mtle-6120: advanced electronic properties of materials...

MTLE-6120: Advanced Electronic Properties of Materials

Band theory of solids

Reading:

I Kasap: 4.1 - 4.5

1

Crystals

I Periodic arrangement of atoms (ions: nuclei + core electrons)

I Previously: classical motion of electrons in crystal

I What was the role of the periodicity? None!

I Next: quantum motion of electrons in crystal

I Bravais lattice: regular grid of points

I Lattice vectors ~a1,~a2, ~a3I Any point on grid n1~a1 + n2~a2 + n3 ~a3I Unit cell: repeat at grid points to fill space

I Multiple atoms possible per unit cell

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2

Classical example: one oscillator

I Mass m attached to a spring with constant k

I Equation of motion in displacement x:

mx = −kx

I In terms of ω0 =√k/m,

x = −ω20x

I Oscillatory solutions e±iω0t with angular frequency ω0

3

Classical example: two oscillators

I Two masses m each attached to a spring with constant k

I Equation of motion in displacements x1 and x2:

mx1 = −kx1mx2 = −kx2

I Solutions: both x1 and x2 oscillate with ω0 =√k/m

4

Classical example: two coupled oscillators

I Two masses m each attached to a spring with constant kI Attached to each other with another spring with constant KI Equation of motion in displacements x1 and x2:

mx1 = −kx1 −K(x1 − x2)

mx2 = −kx2 −K(x2 − x1)

I Coupled equations, x1 and x2 are not each an oscillatorI Add and subtract

m(x1 + x2) = −k(x1 + x2)

m(x1 − x2) = −(k + 2K)(x1 − x2)

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

√k/m

I (x1 − x2) oscillates with frequency√

(k + 2K)/m

5

Coupled oscillators: eigenvalue problem

I Equation of motion in matrix form:

md2

dt2

(x1x2

)= −

(k +K −K−K k +K

)︸︷︷︸

A

·(x1x2

)

I Coupled equations because of off-diagonal terms in A

I Diagonalize A = V ·D · V −1 using eigenvalue diagonal matrix D andeigenvector diagonal matrix V

md2

dt2V −1 ·

(x1x2

)= −D · V −1 ·

(x1x2

)I V −1 combines x into normal modes x1 + x2 and x1 − x2I Corresponding eigenvalues are mω2 = (k +K)∓K i.e. k and k + 2K

6

Three coupled oscillators

I For three oscillators

A =

k +K −K 0−K k + 2K −K

0 −K k +K

I Eigenvalues mω2: k

k +Kk + 3K

7

N coupled oscillators

I 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

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

eigenvalue mω2 = λ (say)I nth row (n 6= 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 problemI nth row (n 6= 1, N) of eigenvalue equation AX = λX is

Cxn−1 +Dxn + Cxn+1 = λxn

I 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)I 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)

I Connect last spring to first

9

Periodic boundary conditions (Born-von Karman)I 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)

I Canceling common factors:

(eNika)Ce−ika +D + Ceika = λ (n = 1)

Ce−ika +D + Ceika(e−Nika) = λ (n = N)

I These are the same as the other equations when eNika = 1 i.e.k = 2πj/(Na)

I Here, j can be any integer, but for j → j +N , k → k + 2π/a andeika → ei(ka+2π) = eika, so nothing changes

10

N →∞ coupled oscillatorsI Eigenvalues mω2 = λ = D + 2C cos ka = (k + 2K)− 2K cos ka

I k = 2πj/(Na) for j = 0, 1, 2, . . . , N − 1 becomes continuous k ∈ [0, 2π)

I 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

I Eigenvalues mω2 form a ‘band’ with range [k, k + 4K]

I Band center set by diagonal element k + 2K

I Band width set by off-diagonal element K (coupling strength)

I Any interval in k of length 2π/a equivalent; conventionally [−π/a, π/a]

11

1D crystal of δ-atomsI An array of δ-potentials at na i.e. V (x) = −V0

∑n δ(x− na)

I Consider E < 0 case first, with κ =√−2mE/~

I Wavefunctions e±κx in each segment, say

ψ(x) = Ane−κ(x−na) +Bne

κ(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

− qκ

(An +Bnt) = −An +Bnt+An−1t−Bn−1

− qκ

(Ant+Bn) = −An+1 +Bn+1t+Ant−Bn

13

δ-crystal: eliminate B’sI 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

I 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

I 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

I 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)

I Since t ≡ e−κa

cosh(κa)− q

2κsinh(κa) = cos(ka)

