Quinn J.J., Yi K.-S. Solid State Physics: Principles and Modern Applications

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Uncorrected Proof

BookID 160928 ChapID 05 Proof# 1 - 29/07/09

150 5 Use of Elementary Group Theory in Calculating Band Structure

t12.1 Table 5.12. Characters and IRs of cubic group

t12.2 (number of operations) (1) (3) (6) (6) (8) (1) (3) (6) (6) (8)

t12.3 class → E 4

423JJ4

J4 J2 J3

t12.4 representation ↓

t12.5 Γ

) 1111111111

t12.6 Γ

)11−1 −1111−1 −11

t12.7 Γ

) 2200−122 00−1

t12.8 Γ



)3−11−103−11−10

t12.9 Γ



)3−1 −1103−1 −110

t12.10 Γ



) 11111−1 −1 −1 −1 −1

t12.11 Γ



)11−1 −11−1 −111−1

t12.12 Γ



) 2200−1 −2 −2001

t12.13 Γ

)3−11−10−31−110

t12.14 Γ

)3−1 −110−311−10

x,−y,−z

(+):

⎛

⎝

⎞

⎠

→

⎛

⎝

−y

−x

⎞

⎠

, C

x,−y,−z

(−):

⎛

⎝

⎞

⎠

→

⎛

⎝

−z

−x

⎞

⎠

417

6. All of the above operations multiplied by the inversion J, 418

⎛

⎝

⎞

⎠

→

⎛

⎝

−x

−y

−z

⎞

⎠

gives the 24 improper rotations.

419

Characters and irreducible representations of the cubic group are listed in 420

Table 5.12. 421

5.6.2 Face Centered Cubic Lattice 422

The primitive translation vectors of a face centered cubic lattice are given by 423

(ˆx +ˆy), a

(ˆz +ˆx), a

(ˆy +ˆz). (5.48)

The primitive vectors of the reciprocal lattice (including the factor 2π)are

424

2π

(−ˆx − ˆy +ˆz), b

2π

(−ˆx +ˆy − ˆz), b

2π

(ˆx − ˆy − ˆz). (5.49)

An arbitrary vector of the reciprocal lattice can be written as

425

= h

+ h

2π

[(−h

− h

+ h

)ˆx +(−h

+ h

− h

)ˆy +(h

− h

)ˆz] .

(5.50)

Brillouin Zone

426

There are eight shortest and six next shortest reciprocal lattice vectors 427

from the origin of reciprocal space to neighboring points (remember that the 428

reciprocal lattice of an FCC is a BCC). They are given by 429

Uncorrected Proof

BookID 160928 ChapID 05 Proof# 1 - 29/07/09

5.6 Empty Lattice Bands for Cubic Structure 151

Fig. 5.8. The ﬁrst Brillouin zone of FCC lattice

1. The eight vectors

2π

[±ˆx, ±ˆy, ±ˆz] whose length is |G| =

2π

√

3. 430

2. The six vectors

2π

(±2ˆx),

2π

(±2ˆy),

2π

(±2ˆz) whose length is |G| =

2π

·2. 431

The ﬁrst Brillouin zone is the volume enclosed by the planes which are the 432

perpendicular bisectors of these 14 G-vectors. The ﬁrst Brillouin zone of FCC 433

lattice has six square faces perpendicular to (100) and eight hexagonal faces 434

perpendicular to (111). (See Fig. 5.8.) 435

The names of high symmetry points are labelled in Fig. 5.8. Γ is the origin. 436

Arbitrary points along (100), (110), and (111) directions are called Δ, Σ, and 437

Λ, respectively. The special points X, L, K,andW are 438

X =

2π

(1, 0, 0),L=

2π





,K =

2π



, 0



,W =

2π



, 1, 0



The energy of a free electron is given by

439

E =

2ma



+ ξ)

+(l

+ η)

+(l

+ ζ)



(5.51)

440

where 441

= −h

− h

+ h

= −h

+ h

− h

= h

− h

and h

are integers. If we measure energy in units of

2ma

,then 442

E(Γ) = l

+ l

E(X)=(l

+1)

+ l

E(L)=(l

)

+(l

)

+(l

)

(5.52)

One should obtain a table similar to Table 5.13. From the table you con-

443

structed, you can draw the empty lattice band structure, showing the bands 444

going from Γ → X and from Γ → L. 445

Uncorrected Proof

BookID 160928 ChapID 05 Proof# 1 - 29/07/09

152 5 Use of Elementary Group Theory in Calculating Band Structure

t13.1 Table 5.13. Energies for FCC empty lattice E(Γ) ≤ 8. Energy is measured in units

