marksphysicshelp...contents. 1 crystals 4 1.1 lattice basics . . . . . . . . . . . . . . . . . . . ....
TRANSCRIPT
Year 2 Solid State Physics
Derek Lee and Rupert Oulton
Notes compiled by Nicholas Reed
January 27, 2018
Contents
1 Crystals 41.1 Lattice basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Reciprocal space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Summary of Chapter 1: What you need to know/be able to do . . . . . . . . . . . . . . . . . . . 5
2 Crystal Diffraction 62.1 Elastic condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Laue condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.3 Bragg planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.4 Interference view of Bragg Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.5 Bragg diffraction in higher dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.6 Brillouin zones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.7 Summary of Chapter 2: What you need to know/be able to do . . . . . . . . . . . . . . . . . . . 9
3 Lattice Vibrations 103.1 Harmonic approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
3.1.1 More than two-body system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.2 Monatomic chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3.2.1 Equation of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.2.2 Propagating wave solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.2.3 First Brillouin Zone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.2.4 Dispersion relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.2.5 Different vibrational modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3.3 Lattice vibrations in higher dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.3.1 Dispersion relation: simple cubic lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.3.2 Brillouin zones for higher dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.4 Finite Systems and Density of States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.4.1 Finite systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.4.2 Density of states in reciprocal space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.5 Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.5.1 Energy and heat capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.5.2 High-temperature regime . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.5.3 Low-temperature regime . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.6 Summary of Chapter 3: What you need to know/be able to do . . . . . . . . . . . . . . . . . . . 18
4 Free Electron Model 194.1 Modelling electrons in a metal as a Fermi Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.2 Filling energy levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.3 Density of States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204.4 Non-zero temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214.5 Electronic heat capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224.6 Real metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224.7 Summary of Chapter 4: Things you need to know/be able to do . . . . . . . . . . . . . . . . . . 22
5 Electron states in a crystal 235.1 Tight binding model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
5.1.1 Single well: quantum tunnelling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235.1.2 Double well: energy splitting due to quantum tunnelling . . . . . . . . . . . . . . . . . . . 235.1.3 Double well: tight binding approximation (non examinable) . . . . . . . . . . . . . . . . . 245.1.4 One-dimensional array: wave motion and energy bands . . . . . . . . . . . . . . . . . . . 265.1.5 The discrete Schrodinger equation for the one-dimensional array . . . . . . . . . . . . . . 275.1.6 Tight binding model in higher dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
5.2 Bloch’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295.3 Nearly Free Electron Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
5.3.1 Phonon band gap in a diatomic chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305.3.2 Electron band gap in a weak periodic potential . . . . . . . . . . . . . . . . . . . . . . . . 31
5.4 Summary of Chapter 5: What you need to know/be able to do . . . . . . . . . . . . . . . . . . . 32
1
6 Electrons in Solids 336.1 Electron models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
6.1.1 Sommerfeld (free electron) model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336.1.2 Nearly free electron model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336.1.3 Tight binding model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
6.2 Bloch’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346.2.1 Zone schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
6.3 Band theory, simplified . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356.4 Metals, Insulators and Conductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366.5 Dynamics of Bloch Electrons and Holes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
6.5.1 Effective mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376.5.2 Consequence of Bloch’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386.5.3 Holes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
6.6 Counting ‘free’ electrons and holes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396.6.1 Counting electrons in a solid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396.6.2 Number of ‘free’ electrons in a metal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426.6.3 ‘Free’ electrons in other solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426.6.4 Law of mass action . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
6.7 Intrinsic and Extrinsic Semiconductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436.7.1 Hydrogen model of dopants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436.7.2 Donor ionization probability: NON-EXAMINABLE . . . . . . . . . . . . . . . . . . . . . 446.7.3 Extrinsic carrier density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
6.8 Summary of Chapter 6: What you need to know/be able to do . . . . . . . . . . . . . . . . . . . 47
