1 5-6• The equilibrium separation of the atom is given by: Req = 21/6σ ∼ 1.12σ

• The repulsive exponent must be greater than 6 to produce a potential well, and is taken as 12 mainly for sim-plicity.

• The attractive term describe an induced dipole-dipole type of interaction:

dipole electric field E ∼ −dVdr∼ 1r3

induced dipole p ∼ αE where α is the polarisability.

the induced dipole-dipole potential Vp ∼ pE ∼ E2 ∼ 1

r6

2 6-7• The lattice sum involves only one atom per unit cell:

UB =1

2

∑y 6=1

Vxy(Rij)

• In counting, the nearest neighbours (nns) dominate because the potentials avery short range (powers of -6 and-12) - next nns and others make only a small contribution.

• For the FCC structure there are 12 nns.

UB = −2ε

[14.45

(σ

R0

)6

− 12.13

(σ

R0

)12]

(2)

where R0 is found from the minimum energy requirement dUdR = 0⇒ R0 = 1.09σ.

• The cohesive energy is the ground state internal energy of the solid and, knowing its functional form, otherproperties of the solid can be determined.

• The L-J potential is an empiricial potential: ab initio Q.M. total energy calculations can determine the potentialfrom first principles.

1.3 Cohesion in Ionic crystals

• Electron transfer between atoms occurs, creating ionic crystals when:

– one atom has a small ionisation potential eg. metallic elements in GI,II: form cation.

– one atom has a large electron affinity (typically non-metals in G VI,VII): form anion.

• The ion formed has a highly spherical charge distribution, and so we can treat this Coulomb potential as if itwere a point charge located at the centre of mass.

• All Coulomb interactions between point charges have to be summed, and the ion core repulsion has to be takeninto account.

• This is more complicated than the Leynard-Jones potential because the Coulomb potential is long range.

1

• The Coulomb interaction potential:

Vi(r12) =e2

4πε0

q1q2r12

where qi are integers

• For ionic solids the (Pauli) repulsion potential is generally written (for nearest neighbours(nns) only):

VE(r12) = λe−r12/ρ

where λ and ρ are adjustable parameters set from experiment.(Bulk modulus B = −V dP

dV= V

(d2U

dV 2

)T,V=V0

)

The total potential of one ion, i, in the lattice is:

Ui =∑j

′V (rij) where V (rij) = VC(rij) + VR(rij)

where the prime means exclude i from the sum.

Simplify by scaling to nearest neighbour distance Rnn: rij = Rnnaij

Ui(Rnn) = zλe−Rnn/ρ +e2

4πε0

∑j

′ qiqjRnnaij

For a stable structure we generally have an ion with one sign with nearest neighbours of the off-sign.

e2

4πε0

∑j

′ qiqjRnnaij

=−e2|q+||q−|

4πε0RnnA

where we have +1 for ions of opposite charge etc. and A is the (positive) Madelung constant. The Madelungconstant depends on structure only, but the summations are difficult (Ewald helped solved the problem - diffrac-tion theory PY3P03 part 1).

3 Problem 1Calculate reciprocal lattice vectors G100,G110,G111 for the conventional f.c.c. reciprocal lattice(direct lattice is thusb.c.c.) with primitive basis vectors:

a∗ =2π

a(x+ y), b∗ =

2π

a(y + z), c∗ =

2π

a(x+ z)

Solution: Reciprocal space of f.c.c. is b.c.c. and vice versa with a→ 4πa in formulae. G100,G110,G111 can be

determined via the cube edges, length a in real space→ 4πa in reciprocal space.

G110 = 2a∗ =4π

a(x+ y)

G100 =4π

a(x)

G111 =4π

a(x+ y + z)

2

4 Problem 2Show that the plane wave eik·r satisfies the general Laue diffraction condition (2k · G + n|G|2 = 0, n integer) for asimple square lattice with cell side length a, if kx = ±πa , ky = ±πa

Solution: Plane wave k=(kx,xy), unit cell Ghk=(h 2πa , k

2πa

)2k ·G+ n|G|2 = 0⇒ 2

2π

a(hkx + hky) + n

(2π

a

)2

(h2 + k2) = 0

⇒ hkx + hky = −nπa

(h2 + k2)

Consider G10 h = 1, k = 0 kx = −nπa⇒ kx = ±π

afor n = ∓1

G01 h = 0, k = 1 ky = −nπa⇒ ky = ±π

afor n = ∓1

5 16-17Some plane wave properties:

• phase must repeat in space every 2π - k·r′=k·r+2π

⇒ |k| |r′⊥| − |r⊥| = 2π

• neighbouring planes of equal phase (amplitude) are separated by the wavelength:

|r′⊥| − |r⊥| = λ⇒ |k|λ = 2π ⇒ k =2π

λ(8)

• planewaves in crystalline solids: physical properties must have the periodicity of the lattice

• a plane wave whose period matches that of the lattice has the same amplitiude at r and r+R:

eik·(r+R) = eik·r

∴ eik·R = 1

but eiG·R = 1 and G ·R = 2πn (n integer)

so k = G (9) the wavevector is a reciprocal lattice vector.

