rn)

1=0

2m'

\q-q

+ £ f-(2/ +

l)P,(q■

q'mqRMiq'Rn)^'^

&ηΐε(Κη)

(10.31)

Derive Eqs. (10.30) and (10.31).

Pseudopotential for aluminum: Consider the empty core pseudopotential,

shown in Figure

10.1,

and normalized for a unit cell of volume Ω as

V(r) = U

e-^^9(r-R

). (10.32)

Here d is a screening length, and R

is a parameter describing a length within

which the cancellations leading to the pseudopotential are taken to be perfect.

Problems 287

The goal is to determine i/o, R

, and d. The experimental data that make

it possible come from the de Haas-van Alphen effect (Section 16.5.2) and

optical data, as in Problem 2 of Chapter

23.

However, all one needs to know

is that it has been possible to measure the Fourier transform U

defined in

Eq. (7.26), at two reciprocal lattice vectors, and the results for aluminum are

(in)

=0.245 eV (10.33)

^(200) =

°·

762

· (10.34)

The reciprocal lattice vectors are being described by their Miller indices and

are measured in units of 2-π/α.

In addition, the Fermi energy, measured relative to the lowest-energy k state,

£jr = 11.7eV, (10.35)

which can be used in Eq. (10.13).

(a) Take the Fourier transform of

Eq.

(10.32).

(b) Determine i/o, d, and R

by finding a reasonable fit of Eq. (10.32) to the

measured values Eq. (10.33), (10.34), and (10.13). The agreement cannot be

perfect, because one is trying to fit three parameters with two variables. Use

any nonlinear fitting routine, or use a process of trial and error as convenient.

The final values should be close to those listed after Eq. (10.38).

Nanotube conductivity: An (m, n) nanotube was defined in Problem 1.3 to

be a graphene sheet rolled up so that the atom

&Xc

= ma\ +««2 finds itself back

at the origin, where a\ and

Ü2

are primitive vectors of the graphene sheet.

(a) Argue that for an (m, n) nanotube, allowed Bloch vectors k must satisfy

k-C = 2nl. Where / is an integer. (10.36)

(b) Consulting Figures 10.9 and 7.10, note that the nanotube should be expected

to be metallic if symmetry point K is an allowed value of k, and a semicon-

ductor or insulator otherwise. Show for positive m, n, that the condition for K

to be an allowed value of k is

m = n Or m — n = 3j. Where j is an integer. (10.37)

The nanotubes with n = m are in fact all metallic, while those with m

— n

= 3j

are very narrow-band semi-conductors at very low temperatures, and effec-

tively metallic at room temperature. The remaining nanotubes are semicon-

ductors. See Louie (2001) or Charlier et al. (2007).

Plane wave band structure, part I: The following five problems outline the

construction of a primitive plane wave band structure code. The programming

task can be carried out in any language, but the computing environment should

be one that

288 Chapter 10. Realistic Calculations in Solids

• can carry out complex arithmetic, including multiplication and exponentia-

tion of complex numbers, and

• can take the (discrete) Fourier transform of a three-dimensional array of

complex numbers.

The first task is to write a subroutine that takes two arguments. The first is

c(i, j), an N x M-dimensional array of complex numbers. The second is an

integer, i\. The columns of c are orthonormal up to i\

—

\. The task of the

subroutine is to replace the i'ith column of c with a column that is orthogonal

to the preceding i\

—

l columns and normalized to unity. Test the program on

the following numbers:

c(i,l) c(i,2) c(i,3) c(i,4)

(0.258198887,0.516397774) (-0.392356962,-0.249681681) (0.303851753,0.138416067) 0

(0.774596632,0.258198887) (0.321019441,-0.071337625) (-0.221133977,0.082717843) 2

(0..0.) (0.107006453,0.749045134) (0.097649745,-0.087221026) 1

(0.,0.) (0.321019351,0.) (0.889513135,0.156192154) 0

The first three columns are orthonormal. Find c(i, 4).

Band structure, part II: Consider the Fourier transform of the pseudopo-

tential in Figure 10.1

Uoe

dK[*&

+ i] ·

(ia38)

Take d = 0.350, R

= 0.943, and i/o = —31.30 (the units are angstroms and

electron volts). These numbers are appropriate for aluminum and are the sub-

ject of Problem 2. The task is to compute

U(r) = Σ S^Ug, (10.39)

taking K to be the reciprocal lattice vectors of

fee lattice of spacing a = 4.05

(Â).

(a) Find Ug

(b) Use Eq. (2.4) to construct the primitive vectors of the reciprocal lattice.