7 Charge Transport 487.1 The Drude model and carrier mobility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
7.1.1 Charge transport in semiconductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 517.1.2 Charge transport in real metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 527.1.3 Transport Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 527.1.4 Bloch Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 527.1.5 Charge carrier mobility and electrical conductivity . . . . . . . . . . . . . . . . . . . . . . 53
7.2 The Hall Effect and Magnetoresistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 537.2.1 Crystals in a static magnetic field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 537.2.2 Transport equations and the Hall effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . 557.2.3 Magnetoresistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 597.2.4 Scattering mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
7.3 Drift, Diffusion and Einstein’s relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 617.3.1 Carrier drift & diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 617.3.2 Einstein’s Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 627.3.3 Generation and recombination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 637.3.4 Excess carrier density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
7.4 Summary of Chapter 7: What you need to know/be able to do . . . . . . . . . . . . . . . . . . . 66
8 The pn Junction 678.1 Thermodynamics and Electrostatics of a pn Junction . . . . . . . . . . . . . . . . . . . . . . . . . 67
8.1.1 Depletion region and built-in potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 688.1.2 Applied bias . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
8.2 The Shockley Equation and Diode Electronics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 738.2.1 pn-junction devices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
8.3 Summary of Chapter 8: What you need to know/be able to do . . . . . . . . . . . . . . . . . . . 75
9 Solids in Other External Fields 769.1 Optical properties of solids and Optoelectronics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
9.1.1 Radiative and non-radiative absorption and emission . . . . . . . . . . . . . . . . . . . . . 769.1.2 Illuminated p-n junction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 779.1.3 Optical diode operation modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
9.2 Origin of Magnetism in Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 789.2.1 Magnetization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 789.2.2 Change to Hamiltonian for electrons in a solid with external magnetic field . . . . . . . . 79
9.3 Chapter 9: What you need to know/be able to do . . . . . . . . . . . . . . . . . . . . . . . . . . 80
2
Comments: UPDATED
Just a couple brief things:
• The first half of these notes (Chapters 1-5) are from Derek Lee’s course. Content is mostly the same, justwith a lot of his extra words removed to try and make the key information clearer. The summaries at theend of each chapter are from Derek’s slides and from what I think is also useful information.
• Second half (Chapters 6-9) is Oulton’s course. This is a comb-through of his slides, lectures and derivationsto try and piece together the course content in a manner that is more usable than it otherwise felt to me.All of the derivations in boxes in this section are things he wrote out via the visualizer and so probablyneed to be understood & reproducible unless otherwise indicated
• In a couple places I’ve included some non-examinable derivations (mostly the tight-binding model 2 welldiscrete Schrodinger equation derivation in Chapter 5, and the ionization probability derivation in Chapter6) because I think they’re particularly useful for understanding
• I didn’t include the section on electrical conduction from Derek Lee’s Chapter 4 as it’s covered in Oulton’ssection on charge mobility & the Sommerfeld-Drude model
• Pretty much all of the diagrams are just taken from lecturer’s slides. In a couple cases I’ve made someminor modifications to make them clearer, and I’ve tried to explain them as best as I can.
January 2018 updateI’ve tried to do another typo pass so these notes will still be useful for the current set of Year 2 students.Obviously the course may have changed somewhat so in some places where I make references to specific partsof the lecture notes or specific problem sheets for more complex derivation points, these references may not beuseful any more.
Also, as I say when you get there, but the ’proof’ for Bloch’s theorem in Chapter 6 is just wrong and I can’tsee how to make it work. It’s in the form that Rupert Oulton gave it to us in when he taught it, but yeah itseems to just be wrong. If someone works out how to do it without needing to resort to translation operatorsor something like that then let me know but yeah.
If you notice any typos you can email me at [email protected], especially if they’re in something importantlike equations (in this pass I noticed that there was an exponential that didn’t have a unitless exponent, forexample).
3
1 Crystals
1.1 Lattice basics
• Crystals are built of periodic structures.
• These lattices are broken into the smallest possible blocks, unit cells.
• The smallest possible unit cell is the primitive unit cell
• The centres of each primitive unit cell are called lattice points
• A vector joining two lattice points is called a lattice vector
• Any lattice point can be denoted as
R = n1a1 + n2a2 + n3a3 (1.1.1)
where a1, a2 and a3 are primitive lattice vectors (smallest possible lattice vector).
Contents of a unit cell are referred to as the basis of the unit cell, and we construct a Bravais lattice from thepoints that are at the centre of each unit cell.
No choice of unit cell is unique; the only requirements are that they all occupy the same volume, and that theycan make up the total volume occupied by the crystal. Two main choices of unit cell:
1. Unit cell with shape and dimensions given by primitive lattice vectors
2. Wigner-Seitz unit cell: the volume around a lattice point R where all of the points enclosed in the volumeare closest to R and no other lattice point.
