MTLE-6120: Advanced Electronic Properties of Materials Band theory - - PowerPoint PPT Presentation

MTLE-6120: Advanced Electronic Properties of Materials Band theory of solids

Reading:

◮ Kasap: 4.1 - 4.5

1

Crystals

◮ Periodic arrangement of atoms (ions: nuclei + core electrons) ◮ Previously: classical motion of electrons in crystal ◮ What was the role of the periodicity? None! ◮ Next: quantum motion of electrons in crystal ◮ Bravais lattice: regular grid of points ◮ Lattice vectors

a1, a2, a3

◮ Any point on grid n1

a1 + n2 a2 + n3 a3

◮ Unit cell: repeat at grid points to fill space ◮ Multiple atoms possible per unit cell

+ + + + + + + + + + + + + + + + + + + +

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

+ + + + + + + + + + + + + + + + + + + +

2

Classical example: one oscillator

◮ Mass m attached to a spring with constant k ◮ Equation of motion in displacement x:

m¨ x = −kx

◮ In terms of ω0 =

• k/m,

¨ x = −ω2

0x ◮ Oscillatory solutions e±iω0t with angular frequency ω0

3

Classical example: two oscillators

◮ Two masses m each attached to a spring with constant k ◮ Equation of motion in displacements x1 and x2:

m¨ x1 = −kx1 m¨ x2 = −kx2

◮ Solutions: both x1 and x2 oscillate with ω0 =

• k/m

4

Classical example: two coupled oscillators

◮ Two masses m each attached to a spring with constant k ◮ Attached to each other with another spring with constant K ◮ Equation of motion in displacements x1 and x2:

m¨ x1 = −kx1 − K(x1 − x2) m¨ x2 = −kx2 − K(x2 − x1)

◮ Coupled equations, x1 and x2 are not each an oscillator ◮ Add and subtract

m(¨ x1 + ¨ x2) = −k(x1 + x2) m(¨ x1 − ¨ x2) = −(k + 2K)(x1 − x2)

◮ Oscillator differential equations in (x1 ± x2) ◮ (x1 + x2) oscillates with frequency

• k/m

◮ (x1 − x2) oscillates with frequency

• (k + 2K)/m

5

Coupled oscillators: eigenvalue problem

◮ Equation of motion in matrix form:

m d2 dt2

• x1

x2

• = −
• k + K

−K −K k + K

• A

·

• x1

x2

• ◮ Coupled equations because of off-diagonal terms in A

◮ Diagonalize A = V · D · V −1 using eigenvalue diagonal matrix D and

eigenvector diagonal matrix V m d2 dt2 V −1 · x1 x2

• = −D · V −1 ·

x1 x2

• ◮ V −1 combines x into normal modes x1 + x2 and x1 − x2

◮ Corresponding eigenvalues are mω2 = (k + K) ∓ K i.e. k and k + 2K

6

Three coupled oscillators

◮ For three oscillators

A =   k + K −K −K k + 2K −K −K k + K  

◮ Eigenvalues mω2:

  k k + K k + 3K  

7

N coupled oscillators

◮ For N oscillators, get N × N matrix

A =           k + K −K −K k + 2K −K −K k + 2K ... ... ... −K −K k + 2K −K −K k + K          

◮ Consider eigenvector of displacements X = (x1, x2, . . . , xN) with

eigenvalue mω2 = λ (say)

◮ nth row (n = 1, N) of eigenvalue equation AX = λX is

Cxn−1 + Dxn + Cxn+1 = λxn where C = −K is the coupling (off-diagonal term) and D = k + 2K is the diagonal term

8

Tri-diagonal matrix eigenvalue problem

◮ nth row (n = 1, N) of eigenvalue equation AX = λX is

Cxn−1 + Dxn + Cxn+1 = λxn

◮ Try solution of form xn = eik(na) where a is spacing between oscillators

(just so that k has usual dimensions of wavevector) Ceik(n−1)a + Deikna + Ceik(n+1)a = λeikna Ce−ika + D + Ceika = λ D + 2Ccos(ka) = λ (Each spring has phase eika relative to previous one)

