Solid State

Question 1

Position operator

Answer 1

x̂ = x

Question 2

Momentum operator

Answer 2

p̂ₓ = -iħ d/dx

Question 3

Angular momentum operator

Answer 3

L̂ₓ = ŷp̂ᶻ - ẑp̂ᵧ, etc.

In spherical,
L̂ᶻ = −iħ ∂/∂φ

Question 4

Hamiltonian

Answer 4

Ĥ = T̂ + V̂ (kinetic + potential)
= p̂ₓ²/2m + V̂(x̂, t)
= -ħ²/2m d²/dx² + V(x, t) (1D)

Question 5

Wavefunction (r,t)

Answer 5

Ψ(r, t) = Ψ(r) exp(-iEt/ħ)

Question 6

TISE

Answer 6

Ĥ Ψ(r) = E Ψ(r)

Question 7

Expectation value

Answer 7

⟨Â⟩ = ∫ Ψ∗ÂΨ dr / ∫ |Ψ|² dr
= ∫ Ψ∗ÂΨ dr (if Ψ normalised)

Question 8

Hamiltonian for hydrogen-like atoms

Answer 8

Ĥ = -ħ²/2mₑ ∇² - Ze²/4πε₀r
where Z is number of protons (2nd term is Coulomb potential)

Solved by
Eₙ,ₗ,ₘₗ = Eₙ
= −Z²Eᴿ/n²

where Eᴿ is the Rydberg energy
Eᴿ = e²/8πε₀a₀ = 13.6 eV

and a₀ is the Bohr radius
a₀ = 4πε₀ħ²/mₑe² = 0.53 Å

Question 9

Split hydrogen wavefunction into its constituent parts

Answer 9

Ψₙ,ₗ,ₘₗ,ₘₛ = ψₙ,ₗ,ₘₗ(r, θ, φ)χₘₛ

ψₙ,ₗ,ₘₗ(r, θ, φ) = Rₙ,ₗ(r)Yₗ,ₘₗ(θ, φ):
* Rₙ,ₗ(r) = const. × (Zᵣ/a₀)ˡ exp(−Zᵣ/na₀) Σ(0→n−l−1) (−1)ᵏ cₖ (Zr/a₀)ᵏ
* Yₗ,ₘₗ(θ, φ) = Pₗ,ₘₗ(θ)Φₘₗ(φ), spherical harmonic functions
* Pₗ,ₘₗ(θ) = associated Legendre polynomials
* Φₘₗ(φ) = exp(imφ)

χₘₛ denotes spin, up or down. (χ1/2 = χ↑, χ−1/2 = χ↓)

Allowed values:
n = 1, 2, …
l = 0, 1, 2, … , n−1
mₗ = −l, −l + 1, … , l − 1, l
mₛ = ±1/2

Question 10

Degeneracy of an energy level

Answer 10

2n²

n = principle quantum number
2 is spin factor

Question 11

Wavefunction of helium

Answer 11

2 electrons

Ψ(r₁, r₂) = 1/√2’ ψ₁ₛ(r₁)ψ₁ₛ(r₂)(↑↓ − ↓↑)

Question 12

Hund’s rules

Answer 12

• Maximize S (fill all mₗ ‘slots’ with aligned spins before any with anti-aligned)
• Maximize L (fill slots from high mₗ to low, with L = Σmₗ)
• If orbital is less than half full, minimize J = |L − S|.
• If orbital is more than half full, maximize J = L + S.

Question 13

Charge density from wavefunction

Answer 13

Charge density ∝ |Ψ|²

Question 14

Born-Oppenheimer approximation

Answer 14

Nuclear mass much greater than electron mass, treat electron motion separately. In practice, consider electron’s (quantum mechanical) motion within fixed configuration of nuclei.

Question 15

Potential for BO approx. of H₂⁺

Answer 15

V = −e²/4πε₀rₐ − e²/4πε₀rᵇ + e²/4πε₀r

where rₐ = |rₑ−a|, rᵇ = |rₑ−b|

There is strong attraction at protons, rₐ → 0 and rᵇ → 0.

see fig(1.12.1) and fig(1.13.1)

Question 16

What is a LCAO?

Answer 16

Linear combination of atomic orbitals (LCAO). Applies when nuclei are well separated. Since we want to find the ground state, choose the combination of the lowest energy states of the constituents.
A molecular orbital (MO, wave function for one electron in a molecule) is constructed by a LCAO using
ψᵢ(rₑ) = Σₘ cₘᵢ ψₘ(rₘ)
where ψₘ are atomic orbitals located at aₘ with rₘ = rₑ − aₘ, and cₘ are mixing coefficients for lowest energy value. May have many such combinations, as labeled by index i.

Question 17

Graph wavefunctions for H₂⁺ using LCAO

Answer 17

see fig.(1.13.2) and fig(1.14.1)

Two possibilities, 1s orbitals in phase or out of phase.

In phase, 1s orbitals add, out of phase they subtract so halfway probability is zero. Graphs are normalised.

Question 18

Bonding in H₂⁺

Answer 18

ψ⁺⁻(rₑ) = N⁺⁻[ψₐ(rₐ) ⁺⁻ ψᵇ(rᵇ)]

Electron distribution:
|ψ⁺⁻|²

“Even” state ψ⁺:
Electron has high probability in area between the two protons, binds the protons together due to Coulomb attraction. Wavefunction ψ⁺ is a bonding orbital.

“Odd” state ψ⁻:
Electron has low probability in area between two protons, will not bind the nuclei.
Wavefunction ψ⁻ is an anti-bonding orbital.

Question 19

Hamiltonian and energy expectation in H₂⁺

Answer 19

Ĥ = −ħ²/mₑ ∇ₑ² − e²/4πε₀rₐ − e²/4πε₀rᵇ + e²/4πε₀r

Expectation values in states ψ⁺⁻:
E⁺⁻ = ∫ (ψ⁺⁻)∗Ĥψ⁺⁻drₑ / ∫ |ψ⁺⁻|²drₑ
= H±Hₐᵇ/1±S

where
H = Hₐₐ = Hᵇᵇ = ∫ ψₐ∗Ĥψₐdrₑ = ∫ ψᵇ∗Ĥψᵇdrₑ
Hₐᵇ = ∫ ψₐ∗Ĥψᵇdrₑ = ∫ ψᵇ∗Ĥψₐdrₑ
S = ∫ ψₐ∗ψᵇdrₑ

Question 20

Plot expectation energy as a function of internuclear separation in H₂⁺

Answer 20

E⁺⁻(r) = E₁ₛ + e²/4πε₀r + C(r)±K(r) / 1±S(r)
S(r) ∝ exp(−r/a₀)
K(r) ∝ −exp(−r/a₀)
C(r) ∝ −1/r

Bonding state σᵍ₁ₛ, in phase, E⁺/Eᵍ
Anti-bonding state σ∗ᵤ₁ₛ, antiphase, E⁻/Eᵘ

Eᵍ is lower and has a minimum at r₀, which is referred to as bond length.
Eᵘ is higher without a minimum.