15

δ-crystal: eigenvaluesI Energy eigenvalue is E = ~ω = −~2κ2

2m

I For the δ-atom, we got a unique κ = mV0/~2 ≡ q/2I But now, an entire family of solutions for k ∈ (−π/a, π/a):

cosh(κa)− q


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

κ =q sinh(κa)

2(cosh(κa)− cos(ka))

I Large q ⇒ large κ⇒ sinh, cosh extremely large

I For q →∞, κ→ q/2 as in the δ-atom

I For large q, substitute κ = q/2 in RHS to get

κ ≈q sinh qa

2

2(cosh qa2 − cos(ka))

≈ q

2

(1 + 2e−qa/2 cos(ka)

)

16

δ-crystal: tight-binding limitI For large q,

κ ≈ q

2

(1 + 2e−qa/2 cos(ka)

)I Correspondingly, the energy

E = −~2κ2

2m≈ −~2q2

8m︸︷︷︸E0

−~2q2e−qa/2

2mcos(ka)

I Exactly like the coupled springs!

I First term = band center = energy E0 of isolated δ-atom

I Second term ∝ band width and coupling strength ∝ e−qa/2

I Coupling proportional to overlap between wavefunctions at adjacent atoms

17

δ-crystal: free statesI For bound states E = −~2κ2

2m < 0, we derived eigenvalue condition

cosh(κa)− q


I For free states E = ~2K2

2m > 0 (using K since k is taken), substituteκ = iK above:

cosh(iKa)− q

2iKsinh(iKa) = cos(ka)

which simplifies to

cos(Ka)− q

2Ksin(Ka) = cos(ka)

18

δ-crystal: energy conditions

cosh(κa)− q

2κsinh(κa) = cos(ka) or cos(Ka)− q

2Ksin(Ka) = cos(ka)

I Plot LHS as a function of E = ~2K2

2m or −~2κ2

2m ; note |RHS| ≤ 1

-1

0

1

-10 -5 0 5 10 15 20 25 30 35 40

For qa = 5cos(ka)

E

h-2/(ma

2)

E = -h-2κ

2/(2m) < 0 E = h-

2K

2/(2m) > 0

I When LHS magnitude > 1, those κ or K not valid solutions

I Correspondingly, not all E valid: bands!

19

δ-crystal: band structureE

h-2/(ma

2)

k

For qa = 5

-20

-10

0

10

20

30

40

-3π/a -2π/a -1π/a 0π/a 1π/a 2π/a 3π/a

I Now plot with respect to k instead

I Repeats when k → k + 2π/a as for the classical springs

20

δ-crystal: band structure qa = 0E

h-2/(ma

2)

k

For qa = 0

-20

-10

0

10

20

30

40

-3π/a -2π/a -1π/a 0π/a 1π/a 2π/a 3π/a

I Potentials no longer present: free electrons!

21


h-2/(ma

2)

k

For qa = 1

-20

-10

0

10

20

30

40

-3π/a -2π/a -1π/a 0π/a 1π/a 2π/a 3π/a

I Weak potentials: nearly-free electrons

I Potential opens up ‘gaps’ at k = 0,±π/a

22


h-2/(ma

2)

k

For qa = 2

-20

-10

0

10

20

30

40

-3π/a -2π/a -1π/a 0π/a 1π/a 2π/a 3π/a

I Gaps widen as potential gets stronger

I Electrons more tightly bound

23


h-2/(ma

2)

k

For qa = 5

-20

-10

0

10

20

30

40

-3π/a -2π/a -1π/a 0π/a 1π/a 2π/a 3π/a

I Band widths reduce as potential gets stronger

I Band structure resembles tight-binding case (C +D cos ka)

24


h-2/(ma

2)

k

For qa = 10

-20

-10

0

10

20

30

40

-3π/a -2π/a -1π/a 0π/a 1π/a 2π/a 3π/a

I Bands begin to approach discrete energy levels like atoms

I At same strength, higher energies less affected ⇒ wider bands

25

δ-crystal wavefunctions

I Wavefunction of the form

ψ(x) = Ane−κ(x−na) +Bne

κ(x−(n+1)a), na < x < (n+ 1)a

I Solution An = eikna

I 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

I 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

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

ψ(x) =

[e−κ(x−na) +

{2κ


}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

I Relative to start of interval x = na, piece in [ ] identical for each n

I Change from one unit cell to another only in eikna i.e. relative phase eika

I Bloch’s theorem: eigenfunctions of any periodic potential ei~k·~r× periodic

function u~k(~r)

I 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


}eκ(x−(n+1)a)