6 17-18Brillouin Zones (Elliot Ch. 2)

• In real space the Wigner-Seitz cell is the atom centred unit cell of minimum volume.

• In reciprocal space the Wigner-Seitz cell is called the first Brillouin zone.

• Brillouin Zones (BZs) are formed from the perpendiculare bisectors(planes in 3D) of the G vectors.

• The first BZ is formed from the nearest reciprocal lattice points.

3

• The second BZ is formed from the next nearest lattice points(subtracting the first BZ volume) and comprisesseparae pieces(and so on).

• All BZs are the same volume.

• Points and lines of high symmetry in a BZ are labelled because experiments(and theory) usually exploit sym-metry to simplify measurement and interpretation of results.

• The centre of the zone, the point of highest symmetry is the gamma point Γ(0, 0, 0)

7 Problem 3Explain the positions of the 1st BZ boundaries of the real space b.c.c. lattice, and calculate the high symmetry pointdistances in terms of the appropriate G vector.

(see handout for picture)

Solution: For 1st BZ boundaries, bisecting planes are needed for nearest reciprocal lattice point directions

For G100 H =2π

a[1, 0, 0]⇒ |ΓH| = 2π

a=

1

2|G100|

From Problem 1:G100 =

4π

a(x), G110 =

4π

a(x+ y), G111 =

4π

a(x+ y + z)

For G110 : N =2π

1[0.5, 0.5, 0]⇒ [Γ, N ] =

2π

a

√0.25 + 0.25 =

π

a

√2 =

1

4|G1,1,0|

G111 : P =2π

1[0.5, 0.5, 0.5]⇒ [Γ, P ] =

2π

a

√0.25 + 0.25 + 0.25 =

π

a

√3 =

1

4|G1,1,1|

8 21-22V. Atom dynamics in crystals

V.1 Dynamics of continuous media

• At length scales much larger than atomic dimensions solids behave as continuous media.

• The properties may be anisotropic due to the underlying crystals symmetry, but this is treated phenomonologi-cally - the properties are measured in different macroscopic directions of the crystal.

• Sound propagation is a good example because v=fλ, v ≈ 5000ms−1 f ≈ 10kHz⇒ λ ≈ 0.5m

• A sound(acoustic) wave is just an elastic wave propagating through a medium, where a time-dependent stress,σ (force per unit area) produces a strain, e(fractional change in displacement) via the elastic stiffness, c.

σi =

6∑j=1

cijej for i = 1 to 6 and cij = cji (10)

(Elliot 3.4.3 p.187)

• With no crystal symmetry we have 21 independent components, but any symmetry reduces the number dramat-ically, eg. cubic crystals have 3 independent components.

c11(= c22 = c33), c12(= c13 = c23), c44(= c55 = c66)

4

9 22-23• Solids are more complicated because they support shear stresses and have more than one nonzero value of e:

displacements can occur that are not aligned with the stress.

• For an isotropic solid there are two independent values of c-B and µ(the shear modulus) - thus allowing both alongitudinal and 2 transverse waves(polarisation unit vectors, e(7) orthogonal to each other) to propagate.

Vlong =

√B + 4

3µ

ρand Vtrans =

õ

ρ(12) (⇒ Vlong > Vtrans)

• For anisotropic solids the simple division into longitudinal and transverse waves may not be possible eg. k maynot be parallel to e where a longitudinal wave is expected.

• Thus, in general we have plane waves propagating through the solid as consequence of the solid possessingrestoring forces(quantified by the modulus) that respond to the displacement.

• The Debye heat capacity. We can progress further with the continuum model by recognising that the solid isfinite and has boundaries so the wave equation has boundary conditions that restrict the allowed values of k - wehave dealt with this before in PY3P06 XIII Density of states.

10 24-25We now know that, for an isotropic solid we have 1 longitudinal and 2 transverse waves.

D(ω)dω =V

(2πr)34πω2dω =

V

(2π)2

[1

v3long− 2

v3trans

]4πω2dω

D(ω)dω =3V

(2πv0)34πω2dω (13)where v0 is the weighted non-dispersing average.