See fig(1.16.1)

Question 21

What is dissociation energy?

Answer 21

Energy required to separate two bonded atoms.
Difference between bonding orbital energy and individual atomic orbital energy.

For H₂⁺ and H₂:
D = E₁ₛ − Eᵍ for one electron (H₂⁺)
D = 2E₁ₛ − Eᵍ for two electrons (H₂)

Question 22

Splitting MO energy levels

Answer 22

Energy levels in an MO can be split into bonding (lower energy) and antibonding (higher energy). Electrons will fill these levels as expeced.

Question 23

MOs up to 2p

Answer 23

1s and 2s split into 1sσ, 1sσ∗, 2sσ, 2sσ∗.

2p splits into 1pσ, 1pπ, 1pπ∗ 1pσ∗.
π can hold 4 electrons, formed from perpendicular 2pₓ- and 2pᵧ-orbitals.

2sσ∗ is close to 2pᶻσ (and they have the same symmetry), so hybridization/linear combination of these two orbitals occurs. The upper hybrid orbital, 2pσ’, will then be above the π-orbtial.

Question 24

Bond order

Answer 24

A measure of the bond strength of a molecule

bond order = ½(no. bonding e⁻ − no. antibonding e⁻)

Question 25

H₂ electronic excitations

Answer 25

Electrons excited to a higher energy orbital, with energy increased by a few eV

Not populated at room temp.

Question 26

H₂ vibrational excitations

Answer 26

Treat the bond between two nuclei as a spring. Vibrational energy is given by SHO:

Eₙ = (n + ½)ħω

where
n = 0, 1, 2, …
angular frequency ω = √k/µ’

µ is the reduced mass of the two protons, k is the spring constant of the bond.

Typical energy interval
∆E ∼ 50 meV.

A few populated at room temp.

Question 27

H₂ rotational excitations

Answer 27

Eᵣₒₜ = ħ²J(J + 1)/2I
where
J = 0, 1, 2, …

For H₂, estimate ħ²/2I ≈ 8 meV

Highly populated at room temp.

Question 28

What is covalent bonding

Answer 28

Bonding through overlap of orbitals

Question 29

Hybridization in carbon

Answer 29

2s, 2pₓ,ᵧ,ᶻ are all close in energy.

sp² hybridization:
* Combination of 2s and 2pₓ,ᵧ to form three arm orbitals on a plane, 120° from each other.
* These orbitals can overlap with similar orbitals of neighbouring carbon atoms to form σ-bonding (very strong), and extend to form an infinite honeycomb lattice (graphene).
* Remaining 2pᶻ orbital forms a π-orbtal with its neigbours, explaining mobile electrons in graphene.
* Each layer graphene is coupled to the others by (weak) van der Waals force (mainly dipole-dipole interactions) to form graphite.

sp³ hybridization:
* Combination of 2s, 2pₓ,ᵧ,ᶻ to form four arm orbitals in tetragonal directions with 109.47° between them.
* Each arm can overlap with the arms of neighbouring carbon to form strong σ bond, and extend in three dimensions to form the structure of diamond.

Question 30

Energy levels in hydrogen chain Hₙ

Answer 30

Number of molecular orbitals scales with n, with energy level differences getting smaller as n increases until there can be a continuum approximation for ‘allowed bands’. (Fig(1.25.1))

Lowest energy orbital, all H 1s orbitals combine in phase. Highest energy orbital, they combine out of phase. In between are (n − 2) orbitals with varying phases.

Question 31

What is the density of states

Answer 31

g(E)dE = the number of allowed energy levels per unit volume of solid in the energy range E → E + dE

Question 32

What is ionic bonding

Answer 32

Bonding through formation of stable ions by loss or gain of electrons

Question 33

Binding energy of ionic solids (pairwise)

Answer 33

If Uᵢⱼ is the interaction energy between ions i and j, define sum Uᵢ to include all interactions
involving ion i:
Uᵢⱼ = Σj(≠i) Uᵢⱼ

Assume Uᵢⱼ consists of:
* short-range repulsion (Pauli) λ exp(−r/ρ) with empirical positive parameters λ and ρ
* Coulomb potential ∝ ±q²/r

Uᵢⱼ = λexp(−rᵢⱼ/ρ) ± q²/4πε₀rᵢⱼ

introducing dimensionless parameter pᵢⱼ, write lattice distances rᵢⱼ = pᵢⱼR with lattice constant R

Uᵢⱼ = λexp(−R/ρ) ± q²/4πε₀R (nearest neighbors)
OR
Uᵢⱼ = ± q²/4πε₀pᵢⱼR (otherwise)

Question 34

Total lattice energy of a crystal of N molecules (2N ions)

Answer 34

Uₜₒₜ = NUᵢ
= N(zλexp(−R/ρ) − αq²/4πε₀R)

where z is number of nearest neighbors
and α = Σj(≠i) ±1/pᵢⱼ is the Madelung constant

Question 35

Madelung energy/potential

Answer 35

Ionic crystal, at equilibrium separation dUₜₒₜ/dR = 0,
from Uₜₒₜ = N(zλexp(−R/ρ) − αq²/4πε₀R) get

R₀²exp(−R₀/ρ) = ραq²/4πε₀zλ

Total energy
Uₜₒₜ(R₀) = −Nαq²/4πε₀R₀ (1−ρ/R₀)

Madelung energy/potential is
−Nαq²/4πε₀R₀

Question 36

Bonding in molecular solids

Answer 36

If neutral atoms or molecules are stable, solids can be held together by weak Van der Waals forces.

Typical interaction potential between two inert
atoms/molecules can be written as Lennard-Jones potential
U(r) = 4Є[(σ/r)¹² − (σ/r)⁶]
where Є and σ are the parameters.

Repulsive first term approximates short-range potential due to Pauli exclusion principle, second term is Van der Waals potential.

Van der Waals is mainly due to attraction between induced dipoles of atoms (two inert atoms, Coulomb interaction distorts original spherical charge distribution, producing dipole-dipole interaction)

Question 37

Total potential energy of a molecular solid

Answer 37

Ignoring kinetic energy of inert gas atoms, energy given by summing Lennard-Jones potential over all pairs of atoms in the crystal.

N atoms:
Uₜₒₜ = N4Є/2 Σj(≠i) [(σ/pᵢⱼR)¹² − (σ/pᵢⱼR)⁶]
where rᵢⱼ = pᵢⱼR is separation between i and j atoms with the dimensionless parameter pᵢⱼ . Factor of ½ is to avoid double counting.