2ma

t13.2 h

E(Γ) E(X) E(L)

t13.3 0 0 0 0 0 0 0 1

t13.4 1 0 0 −1 −11 3 2

t13.5 −10 0 1 1−13 6

t13.6 0 1 0 −11−13 2

t13.7 0 −101−11 3 6

t13.8 0 0 1 1 −1 − 13 6

t13.9 0 0 −1 −111 3 2

t13.10 1 1 1 −1 −1 −13 2

t13.11 −1 −1 −1111 3 6

t13.12 1 1 0 −200 4 1

t13.13 −1 −10200 4 9

t13.14 1 0 1 0 −20 4 5

t13.15 −10−1020 4 5

t13.16 0 1 1 0 0 −24 5

t13.17 0 −1 − 1002 4 5

t13.18 1 −100−22 8 9

t13.19 −11 0 0 2−28 9

t13.20 1 0 −1 − 202 8 5

t13.21 −10 1 2 0−28 13

t13.22 0 1 −1 − 220 8 5

t13.23 0 −112−20 8 13

t13.24 2 1 1 −2 −20 8 5

t13.25 −2 −1 −1220 8 13

t13.26 1 2 1 −20−28 5

t13.27 −1 −2 −1202 8 13

t13.28 1 1 2 0 −2 −28 9

t13.29 −1 −1 −2022 8 9

Uncorrected Proof

BookID 160928 ChapID 05 Proof# 1 - 29/07/09

5.6 Empty Lattice Bands for Cubic Structure 153

19/4

(4)

(8)

(6)

(2)

(6)

Fig. 5.9. Empty lattice band of the FCC lattice. The energy E is measured in units

2ma

and plotted as a function of k. Each band is schematically represented by a

straight line going from E

(Γ) to E

(X) or E

(L) even though the bands really have

a more complicated (quadratic form) dependence of the Bloch wave vector k.The

set of integers (l

) is indicated for each band

This empty lattice lattice band structure is shown in Fig. 5.9. Note that k = 446

2π

ˆx at X and k =

(ˆx +ˆy +ˆz)atL.TheenergyE is sketched as a function 447

of k. 448

5.6.3 Body Centered Cubic Lattice 449

The primitive translations of the reciprocal lattice (including 2π)are 450

2π

(ˆx +ˆz), b

2π

(−ˆx +ˆy), b

2π

(−ˆy +ˆz). (5.53)

Therefore a general reciprocal lattice vector G

is given by 451

2π

+ h

]

2π

[(h

− h

)ˆx +(h

− h

)ˆy +(h

+ h

)ˆz] . (5.54)

Uncorrected Proof

BookID 160928 ChapID 05 Proof# 1 - 29/07/09

154 5 Use of Elementary Group Theory in Calculating Band Structure

Fig. 5.10. (a)First Brillouin zone of the BCC lattice, (b) cross section of the four

planes bisecting the G = ±

2π

(ˆx±ˆy) perpendicular to the z-axis of the ﬁrst Brillouin

zone of a BCC lattice

The 12 shortest reciprocal lattice vectors are 452

2π

(ˆx ± ˆy) , ±

2π

(ˆy ± ˆz) , ±

2π

(ˆx ± ˆz) .

They have length

2π

√

2. The ﬁrst Brillouin zone is formed by the 12 planes 453

that bisect these 12 shortest reciprocal lattice vectors.(See Fig. 5.10a.) 454

Figure 5.10b shows the cross section of the four planes that bisect the 455

shortest G = ±

2π

(ˆx ± ˆy) perpendicular to the z-axis of the ﬁrst Brillouin 456

zone of the BCC lattice. The empty lattice energy bands can be written by 457

2ma



+ ξ)

+(l

+ η)

+(l

+ ζ)



(5.55)

458

where 459

= h

− h

= h

− h

= h

+ h

and h

are integers. We use the symbols of k =

2π

(ξ,η, ζ), H=

2π

(1, 0, 0), 460

2π

(

), and N=

2π

(

, 0). Thus, we have, in units of

2ma

461

(Γ) = l

+ l

(H)=(l

+1)

+ l

(P )=(l

)

+(l

)

+(l

)

(N)=(l

)

+(l

)

+ l

(5.56)

Since l

= h

− h

, l

= h

− h

, l

= h

+ h

,wecanwrite 462

E(Γ) = (h

− h

)

+(h

− h

)

+(h

+ h

)

E(H)=(h

− h

+1)

+(h

− h

)

+(h

+ h

)

E(P )=(h

− h

)

+(h

− h

)

+(h

+ h

)

E(N)=(h

− h

)