]eikna

I Assume large enough V0 and hence q and κ that e−κa � 1. Also, in thislimit, κ ≈ q/2, so:

ψ(x) ≈ e−κ(x−na)eikna + eκ(x−(n+1)a)eik(n+1)a

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

ψ(x) ≈∑n

e−κ|x−na|eikna (all x)

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

(analogous to the spring oscillations)

28

δ-crystal wavefunctions: nearly-free limit

I Consider from a different perspective

I Free electron eiKx incident on atom at x = 0

I Reflected wave Re−iKx

I Free electron eiKx incident on atom at x = a

I Reflected wave eiKaRe−iK(x−a) = Re−iKx · e2iKa

I Neighbouring atoms reflect with relative phase e2ika

I Strong reflections at 2Ka = 2nπ ⇒ K = nπ/a

29


h-2/(ma

2)

k

For qa = 1

-20

-10

0

10

20

30

40

-3π/a -2π/a -1π/a 0π/a 1π/a 2π/a 3π/a

I Hence strongest effect of potential at K = nπ/a (which is k = 0,±π/a)

I Therefore, potential opens up ‘gaps’ at k = 0,±π/a

30

δ-crystal wavefunctions: general example 1E

h-2/(ma

2)

k

-20

-10

0

10

20

30

40

-1π/a 0π/a 1π/a

For qa = 5Re Ψ(x)

0a 1a 2a 3a 4a 5a

I Bound state with k ≈ 0

I Similar phase at x = na

31


h-2/(ma

2)

k

-20

-10

0

10

20

30

40

-1π/a 0π/a 1π/a

For qa = 5Re Ψ(x)

0a 1a 2a 3a 4a 5a

I Bound state with k ≈ π/(2a)

I Phase at x = na has period ∼ 4 unit cells

32


h-2/(ma

2)

k

-20

-10

0

10

20

30

40

-1π/a 0π/a 1π/a

For qa = 5Re Ψ(x)

0a 1a 2a 3a 4a 5a

I Bound state with k ≈ π/aI Phase at x = na has period ∼ 2 unit cells (alternates sign)

I Maximum probability at attractive δ-potential: lower energy

33


h-2/(ma

2)

k

-20

-10

0

10

20

30

40

-1π/a 0π/a 1π/a

For qa = 5Re Ψ(x)

0a 1a 2a 3a 4a 5a

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

I Minimum probability at attractive δ-potential: higher energy

I Jump between this and previous case causes jump in energy ⇒ band gap

34


h-2/(ma

2)

k

-20

-10

0

10

20

30

40

-1π/a 0π/a 1π/a

For qa = 5Re Ψ(x)

0a 1a 2a 3a 4a 5a

I Free state with k ≈ π/(2a)


35


h-2/(ma

2)

k

-20

-10

0

10

20

30

40

-1π/a 0π/a 1π/a

For qa = 5Re Ψ(x)

0a 1a 2a 3a 4a 5a

I Free state with k ≈ 0

I Similar phase at x = na

36


h-2/(ma

2)

k

-20

-10

0

10

20

30

40

-1π/a 0π/a 1π/a

For qa = 5Re Ψ(x)

0a 1a 2a 3a 4a 5a

I Free state with k ≈ 0, but higher K

I Similar phase at x = na, but more oscillations within each unit cell

37


h-2/(ma

2)

k

-20

-10

0

10

20

30

40

-1π/a 0π/a 1π/a

For qa = 5Re Ψ(x)

0a 1a 2a 3a 4a 5a

I Free state with k ≈ π/(2a)


38


h-2/(ma

2)

k

-20

-10

0

10

20

30

40

-1π/a 0π/a 1π/a

For qa = 5Re Ψ(x)

0a 1a 2a 3a 4a 5a

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

39

Many electrons

I So far: energy of a single electron in a periodic potential

I Reality: many interacting electrons

I DFT picture: eigen-states of effective potential VKS(~r), fill up electrons inascending order of energy

I Assume δ-crystal potential was such an effective potential

I To fill electrons, need to know how many states in a band?