Summations are dominated by the first nearest-neighbor terms. For fcc and hcp structures:
have
Σj(≠i) 1/pᵢⱼ¹² = 12.13
Σj(≠i) 1/pᵢⱼ⁶ = 14.45

Equilibrium lattice constant R₀ determined by
dUₜₒₜ/dR = 0

For fcc structure of all inert gas atom solids, we find
R₀/σ = 1.09

Question 38

Metallic bonding (basic)

Answer 38

Characterized by delocalization of valence electrons (free electrons, or conducting electrons) throughout the solid, usually one or two electrons per atom.

Metal crystals are non-directional, they are mostly close packed structures.

Interaction of ion cores with free electrons makes a large contribution to the binding energy, but the characteristic feature of metallic binding is lowering the energy of free electrons.

Question 39

2D lattice translation vector

Answer 39

Rₙ,ₘ = na + mb
where n and m are integers
and a and b are non-collinear lattice vectors, with some angle ϕ between them.

Question 40

What are the 2D Bravais lattices? Draw them.

Answer 40

Square: |a| = |b|, φ = 90°, 4 mirror planes, 4-fold rotational symmetry
Rectangular: |a| ≠ |b|, φ = 90°, 2 mirror planes, 2-fold rotational symmetry
Hexagonal: |a| = |b|, φ = 120°, 6 mirror planes, 6-fold rotational symmetry
Centred Rectangular: |a| ≠ |b|, φ ≠ 90°, 2 mirror planes, 2-fold rotational symmetry
Oblique: |a| ≠ |b|, φ ≠ 90°, 0 mirror planes, ?-fold rotational symmetry

Fig(2.3.1)

Question 41

What is crystal structure?

Answer 41

A copy of a basis attatched to each point defined by the lattice’s translation vectors.

Question 42

What is a unit cell?

Answer 42

The unit cell is defined by a parallelogram with the lattice vectors (a, b) forming two sides.

Primitive unit cells are the simplest cell that can be chosen. They contain one copy of the basis and one lattice point.

Sometimes a more complex unit cell, which contains more than one lattice point and more than one copy of the basis is chosen, usually to reflect the symmetry of the lattice.

Question 43

3D lattices

Answer 43

Defined by three non-coplanar lattice vectors (a, b, c) separated by angles (α, β, γ).

Question 44

(Examinable) 3D Bravais lattices

Answer 44

Simple/primative cubic: Point at each corner
Body-centred cubic: Point at each corner, one in centre
Face-centred cubic: Points at each corner and at centre of each face
Hexagonal: Layers of 2D hexagonal

Question 45

Unit cells for cubic lattices

Answer 45

Simple:
* Primitive unit cell same as adopted unit cell
* 6 nearest neighbours

Body-centred (bcc):

Primitive cell:
* a’ = a/2 (x̂ + ŷ − ẑ)
* b’ = a/2 (ŷ + ẑ − x̂)
* c’ = a/2 (ẑ + x̂ − ŷ)
* Angles α’ = β’ = γ’ = cos⁻¹(−1/3)= 109°29’

Adopted cell:
* a = b’ + c’
* b = c’ + a’
* c = a’ + b’
* Angles α = β = γ = π/2
* Contains 2 basis
* 8 nearest neighbours

Face-centred (fcc):

Primitive cell:
* a’ = a/2 (x̂ + ŷ)
* b’ = a/2 (ŷ + ẑ)
* c’ = a/2 (ẑ + x̂)
* Angles α’ = β’ = γ’ = π/3

Adopted cell:
* a = a’ + b’ − c’
* b = b’ + c’ − a’
* c = c’ + a’ − b’
* Angles α = β = γ = π/2
* Contains 4 basis
* 12 nearest neighbours

Question 46

Miller index

Answer 46

A convention for labeling planes within a structure.
* Find the intercepts of the plane on the a, b, c axes in terms of lattice constants. (lattice can be primitive or non-primitive.)
* Take the reciprocals of these numbers (removes ∞’s).
* (Usually) reduce to the smallest three integers having the same ratio. Express as (hkl).

The Miller indices (hkl) defines a set of planes equally spaced from each other, with the first plane going through the origin (0, 0, 0), and the next intersecting unit cell axes at the points (a/h, b/k, c/l).

If a plane cuts an axis on the negative side of the origin, corresponding index is negative, indicated by placing a minus sign above the index: (hk̄l)

Question 47

Measuring lattices with x-ray diffraction

Answer 47

Cubic lattice, (hkl) planes separated by distance
dₕₖₗ = a/√h² + k² + l²’

To measure interplanar spacing, dₕₖₗ, must satisfy two conditions:
1. The Bragg diffraction law,
2. The planes must be correctly aligned w.r.t. incident and scattered beams (usually x-rays). (fig 2.14.1)

Bragg’s law for (constructive) diffraction of x-ray with wavelength λ is given by:
nλ = 2dsinθ
with integer n
where d is distance between layers
Total scattering angle will be 2θ

Question 48

Estimating the density of a solid from x-ray diffraction.

Answer 48

• Use X-rays to measure the lattice parameters (size of the lattice vectors/unit cell)
• Add up the mass of atoms in the basis
• Divide the mass by the volume of the primitive unit cell
• (Sometimes it’s simpler to use conventional unit cell then remember that it contains multiple copies of the basis.)
• Resulting X-ray density is often an overestimate, as real solids have vacancies
  (missing atoms/ions etc.)

Question 49

Modes of vibration in a solid

Answer 49

At every wavevector k, there are one longitudinal (forwards-backwards) and two transverse (up-down and left-right) modes.

Question 50

EoM for longitudinal vibrations in a monatomic 1D chain

Answer 50

Considering the nth atom, displacement uₙ:
* Force to the left is Fₗ = K(uₙ − uₙ₋₁)
* Force to the right is Fᵣ = K(uₙ₊₁ − uₙ)
EoM is
Müₙ = Fᵣ − Fₗ = K (uₙ₊₁ − 2uₙ + uₙ₋₁)

Spring const. K

Question 51

Derive dispersion relation of longitudinal wave in 1D monatomic chain

Answer 51

Equation of motion:
Müₙ = K(uₙ₊₁ − 2uₙ + uₙ₋₁)

Let all atoms oscillate with the same max. amplitude A, assume a solution of the form:
uₙ = Aexp[ i(kxₙ⁰ − ωt)]

where A is a constant and equilibrium position of nth atom is xₙ⁰ = na

Sub into EoM:
−ω²MAexp(i(kna−ωt) = KA (exp(i[k(n+1)a−ωt]) − 2exp(i(kna−ωt)) + exp(i[k(n−1)a−ωt]))

Cancel Aexp(ikna):
−ω²M = K(exp(ika) − 2 + exp(−ika))
= 2K[cos(ka) − 1]

Dispersion relation ω(k):
ω²M = 4K sin²(ka/2)
ω(k) = 2√K/M’ |sin(ka/2)|

ω(k) is periodic, all possible ω values contained in range −π/a ≤ k ≤ π/a
This is the First Brillouin zone.