Obtain a routine capable of performing the transform forN xN xN arrays of

numbers; the problems will use

= 4, 5, 8, and 12. If such a routine cannot

be located, the problems can be performed for

= 4, 8 only.

(d) For a given value of N, for what values of r will a fast Fourier transform

routine compute U(r), and how will they be indexed?

(e) Use the routine to evaluate Eq. (10.39), taking N

—

5, and printing the 125

values of U(r) and r in the Wigner-Seitz cell. Here are some of the values.

The first three numbers are x, y, and z coordinates of r in angstroms, and the

final number is the value of the potential U(r) in Rydbergs.

Problems 289

405

0.405

0.810

1.215

1.620

0.405

0.810

1.215

0.405

0.810

1 .215

620

0.405

0.810

1.215

620

0.201763183,0

-0.201839209,0

-0.757843018,0

-0.201839209,0

-0.201839194,0

-0.658886611,0

-0.718343496,0

-0.734894872,0

If only

= 4 is possible, then U does not come out to be real; the first values,

in Rydbergs, are

50625

0.50625

1.01250

1.51875

0.50625

1.01250

0.50625

1.01250

1.51875

0.50625

1.01250

1.51875

0.166458577, 0

-0.265354961,

-0.911596537,

-0.265354961,-0

-0.265354931,

-0.591810703,-0

-0.781587422,

)

240282357)

)

240282357)

240282387)

186816186)

103971817)

6. Plane wave band structure, part

III:

Consider Schrödinger's equation in the

form of Eq. (7.33).

The task is to find the lowest-lying eigenvalue δ and eigenfunction ψ solving

this equation with the potential given by Eq. (10.38). Look for a solution with

Bloch index k = 0; essentially, this means that one sets q = 0 in Eq. (7.33).

Restrict all calculations to the 5 x 5 x 5 set of K considered in the previous

portion of the problem.

(a) Define a

x 5 x 5 complex array ψ(Κ). Let £

max

be £^ , where

max

is the

largest value of K under consideration. Write a routine to compute

^(£) = (£|-£

max

)^)

Ç^(^-^)·

Κ'

(10.40)

The hard part of the computation is the convolution ]Γ^, Ug,ip(K

—

K'). To

perform the convolution, use the fact that

U^(K-K')

= ^[Γ'^χϊ-'^ί)]]; (10.41)

that is, take the inverse Fourier transforms of U and i/>, form a new array by

multiplying together each element of U and ψ, take the Fourier transform of

the result, and divide by

This procedure is vastly faster than performing

the sums described by the convolution directly.

(b) To find the ground state of

Eq.

(7.33),

i. Choose some random initial normalized ψ(Κ), such as ψ(0) = 1, with all

other components zero.

ii.

Find

-φ'

according to Eq. (10.40).

iii.

Normalize ψ'.

iv. Put

-ψ'

back into the right-hand side of

Eq.

(10.40), and find ψ".

v. Normalize ψ".

290

Chapter 10. Realistic Calculations in Solids

vi.

Continue in this fashion until the process converges. One should be left

with the lowest-energy eigenstate of Schrödinger's equation.

vii.

What is the eigenvalue £ of this state from Eq. (7.33)?

Plane wave band structure, part IV: Consider again the potential given in

Eq. (10.38). The new task is to compute the lowest-lying six eigenvalues £ ^

(ε

-ε)ψ(ξ)+Σ

U^iq-'K')

(10.42)

for

k =

[2π/α](Λ .4 .4), and a

4.05Â. Combine the routine for comput-

ing action of the Hamiltonian on wave functions with the orthogonalization

routine Problem 4. Use an 8 x 8 x 8 grid for K.

For the first four eigenvalues, on 8 x 8 x 8 and 12 x 12 x 12 grids one has the

following in Rydbergs:

8x8x8

-0.27

0.13

1.17

12x 12x 12

-0.27

0.13

1.17

8. Plane wave band structure, part V:

(a) By adding an outer loop to the program produced in the previous problem,

calculate the band structure of aluminum. Use a 4 x 4 x 4 grid. Carry out the

calculation along the trajectory L to Γ, Γ to X, X to U, U to Γ. Use about ten

k points for each leg of the trajectory. Plot the first six bands. Do not expect

the results to be perfect; the energy levels do not always appear in the correct

order, and sometimes the convergence is slow. With a cutoff of around 10,000

on iterations for each energy level, one obtains the following figure:

References

291

The task is to produce a graph of roughly similar quality, with energies in-

dicated on the y axis in electron volts. Compare with the energy bands of

aluminum shown in Figure 10.6.

(b) Suppose that energies in Figure 10.6 were reported relative to the Fermi

level. Outline in words the calculations needed to determine the Fermi level.

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