1.2 Reciprocal space
The lattice vectors described so far are in regular space, but we will want to deal with plane waves. Thesetypically propagate with a wavevector k, and thus propagate in reciprocal space, which has units of m−1.
• Propagating waves typically have a k · r exponent so we wish to define a coordinate system such that thisdot product is simple.
• We define the wavevector k of the propagating wave as
k = k1b1 + k2b2 + k3b3 (1.2.1)
for dimensionless k constants.
Could have expressed k in terms of the primitive lattice vectors, but things become difficult if the lattice has anon-orthogonal basis. Instead insist that the vectors b1,2,3 are such that the dot product cross terms will be 0.
• For simplicity, we will require that
k · r = 2π(k1r1 + k2r2 + k3r3) (1.2.2)
• As a result of the cross term condition, we require that
ai · bj = 2πδij (1.2.3)
We wish to construct all the b vectors to form an orthogonal basis, such that b1 is perpendicular to both a2,3,and so forth for the other b vectors. To ensure this we define these vectors as
b1 = 2πa2 × a3
a1 · (a2 × a3), b2 = 2π
a3 × a1
a1 · (a2 × a3), b3 = 2π
a1 × a2
a1 · (a2 × a3)(1.2.4)
The factor of 2π ensures that the conventional condition is satisfied. The term in the denominator is the volumeof the unit cell in real space.
4
1.3 Summary of Chapter 1: What you need to know/be able to do
• Know the meanings of all the terms in bold in this section.
• Write down the definition of and give an explanation of the meaning of the b vectors (& be able to calculatethem).
5
2 Crystal Diffraction
2.1 Elastic condition
• We use electromagnetic waves to investigate crystal structures, typically collimated beams of monochro-matic X-rays.
• An incident beam can be characterised by a wavevector kin, and the outgoing scattered beam by kout.
• Assume photons are elastically scattered by interactions (and thus ignore any absorption and re-emissionprocesses).
For elastic scattering, require that|kin| = |kout| (2.1.1)
Can define a scattering wavevector q asq = kout − kin (2.1.2)
Clearly, these three wavevectors will form a triangle in reciprocal space. Can then write the elastic condition(2.1.1) as
k2out = (kin + q)2 = k2
in (2.1.3)
Expanding the bracket gives that k2in + 2kin · q + q2 = k2
in and thus we reach that
2kin · q + q2 = 0 (2.1.4)
2.2 Laue condition
• In a crystal with primitive lattice vectors a1,2,3, the scattering wavevector must take the form of one ofthe reciprocal lattice vectors ∴ say that
q = G = h1b1 + h2b2 + h3b3 (2.2.1)
This is known as the Laue condition.
• Combining the elastic and Laue conditions tells us that the possible momentum changes for a photon withincident wavevector kin and scattered wavevector kout are equal to ~G.
• Can alternatively sub Laue condition into Equation 2.1.4 to say that
2kin ·G + G2 = 0 (2.2.2)
Can also attempt to satisfy these conditions with a geometrical method, known as the Ewald method. Putthe tip of the incoming wavevector at a reciprocal lattice point and draw a circle (or sphere in 3D lattices)around its tail that has radius equal to the wavevector’s length. A possible outgoing wavevector is found wherethis surface intersects another reciprocal lattice point (see Figure 1)
6
Figure 1: A demonstration of the Ewald method. The vector G is a reciprocal lattice vector and is the scatteringvector between the incident wavevector kin and the outgoing wavevector kout. This diagram allows us to findthe possible kout vectors.
2.3 Bragg planes
For reciprocal lattice vector G and lattice point R we can write that
G ·R = 2πm (2.3.1)
for m ∈ Z, by the definitions from Section 1. This produces a plane of regular lattice sites (not reciprocal lattice
sites) for each m, normal to G and a distance2πm
|G|from the origin (see expressions for planes in the 1st Year
Linear Algebra course). These parallel planes are referred to as the Bragg planes.
Can also express Equation 2.3.1 ase−iG·R = 1 (2.3.2)
for all R, as em2πi = 1 for all m ∈ Z
• Incoming, outgoing and scattering wavevectors must form an isosceles triangle, so wavevector must reflectoff of Bragg plane at an angle equal to its angle of incidence θ.