Question 52

Limit of longitudinal waves in FBZ (monatomic)

Answer 52

ω(k) = 2√K/M’ |sin(ka/2)|

Small k (long wavelength)
ω = ka√K/M’
ω → 0 as k → 0.

In this limit, usually write
ω = vₚk
vₚ = a√K/M’
where vₚ is the speed of sound.
Hence longitudinal vibration is also known as the ‘acoustic mode’.

When k = 0, all atoms move in phase, i.e. a bulk translation.

Large k
dω/dk → 0, as k →π/a

i.e. the group velocity vᵍ → 0
At Brillouin zone boundaries, k = ±π/a, stationary wave.

Max. frequency is given by ω = 2√K/M’, known
as the cut-off frequency of the lattice.

Question 53

Dispersion relation for longitudinal vibrations in a 1D diatomic chain

Answer 53

Two types of atom/ion, masses M and m. Basis is 2 atoms. Assume atoms are equally spaced, a/2 apart. Only nearest-neighbour interactions, equations of motion for nth and (n−1)th atoms:
Müₙ = K (uₙ₊₁ − 2uₙ + uₙ₋₁)
müₙ₋₁ = K (uₙ − 2uₙ₋₁ + uₙ₋₂).

Mass M, assume solution of the form:
uₙ = Aexp[ i(kx⁰ₙ − ωt)]
where undisplaced position of nth atom x⁰ₙ = na/2.

Mass m, assume solution of the form:
uₙ₋₁ = αAexp[ i(kx⁰ₙ₋₁ − ωt)]
where α describes amplitude and phase of mass m atoms relative to mass M.

Sub into the equations of motion:
−αω²m = K [exp(ika/2) + exp(−ika/2) − 2α]
= 2K (cos(ka/2) − α)

−ω²M = 2K(α cos(ka/2) − 1)

Rearrange:
α = 2K−ω²M / 2Kcos(ka/2)
and
α = 2Kcos(ka/2) / 2K−ω²m

Equate:
mMω⁴ − 2K(M+m)ω² + 4K²sin²(ka/2) = 0

Quadratic:
ω² = K(M+m)/Mm ± K√(M+m/Mm)² − 4/Mm sin²(ka/2) ‘
where + and − signs give two solutions/modes

For small k,
ω² → K(M+m)/Mm ± K(M+m)/Mm

Question 54

Longitudinal modes for diatomic 1D chain

Answer 54

ω² = K(M+m)/Mm ± K√(M+m/Mm)² − 4/Mm sin²(ka/2) ‘

Lower branch (−) is longitudinal acoustic mode (LA), since ω → 0 as k → 0.
At zone centre (k = 0)
α = 2Kcos(ka/2)/2K−ω²m
= 2K−ω²M/2Kcos(ka/2) = 1
Oscillations of the two types of atom have same amplitude and are in phase.

Upper branch (+) is longitudinal optical mode (LO).
At the zone centre (k → 0)
ω² → 2K(M+m)/Mm

α = 2K/2K−ω²m
= 2K−ω²M/2K = −M/m

Oscillations of the two types of atoms are antiphase, but because magnitude of the phase term is given by the ratio of the masses, the centre of mass for the two atoms remains fixed.

Both
At BZ boundaries k = ±π/a

ω² → K(M+m)/Mm ± K(M−m)/Mm

ω² = 2K/m (+branch)
ω² = 2K/M (−branch)

and
α = ∞ (+branch)
α = 0 (−branch)

The + solution corresponds to m oscillating and M at rest, and the reverse for the − solution

• Lower branch, longitudinal acoustic (LA) mode, both atoms/ions in basis oscillate in phase, entire basis displaced. Long wavelength sound wave as k → 0. Typical frequency of order 1012 Hz.
• Upper branch, longitudinal optical (LO) mode, atoms/ions vibrate against each other, out of phase, fixed centre of mass. Large induced dipole moment parallel to k, which interacts strongly with EM radiation. Photons can be absorbed or emitted, hence longitudinal optical (LO) mode. Frequencies ∼ 1013Hz in IR for an ionic crystal.

Question 55

Why no transverse waves in liquids/gas?

Answer 55

Transverse waves in a solid are supported by shear strength of the solid. In liquids or gases, there is no/very weak shear strength, so transverse waves can’t exist.

Question 56

Number of modes in 3D

Answer 56

p atoms per unit cell, 3p modes.
p modes per direction, so p will be longitudinal and 2p will be transverse.

Question 57

What is a phonon?

Answer 57

Normal modes of lattice vibration are quantized by phonons. They are spinless Bosons and are often refered to as (quasi)particles.

A phonon describes the wave-like excitations of displacements in a lattice.
The allowed energies of a normal mode, s, with angular frequency ωₛ(k) are given by:
Eₛ(k) = (nₖ,ₛ + ½) ħωₛ(k)
where nₖ,ₛ = 0, 1, 2, …

Looks like QM SHO, but rememebr this is collective oscillations of hella atoms.

We say that there are nₖ,ₛ phonons of type s with wave vector k present in the crystal, so nₖ,ₛ is the number of phonons of a particular type present. If n = 0 there are no phonons
in that mode of vibration.

Question 58

Exitation of phonons due to interaction with light

Answer 58

Occurs when EM dispersion relation ω=cₘk overlaps with optical branch ω. (not possible for acoustic mode as cₘ⟩⟩vₚ)

Only happens very close to k=0, and freq. of optical mode here is
ω² = 2K(M+m)/Mm

fig(3.14.1)

Question 59

Main models for valence electrons in solids.

Answer 59

1. Free-Electron Model (FEM): Assumes the valence electrons are free to move throughout the solid (sea of electrons/ electron gas) and effectively ignore the interaction between them and the atomic cores (set potential of the cores to a constant or zero). This model applies to simple metals (outer shells are s- or p-orbitals) and some properties of other metals.
2. Nearly-Free-Electron Model (NFEM): Assumes the electrons are nearly free, but there is a weak interaction between them and the atomic cores (ions).
3. Tight-Binding Model (TBM): Assumes the valence electrons are quite tightly bound to individual cores and retain some atomic character in the solid. Electrons can “hop” from one site to another to conduct electric current. This model applies to graphene, for example.

Question 60

Energy levels for FEM

Answer 60

Assume valence electrons are so weakly held that the atomic orbitals, we have a sea or gas of electrons.
We use the electron-in-a-box model for a 3D infinite potential well:
V(x, y, z) = 0 if 0 ≤ x, y, z ≤ L, = ∞ otherwise.