◮ All rows satisfy this condition except first and last

eika(D − K) + Ce2ika = λeika (n = 1) Cei(N−1)ka + (D − K)eNika = λeNika (n = N)

◮ Connect last spring to first

9

Periodic boundary conditions (Born-von Karman)

◮ Edge equations after connecting last spring to first

CeNika

from n = N

+eikaD + Ce2ika = λeika (n = 1) Cei(N−1)ka + DeNika + Ceika

from n = 1

= λeNika (n = N)

◮ Canceling common factors:

(eNika)Ce−ika + D + Ceika = λ (n = 1) Ce−ika + D + Ceika(e−Nika) = λ (n = N)

◮ These are the same as the other equations when eNika = 1 i.e.

k = 2πj/(Na)

◮ Here, j can be any integer, but for j → j + N, k → k + 2π/a and

eika → ei(ka+2π) = eika, so nothing changes

10

N → ∞ coupled oscillators

◮ Eigenvalues mω2 = λ = D + 2C cos ka = (k + 2K) − 2K cos ka ◮ k = 2πj/(Na) for j = 0, 1, 2, . . . , N − 1 becomes continuous k ∈ [0, 2π) ◮ Spacing in k = 2π/(Na) = 2π/L, where L = Na is length of system k k+2K k+4K

• 3π/a
• 2π/a
• 1π/a

0π/a 1π/a 2π/a 3π/a k mω2 ◮ Eigenvalues mω2 form a ‘band’ with range [k, k + 4K] ◮ Band center set by diagonal element k + 2K ◮ Band width set by off-diagonal element K (coupling strength) ◮ Any interval in k of length 2π/a equivalent; conventionally [−π/a, π/a]

11

1D crystal of δ-atoms

◮ An array of δ-potentials at na i.e. V (x) = −V0

• n δ(x − na)

◮ Consider E < 0 case first, with κ =

√ −2mE/

◮ Wavefunctions e±κx in each segment, say

ψ(x) = Ane−κ(x−na) + Bneκ(x−(n+1)a), na < x < (n + 1)a

12

δ-crystal: matching conditions

Continuity at x = na and (n + 1)a: An + Bne−κa = An−1e−κa + Bn−1 Ane−κa + Bn = An+1 + Bn+1e−κa Derivative matching conditions at x = na and (n + 1)a: −2mV0 2

• An + Bne−κa

= κ

• −An + Bne−κa

− κ

• −An−1e−κa + Bn−1
• −2mV0

2

• Ane−κa + Bn
• = κ
• −An+1 + Bn+1e−κa

− κ

• −Ane−κa + Bn
• Simplify using q ≡ 2mV0/2 and t ≡ e−κa:

An + Bnt = An−1t + Bn−1 Ant + Bn = An+1 + Bn+1t − q κ(An + Bnt) = −An + Bnt + An−1t − Bn−1 − q κ(Ant + Bn) = −An+1 + Bn+1t + Ant − Bn

13

δ-crystal: eliminate B’s

◮ Add first and third equations, subtract fourth from secondequation:

(1 − q/κ)(An + Bnt) = 2An−1t − An + Bnt (1 + q/κ)(Ant + Bn) = 2An+1 − Ant + Bn

◮ Collect all A terms to left side and B to right side:

(2 − q/κ)An − 2An−1t = (q/κ)Bnt (2 + q/κ)Ant − 2An+1 = −(q/κ)Bn

◮ Eliminate Bn:

−An−1t +

• 1 + t2 + (t2 − 1)q

2κ

• An − An+1t = 0

14

δ-crystal: eigenvalue condition

−An−1t +

• 1 + t2 + (t2 − 1)q

2κ

• An − An+1t = 0

◮ Substitute An = eikna:

−teik(n−1)a +

• 1 + t2 + (t2 − 1)q

2κ

• eikna − teik(n+1)a = 0

−te−ika +

• 1 + t2 + (t2 − 1)q

2κ

• − teika = 0

1 + t2 + (t2 − 1)q 2κ = 2t cos(ka) 1/t + t 2 − q 2κ · 1/t − t 2 = cos(ka)

◮ Since t ≡ e−κa

cosh(κa) − q 2κ sinh(κa) = cos(ka)

15

δ-crystal: eigenvalues

◮ Energy eigenvalue is E = ω = − 2κ2 2m ◮ For the δ-atom, we got a unique κ = mV0/2 ≡ q/2 ◮ But now, an entire family of solutions for k ∈ (−π/a, π/a):

cosh(κa) − q 2κ sinh(κa) = cos(ka)

◮ First consider limit of deep potential i.e. large q; rearrange:

κ = q sinh(κa) 2(cosh(κa) − cos(ka))

◮ Large q ⇒ large κ ⇒ sinh, cosh extremely large ◮ For q → ∞, κ → q/2 as in the δ-atom ◮ For large q, substitute κ = q/2 in RHS to get

κ ≈ q sinh qa

2

2(cosh qa

2 − cos(ka)) ≈ q

2

• 1 + 2e−qa/2 cos(ka)
• 16

δ-crystal: tight-binding limit

◮ For large q,

κ ≈ q 2

• 1 + 2e−qa/2 cos(ka)
• ◮ Correspondingly, the energy

E = −2κ2 2m ≈ −2q2 8m

E0

−2q2e−qa/2 2m cos(ka)

◮ Exactly like the coupled springs! ◮ First term = band center = energy E0 of isolated δ-atom ◮ Second term ∝ band width and coupling strength ∝ e−qa/2 ◮ Coupling proportional to overlap between wavefunctions at adjacent atoms

17

δ-crystal: free states

◮ For bound states E = −2κ2 2m

< 0, we derived eigenvalue condition cosh(κa) − q 2κ sinh(κa) = cos(ka)

◮ For free states E = 2K2 2m

> 0 (using K since k is taken), substitute κ = iK above: cosh(iKa) − q 2iK sinh(iKa) = cos(ka) which simplifies to cos(Ka) − q 2K sin(Ka) = cos(ka)

18

δ-crystal: energy conditions

cosh(κa)− q 2κ sinh(κa) = cos(ka)

• r

cos(Ka)− q 2K sin(Ka) = cos(ka)

◮ Plot LHS as a function of E = 2K2 2m

• r −2κ2

2m ; note |RHS| ≤ 1

• 1

1

• 10
• 5

5 10 15 20 25 30 35 40 For qa = 5 cos(ka) E h

• 2/(ma2)

E = -h

• 2κ2/(2m) < 0

E = h

• 2K2/(2m) > 0

◮ When LHS magnitude > 1, those κ or K not valid solutions ◮ Correspondingly, not all E valid: bands!

19

δ-crystal: band structure

E h

• 2/(ma2)

k For qa = 5

• 20
• 10

10 20 30 40

• 3π/a
• 2π/a
• 1π/a

0π/a 1π/a 2π/a 3π/a

◮ Now plot with respect to k instead ◮ Repeats when k → k + 2π/a as for the classical springs

20

δ-crystal: band structure qa = 0

E h

• 2/(ma2)

k For qa = 0

• 20
• 10

10 20 30 40

• 3π/a
• 2π/a
• 1π/a

0π/a 1π/a 2π/a 3π/a

◮ Potentials no longer present: free electrons!

