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Lecture VIII

Band theory

dr hab. Ewa Popko

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Band TheoryThe calculation of the allowed electron states in a solid is referred to as band theory or band structure theory.

To obtain the full band structure, we need to solve Schrödinger’s equation for the full lattice potential. This cannot be done exactly and various approximation schemes are used. We will introduce two very different models, the nearly free electron and tight binding models.

We will continue to treat the electrons as independent, i.e. neglect the electron-electron interaction.

3

Bound States in atoms

r4qe = )r(V

o

2

Electrons in isolated atoms occupy discrete allowed energy levels E0, E1, E2 etc. .

The potential energy of an electron a distance r from a positively charge nucleus of charge q is

-8 -6 -4 -2 0 2 4 6 8-5

-4

-3

-2

-1

0

F6 F7 F8 F9

r

V(r)E2

E1

E0

r

0

Increasing Binding Energy

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Bound and “free” states in solids

-8 -6 -4 -2 0 2 4 6 8-5

-4

-3

-2

-1

0

F6 F7 F8 F9

r-8 -6 -4 -2 0 2 4 6 8

-5

-4

-3

-2

-1

0

F6 F7 F8 F9

r-8 -6 -4 -2 0 2 4 6 8

-5

-4

-3

-2

-1

0

F6 F7 F8 F9

r

V(r)E2

E1

E0

The 1D potential energy of an electron due to an array of nuclei of charge q separated by a distance a is

Where n = 0, +/-1, +/-2 etc.

This is shown as the black line in the figure.

n

2

naro4qe = rV

)(

r

0

0+ + + + +

aNuclear positions

V(r) lower in solid (work function).

Naive picture: lowest binding energy states can become free to move throughout crystal

V(r)Solid

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Energy Levels and Bands Isolated atoms have precise allowed energy levels.

In the presence of the periodic lattice potential bands of allowed states are separated by energy gaps for which there are no allowed energy states.

The allowed states in conductors can be constructed from combinations of free electron states (the nearly free electron model) or from linear combinations of the states of the isolated atoms (the tight binding model).

+E

+ + + +position

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Influence of the lattice periodicityIn the free electron model, the allowed energy states are

 where for periodic boundary conditions

 nx , ny and ny positive or negative integers.

)(2

2222

zyx kkkm

E

Lnk

Ln

kLnk z

zy

yx

x 2;

2;2

E

k0

-5

-4

-3

-2

-1

0

r

E

Periodic potentialExact form of potential is complicated

Has property V(r+ R) = V(r) where

R = m1a + m2b + m3c

where m1, m2, m3 are integers and a ,b ,c are the primitive lattice vectors.

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Waves in a periodic latticeRecall X-ray scattering in Solid State:

n= 2asin

Consider a wave, wavelength moving through a 1D lattice of period a.

Strong backscattering for n= 2a

Backscattered waves constructively interfere.

Wave has wavevector k = 2

a

Wave moving to right

Scattered waves moving to left

Scattering potential period a

1D lattice: Bragg condition is k = n/a (n – integer)

3D lattice: Scattering for k to k' occurs if k' = k + G

where G = ha1 + ka2 + la3 h,k,l integer and a1 ,a2 ,a3

are the primitive reciprocal lattice vectors

k

k'

G

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Real and Reciprocal Lattice Spaces • R for a crystal can be expressed in general as:

R=n1a1+n2a2+n3a3 where a1, a2 and a3 are the primitive lattice vectors and n1,n2 and n3 are integers

• Corresponding to a1, a2 and a3 there are three primitive reciprocal lattice vectors: b1, b2 and b3 defined in terms of a1, a2 and a3 by:

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Bragg scattering & energy gaps1D potential period a. Reciprocal lattice vectors G = 2n /a

A free electron of in a state exp( ix/a), ( rightward moving wave) will be Bragg reflected since k = /a and a left moving wave exp( -ix/a) will also exist.

In the nearly free electron model allowed un-normalised states for k = /a are

ψ(+) = exp(ix/a) + exp( - ix/a) = 2 cos(x/a)ψ(-) = exp(ix/a) - exp( - ix/a) = 2i sin(x/a)

+E

+ + + +position a

N.B. Have two allowed states for same k which have different energies

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Reciprocal lattice

Use of reciprocal lattice space: Wave vectors k for Bloch waves lie in the reciprocal lattice space. Translation symmetry=> a Bloch wave can be characterized by two wavevectors (or wavelengths) provided they differ by a reciprocal lattice vector! Example in 1D:

Suppose k’=k+(2/a) then k(x)=exp(ikx)u(x)

and k’(x)=exp(ik’x)u(x)=exp(ikx)exp(i2x/a)u(x) =exp(ikx)u’(x)

essentially have the same “wavelength”

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Cosine solution lower energy than sine solutionCosine solution ψ(+) has maximum electron probability density at minima in potential.