Solve TISE with V = 0, using boundary conditions ψ = 0 when x = 0, L etc.
Resulting wavefunctions
ψₙₓ,ₙᵧ,ₙᶻ(x, y, z) = (2/L)³’² sin(nₓπx/L) sin(nᵧπy/L) sin(nᶻπz/L)
with nₓ, nᵧ, nᶻ = 1, 2, 3, …

Wave vector k = (kₓ, kᵧ, kᶻ) given by
kₓ = nₓπ/L etc.
and energy levels given by
Eₓ,ₙᵧ,ₙᶻ = ħ²k²/2m
= ħ²/2m (kₓ² + kᵧ² + kᶻ²)
= π²ħ²/2mL² (nₓ² + nᵧ² + nᶻ²)

Each energy level can be described by combinations of (nₓ, nᵧ, nᶻ), which are effectively the number of half-waves that fit into the x, y, z directions of the cube.

For a box representing a real crystal, which may be large enough to accommodate 10²⁰ or more atoms, energy intervals are so small that allowed states exist as a continuum in energy.

The Free Electron Model predicts that E(k) varies with k in a parabolic way (essentially all energy is kinetic, V is zero or constant)

Question 61

Periodic boundary conditions (PBC)

Answer 61

Introduced when dealing with electrons moving in a macroscopic size (mm, rather than Å).

Crystal structure is periodic, so divide the solid into boxes of length L, and require that the wave functions are periodic in x, y, z, with
period L:
ψ(x+L, y, z) = ψ(x, y, z), etc.

Particle moving in a box with periodical boundary conditions, solutions are exponential functions:
ψ(r) = Aexp(ik·r) with A const.

PBC is that:
exp(ik·L) = 1

This means:
kₓ = 2πnₓ/L
kᵧ = 2πnᵧ/L
kᶻ = 2πnᶻ/L
with nₓ, nᵧ, nᶻ = 0, ±1, ±2, …

Question 62

Derive density of states

Answer 62

From PBC:
kₓ = 2πnₓ/L
kᵧ = 2πnᵧ/L
kᶻ = 2πnᶻ/L
with nₓ, nᵧ, nᶻ = 0, ±1, ±2, …

Wave vector/reciprocal/k- space defined by points of triple k-values, (kₓ, kᵧ, kᶻ).

In 3D, each point occupies a volume element in reciprocal space:
(2π/L)³ = 8π³/V, with V = L³

To find number of states with k between k and k+dk:
* Thin spherical shell, volume of k-space between two spheres is
4πk² dk.
* Number of states in this k-space volume is
(4πk²dk) / (8π³/V) = Vk²dk/2π²
* Number of states per volume is
dn = k²dk/2π²
* Electrons have spin degree of freedom, so multiply a factor of 2:
dn = 2 × k²dk/2π²
* Convert to energy E(k) using
E = ħ²k²/2m
dE = ħ²/m kdk
k = √2mE/ħ²’
* Number of possible electron states with energy from E to E +dE, per unit volume V is
dn = k/π² kdk
= 1/π² √2mE/ħ²’ m/ħ² dE
= √2’m³’²/π²ħ³ √E’ dE
= g(E) dE
* g(E) is density of states

Question 63

Density of states eq in 3D and 2D

Answer 63

3D: g(E) = √2’m³’²/π²ħ³ √E’
2D: m/πħ² (const.)

Question 64

Fermi quantities

Answer 64

Fermi energy: Eₘₐₓ = EF, highest occupied energy level
Fermi wavevector: kF = √2mEF/ħ²’, from EF = ħ²kF²/2m
Fermi momentum and velocity: PF = mvF = ħkF
Fermi temperature: TF = EF/kᴮ

Question 65

Effects of interatomic spacing on potential

Answer 65

• Large increase in KE as interatomic spacing is reduced. This is the dominant source of repulsion in the region of equilibrium separation.
• Coulomb interactions, including dipole-dipole attractive interactions and nuclear repulsion.
• Exchange and (many-body) correlation energy (dont need to know)

Question 66

Sketch potential for NFEM 1D chain.

Answer 66

Potential wells at each ion
Fig(4.11.1)

Question 67

What are Bloch functions?

Answer 67

Approximating the form of the wave function for an electron moving along the row, the lattice will change the free electron wave function exp(ikx) so that its amplitude varies with the periodicity of the lattice.
ψ(x) = exp(ikx)u(x)
where the function u(x) has the property
u(x + a) = u(x)

This is a Bloch function, a function which is periodic with the lattice. The wavelength is still given by λ = 2π/k.
The energy is no longer entirely kinetic, because of the lattice potential.

Question 68

What causes energy gaps in NFEM

Answer 68

At Brillouin zone boundaries, k = ±nπ/a, there are two allowed energies for each k value. This means an energy gap is created at k = ±nπ/a, where diffraction blocks propagation of the electron wavefunction. The free electron dispersion curve is interrupted by energy gaps.

Question 69

reduced zone scheme

Answer 69

Plot energy with band gaps as a function of k, fold into first Brillouin zone. Therefore can see many zones on one graph.

Question 70

Classification of solids

Answer 70

The division of solids into conductors, semiconductors and insulators depends on the occupation of their bands and the size of the gap between them.
From zone volume/state volume, each BZ can contain N discrete k values, so 2N possible electron states.
If an element has only one valence electron per atom, the band is half full, hence metallic conduction can occur.
With two valence electrons per atom (divalent), the first zone is completely full, hence no conduction can occur (semiconductor/insulator).
Trivalent elements contribute 3 electrons, half-filling the second zone, hence again metallic.

Note that there are plenty of divalent metals (alkaline earths, Mg, Ca etc), NFEM can only explain the metallic conductivity by assuming that the 1st and 2nd zones overlap.

Question 71

Crystal orbitals

Answer 71

LCAO method extended to a crystal.
Ψᵢ = Σₘ cᵢₘψₘ
where ψₘ is an atomic orbital, Ψᵢ is a crystal orbitals, and cᵢₘ is a mixing coefficient.
Resulting crystal orbitals must behave like Bloch functions, repeating with the periodicity of the lattice.

Question 72

Wavefunction for the electron in tight-binding model for a 1D solid of lattice spacing a.

Answer 72

Wave function for atomic orbital on atom n on the chain, φₙ(x)

Crystal orbital:
ψ(x) = Σₙ cₙφₙ(x)

ψ(x) must behave like a Bloch function:
ψ(x) = exp(ikx) u(x), where u(x+a) = u(x)

Mixing coefficients, equate the two expressions for ψ(x), dist. along the chain to n is na:
cₙ = exp(ikna)

k = 2πm/L
where L = N a
m = 0, ±1, ±2, ±3, …
and only N of the solutions generate distinct crystal orbitals; 1 BZ contains N allowed values of k.