21

δ-crystal: band structure qa = 1

E h

• 2/(ma2)

k For qa = 1

• 20
• 10

10 20 30 40

• 3π/a
• 2π/a
• 1π/a

0π/a 1π/a 2π/a 3π/a

◮ Weak potentials: nearly-free electrons ◮ Potential opens up ‘gaps’ at k = 0, ±π/a

22

δ-crystal: band structure qa = 2

E h

• 2/(ma2)

k For qa = 2

• 20
• 10

10 20 30 40

• 3π/a
• 2π/a
• 1π/a

0π/a 1π/a 2π/a 3π/a

◮ Gaps widen as potential gets stronger ◮ Electrons more tightly bound

23

δ-crystal: band structure qa = 5

E h

• 2/(ma2)

k For qa = 5

• 20
• 10

10 20 30 40

• 3π/a
• 2π/a
• 1π/a

0π/a 1π/a 2π/a 3π/a

◮ Band widths reduce as potential gets stronger ◮ Band structure resembles tight-binding case (C + D cos ka)

24

δ-crystal: band structure qa = 10

E h

• 2/(ma2)

k For qa = 10

• 20
• 10

10 20 30 40

• 3π/a
• 2π/a
• 1π/a

0π/a 1π/a 2π/a 3π/a

◮ Bands begin to approach discrete energy levels like atoms ◮ At same strength, higher energies less affected ⇒ wider bands

25

δ-crystal wavefunctions

◮ Wavefunction of the form

ψ(x) = Ane−κ(x−na) + Bneκ(x−(n+1)a), na < x < (n + 1)a

◮ Solution An = eikna ◮ From matching conditions

(2 + q/κ)Ant − 2An+1 = −(q/κ)Bn where t = e−κa, yielding Bn =

• (2κ/q)(eika − e−κa) − e−κa

eikna

◮ Substitute:

ψ(x) =

• e−κ(x−na) +

2κ q (eika − e−κa) − e−κa

• eκ(x−(n+1)a)
• eikna

26

δ-crystal wavefunctions: Bloch’s theorem

◮ Same form for E > 0 with κ → iK, so all wavefunctions of form

ψ(x) =

• e−κ(x−na) +

2κ q (eika − e−κa) − e−κa

• eκ(x−(n+1)a)
• eikna

ψ(x) =

• e−iK(x−na) +

2iK q (eika − e−iKa) − e−iKa

• eiK(x−(n+1)a)
• eikna

◮ Relative to start of interval x = na, piece in [

] identical for each n

◮ Change from one unit cell to another only in eikna i.e. relative phase eika ◮ Bloch’s theorem: eigenfunctions of any periodic potential ei k· r× periodic

function u

k(

r)

◮ In this case, periodic part (called Bloch function) is

uk(x) = e−(κ+ik)(x−na)+ 2κ q (1 − e−(κ+ik)a) − e−(κ+ik)a

• e(κ−ik)(x−(n+1)a)

(and similarly for κ → iK)

27

δ-crystal wavefunctions: tight-binding limit

ψ(x) =

• e−κ(x−na) +

2κ q (eika − e−κa) − e−κa

• eκ(x−(n+1)a)
• eikna

◮ Assume large enough V0 and hence q and κ that e−κa ≪ 1. Also, in this

limit, κ ≈ q/2, so: ψ(x) ≈ e−κ(x−na)eikna + eκ(x−(n+1)a)eik(n+1)a

◮ Same form to the left and right of each na, only with κ ↔ −κ ⇒

ψ(x) ≈

• n

e−κ|x−na|eikna (all x)

◮ Exactly the δ-atom wavefunction at each na, combined with phase eikna

(analogous to the spring oscillations)

28

δ-crystal wavefunctions: nearly-free limit

◮ Consider from a different perspective ◮ Free electron eiKx incident on atom at x = 0 ◮ Reflected wave Re−iKx ◮ Free electron eiKx incident on atom at x = a ◮ Reflected wave eiKaRe−iK(x−a) = Re−iKx · e2iKa ◮ Neighbouring atoms reflect with relative phase e2ika ◮ Strong reflections at 2Ka = 2nπ ⇒ K = nπ/a