• The entire beam is therefore deflected by 2θ.
• Can write that kin ·G = −|kin||G| sin θ.
Bragg planes are illustrated in Figure 2
Figure 2: A set of Bragg planes with a reflected wavevector. The Bragg planes are normal to the reciprocallattice vector (and scattering vector) G.
7
2.4 Interference view of Bragg Diffraction
Consider a one-dimensional array of scatterers (e.g. atoms) that has the following properties:
• Scatterer spacing a
• Scatterers placed from x = -Na to x = Na (2N + 1 scatterers)
Waves entering this array:
• For wavevector kin = k > 0 the wave travels to the right, incident on lattice from the left
• Part of this wave is transmitted and exits to the right
• The rest is reflected by the crystal and exits to the left with wavevector kout = -k < 0
• Scattering wavevector is therefore q = -2k.
• Total scattered wave is a superposition of reflected waves from every scatterer
Consider one wave reflected from the origin x = 0, given as ψ0, and one from the scatterer at x = na, given asψn. These are reflected waves and propagate to the left, and thus take the form of e−ikx. We then reach thefollowing points:
– The path difference between ψ0 and ψn when they leave the lattice is ∆xn = 2na.
– Constructive interference occurs for integer wavelength path difference, i.e. if, for m ∈ Z,
∆xn = 2na = mλ (2.4.1)
– This can be related to the phase difference between the waves, by
∆φn = 2π∆x
λ= k∆xn (2.4.2)
– Constructive interference condition is therefore given as
∆φn = 2nka = 2πm (2.4.3)
– Must hold for every scatterer in the chain
– This requires every phase change between scatterers to be 2π
– A discrete set of constructively interfering wavevectors is therefore produced by these conditions
The incident wavevectors that are strongly back-scattered have kin =π
am for m ∈ Z (find explicitly as (2.4.3)
must hold for n = 1) and m in ± pairs, i.e. m = 0, ±1, ±2, etc. These wavevectors will form the reciprocallattice in this one dimensional array.
2.5 Bragg diffraction in higher dimensions
• The phase at a position r can be expressed as kin · r
• The phase difference between a wave arriving at the origin and a wave arriving at r = R is given as
∆φin = φin(R)− φin(0) = kin ·R (2.5.1)
• The phase difference at any point between a wave propagating from the origin and from R, if they areinitially in phase, is given as
∆φout = −kout ·R (2.5.2)
8
• The total phase difference for waves scattering from the unit cell at R and the origin is therefore
δφR = (kin − kout) ·R = −q ·R (2.5.3)
• Need this to be true for all R and thus equal to a multiple of 2π for all R
• Thus require e−iq·R = 1 for all R
• Same property as defines a reciprocal lattice vector =⇒ all scattering vectors must be reciprocal latticevectors
2.6 Brillouin zones
• Want to define a unit cell for the reciprocal lattice
• Define the first Brillouin zone as the Wigner-Seitz unit cell that contains the origin k = 0.
• Any wavevector k′ can be related to one in the first Brillouin zone by k = k′ + G for reciprocal latticevector G (wavefunctions are unchanged by adding an integer multiple of G to k by periodicity, will beproved later).
• If we only perform lattice translations, waves with wavevector k and with wavevector k′ are indistinguish-able.
2.7 Summary of Chapter 2: What you need to know/be able to do
• Know and be able to apply elastic and Laue conditions
• Know Equation 2.2.2 & be able to apply it
• Make arguments about phase differences in 3D to make statements about the scattering vector (Section 2.5)
• Understand the concept of a Brillouin zone
9
3 Lattice Vibrations
Understanding how a lattice vibrates can allow us to examine properties of the crystal, and ultimately derivesome thermodynamic quantities.
For a crystalline material, we expect:
1. Attractive force at large separation
2. Repulsive force when atoms or ions are close
Begin with a two body system, then generalise to more complicated systems. The interatomic potential expe-rienced by both bodies in this system will be a function of relative position r12 = |r1 − r2|.
3.1 Harmonic approximation
• Two particles at rest in equilibrium have minimum of potential energy
• Consider changes near this equilibrium such that we can expand in a quadratic form, as
V (r1 − r2) = V (reqm12 ) +1
2
∑i,j=x,y,z
κij ui12 u
j12 + . . . (3.1.1)
Note: This is just a Taylor expansion, but the linear term goes to zero as the first derivative is 0 at aminimum of potential, which is our point of expansion.