Sine solution ψ(-) has maximum electron probability density at maxima in potential.

Cos(x/a) Sin(x/a)

Cos2(x/a)

Sin2(x/a)

In a periodic lattice the allowed wavefunctions have the property

where R is any real lattice vector.

22 )()( rRr ψψ

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Magnitude of the energy gapLet the lattice potential be approximated by Let the length of the crystal in the x-direction to be L. Note that L/a is the number of unit cells and is therefore an integer. Normalising the wavefunction ψ(+) = Acos(x/a) gives 

so Solving Schrödinger’s equation with  

 

)/2(cos)( 0 axVxV

1)/(cos22

0 dxaxA

L

21

2

LA

)()( ψψ EH

)()()/2cos(2 02

22

EaxVxm

220

22 VmkE

)/2(cos)( 0 axVxV

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Gaps at the Brillouin zone boundaries

At points A ψ(+) = 2 cos(x/a) and E=(k)2/2me - V0/2 .

At points B ψ(-) = 2isin(x/a) and E=(k)2/2me + V0/2 .

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Bloch StatesIn a periodic lattice the allowed wavefunctions have the property

where R is any real lattice vector.

Therefore where the function (R) is real, independent of r, and dimensionless.  Now consider ψ(r + R1 + R2). This can be written Or

Therefore (R1 + R2) = (R1) + (R2)

 

(R) is linear in R and can be written (R) = kxRx + kyRy + kzRz = k.R. wherekx, ky and kz are the components of some wavevector k so

(Bloch’s Theorem) 

22 )()( rRr ψψ

)()( . rRr Rk ψψ ie

)()( )(21

21 rR ψψ RRRr ie

)()( )( rRr Rψψ ie

)()()( )()(2

)(21

211 rRrRRr RRR ψψψ iii eee

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(Bloch’s Theorem)

For any k one can write the general form of any wavefunction as

 where u(r) has the periodicity ( translational symmetry) of the lattice. This is an alternative statement of Bloch’s theorem.

)2()()( . rr rk ueiψ

)1(ψ(r)eR)ψ(r ik.R

Alternative form of Bloch’s Theorem

Re [ψ(x)]

x

Real part of a Bloch function. ψ ≈ eikx for a large fraction of the crystal volume.

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Bloch Wavefunctions: allowed k-statesψ(r) = exp[ik.r]u(r)

kp

Periodic boundary conditions. For a cube of side L we require

ψ(x + L) = ψ(x) etc.. So

but u(x+L) = u(x) because it has the periodicity of the lattice therefore

Therefore i.e. kx = 2 nx/Lnx integer.

Same allowed k-vectors for Bloch states as free electron states.

Bloch states are not momentum eigenstates i.e.

The allowed states can be labelled by a wavevectors k.

Band structure calculations give E(k) which determines the dynamical behaviour.

L)u(xeL)u(xe xikL)(xik xx

xikL)(xik xx ee

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Nearly Free ElectronsNeed to solve the Schrödinger equation. Consider 1D

write the potential as a Fourier sum

where G = 2n/a and n are positive and negative integers. Write a general Bloch function

where g = 2m/a and m are positive and negative integers. Note the periodic function is also written as a Fourier sum

Must restrict g to a small number of values to obtain a solution. For n= + 1 and –1 and m=0 and 1, and k ~ /a

E=(k)2/2me + or - V0/2

(x) E = (x) (x) V + x

2m

- 2

22

ψψ

G

iGxGeVxV )(

g

igxg

ikxikx eAeruex )()(ψ

Construct Bloch wavefunctions of electrons out of plane wave states.

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Tight Binding ApproximationNFE Model: construct wavefunction as a sum over plane waves.

Tight Binding Model: construct wavefunction as a linear combination of atomic orbitals of the atoms comprising the crystal.

Where (r)is a wavefunction of the isolated atom

rj are the positions of the atom in the crystal.

) ( c = )( j j

jr-r r ψ

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Molecular orbitals and bondingConsider a electron in the ground, 1s, state of a hydrogen atom

The Hamiltonian is

Solving Schrödinger’s equation :

E = E1s = -13.6eV

o

2

4e = where

RadiusBohr theis a where a 1 = (r) i.e. oo e ar/-3/2 o

r

- 2m

- = H22

+E1s

V(r)

(r)

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Hydrogen Molecular IonConsider the H2

+ molecular ion in whichone electron experiences the potential of two protons. The Hamiltonian is

We approximate the electron wavefunctions as

and

|R - r|-

r -

2m- = )rU( +

2m- = H

2222

] + A[ |)] R - r(| + )r([ A = )r( 21 ψ

] B[ |)]R - r(| )r([ B = )r( 21 ψ

p+ p+

e-

R

r

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Bonding andanti-bonding states Solution:E = E1s – (R) for

E = E1s + (R) for

(R) - a positive function

Two atoms: original 1s stateleads to two allowed electron states in molecule.