29

δ-crystal: band structure qa = 1

E h

• 2/(ma2)

k For qa = 1

• 20
• 10

10 20 30 40

• 3π/a
• 2π/a
• 1π/a

0π/a 1π/a 2π/a 3π/a

◮ Hence strongest effect of potential at K = nπ/a (which is k = 0, ±π/a) ◮ Therefore, potential opens up ‘gaps’ at k = 0, ±π/a

30

δ-crystal wavefunctions: general example 1

E h

• 2/(ma2)

k

• 20
• 10

10 20 30 40

• 1π/a

0π/a 1π/a For qa = 5 Re Ψ(x) 0a 1a 2a 3a 4a 5a

◮ Bound state with k ≈ 0 ◮ Similar phase at x = na

31

δ-crystal wavefunctions: general example 2

E h

• 2/(ma2)

k

• 20
• 10

10 20 30 40

• 1π/a

0π/a 1π/a For qa = 5 Re Ψ(x) 0a 1a 2a 3a 4a 5a

◮ Bound state with k ≈ π/(2a) ◮ Phase at x = na has period ∼ 4 unit cells

32

δ-crystal wavefunctions: general example 3

E h

• 2/(ma2)

k

• 20
• 10

10 20 30 40

• 1π/a

0π/a 1π/a For qa = 5 Re Ψ(x) 0a 1a 2a 3a 4a 5a

◮ Bound state with k ≈ π/a ◮ Phase at x = na has period ∼ 2 unit cells (alternates sign) ◮ Maximum probability at attractive δ-potential: lower energy

33

δ-crystal wavefunctions: general example 4

E h

• 2/(ma2)

k

• 20
• 10

10 20 30 40

• 1π/a

0π/a 1π/a For qa = 5 Re Ψ(x) 0a 1a 2a 3a 4a 5a

◮ Free state with k ≈ π/a ◮ Phase at x = na has period ∼ 2 unit cells (alternates sign) ◮ Minimum probability at attractive δ-potential: higher energy ◮ Jump between this and previous case causes jump in energy ⇒ band gap

34

δ-crystal wavefunctions: general example 5

E h

• 2/(ma2)

k

• 20
• 10

10 20 30 40

• 1π/a

0π/a 1π/a For qa = 5 Re Ψ(x) 0a 1a 2a 3a 4a 5a

◮ Free state with k ≈ π/(2a) ◮ Phase at x = na has period ∼ 4 unit cells

35

δ-crystal wavefunctions: general example 6

E h

• 2/(ma2)

k

• 20
• 10

10 20 30 40

• 1π/a

0π/a 1π/a For qa = 5 Re Ψ(x) 0a 1a 2a 3a 4a 5a

◮ Free state with k ≈ 0 ◮ Similar phase at x = na

36

δ-crystal wavefunctions: general example 7

E h

• 2/(ma2)

k

• 20
• 10

10 20 30 40

• 1π/a

0π/a 1π/a For qa = 5 Re Ψ(x) 0a 1a 2a 3a 4a 5a

◮ Free state with k ≈ 0, but higher K ◮ Similar phase at x = na, but more oscillations within each unit cell

37

δ-crystal wavefunctions: general example 8

E h

• 2/(ma2)

k

• 20
• 10

10 20 30 40

• 1π/a

0π/a 1π/a For qa = 5 Re Ψ(x) 0a 1a 2a 3a 4a 5a

◮ Free state with k ≈ π/(2a) ◮ Phase at x = na has period ∼ 4 unit cells

38

δ-crystal wavefunctions: general example 9

E h

• 2/(ma2)

k

• 20
• 10

10 20 30 40

• 1π/a

0π/a 1π/a For qa = 5 Re Ψ(x) 0a 1a 2a 3a 4a 5a

◮ Free state with k ≈ π/a ◮ Phase at x = na has period ∼ 2 unit cells (alternates sign)