- The Debye density of vibrational states (∼ ω2)We introduce atoms at this stage because we know that this oscillating system and 3N-6 degees of freedom and

thus can only have this number of modes ('3N for real systems) - this means that there is a maximum frequency:

3N =

∫ ωD

0

D(ω)dω =3V

(2πv0)3

∫ ωD

0

4πω2dω

3N =4πω3

DV

(2πv0)3⇒ ωD = v0

(6π2N

V

) 13

(14)

- this maximum frequency is called the Debye frequency, ωD.

V.2 The Debye heat capacity Einstein (1907) assumed that each atom in the solid oscillated at a single frequency,ωE ,characteristic of the solid (PYP306)

CV =

(∂U

∂T

)N,V

= 3Nk

(θET

)2eθET(

eθET − 1

)2 (55)

5

where θE =(~ωE

k

)is called the Einstein characteristic temperature.

Debye model: the thermal average value of X is

< X >=

∫D(ε)f(ε, T, µ)Xdε (40) with phonon f(ε) =

1

exp( εkT − 1)

(54)

U =< E >=3V

(2πv0)3

∫ ωD

0

~ω4ω2dω

e~ωkT − 1

Using x = ~ωkT , we have:

U =3V (kT )4

2~3x2v30

∫ D

0

x3dx

ex − 1

But ωD = V0

(6π2N

V

) 13

⇒ V 30 = ω3

D

V

6π2N=

(xDkT )3

~3V

6π2N

U =9NkT

x3D

∫ xD

0

x3dx

ex − 1(15) neglecting zero-point energy

In general , this integral has to be evaluated numerically because of the upper limit, but particular cases can be treatedanalytically.

1. Low temperature kt ~ωx 1⇒ xD →∞ at upper limit of integral, and the (standard) integral is π4

15

We define a Debye temperature, kθD = ~ωD and xD = θDT

U(T ) =3π4NkT 4

5θ3Dand CV (T ) =

∂U

∂T=

12π4Nk

5

(T

θD

)3

(16)

Debye formula for the low temperature heat capacity.So CV ∼ T 3 for non-metallic solids at low temperatures, in agreement with experiment. Note:

• as xD = θDT , a high Debye temperature means a long temperature range over which (16) is applicable.

• High θD comes from strong bonds and light atoms:

θD =~ωmaxk

and harmonic oscillator ω =

√k2

m

• For the elements, highest is diamond at 2230K, lowest is caesium at 38K.

2. High temperature behaviour: kT ~ω ⇒ 1∫ xD

0

x3dx

ex − 1=

∫ xD

0

x3dx

(x+ x2

2! + . . . )=

∫ xD

0

x2dk

1 + x2! + . . .

=1

3x3D

From(15): U =3NkT

x3D

x3D3

= 3Nkt

CV =∂U

∂T= 3Nk = 3R Dulong and Petit’s law.

3. Between these limits, numerical methods are used - note that plotting the reduced variable, TθD

, results in auniversal curve, to within about 15%.

6

11 27-28V.4 The 1D atomic chain (similar to beads on a string: SF2003)

• Periodic boundary conditions - Born-von Karmann conditions - for a periodic arroy of N atoms, separation a, oftotal length L = Na:

U(r) = U(r + L) (17)

- imagine the first and last atoms linked in some way.

• The solutions become running waves with an integer number of wavelengths fitting into L:

kri = 0± ni(

2π

L

)±, . . . ,±

(Nπ

L

)i = (x, y, z) (18)

The separation of the k values is twice that of the standing waves, but there are twice as many allowed values,so the density of states is the same.

• Running wave solutions are useful because net transfer of energy occurs (contrast standing waves).

12 30-31The 1D chain: Springs (or elastic strings) have return forces analogous to interatomic potential(bottom of potentialwell is parabolic for small displacement)-look at longitudinal motion.Masses are analogous to ion cores - apply F = d2x

dt2 m to each mass (a is unit cell dimensions). Model reveals the"essential physics" of atom dynamics of 3D solids:»INSERT FIGURE 12«spring forces n-2 n n+1 n+2th particlesnth atom at equilibrium : x0n = nanth atom displaced: xn = na+ an (and ( ∂∂x = ∂

∂u ))If the interatomic potential is φ(a), and only nearest neighbour interactions matter, the total potential energy, U=Nφ(a).Use Taylor series expansion, summing over all atoms

(f(x)=f(a)+(x-a)f’(a)+...)

U(x) = Nφ(a) +∑m1

1

m!