Find for N atoms in a solid have N allowed energy states

)r(ψ

)r(ψ)r(ψ

-6 -4 -2 0 2 4 6

-1.4

-1.2

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

r

-6 -4 -2 0 2 4 6

-1.4

-1.2

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

r

V(r)

2)r(ψ

)r(ψ

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The tight binding approximation for s states

)aos(kc 2- -)e ae a( - - = k)( E xxx k i-k i

+ + + + +aNuclear positions

Solution leads to the E(k) dependence!!

1D:

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E(k) for a 3D lattice Simple cubic: nearest neighbour atoms at 

So E(k) = 2(coskxa + coskya + coskza)

Minimum E(k) = 6for kx=ky=kz=0

Maximum E(k) = 6for kx=ky=kz=+/-/2

Bandwidth = Emav- Emin = 12

For k << acos(kxx) ~ 1- (kxx)2/2 etc.

E(k) ~ constant + (ak)2/2c.f. E = (k)2/me

),,();,,();,,( a000a000a

-4 -2 0 2 4-18

-16

-14

-12

-10

-8

-6

-4

-2

0

F1

k [111] direction/a/a

E(k)

Behave like free electrons with “effective mass” /a2

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Each atomic orbital leads to a band of allowed states in the solid

Band of allowed states

Band of allowed states

Band of allowed states

Gap: no allowed states

Gap: no allowed states

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Independent Bloch states

Bloch states

Let k = k ́ + G where k is in the first Brillouin zoneand G is a reciprocal lattice vector.

But G.R = 2n, n-integer. Definition of the reciprocal lattice. So

k is exactly equivalent to k.

)()( rRr k.Rψψ ie

)(ee)( Gi rRr .Ri.Rk ψψ

)(e)( and 1e iiG rRr .Rk.R ψψ .Rkk.R ii ee

-4 -2 0 2 4-18

-16

-14

-12

-10

-8

-6

-4

-2

0

F1

k [111] direction

/a/a

E(k)

The only independent values of k are those in the first Brillouin zone.

Solution of the tight binding model is periodic in k. Apparently have an infinite number of k-states for each allowed energy state.

In fact the different k-states all equivalent.

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Reduced Brillouin zone schemeThe only independent values of k are those in the first Brillouin zone.

Results of tight binding calculation

Results of nearly free electron calculation

Discard for |k| > /a

Displace into 1st B. Z.

Reduced Brillouin zone scheme

-2/a2/a

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Extended, reduced and periodic Brillouin zone schemes

Periodic Zone Reduced Zone Extended Zone

All allowed states correspond to k-vectors in the first Brillouin Zone.

Can draw E(k) in 3 different ways

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The number of states in a bandIndependent k-states in the first Brillouin zone, i.e. kx < /a etc.

Finite crystal: only discrete k-states allowed

Monatomic simple cubic crystal, lattice constant a, and volume V.

One allowed k state per volume (2)3/V in k-space.

Volume of first BZ is (2/a)3

Total number of allowed k-states in a band is therefore

.etc,....2,1,0,2

xx

x nLn

k

NaV

Va

3

33 22

Precisely N allowed k-states i.e. 2N electron states (Pauli) per band

This result is true for any lattice: each primitive unit cell contributes exactly one k-state to each band.

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Metals and insulatorsIn full band containing 2N electrons all states within the first B. Z. are occupied. The sum of all the k-vectors in the band = 0.

A partially filled band can carry current, a filled band cannot

Insulators have an even integer numberof electrons per primitive unit cell.

With an even number of electrons perunit cell can still have metallic behaviourdue to band overlap.

Overlap in energy need not occurin the same k direction

E

k0 a

EF

Metal due to overlapping bands

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Full Band

Empty Band

Energy Gap

Full Band

Partially Filled Band

Energy GapPart Filled Band

Part Filled Band EF

INSULATOR METAL METAL or SEMICONDUCTOR or SEMI-METAL

E

k0 a

EF

E

k0 a

E

k0 a

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Bands in 3D In 3D the band structure is much more complicated than in 1D because crystals do not have spherical symmetry.

The form of E(k) is dependent upon the direction as well as the magnitude of k.

Figure removed to reduce file size

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