Rewrite wave function:
ψₖ(x) = Σₙ exp(ikna)φₙ(x)

Question 73

Energy in crystal orbitals (TBM)

Answer 73

ψₖ(x) = Σₙ exp(ikna) φₙ(x)

Hamiltonian expectation value:
E = ∫ ψₖ∗Ĥψₖdτ / ∫ |ψₖ|²dτ
where dτ = dx in 1D.

Sub in ψₖ(x):
∫ ψₖ∗Ĥψₖ dτ = Σₙ,ₘ(1→N) exp[i(n−m)ka] ∫φₘ∗Ĥφₙ dτ
∫ ψₖ∗ψₖ dτ = Σₙ,ₘ(1→N) exp[i(n−m)ka] ∫φₘ∗φₙ dτ

Taking atomic orbitals to be normalized, and neglecting overlap:
∫ φₘ∗φₙ dτ = δₘₙ = 0, if m≠n, = 1 if m=n
1, if n = m

∫ ψₖ∗ψₖ dτ = N

Assume only interactions are between two adjacent orbitals in the chain
∫ φₙ∗Ĥφₙ dτ = α, energy of an isolated atomic orbital
∫ φₘ∗Ĥφₙ dτ = β if n and m are nearest neighbours, = 0 otherwise.

∫ ψₖ∗Ĥψₖ dτ = N[α + β(exp(-ika)+exp(ika)]
= N[α + 2βcos(ka)]

Energy per orbital:
Eₖ = α + 2β cos(ka)

Note that β (interaction energy) is a stabilizing effect, so is inherently negative. At k=0, E = α+2β, which corresponds to a bonding array of 1s orbitals.
At k = ±π/a, E = α − 2β, which corresponds to an antibonding array of 1s orbitals.
See fig(4.19.1)

p-orbital band disperse in the opposite direction, see fig(4.20.2)

Energy gaps come from the atomic energy levels.

Question 74

Number of crystal orbitals

Answer 74

n atomic orbitals = n crystal orbitals, which can take 2 electrons each

Question 75

Why does carbon (2 valence electrons) act like it has 4 valence electrons

Answer 75

In C, 2s and 2p levels are close in energy (only ~4 eV apart)
Two possible processes:
* Promotion of a 2s electron to 2p; 4 electrons are now available. Energy needed for promotion compensated by the extra bond energy when 4 bonds are formed.
* Hybridization of the atomic orbtials involved to form 4 equivalent hybrid AOs; the four electrons can now form 4 equivalent bonds. There are two types of hybridization for carbon: sp³ and sp² hybridizations.

see fig(4.22.2)

Question 76

sp³ hybridisation energy and contribution

Answer 76

Hybrid state:
ψ = A(γφ₂ₛ + φ₂ₚ)
where A is a normalization factor and γ² is the s-to-p state mixing ratio.

Mixing ratio is related to bond angle as:
γ² = −cosθ = 0.33
for sp³ hybrids with angle θ = 109°28’

Normalizing the hybrids:
∫ ψ∗ψdτ = 1 = A²(γ² + 1)
A = 1/√γ²+1’

General energy of a hybrid state:
Eₕᵧ = ∫ ψ∗Ĥψdτ / ∫ ψ∗ψdτ
= γ²Eₛ+Eₚ/γ²+1

For sp³:
Eₕᵧ = 0.33Eₛ+Eₚ / 0.33+1
= Eₛ+3Eₚ/4

2s contribution to each hybrid:
γ²/γ²+1 = 0.25, 25%.

Question 77

sp² hybridisation energy and contribution

Answer 77

Hybrid state:
ψ = A(γφ₂ₛ + φ₂ₚ)
where A is a normalization factor and γ² is the s-to-p state mixing ratio.

Mixing ratio is related to bond angle as:
γ² = −cosθ = 0.5
for sp² hybrids with angle θ = 120

Normalizing the hybrids:
∫ ψ∗ψdτ = 1 = A²(γ² + 1)
A = 1/√γ²+1’
​
General energy of a hybrid state:
Eₕᵧ = ∫ ψ∗Ĥψdτ / ∫ ψ∗ψdτ
= γ²Eₛ+Eₚ/γ²+1
​
For sp²:
Eₕᵧ = 0.5Eₛ+Eₚ / 0.5+1
= Eₛ+2Eₚ/3
​
2s contribution to each hybrid:
γ²/γ²+1 = 0.33, 33%.

Question 78

What causes the honeycomb bonding of graphene

Answer 78

Valence orbitals at each C are hybridised, forming 3 sp² orbitals and leaving one pᶻ orbital at each site unhybridised and perpendicular to the 2D plane.
The sp² hybrids from each C atom form 3 σ-bonding orbitals (and 3 σ∗-antibonding orbitals) with the sp² orbitals of the 3 neighbouring C atoms. Bonding orbital associated with each σ-bond is occupied by 2 electrons (↑↓), so full. Corresponding σ∗-antibonding orbitals are empty.
There is one electron per C atom left in the 2pᶻ (π) orbital: this is the origin of the remarkable properties of graphene.

Question 79

Heat capacity in limit T→0

Answer 79

C = dQ/dT
where Cᵥ ≈ Cₚ

Question 80

Dulong and Petits Law

Answer 80

Around 50% of solid monoatomic elements (one atom per basis) at high temperature have a molar heat capacity ∼25 JK⁻¹mol⁻¹ ≈ 3R, with R = 8.31 JK⁻¹mol⁻¹.
This suggests a common origin: lattice vibrations.
(Universal gas constant R = kᴮNᴬ)

Question 81

Classical energy for 3D vibration of atom of mass m

Answer 81

Only considering nearest-neighbour interactions, and assuming spring constant K for the restoring force in all x, y, z directions.

Total energy:
u = ½mv² + ½K|r−r₀|²
where r₀ is the equilibrium position.

Question 82

Equipartition principle

Answer 82

The average energy in each accessible degree of freedom of a system in thermal equilibrium is ½kᴮT.
Accessible degrees of freedom of a system: (pₓ, pᵧ, pᶻ) from kinetic energy, and (x, y, z) from the vibrational potential energy, total 6.

Question 83

Classical total internal energy of one mole with p atoms per basis

Answer 83

U = U₀ + pNᴬ × 6 × kᴮT/2
= U₀ + 3pRT
where U₀ is the cohesive energy at T = 0.
If p = 1, molar heat capacity is
C = dU/dT = 3R = 24.93 JK⁻¹mol⁻¹

This model predicts Cᵥ is independent of temperature, clearly not the case at low temperature. Need QM for explanation of low-temperature behaviors.