+(h

− h

)

+(h

+ h

)

(5.57)

Uncorrected Proof

BookID 160928 ChapID 05 Proof# 1 - 29/07/09

5.7 Energy Bands of Common Semiconductors 155

5.7 Energy Bands of Common Semiconductors 463

Many common semiconductors which crystallize in the cubic zincblende struc- 464

ture have valence–conduction band structures that are quite similar in gross 465

features. This results from the fact that each atom (or ion) has four electrons 466

outside a closed shell and there are two atoms per unit cell. For example, 467

silicon has the electron conﬁguration [Ne]3s

, i.e., two 3s electrons and 468

two 3p electrons outside a closed neon core. With two silicon atoms per 469

unit cell, this gives eight electrons per unit cell. The empty lattice has a 470

single Γ

band at E

= 0 and 8-fold degenerate bands at E

=3.The471

eightfold degeneracy is lifted by the periodic potential, so the valence and 472

conduction bands at Γ will arise from these eight bands. Germanium has 473

the electron conﬁguration of [Ar]3d

, and III–V compounds like GaAs 474



[Ar]3d





[Ar]3d

.

look just like Ge if one 4p electron

475

transfersfromAstoGaleavingasomewhationicGa

−

molecule in the 476

unit cell instead of two Ge atoms. The same is true if any III–V elements 477

replace a pair of Si atoms or Ge atoms in a zincblende structure. 478

A nice example of the use of group concepts in studying energy band 479

structure is a simple nearly free electron type model used to give a rather good 480

description of the valence–conduction band semiconductors with zincblende 481

structures. We will give a rough sketch of the calculation, referring the reader 482

to an article by D. Brust

. To describe the band structures of Si and Ge, Brust 483

use the following 15 plane-wave wave functions corresponding to the 15 bands 484

at Γ which have energy E ≤ 4. (See the 15 bands at Γ in Fig. 5.9.) We can 485

write these 15 plane waves as w

,withi =1, 2, 3,...,15 deﬁned by 486

=1E

(Γ) = 0,

= w

∗

2πi

(x+y+z)

(Γ) = E

(Γ) = 3,

= w

∗

2πi

(x−y−z)

(Γ) = E

(Γ) = 3,

= w

∗

2πi

(−x+y−z)

(Γ) = E

(Γ) = 3,

= w

∗

2πi

(−x−y+z)

(Γ) = E

(Γ) = 3,

= w

∗

2πi

(Γ) = E

(Γ) = 4,

= w

∗

2πi

(Γ) = E

(Γ) = 4,

= w

∗

2πi

(Γ) = E

(Γ) = 4,

From these 15 functions w

,...,w

, one can construct linear superposi- 487

tions belonging to IRs of the group of the wave vector Γ, X, L, etc. Some 488

examples are 489

D. Brust, Phys. Rev. 134, A1337 (1964).

Uncorrected Proof

BookID 160928 ChapID 05 Proof# 1 - 29/07/09

156 5 Use of Elementary Group Theory in Calculating Band Structure

√

belongs to Γ

√

− w

+ w

− w

] belongs to Γ

√

+ w

− w

+ w

− w

+ w

− w

] belongs to Γ

√

+ w

− w

] belongs to Γ



√

+ w

] belongs to Γ

If you use these combinations of plane waves, the Schr¨odinger equation breaks

490

up into a block diagonal 15 × 15 matrix as shown in (5.58). 491





[3 × 3]



[2 × 2]

[3 × 3]



=0. (5.58)

Here Γ

and Γ



are 1×1, Γ

is a 2×2, and Γ

,Γ

,andΓ



are 3×3 matrices, 492

respectively. Oﬀ diagonal elements and bands in higher empty lattice states 493

are treated by standard nondegenerate perturbation theory. 494

Now the question arises “What do we use for the periodic potential 495

V (r)=



·r

?” Brust simply treated the parameters V

as 496

phenomenological coeﬃcients to be obtained by ﬁtting band gaps and eﬀective 497

masses measured experimentally or ﬁtting more detailed ﬁrst-principles band 498

structure calculations. He found that he could obtain a reasonably satisfactory 499

ﬁt by keeping only three parameters: 500

V (3) = V

when l

+ l

=3,

V (8) = V

when l

+ l

=8, (5.59)

V (11) = V

when l

+ l

=11.