∂mφ

∂um

∑n=1

N(un − un−1)m (22)

Now a is the equilibrium separation, so there is no net force on the atom⇒ ( ∂φ∂U )U=0 = 0.The dominant term is the quadratic term, and if we neglect higher terms, we have the harmonic appriximation - simplespring, but also small atomic displacements.

From(22) Uharm =1

2k

N∑n=1

(un − un+1)2 with k =

(∂2U

∂a2

)u=0

The net force on the nth atom is

Fn = k(un+1 − un)− k(un − un−1 = k(un+1 − 2un + un−1)

Equation of motion for the nth atom is:

m∂2u

∂t2= k(un+1 − 2un + un−1

(= −∂Uharm

∂un

)(23)

7

13 31-32We know that the most general solution for this periodic system is a Bloch wave (20)

un(x0n, t) = A exp[i(kx0n − ωkt)]

where A ≡ u0k(r) = u0k(r + a), the unit cell function.

• For discrete values of k corresponding to running waves with periodic boundary conditions;

• Using x− n0 = na we have un = A exp[i(kna− ωkt)] andun = −ω2

kA exp[i(kna− ωkt)]

• Substituting into the E.o.m. and canceling

−mω2k = k[exp(ika)− 2 + exp(−ika)] = 2k[cos(ka)− 1]

ωk = 2

√k

m| sin(

1

2ka)| (24)

• There is a maximum cut-off value ωmaxk = 2√

mm for k = nπ

a n = (±1,±3, . . . )

• The frequency is not proportional to k, as in elastic media, but has a periodic, sinusoidal behaviour.

• The periodicity is that of a reciprocal lattice 2πa .

14 33-34Phase velocity and group velocity

• The velocity of points of constant phase in a wave is called the phase velocity, vφ.

φ(x, t) = kx− ωt anddφ

dt=∂φ

∂x

dx

dt+∂φ

∂t= k

dx

dt= −ω

Constant phasedφ

dt= 0⇒ k

dx

dt− ω = 0 and

dx

dt= vφ =

ω

k(25)

• Where vφ = f(x), waves of different frequency have different phase velocites:

– A pulse or wave packet can be represented as a Fourier sum of sinusoidal components.

– The wave packet will travel at the group velocity but will broaden due to the dispersion of the phasevelocities of the sinusoidal components eg. 2 waves similar in ω, k:

ω1 = cos(kx− ωt) and ω2 = cos([x+ 24 k]x− [ω + 24 ω]t)

ω1 + ω2 = 2 cos([k +4k]x− [ω +4ω]t)∗cos([4k]x− [4ω]t)

beat amplitude

modulated by amplitude factor.

8

15 34-35The phase and group velocities of the vibrational waves are then:

vφ =ω

k=

2

k

√k

msin

(1

2ka

)vg =

dω

dk= a

√k

mcos

(1

2ka

)It is clear that the group velocity is zero at the BZ boundaries ±πa (gradient is 0) so we have standing waves there -interference between waves travelling in opposite directions after Bragg reflection from the periodic array of atoms

nλ = 2d sin θ ⇒ n

(2π

k

)= 2a⇒ k = n

(πa

)with 2θ = 180

(Note! This is a general result for wave in a solid - applied to electrons in section VI)In the long wavelength limit(zone centre) ka 1, and we exppect to regain the continuos model:

vφ =ω

k=

2

k

√k

m(1

2ka) = a

√k

m= vg (no dispersion)

16 40-41V.6. Three-dimensional crystals

• The main additional contributions come from transverse (shear) waves.

• Transverse Acoustic (TA) modes will generally have differente frequencies to LA modes because of their dif-ferent velocities(V.I)

• Modes and branches for N unit cells, each cell with p atoms

– 3pN degrees of vibrational freedom (we just ignore the 6)

– Each branch has N modes, so there are 3p branches

– There can only be one LA and two TA branches so there are 3(p-1) optic branches

– There are (p-1) LO and 2(p-1) TO branches (only 1 LO but 2 orthogonal TO displacements)

• Calculation of the vibrational behaviour of 3D crystals starts by calculating the interatomic potential usingthe Born-Oppenheimer approximation and varying the interatomic separation - matrix methods are needed andsymmetry relations are used to simplify the problem

• Theory is compared with experimental results form, most commonly, inelastic neutron scattering studies alonghigh symmetry direction of the BZ

17 43-44V.7. Phonons and crystals momentum We have used quanutm theory to derive/devise the Block theorem butotherwise classical mechanics.