Question 84

Quantum/Einstein model of 3D vibration

Answer 84

Atoms are vibrating independently as QM SHO in each direction x, y, z. Allowed energies of vibration (in each direction) given by:
E = (nₓ + nᵧ + nᶻ + ³⁄₂) ħω
ω = √K/m’
nₓ, nᵧ, nᶻ = 0, 1, 2, 3, …

Question 85

Einstein temerature

Answer 85

Allowed energies of vibration:
E = (nₓ + nᵧ + nᶻ + ³⁄₂) ħω

Expect a reduction in accessibility of quantum states when kᴮT < ħω (as ħω is difference in energy between quantum states), i.e. when:
T < Θᴱ = ħ/kᴮ √K/m’

Einstein temperature, Θᴱ, is the limiting temperature below which the first excited vibrational state becomes inaccessible (is weakly occupied). The theory correctly predicts that the heat capacity of solids becomes smaller below Θᴱ (which varies from one solid to the next dependent on K and m).

Question 86

Einstein model internal energy

Answer 86

Use the density of states and an occupation function:
g(E) × f(E, T)dE = D(E, T)dE
= average number of particles occupying quantum states with energies between E and E+dE.
D(E, T) is the distribution function.
Integrating ED(E, T) over all energies gives an expression for the energy U.

Assume all thermal excitations have the same excitation energy,
ħω = kᴮΘᴱ

In one mole of elemental solid, the density of
vibrational states is:
g(E) = 3Nᴬδ(E−kᴮΘᴱ)
as there are 3Nᴬ SHO’s.
The δ function guarantees that all vibration states have the same energy.

Since lattice vibrations/phonons are bosons, use the Bose-Einstein distribution function:
f(E, T) = 1/exp(E/kᴮT)−1
where chemical potential of phonon µ = 0 (no conservation of phonon number)

Internal energy, replacing E with kᴮΘᴱ:
U = ∫∞ E × f(E, T) × g(E)dE
= 3NᴬkᴮΘᴱ/exp(Θᴱ/T−1)

Question 87

Einstein model heat capacity

Answer 87

Internal energy:
U = ∫∞ E × f(E, T) × g(E)dE
= 3NᴬkᴮΘᴱ/exp(Θᴱ/T)−1

Heat capacity:
Cᵥ = dU/dT
= 3RΘᴱ²exp(Θᴱ/T)/T²(exp(Θᴱ/T)−1)²

High T limit, T⟩⟩Θᴱ
exp(Θᴱ/T)−1 ≈ Θᴱ/T
Cᵥ = 3R, classical result

Low T limit, T→0
exp(Θᴱ/T)−1 → exp(Θᴱ/T)
Cᵥ ∝ exp(−Θᴱ/T)
Real solids, heat capacity at lowest temperatures ⟩ Einstein prediction

Question 88

Debye model

Answer 88

Einstein model for specific heat, all atoms vibrate independently in the QM SHO with the same single excitation frequency. Phonons actually have a spread of frequencies, caused by coupling of atomic vibrations.

Debye model for specific heat of a solid:
Lattice vibrations are quantum quasiparticles called phonons, with following assumptions:
* No minimum phonon energy; ω → 0 continuously for acoustic modes (long wavelength).
* There is a maximum phonon energy, there can be no phonon with a wavelength less than the lattice spacing. Maximum (cut-off) frequency is the Debye frequency, ωᴰ.

Use boson distribution for phonons:
f(E, T) = 1/exp(E/kᴮT−1)

Density of states, use:
g(E) = AE² if E≤ħωᴰ, = 0 if E>ħωᴰ
where
A = 9R/kᴮ⁴Θᴰ³ per mole,
Θᴰ = ħωᴰ/kᴮ

Predicts Cᵥ ∝ T³ at low T, correcting faliure of Einstein model

Question 89

What is density of states of 3D phonons proportional to?

Answer 89

g(E) ∝ E²
Can be shown using periodic boundary conditions, finding allowed wavevectors, finding number of states in interval dk (in spherical), using acoustic energy wavevector relation E=ħω(k)=ħvₚk

Question 90

Fermi-Dirac distribution

Answer 90

f(E, T) = 1/exp[(E−EF)/kᴮT]+1
When E = EF, f = ½
In general, EF = µ

Question 91

Volume and number of electrons in Fermi sphere

Answer 91

Volume:
VF = ⁴⁄₃πkF³

Number of electrons:
N = 2 × ⁴⁄₃ πkF³/(2π/L)³
= VkF³/3π²

Question 92

Energy and density of states in Fermi sphere

Answer 92

E = π²ħ²/2mL² (nₓ² + nᵧ² + nᶻ²)

g(E)dE = 1/π² √2mE/ħ²’ mdE/ħ²
= √2’m³’²/π²ħ³ √E’ dE

Question 93

Fermi energy, Fermi velocity, Fermi momentum, and Fermi temperature from electron density

Answer 93

N = VkF³/3π²

kF = (3π²N/V)¹’³

EF = ħ²/2m (3π²N/V)²’³

vF = ħ/m (3π²N/V)¹’³

pF = ħkF

TF = EF/kᴮ, point of classical behaviour, usually HELLA BIG

Question 94

Range of electrons that can excite in a Fermi gas

Answer 94

Only electrons in quantum states within energy range ≈ ±kᴮT of EF can change states. Exclusion principle prevents electrons with energy much more than ≈ kᴮT below EF from accepting energy because there are no vacant states for these electrons to move into.

Question 95

Quantum internal energy of a mole of electrons

Answer 95

U = U₀ + (no electrons that can accept energy) × 3 × kᴮT/2

U₀ is energy at T = 0K
No. electrons able to accept energy given by the distribution function
D(E, T) = g(E) × f(E, T)
with a range ±kBT of EF

Necessary integrals at real temperatures are quite difficult, but can make a good estimate:
* We know f(EF, T) = 1/2, so D(EF, T) = g(EF)/2.
* Considering graph of D(EF,T) at T=0 and T⟩0, need to assess two triangular areas. (Fig(5.16.1))
* Each triangle height g(EF)/2 and base kᴮT.
* Integrating areas under distribution function gives number density of electrons, n = N/V
* So the number density of electrons that can have energies increased when temperature raised from T = 0K to T = TK is:
n’ = ½ g(EF) kᴮT

Assume each electron can accept energy of order (3kᴮT)/2, i.e. electrons with energies ~EF = kᴮTF are behaving classically.

Internal energy per unit volume:
U(T) = U₀ + n’ × ³⁄₂kᴮT
= U₀ + ¾ g(EF)kᴮ²T²

Close to Fermi energy, density of states is:
g(EF) = 3n/2EF

U(T) = U₀ + ⁹⁄₈kᴮ²T² n/EF

Question 96

Quantum electronic heat capacity at constant volume, per unit
volume

Answer 96

U(T) = U₀ + ⁹⁄₈kᴮ²T² n/EF

Cᵥᵉ = dU/dT ≈ ⁹⁄₄ kᴮ²T n/EF

This is reasonably close to the full analysis result
Cᵥᵉ = π²/2 kᴮ²T n/EF

Note difference: in classical, all electrons of the system participate in thermal excitations hence contribute to the heat capacity; but in QM, only electrons near the Fermi surface are
thermally excited and contribute to the heat capacity of metals.