I Periodic boundary conditions in length L, spacing in k is 2π/L

I Range of k is from −π/a to +π/a with length 2π/a

I Therefore, number of k is (2π/a)/(2π/L) = L/a = number of unit cells

I So one electron per unit cell ⇔ one set of ~k ∈ [−π/a, π/a]

I Except spin! Two electrons per unit cell ⇔ one set of ~k ∈ [−π/a, π/a]

40

Filling up electrons

k

Eh-2/(ma2)

0

10

20

30

40

-π/a 0 π/a

1/cell2/cell

3/cell

4/cell

5/cell

6/cell

I Filled state with maximum energy:Highest Occupied Molecular Orbital(HOMO)

I Empty state with minimum energy:Lowest Unoccupied Molecular Orbital(LUMO)

I Each electron/cell fills half a band

I Odd electrons/cell: HOMO = LUMOin middle of band ⇒ metal

I Even electrons/cell: HOMO at bandmaximum, gap to LUMO at nextband minimum ⇒ semiconductor /insulator

I Two types of semiconductors in 1D:direct gap at Γ i.e. k = 0 ordirect gap at zone boundary k = ±π

a

41

Beyond one dimension

I So far: 1D crystal, unit cell length a

I Electron energies depend on Bloch wave-vector k periodic with 2π/a

I Periodicity because phase between unit cells eika unchanged whenk → k + 2π/a

I Therefore we could represent all properties in interval [−π/a, π/a]

I In 3D, crystal represented by lattice vectors ~a1, ~a2 and ~a3I Electron energies depend on Bloch wave-vector ~k

I What is the periodicity in ~k?

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

I 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

I Consider vectors ~bi defined such that ~bi · ~aj = 2πδij

I Suppose ~G =∑imi

~bi ≡ m1~b1 +m2

~b2 +m3~b3 with mi integers

I Then ~G · ~aj =∑imi

~bi · ~aj =∑imiδij = mj (an integer)

I So all the ~G that leave ei~k·~aj unchanged also a lattice with basis vectors

~b1, ~b2, ~b3I Let A = (~a1,~a2,~a3) and B = (~b1,~b2,~b3)

(3x3 matrices with vectors in columns)

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

(generalization of 2π/a)

I 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)

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

crystals because constructive interference when ei~G·~aj = 1

43

Brillouin zone

I Unit cell of the reciprocal lattice (any shape possible, only volume matters)

I Fundamental Brillouin zone: set of ~k closer to ~G = 0 than any other~G =

∑imi

~biI In 1D, reduces to [−π/a,+π/a] (i.e. the set of k such that |k| < π/a

I For cubic crystals with spacing a, a cube with |kx|, |ky|, |kz| ≤ π/aI For general lattice, shapes can be quite complex (up to 6 edges in 2D, 14

facets in 3D)

44

Example: FCC latticeI Face centered cubic lattice with cubic length a:

(~a1,~a2,~a3) =a

2

0 1 11 0 11 1 0

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

(~b1,~b2,~b3) =4π/a

2

−1 1 11 −1 11 1 −1

I Brillouin zone is a truncated octahedron

I 8 hexagonal faces: nearest neighbours

I 6 square faces: next-nearest neighbours

L

W

Γ

K

X

45

Band structure of gold (FCC metal)

I Band structure E(~k) for ~k ∈BZ: need 4D plot!

I Paths connecting special high-symmetry points show important features

-6

-4

-2

0

2

4

6

Γ X W L Γ K

Ener

gy [e

V]

PBEsol+UGLLBscQSGW

Expt

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

I Note complexity compared to 1D:many bands overlapping in energy

I E = 0 typically set to HOMO level

I Valence electrons/cell = 11 (odd), configuration: 5d106s1

L

W

Γ

K

X

46

Band structure of platinum (FCC metal)

-10

-5

0

5

10

Γ X W L Γ K

E -

EF [

eV

]

I HOMO = Valence Band Maximum (VBM) with energy Evand LUMO = Conduction Band Minimum (CBM) with energy Ec

I Valence electrons/cell = 10 (even), configuration: 5d96s1

I Even electrons / cell can be metallic in 3D!

47

Band structure of platinum at a = 20 A

5d6s

6p

48


-10

-5

0

5

10

Γ X W L Γ K

ε -

εF [eV

]

49


-10

-5

0

5

10

Γ X W L Γ K

ε -

εF [eV

]

50

Band structure of platinum at a = 3.92 A

-10

-5

0

5

10

Γ X W L Γ K

ε -

εF [eV

]

51

Band structure of silicon (diamond-cubic semiconductor)

E in e

V

6

0

−12L Λ Γ Δ Χ Σ Γ

EvEc

−6

I HOMO = Valence Band Maximum (VBM) with energy Evand LUMO = Conduction Band Minimum (CBM) with energy Ec

I HOMO-LUMO gap Eg = Ec − Ev ≈ 1.1 eV

I HOMO and LUMO at different ~k ⇒ indirect band gap

I Diamond: similar band structure, much larger gap (≈ 5.5 eV) ⇒ insulator

I Valence electrons/cell = 8 (even), configuration: 3s23p2 (two Si/cell)