39

Many electrons

◮ So far: energy of a single electron in a periodic potential ◮ Reality: many interacting electrons ◮ DFT picture: eigen-states of effective potential VKS(

r), fill up electrons in ascending order of energy

◮ Assume δ-crystal potential was such an effective potential ◮ To fill electrons, need to know how many states in a band? ◮ Periodic boundary conditions in length L, spacing in k is 2π/L ◮ Range of k is from −π/a to +π/a with length 2π/a ◮ Therefore, number of k is (2π/a)/(2π/L) = L/a = number of unit cells ◮ So one electron per unit cell ⇔ one set of

k ∈ [−π/a, π/a]

◮ Except spin! Two electrons per unit cell ⇔ one set of

k ∈ [−π/a, π/a]

40

Filling up electrons

k E h

• 2/(ma2)

10 20 30 40

• π/a

π/a

1/cell 2/cell 3/cell 4/cell 5/cell 6/cell

◮ Filled state with maximum energy:

Highest Occupied Molecular Orbital (HOMO)

◮ Empty state with minimum energy:

Lowest Unoccupied Molecular Orbital (LUMO)

◮ Each electron/cell fills half a band ◮ Odd electrons/cell: HOMO = LUMO

in middle of band ⇒ metal

◮ Even electrons/cell: HOMO at band

maximum, gap to LUMO at next band minimum ⇒ semiconductor / insulator

◮ Two types of semiconductors in 1D:

direct gap at Γ i.e. k = 0 or direct gap at zone boundary k = ±π

a

41

Beyond one dimension

◮ So far: 1D crystal, unit cell length a ◮ Electron energies depend on Bloch wave-vector k periodic with 2π/a ◮ Periodicity because phase between unit cells eika unchanged when

k → k + 2π/a

◮ Therefore we could represent all properties in interval [−π/a, π/a] ◮ In 3D, crystal represented by lattice vectors

a1, a2 and a3

◮ Electron energies depend on Bloch wave-vector

k

◮ What is the periodicity in

k?

◮ Phase between unit cells ei k· aj along jth lattice direction (j = 1, 2, 3) ◮ If

k → k + G such that G · aj = 2πnj, then ei

k· aj → ei( k+ G)· aj = ei k· ajei2πnj = ei k· aj (unchanged)

42

Reciprocal lattice

◮ Consider vectors

bi defined such that bi · aj = 2πδij

◮ Suppose

G =

i mi

bi ≡ m1 b1 + m2 b2 + m3 b3 with mi integers

◮ Then

G · aj =

i mi

bi · aj =

i miδij = mj (an integer) ◮ So all the

G that leave ei

k· aj unchanged also a lattice with basis vectors

• b1,

b2, b3

◮ Let A = (

a1, a2, a3) and B = ( b1, b2, b3) (3x3 matrices with vectors in columns)

◮ Defining condition written as BT · A = 2πI ⇒ B = 2πA−T

(generalization of 2π/a)

◮ Specifically in 3D, this inverse can be written explicitly as

• b1 = 2π(

a2 × a3)

• a1 · (

a2 × a3),

• b2 = 2π(

a3 × a1)

• a2 · (

a3 × a1), and

• b3 = 2π(

a1 × a2)

• a3 · (

a1 × a2)

◮ Reciprocal lattice vectors relevant for (X-ray / electron) diffraction from

crystals because constructive interference when ei

G· aj = 1

43

Brillouin zone

◮ Unit cell of the reciprocal lattice (any shape possible, only volume matters) ◮ Fundamental Brillouin zone: set of

k closer to G = 0 than any other

• G =

i mi

bi

◮ In 1D, reduces to [−π/a, +π/a] (i.e. the set of k such that |k| < π/a ◮ For cubic crystals with spacing a, a cube with |kx|, |ky|, |kz| ≤ π/a ◮ For general lattice, shapes can be quite complex (up to 6 edges in 2D, 14

facets in 3D)