Question 97

Overall heat capacity from electrons and lattice

Answer 97

Contribution to the heat capacity of a solid from its electrons, Cᵥᵉ, is proportional to T.
Taking unit volume as the molar volume and assuming each atom donates one electron to the free electron gas, heat capacity per mole of atoms is:
Cᵥᵉ = π²/2 Nᴬkᴮ T/TF
= π²RT/2TF

This is very small compared with 3R, so will only become observable at very low temperature, when lattice contribution drops out.

At low temperatures, for T ⟨⟨ TD, expect lattice contribution to be ∼ T³

Overall:
Cᵥᵗᵒᵗ = Cᵥᵉ + Cᵥˡᵃᵗᵗᶦᶜᵉ = γT + αT³

Question 98

The Wiedemann-Franz Law

Answer 98

For a metal at room temperature/high temperature
κ/σ = LT
where κ is thermal conductivity, σ is electrical conductivity.
L is a constant (Lorenz number) and T is the absolute temperature.

Question 99

Assumptions in the Drude model

Answer 99

• Electrons behave as a free electron gas.
• The electrons are free to move between the positively charged ionic cores.
• The electrons collide with/ are scattered by the ionic cores only and not with each other; these are instantaneous collisions which change the electron velocity.
• Between collisions, the electrons do not interact with the ionic cores or each
  other (no long-range interactions between electrons, or between electrons and ions).
• Electrons achieve thermal equilibrium with their surroundings only through
  collisions.

Question 100

Derive electrical conductivity in the Drude model

Answer 100

1. In absence of an electric field, E = 0, motion of electrons is random. Define average time between collisions, τ꜀.
2. Apply electric field, in between collisions electrons will experience force due to electric field E, there will be a progressive drift in one direction.
3. In time τ꜀, electron experiences force F = −eE, and acquires drift velocity in the opposite direction to the field:
   F = −eE = mv/τ꜀
   hence v = −eτ꜀E/m
4. Constant of proportionality between v and −E is the mobility, µ:
   µ = eτ꜀/m
5. Current density, J, due to electrons of average velocity v and electron density n is:
   J = −nev = σE
   where conductivity σ is:
   σ = ne²τ꜀/m

Its inverse is resistivity ρ = 1/σ = m/(ne²τ꜀). Most metals, ρ ~ 1−20 mΩcm at room temp, with n ~ 10²²−10²³ cm⁻³, and τ꜀ ~ 1−10 fs

Question 101

Internal energy and heat capacity in Drude model

Answer 101

Electrons are particles of a classical gas, so each possess mean kinetic energy given by equipartition.
½m⟨v²⟩ = ³⁄₂ kᴮT

Potential is ignored, so energy of electrons is all KE, internal energy, U (per volume) given by:
U = ³⁄₂ nkᴮT

Heat capacity:
Cᵥᵉ = dU/dT = ³⁄₂ nkᴮ

Question 102

Derive thermal conductivity, and therefore the Wiedemann-Franz law, using the Drude model

Answer 102

Heating a metal rod, once dynamic equilibrium is established, there is a temperature gradient, ∆T/∆x, along the rod.

Thermal conductivity, κ, defined by:
Jq = 1/A dQ/dt = −κ dT/dx
where Jq is flux of heat (energy per second per unit area).

Consider planes at x and x+dx along rod. Change in energy over distance dx:
−³⁄₂ kᴮdT
Work done:
Fdx = −³⁄₂ kᴮdT

Drude model, electrons scattered as they diffuse down rod. Assume same average time between collisions as electrical conductivity (τ꜀), and drift velocity given by:
v = Fτ꜀/m = −3τ꜀kᴮ/2m dT/dx

Equipartition, each electron carries heat energy ³⁄₂ kᴮT, so flux of heat energy, Jq, is given by:
Jq = n × ³⁄₂ kᴮT × v
= − 3²nτ꜀kᴮ²T/2²m dT/dx

so thermal conductivity κ is:
κ = −Jq/(dT/dx) = 3²nτ꜀kᴮ²T/2²m

Ratio with electric conductivity:
κ/σ = (3²nτ꜀kᴮ²T/2²m)/(ne²τ꜀/m)
= (3kᴮ/2e)²T
Wiedemann-Franz law.

Lorenz number, L, in this Drude model,
Lᴰ = (3kᴮ/2e)²

Question 103

Magnitude of drift velocity for electrons in a current density

Answer 103

v = j/ne

Remember that drift velocity is in the opposite direction of conventional direction of current flow.

Question 104

Quantum correction to electronic conductivity

Answer 104

QM free electron model, electrons occupy traveling wave quantum states inside Fermi surface in k-space. No electrical field, E = 0, all k states equally likely, average electron velocity = 0.
In presence of an electrical field, there is force F = −eE on each electron, accelerating it in opposite direction to the field. Distribution of electrons becomes unbalanced, now more electrons traveling in one direction than the other. Shifts the whole Fermi sphere by a very small amount, ∆k, in opposite direction to the field.

Group velocity of traveling wave quantum state of momentum k:
v = dω/dk = 1/ħ dEₖ/dk = ħk/m
Field is applied momentum shift of electron:
∆k = m/ħ ∆v
This small ‘drift’ velocity ∆v is what allows for electronic conductivity.

Question 105

Why is current constant under a sustained electric field?

Answer 105

Net drift velocity is very small (∼10⁻³ ms⁻¹, compared with the Fermi velocity vF ∼0.01c.), so Fermi sphere moved by a tiny amount. If field maintained, electrons should continue to accelerate, Fermi sphere should continue to shift, velocity and current should rise indefinitely.

Instead, a constant current is reached, so there must be a scattering mechanism.

Constant drift velocity achieved because electrons near Fermi surface are scattered to vacant states of lower |k|-value and energy, whose velocity is in the opposite direction to drift velocity.
∆k = m/ħ ∆v = −eτ/ħ E
∆v isdrift velocity, τ is time between scattering
events.
Change in momentum of order ~2ħkF, diameter of fermi sphere. Shift can only occur in partially full band, so only solids with partly occupied energy bands are good electronic conductors

Question 106

What causes electron scattering in solids?

Answer 106

Electron scattering is associated with imperfections in the periodicity of the lattice,:
* Electron-phonon scattering: thermal vibration of the lattice prevents atoms being on perfectly periodic sites at the same time. Corresponding collision time is denoted as τₚ, which is clearly temperature dependent.
* Electron-impurity scattering: presence of impurity atoms and other defects
upsets lattice periodicity. Collision time τᵢ is, in general, independent of temperature.

For electric conductivity, we still adopt the Drude model but identify two characteristic collision times: τₚ due to collisions with thermal phonons (dependent on temperature), and τᵢ due to collisions with impurities (independent of temperature).