Remember, by cubic symmetry, V

1,1,1

= V

1,1,−1

= V

1,−1,−1

= V

−1,−1,−1

etc. 501

For Si, Brust found that V (3) −0.21 Ry,V(8)  0.04 Ry,V(11)  0.08 Ry. 502

For Ge he found that V (3) −0.23 Ry,V(8)  0.00 Ry,V(11)  0.06 Ry. 503

The band structure obtained for Si is illustrated in Fig. 5.11. This diagram 504

shows 11 bands at Γ out of the 15 bands we put into the calculation. Since 505

there are two atoms per unit cell and four valence electrons per atom, we 506

have eight electrons per unit cell or enough to ﬁll four bands. Thus the Γ



507

state is the top of the valence band. The conduction band is Γ

at Γ, but 508

the minimum is at near the X-point, so the conduction band minimum has 509

six valleys, each very near one of the six X points. 510

Uncorrected Proof

BookID 160928 ChapID 05 Proof# 1 - 29/07/09

5.7 Energy Bands of Common Semiconductors 157

(1)

(2)

(3)

Fig. 5.11. Simple band structure of Si

Uncorrected Proof

BookID 160928 ChapID 05 Proof# 1 - 29/07/09

158 5 Use of Elementary Group Theory in Calculating Band Structure

Problems 511

5.1. At E(X) = 0.25

2ma

there are two degenerate bands. At E(X) = 512

1.25

2ma

there are four. Determine the linear combinations of degenerate 513

states at these points belonging to IR’s of G

.DothesameforE(Γ) = 514



2ma



and 2



2ma



515

5.2. Make a table, with values of h

, of the resulting l

and E(Γ), 516

E(X), E(L) for all bands with E(Γ) ≤ 4 for an FCC lattice. 517

5.3. Use your knowledge of the irredicible representations at E(Γ) = 0, 1, 2(in 518

units of

2ma

)andatE(X)=0.25 and 1.25, together with the compatibility 519

relations to determine the irreducible representations for each of these bands 520

along the line Δ. 521

5.4. Tabulate E(Γ),E(H),E(P ) for all bands that have E(Γ) ≤ 4. Then 522

sketch E vs. k along Δ(Γ → H) and along Λ(Γ → P ). 523

5.5. Do the same as above in part 4 for simple cubic lattice where 524

2ma



+ ξ)

+(l

+ η)

+(l

+ ζ)



for Γ, X =

(1, 0, 0), and R =

(1, 1, 1). Sketch E vs. k along Γ → X and 525

along Γ → R for all bands having E(Γ) ≤ 4. 526

5.6. Use the irreducible representations at E(X) = 0.25 to evaluate 527



(0.25) | V (r) | Ψ

(0.25)



where Ψ

(0.25) is the wave function at E(X) = 0.25 belonging to the 528

irreducible representation X

. 529

(a) Show that V

=0ifi = j. 530

(b) Show that the diagonal matrix elements give the same energies (and 531

band gap) as obtained by degenerate perturbation theory with the 532

original plane waves. 533

Uncorrected Proof

BookID 160928 ChapID 05 Proof# 1 - 29/07/09

5.7 Energy Bands of Common Semiconductors 159

5.7. A two dimensional rectangular lattice has a reciprocal lattice whose 534

primitive translations, including the 2π,areb

2π

ˆx and b

2π

√

ˆy. 535

2π

536

(a) List the operations belonging to G

. 537

(b) Do the same for G

and G

. 538

(k, r)=e

i(k+K

)·r

and E

(k)=

¯h

(k + K

)

. Here, K

= l

+ 540

,andl

and l

are integers. Tabulate the energies at Γ and at X 541

for (l

) =(0,0), (0, ±1), (−1,0), (1,0), and (−1, ±1). 542

(d) Sketch (straight lines are OK) E vs. k along the line Δ (going from Γ 543

to X) for these bands. 544

(e) Two degenerate bands at the point E(Γ) = 0.5 connect to E(X) = 0.75. 545

Write down the wave functions for an arbitrary value of k

for these 546

two bands. 547

(f) From these wave functions, construct the linear combinations belonging 548

to irreducible representations of G

. 549

5.8. Graphene has a two-dimensional regular hexagonal reciprocal lattice 550

whose primitive translations are b

2π

(1, −

√

)andb

2π

(0,

√

). 551

(a) List the operations belonging to G

. 552

(b) Do the same for G

and G

.Notethatk

2π

(

, 0) and k

= 553

2π

(

√

). 554

(k, r)and 555

(k)atΓandK. 556

(d) Tabulate the energies at Γ and at K for (l

) =(0,0), (0, ±1), (±1, 0), 557

(−1, −1), and (1, 1). 558

(e) Sketch E vs. k along the line going from Γ to K for these bands. 559

(f) Write down the wave functions for the three fold degenerate bands at 560

the energy E(K)=4/9. 561