• The energy (α(amplitude)2) of a classical normal mode can take any value

• The quantum oscillator has certain allocated eigenvalues of energy, characterised by integer n(k, p) where p isthe branch index

ε(k, p) =

[n(k, p) +

1

2

]~ωk(φ) (31)

9

• The quantised oscillation is called a phonon and nis the number of phonons excited in the system

• The solid has energy even with no phonons excited, the zero point energy:∑k,p

1

2~ωkp

• The total vibrational energy of the solid is:

E =∑k,p

ε(k, p)

• Phonons are spin zero, and are not conserved, so they are like photons and have the same statistical distributionfunction, as we showed in PY3P06.

18 43-44From PY3P06, the thermal average phonon occupation number per mode per branch

n(k, p) =< n >=1

exp[~ωk(p)kT

]− 1

Planck Dist. law PY3P06 used in V.2 Debye theory

Also Z =

∞∑n=0

exp

[−(n+

1

2)~ωkkT

]=

exp[− 12~ωk/kT ]

1− exp[−~ωk/kT ]

And U = kT 2 ∂(lnZ)

∂T=

1

2~ωk +

~ωkexp[~ωk/kT ]− 1

(32)

Crystal momentum

• Quantum mechanical momentum of free particle

p = ~k from momentum operator

−i~5 (Aeikr) = ~k(Aeikr)

This applies to a phonon also, except that k is only defined modulo G: k′ = k +G

• We call ~k the crystal momentum which behaves like a real momentum in many situations.

• The total wavevector of waves interacting in a crystal are conserved modulo G.

• Measurement of phonon dispersion curves by inelastic neutran scattering

• If an external quantum particle, initial wavevector Ki, scatters inelastically off a solid, gaining (losing) energyby absorbing (creating) a phonon, wavevector k, crystal momentum conservation gives:

Kf −Ki = ±k +G (33)

where Kf is the final wavevector of the scattered particle for each branch.

10

• Absorption of a phonon by the neutron is generally more important at normal temperatures, giving, with energyconservation:

Ef − Ei = ~ωk ⇒p2f

2mn− p2i

2m2= ~ω (pf−pi)

~

where mn is the neutron mass and p = ~k

• The velocity and direction of the incoming neutron beam is selected and, typically, the detector is placed in aparticular direction and the energy of the scattered neutrons analysed; thus we can find the energy and the crystalmomentum of the phonon:

~ωk = (Ef − Ei) and k ±G =(pf − pi)

~

• This is repeated for different energies and directions, and different crystal orientiations to map out the dispersioncurves.

• Thermal neutrons are used because their energy is similar to the phonons, making the change of energy onabsorption large and thus easily detectable and, because of their large mass, their wavevector is larger than the1st BZ, making the whole zone easily accessible by neutron scattering. The neutron wavevector:

kn =

√2mnE

~=

√3mnkT

~∼ 40nm−1 (at 300K)

[Tyical real space unit cell ∼ 0.5mm, so 2πa ∼ 12nm−1]

19 Problem 4(See Hook & Hall 2nd ed. p.351 or your notes for images)Explain why:

1. the TA and LA modes are degenerate when k = 0.5G100

2. the right-hand end of (a) matches the left-hand end of (b)

3. the slopes of the curves do not vanish at the 1st BZ boundary in (b)

Solution: (Remember problems 1 and 3)

19.1Adding a reciprocal lattice vector (rlv) does not change the system: we used a rlv that turns the LA vector into a TAvector. Our LA vector was: k = 2π

a x, and the corresponding TA vector would be y or z with the same displacement.

a∗ =2π

a(x+ y)

⇒ k − a∗ =2π

a(x− x− y) = −2π

ay

subtracting rlv leaves atomic displacements unchanged, but propagation vector is now orthogonal to LA so must beTA.

11

19.2Same approach for RHE(a) LHE(b)

(a) k = 0.5G100 =2π

ax

(b) k′ = 0.5G111 =2π

a(x+ y + z)

(k′ − k) =2π

a(y + z) = b∗

⇒ these are equivalent because they differ by a rlv

19.3BZ boundary at 0.25G100 (Remember problem 3). The [1,1,1] direction does not intersect the BZ boundary at 90. Itis only the component of Vkω ⊥ to BZ boundarry that must vanish to give zero group velocity.

20 Problem 5(again see notes or H&H) The figure below shows phonon dispersion curves for potassium, which has a b.c.c. unitcell with a = 0.523 nm. Find the velocity of the L and T1 modes in the [1,1,0] direction near the Γ point, if the initialslopes are 1

|G110|1.50× 1013 ms−1 and 1|G110]|

1.00× 1013ms−1, respectively.