52

Band structure of GaAs (zinc-blende semiconductor)

I HOMO-LUMO gap Eg = Ec − Ev ≈ 1.4 eV

I HOMO and LUMO at same ~k (Γ) ⇒ direct band gap

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

53

Electron motion in a crystal

I Crystal causes electronic band structure E(~k)

I Apply force F to electron

~F =d~p

dt= ~

d~k

dt

I 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

I Quantum mechanically d~vdt = ∇~k

(∇~kE(~k)/~2 · ~F

)I Compare with classical d~v/dt = ~F/m

I ∇~k(∇~kE(~k)/~2·

)effectively like inverse mass

I In general, effective mass tensor m∗ij(~k) = ~2inv

(∂ki∂kjE(~k)

)I Depends on band and ~k in general, not a single constant!

I For metals, typically report average m∗ for all HOMO levels (Fermi surface)

I For semiconductors, typically report average m∗ each for VBM and CBM

55

Effective mass: typical values

Materialm∗/me

Fermi surface VBM CBMCu 1.01Ag 0.99Au 1.10Pt 13Si -0.16,-0.49 0.98,0.19Ge -0.04,-0.28 1.64,0.08

GaAs -0.082 0.067

I Multiple values in column ⇒ anisotropic (longitudinal / transverse)

I Recall mobility µ = eτm →

eτm∗

I Semiconductors: typically lower effective mass ⇒ higher mobility

I Noble metals: free-electron like because m∗ ≈ me

I d-metals like Pt: flat bands, low curvature (∂2E/∂k2), high m∗ (poormobility)

56

Negative effective mass: holes

k

Eh-2/(ma2)

0

10

20

30

40

-π/a 0 π/a

I Top of bands: negative∂2E/∂k2 ⇒ m∗ < 0

I Apply force: electron moves inopposite direction!

I If band filled except for one electron,missing electron behaves like positivecharge (moves in direction of force)

I Semiconductor at finite T : fewelectrons in empty conduction bands,few holes in filled valence bands; bothconduct

I But why do all the electrons in thefilled band not conduct?

57

Conductivity due to a filled band in 1DI Consider a single filled band E(k) with applied electric field EI Drift velocity of electrons in state k is v(k) = −eEτ/m∗(k)I Average drift velocity of all electrons in band is:

vd = 〈v(k)〉k ≡a

2π

∫ π/a

−π/adkv(k) =

a

2π

∫ π/a

−π/adk (−eEτ/m∗(k))

= −eEτ a2π

∫ π/a

−π/adk

1

m∗(k)= −eEτ a

2π

∫ π/a

−π/adk

∂2E

~2∂k2

= −eEτ a2π

[∂E

~2∂k

]π/a−π/a

= −eEτ a2π

(∂E

~2∂k

∣∣∣∣k=π/a

− ∂E

~2∂k

∣∣∣∣k=−π/a

)= 0

because of the periodicity of E and hence ∂E/∂k.I Filled band does not conduct: positive and negative mass contributions

cancel exactly!I Same proof extends to 3D with some vector calculus

58

Density of statesI Most properties depend on the number of states available at a given energy

I We know distribution of states (per unit volume, including spin) with ~k:∫2dk

2πin 1D,

∫2d~k

(2π)2in 2D,

∫2d~k

(2π)3in 3D

I We used this to count states in frequency intervals for light ω = ck

I In general, given En(~k) for many bands indexed by n, number of states perenergy interval dE:

g(E) =∑n

∫2d~k

(2π)dδ(E − En(~k)) (any dimension)

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

g(E) =

(√2m

2π~

)dΘ(E)

2/√E, d = 1

2π, d = 2

4π√E, d = 3

59

Density of states: free electrons

0

5

10

15

0 1 2 3 4 5

g(E

) [e

V-1

nm

-d]

E [eV]

1D2D3D

I Singularity at band edge in 1D (∝ 1/√E)

I Constant and abruptly drops to zero at band edge in 2D

I Goes to zero smoothly at band edge in 3D (∝√E)

60

Density of states: parabolic-band semiconductor

0

1

2

3

4

-3 -2 -1 0 1 2 3 4 5

E [eV]

g(E) [eV-1

nm-3

]Valence band, m

* = -0.3

Conduction band, m* = 0.5

I Parabolic bands near each band edge, with different effective masses

I Overall DOS reduces with reduced effective mass magnitude

61

Density of states: silicon

I Can calculate numerically from band structure

I Parabolic band approximation valid only for narrow energy range near gap

62

mtle-6120: advanced electronic properties of materials...

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