44

Example: FCC lattice

◮ Face centered cubic lattice with cubic length a:

( a1, a2, a3) = a 2   1 1 1 1 1 1  

◮ Reciprocal lattice is body-centered with cubic length 4π/a:

( b1, b2, b3) = 4π/a 2   −1 1 1 1 −1 1 1 1 −1  

◮ Brillouin zone is a truncated octahedron ◮ 8 hexagonal faces: nearest neighbours ◮ 6 square faces: next-nearest neighbours L W Γ K X

45

Band structure of gold (FCC metal)

◮ Band structure E(

k) for k ∈BZ: need 4D plot!

◮ Paths connecting special high-symmetry points show important features

• 6
• 4
• 2

2 4 6 Γ X W L Γ K Energy [eV] PBEsol+U GLLBsc QSGW Expt

(DFT vs. ARPES [Nature Comm. 5, 5788 (2014)])

◮ Note complexity compared to 1D:

many bands overlapping in energy

◮ E = 0 typically set to HOMO level ◮ Valence electrons/cell = 11 (odd), configuration: 5d106s1 L W Γ K X

46

Band structure of platinum (FCC metal)

• 10
• 5

5 10 Γ X W L Γ K E - EF [eV]

◮ HOMO = Valence Band Maximum (VBM) with energy Ev

and LUMO = Conduction Band Minimum (CBM) with energy Ec

◮ Valence electrons/cell = 10 (even), configuration: 5d96s1 ◮ Even electrons / cell can be metallic in 3D!

47

Band structure of platinum at a = 20 ˚ A

5d 6s 6p

48

Band structure of platinum at a = 10 ˚ A

• 10
• 5

5 10 Γ X W L Γ K ε - εF [eV]

49

Band structure of platinum at a = 6 ˚ A

• 10
• 5

5 10 Γ X W L Γ K ε - εF [eV]

50

Band structure of platinum at a = 3.92 ˚ A

• 10
• 5

5 10 Γ X W L Γ K ε - εF [eV]

51

Band structure of silicon (diamond-cubic semiconductor)

E in eV

6 −12 L Λ Γ Δ Χ Σ Γ Ev Ec −6

◮ HOMO = Valence Band Maximum (VBM) with energy Ev

and LUMO = Conduction Band Minimum (CBM) with energy Ec

◮ HOMO-LUMO gap Eg = Ec − Ev ≈ 1.1 eV ◮ HOMO and LUMO at different

k ⇒ indirect band gap

◮ Diamond: similar band structure, much larger gap (≈ 5.5 eV) ⇒ insulator ◮ Valence electrons/cell = 8 (even), configuration: 3s23p2 (two Si/cell)

52

Band structure of GaAs (zinc-blende semiconductor)

◮ HOMO-LUMO gap Eg = Ec − Ev ≈ 1.4 eV ◮ HOMO and LUMO at same

k (Γ) ⇒ direct band gap

◮ Valence electrons/cell = 8 (even), configuration: Ga(4s24p1), As(4s24p3)

53

Electron motion in a crystal

◮ Crystal causes electronic band structure E(

k)

◮ Apply force F to electron

• F = d

p dt = d k dt

◮ Acceleration of electron

d v dt = d dt

• ∇

kω(

k)

• = d

dt ∇

kE(

k)

• = ∇

k

• ∇

kE(

k)1 · d k dt

• = ∇

k

• ∇

kE(

k) 1 2 · F

• 54

Effective mass

◮ Quantum mechanically d v dt = ∇ k

• ∇

kE(

k)/2 · F

• ◮ Compare with classical d

v/dt = F/m

◮ ∇ k

• ∇

kE(

k)/2·

• effectively like inverse mass

◮ In general, effective mass tensor m∗ ij(

k) = 2inv

• ∂ki∂kjE(

k)

• ◮ Depends on band and

k in general, not a single constant!