Solution:v =

ω

k=

2πf

k∴ v = slope× 2π

|G110| =4π

a

√2 = 3.40× 1010m−1

LA velocity: v =1.5× 1013 × 2π

|G110|= 2770ms−1

TA velocity: v =1.0× 1013 × 2π

|G110|= 1850ms−1

21 46-47Electrons in crystals

VI.I The free-electron gas

• From our basic ion core + valence electron model, the simplest assumption is that the valence electrons from agas that is free to move throughout the volume of the solid, experiencing a constant potential everywhere, exceptat the surface of the solid where a potential barrier prevents escape.

• The Drude model treats these electrons as point particles( cf. kinetic theory) using classical statistics and,although some useful results are obtained(see later), FD statistics are really needed.

12

• We return to the 3D square well particle-in-a-box problem.

Electron wavefunction ψ(r) = A exp(ik · r) [Note: plane wave]

nodes at surfaces⇒ standing waves.

ψ(r) =

√8

L3sin(n1πx

L

)sin(n2πy

L

)sin(n3πz

L

)where the ni quantum number are non-negative integers.Note: vibration in elastic sold - similar result from antinodes at surface.

• In k-space we have

k1 = n1π

Land ε(k)

h2k2

2m2(34)

• From the Pauli exclusion principle, each allowed k-state can contain two electrons of opposite spin(no magneticfield), and we can work out the DOS, Fermi energy, etc. at 0K.

εF =~2k2F2me

=~2

2me

(3π2N

V

)(60) PY3P06

• For metals εF >2.5eV and kT ∼ 0.025eV at room temperature.⇒ µRT ' εF and the Fermi surface is quite sharp.

Fermi wavevector, kF =

(3π2N

V

)1/3

(35)

22 54-55Solutions of the Schrödinger equation in the weak periodic potential of the lattice are Bloch functions:

ψk(r) = exp(ik · r)ukr and uk(r +R) = uk(r), k real (20)

We make progress by using the symmetry of the system to expand the Schrödinger equation as a Fourier series ink-space(plane wave expansion of potential and wavefunction).

Complex Fourier Series

f(x) =

∞∑n=−∞

cn exp

(i2πnx

a

)( n integer; period a)

V (r) =

∞∑n=−∞

∞∑k=−∞

∞∑l=−∞

Vhkl exp (iGhkl · r) =

=∑G

VG exp (iG · r) period is r

and VG =1

Ω

∫uni tcell

dr V (r) exp(−iG · r)

(Physically distinct values li within the unit cell)Can show mathematically be proving: ∫

unit celldr exp(−iG · r) = 0

13

Schrödinger equation:

− ~2

2me52 ψ(r) + V (ψ(r)) = εψ(r) and V (r +B) = V (r)

Now we expand.V (r) =

∑G

VGeiG−r and ψk(r) =

∑k

Ckeik·r

∑k

~2

2mek2Cke

(ik·r) +∑G,k′

Ck′VGei(k′+G)·r) = ε

∑k

Ckeik·r

Choosing k′ = k −G:

⇒∑k

eik·r

[(~2k2

2me− ε)Ck +

∑G

VGCk−G

]= 0

Thus for all k: (~2k2

2me− ε)Ck +

∑G

VGCk−G = 0 (37)

Note that the expansion coefficients in (37) differ only by G, as a result of the periodicity. Using these expansions, wecan easily show:

Uk(r) =∑G

Ck−Ge−iG·r (38)

ψk+G(r) = ψk(r) , ε(k +G) = εk

We can now carry over the results on BZ’s from section V-4, as the structure and arguments are the same, with ωbecoming ε. The ε vs. k plots are called the bond structure.

The simplest energy bond scheme is that of the empty lattice, where the ion core potential tends to zero and thus theFourier coefficients at the potential tends to zero - in the limit, this should be the free-electrion gas parabola in k-space.

from (37) =

(~2k2

2me

)Ck = 0 and Uk(r) = 1

ε =~2k2

2meand φk(r) = e(ik · r)

23 55-56We now introduce our weak pseudopotential and take the first term in potential. This is the IVFE model. We expectdifferences of the 1st BZ boundaries, where we have k = ±G/2, (37) becomes:(

~2(G2 )2

2me− ε

)CG/2 + VGC−G/2 = 0

(~2(G2 )2

2me− ε

)C−G/2 + V−GCG/2 = 0

a pair of simultaneous equations.

det

(~2(G2 )2

2me− ε)

VG

V−G

(~2(G2 )2

2me− ε) = 0

14

and solving:

ε± =~2(G2 )2

2me± |VG| (38)

ε = 2|VG| ,C−G/2

CG/2= ±1

Note:V(r) is real⇒ VGeiG·r + V−Ge

iG·r is real.

⇒[VGe

eG·r]∗ = V−GeiG·r

⇒ V ∗G = V−G

24 Problem 6 - Exam?<!>Alkali metals adopt the b.c.c. structure, the 1st BZ of which has special points Γ((0,0,0), H(1,0,0), N( 12 , 12 ,0) andP( 12 , 12 , 12 ). Sodium has a lattice constant of 0.429nm, a density of 970 kg/m3, and 1 mole is 0.023 kg:

• Show that the closest BS surface is at a distance of 0.707(2π/a) from the zone centre.

• Show that the Fermi vector for alkali metals is given by kF = 0.62(2π/a), where a is the lattice constant.

• find how close the Fermi vector approaches the nearest BZ surface.

Solution:

24.1distance = |j|=

√k2x + k2y + k2z see problem 3

ΓN = 0.7072π

a

24.2

kF =

(3π2N

V

)1/3

(35)

N

V= nc =

NAzρ

M=

6.23× 1023 × 970

0.023= 2.54× 1028m−3

⇒ kF = (3π2 × 2.54× 1028)1/3 = 9.09× 109m−1 = 0.6212π

a

24.3kF|k|

=0.62

0.707= 0.877

The Fermi vector is far enough away that pertubation by the zone boundary is negligible, resulting in a Fermi sphereand free electron behaviour.

15

25 68-69VI.7 Beyond the NFE model

• The NFE model is not appropriate for the following types of solids:

– insulators with wide band gaps, because a wide band gap implies strong lattice potential– transition metals, lanthanides & actinides where d- and f- electrons that are more tightly bound to the ion

core are important– covalent solids with directional bonds formed from the combination (and hybridisation (overlap of wave-

functions, Hybrid orbitals)) of directed atomic orbitals.

• Theory may start from atomic orbitals and examine the interaction of neightbouring atoms - the tight-binding orlinear combination of atomic orbitals (LCAO) approach: this closely resembles the simple picture of interactionleading to band formation(Fig 2), and is widely used for molecules.

• The independent electron picutre is also limited, in that electron-electron interactions are important. The ap-proximation works reasonably because of the screening, and because the electron and its exchange-correlationhole, taken together, behave as independent quasi-particles.

• The simplest approach to screening is the Thomas-Fermi model which produces a screened Coulomb potentialthat falls exponentially with distance from the nucleus.

26 69-70VII. Electron dynamics in crystals [Elliot. Ch. 6]

VII.1 Electrical conductivity by electron mobility.

• The electrical conductivity is defined byJ = σE (40)

where J is the current density, E is the electric fiend and σ is the second rank electrical conductivity tensor(which is symmetric, σij = σji, and reduces to a scalar for isotropic systems).

• The e.o.m. of a free electron in an applied E and B field is

F = medv

dt= −e(E + v ×B) (41) (we set B = 0 from now on)

• We can also apply (41) to the group velocity of a wave packet (superposition of plane wave states), provided itsspatial extent is smaller than any spatial variation in E and B:

Vg = 5kω =

(1

~

)4k ε (42)

• The simplest Drude model is based on kinetic theory applied to the free electron gas (point particles; relaxationtime τ between inelastic collisions with ion cores - independent of position and velocity; emerge from collisionin random direction with energy determined by the local temperature).

• If the no. density of electrons per unit volume is nc and the electrons move a distance vδt in the direction of v,then nc(vδt) will cross areaA ⊥ to v, so the charge flow per unit area will be−enc(vδt) and the current densityis:

J = −ncev (43) [N.B. n is often used in solid state books, instead of nc]

16

• With no field, the electrons move randomly so v averages to zero and there is no net current.

• In the presence of an E field, a typical electron at a time t since its last collision, when it acquired velocity v0,will have a velocity:

v0 −(eEt

me

) [force× time

mass=velocity

]• Finally, when we average over all the electrons in the time τ between collisions, v0 averages to zero, leaving the

drift velocity, vd:

vd =−eEτme

= −µeE where the electron mobility µe =eτ

me(44)

J =nce

2τ

meE ⇒ σ =

nce2τ

me= nceµe (45)

• We can check that this is a reasonable expression for the static or D.C. conductivity by using experimental valuesof σ to calculate τ :

τ =meσ

nce2=

meσ

(zρNA/m)e2(46)

where z=valence and ρ=density.

• For Cu at 273K, σ = 6.41× 107(Ωm)−1, ρ = 8920kgm−3, m = 0.0635kg, τ = 2.7× 10−14s.