◮ For metals, typically report average m∗ for all HOMO levels (Fermi surface) ◮ For semiconductors, typically report average m∗ each for VBM and CBM

55

Effective mass: typical values

Material m∗/me Fermi surface VBM CBM Cu 1.01 Ag 0.99 Au 1.10 Pt 13 Si

• 0.16,-0.49

0.98,0.19 Ge

• 0.04,-0.28

1.64,0.08 GaAs

• 0.082

0.067

◮ Multiple values in column ⇒ anisotropic (longitudinal / transverse) ◮ Recall mobility µ = eτ m → eτ m∗ ◮ Semiconductors: typically lower effective mass ⇒ higher mobility ◮ Noble metals: free-electron like because m∗ ≈ me ◮ d-metals like Pt: flat bands, low curvature (∂2E/∂k2), high m∗ (poor

mobility)

56

Negative effective mass: holes

k E h

• 2/(ma2)

10 20 30 40

• π/a

π/a

◮ Top of bands: negative

∂2E/∂k2 ⇒ m∗ < 0

◮ Apply force: electron moves in

• pposite direction!

◮ If band filled except for one electron,

missing electron behaves like positive charge (moves in direction of force)

◮ Semiconductor at finite T: few

electrons in empty conduction bands, few holes in filled valence bands; both conduct

◮ But why do all the electrons in the

filled band not conduct?

57

Conductivity due to a filled band in 1D

◮ Consider a single filled band E(k) with applied electric field E ◮ Drift velocity of electrons in state k is v(k) = −eEτ/m∗(k) ◮ Average drift velocity of all electrons in band is:

vd = v(k)k ≡ a 2π π/a

−π/a

dkv(k) = a 2π π/a

−π/a

dk (−eEτ/m∗(k)) = −eEτ a 2π π/a

−π/a

dk 1 m∗(k) = −eEτ a 2π π/a

−π/a

dk ∂2E 2∂k2 = −eEτ a 2π ∂E 2∂k π/a

−π/a

= −eEτ a 2π

• ∂E

2∂k

• k=π/a

− ∂E 2∂k

• k=−π/a
• = 0

because of the periodicity of E and hence ∂E/∂k.

◮ Filled band does not conduct: positive and negative mass contributions

cancel exactly!

◮ Same proof extends to 3D with some vector calculus

58

Density of states

◮ Most properties depend on the number of states available at a given energy ◮ We know distribution of states (per unit volume, including spin) with

k: 2dk 2π in 1D,

• 2d

k (2π)2 in 2D,

• 2d

k (2π)3 in 3D

◮ We used this to count states in frequency intervals for light ω = ck ◮ In general, given En(

k) for many bands indexed by n, number of states per energy interval dE: g(E) =

• n
• 2d

k (2π)d δ(E − En( k)) (any dimension)

◮ For a free electron E(k) = 2k2/(2m),

g(E) = √ 2m 2π d Θ(E)      2/ √ E, d = 1 2π, d = 2 4π √ E, d = 3

59

Density of states: free electrons

5 10 15 1 2 3 4 5 g(E) [eV-1nm-d] E [eV] 1D 2D 3D ◮ Singularity at band edge in 1D (∝ 1/

√ E)

◮ Constant and abruptly drops to zero at band edge in 2D ◮ Goes to zero smoothly at band edge in 3D (∝

√ E)

60

Density of states: parabolic-band semiconductor

1 2 3 4

• 3
• 2
• 1

1 2 3 4 5 E [eV] g(E) [eV-1nm-3] Valence band, m* = -0.3 Conduction band, m* = 0.5 ◮ Parabolic bands near each band edge, with different effective masses ◮ Overall DOS reduces with reduced effective mass magnitude

61

Density of states: silicon

◮ Can calculate numerically from band structure ◮ Parabolic band approximation valid only for narrow energy range near gap

62