• If we now calculate the mean free path between collisions, using equipartition of energy to estimate the meanvelocity:

1

2mev

20 =

3

2kτ ⇒ v0 =

√3kτ

me= 1.11× 105ms−1

mean free path l = v0τ = 3.01× 10−9m ∼ 3nm. This seems reasonable, as it is comparable to the interatomicspacing(∼ 10 times longer).

• but, this is WRONG: FD statistics, and electrons do not scatter inelastically from ion cores (VII, VIII).

VII.2 Electron Dynamics in a periodic potential

• An electron in a periodic potential experiencing an external force( eg. from an electric field) will have anacceleration arising from both the lattice potential and the external force.

• The effective mass,m∗e , is defined to include the effect of the lattice potential on the respense, and the formalismis then convenient for isolating the effect of the external force.

• If we apply a D.C. electric field (B=0) to a solid for a time δt, the energy and momentum of the electron wavepacket will increase:

~dk

dt= −eE (47) (rate of change of momentum = force, p = ~k)

• From (41):

F = medv

dt= −eE ⇒ dv

dt= − eE

me(48)

17

• Now, from (42)dvgdt

=1

~d

dt(5kε) =

1

~5k (5kε)

dk

dt

and,

~dk

dt= −eE (47)

⇒ dvgdt

=1

~25k (5kε)(−eE)

• From (48):1

m∗e=

1

~25k (5kε)

(m∗e)ij =~2∂2εi∂ki∂kj

=~

∂(vg)i∂kj

(49)

The effective mass is inversely proportional to the curvature of the band at k.

• The effective mass replaces electron mass in (48), and it can be seen that, in general, the direction of the E fieldand the acceleration of the electron may be different.

27 77-78• Electron-phonon scattering and umklapp processes [Elliot 6.3.1.3]

• Only longitudinal phonons couple to electrons (via Ve-phαu.q)[q is the phonon wavelength, u=displacement]

• LA phonons are the moste important, because they have lower energies

• The temperature dependence of the e-ph relaxation time, τph, has different regimes:At high T ( θD), no. of phonons per mole:

< n(ωq, T ) >= fph(ωq, T ) ≈ kT

~ωq

⇒ e-ph collision rate α T and

Te-ph α T−1(T θD) (51)

At low T, the energy requirement that ~ωq kT means that electrons scatter to and from states very close tothe Fermi surface, and also ωq = vq , so the allowed surface area in k-space α q2 α T 2, in addition, scatteringrate α Ue-ph α q.⇒ e-ph collision rate α T3 and

Te-ph α T−3 (T θD) (52)

• This low T situation where the phonon momentum is small can be assisted by the umklapp process, where thephonon adds momentum from a reciprocal lattice vector, G (except at very low T - see Elliot).

18

28 83-84VIII.1 Lattice contribution to thermal conductivity of insulators

• We know that insulators conduct heat much less well than metals, so we expect the lattice contribution ofmetallic solids to be small.

• from (54), if we follow Debye and use the velocity of sound (5010 ms−1, compared to 1.1×105ms−1 for elec-trons as classical particles), we get a value of κ ∼500 times smaller, assuming the relaxation time is comparable- this is a bad estimate of κ.

• To go further, we adapt (54) for phonons, where we introduce the group velocity and sum over phonon branch pand wavevector k:

κ =1

3V

∑k,p

Cv(k, p)v2g(k, p)Tk,p (55)

where Cv(v, p) is the mode heat capacity of phonon k in branch p

• Recalling section V.5 near the BZ centre ( small k or longe wavelength), the group velocity can be replaced bythe phase velocity i.e. the velocity of sound.

• The T dependence of the thermal conductivity is then determined by the behaviour of the heat capacity & therelaxation time.

29 84-85VIII.2 Electron contribution to the thermal conductivity in metallic systems

• Modify the Drude expression to account for FD statistics and the Fermi surface from (54)

K =Cv3V

v2τ ⇒ κ =Cv3V

v2F τ (56)

because only electrons very close to the Fermi energy contribute to transport ( this expression assume that theFermi surface is spherical).

• Applying this approach to the Wiedmann-Franz expression:

Cv =π2Nk62T

2εF(36)

σ =nce

2τ

me(electronic conductivity) and v2F =

2εFme

⇒ κ

σT=π2

3

(k

e

)2

= 2.443× 10−8WΩK−2

• Better agreement than the Drude model, which has 32 - from (55) - see table.

• The Drude Cv = 32NK is too large by 0.3

(εFKT

), but this is compensated by a squared classical velocity that is

too small by 0.67 × same factor - hence the very good agreement that Drude obtained.

19