• We define κij as
κij =∂2V (r)
∂ri∂rj
∣∣∣∣eqm
(3.1.2)
• rx, y, z are the Cartesian coordinates of r
• ux, y, z12 are the coordinates of u12 where u12 = r12 − reqm12 (effectively displacements relative to the equi-librium position).
We call this assumption of a quadratic interaction the harmonic approximation.
3.1.1 More than two-body system
Now generalise to a crystal, an array of identical atoms.
• The entire interaction is the sum of the interaction between each pair of atoms, so we write that
Vtot(r1, r2, . . . ) =∑m<n
V (rm − rn) (3.1.3)
where the restriction on the sum prevents us from counting any interaction twice
• Particles are at equilibrium positions in lattice sites, so Rn = reqmn
• Only relative displacements generate interaction energy and thus a force, so only wish to study theun = rn(t)−Rn.
• Can write potential as a function of relative displacement, i.e.
V (rm − rn) = V (Rm −Rn + umn(t))
for umn = um − un, the relative displacement between atoms m and n
• Restrict to a regime when displacements are small compared to lattice spacing. This is an approximationfor low temperatures when little thermal energy is available for oscillations away from the equilibriumposition
10
• In this regime, can perform a Taylor expansion of the interaction potential around umn = 0, giving aquadratic potential
Vtot ≈∑i,j
∑m<n
κmnij2uimn u
jmn (3.1.4)
κmnij =∂2Vtot
∂rim∂rjn
∣∣∣∣eqm
(3.1.5)
We neglect the constant term from the expansion as it will only provide an offset and won’t affect thephysics we see.
3.2 Monatomic chain
To begin with, we will look at a one-dimensional, monatomic chain. We have a chain of masses connected bythe harmonic interaction potential, which we can therefore consider as a series of springs. Only longitudinaldisplacement will be considered in the one-dimensional case.
3.2.1 Equation of Motion
Consider an infinite chain of particles of mass m:
• Label particles by an index n
• Each particle has position xn(t) at a time t
• Equilibrium spacing is a such that we can write equilibrium positions as xeqmn = na
• The displacement from equilibrium is given as
un(t) = xn(t)− na (3.2.1)
For the purposes of the harmonic approximation, we are only considering the nearest neighbour interactions.This means each atom is effectively only interacting with the adjacent two. We will therefore write the potentialenergy of the system as
Vtot =κ
2
∞∑n=−∞
(un+1 − un)2 (3.2.2)
This is effectively masses connected by springs that obey Hooke’s law, with spring constant κ.
• For two masses joined by a single spring the force on the n = 1 mass is
− ∂V∂x1
= − ∂V∂u1
= κ(u2 − u1) (3.2.3)
• Obviously the partial derivatives are the same as xn is just a shift of un.
• The force on the other mass is equal and opposite.
• We can show that the natural frequency of oscillation for two masses is
√2κ
m.
• We then define the Debye frequency as
ωD =
√κ
m(3.2.4)
• Also define Debye temperature as
TD =~ωDkB
(3.2.5)
11
For the entire chain of atoms, the net force Fn on the nth mass is the sum of the forces from the two springsconnected to that mass. We can therefore write that
Fn = −∂Vtot∂xn
By substituting in equation 3.2.2, we can then say that
Fn = −κ2
∂
∂un
[(un+1 − un)2 + (un − un−1)2
]By then evaluating the derivative and collecting terms we get that
Fn = κ(un+1 + un−1 − 2un) (3.2.6)
The positive direction is to the right. From this the equation of motion for the nth mass is
md2undt2
= κ(un+1 + un−1 − 2un) (3.2.7)
3.2.2 Propagating wave solutions
• Any arbitrary solution of this equation will be a sum of specific solutions
• For lattice spacing a, try a trial solution of
un(t) = ukei(nka−ωt) (3.2.8)
• Solutions are wave-like in space with a single frequency such that we have simple harmonic motion
• All of the masses oscillate collectively for this solution
• The general solution is given as a linear superposition of these wave-like solutions, as
un(t) = Re
(∑k
ukei(nka−ωkt)
)(3.2.9)
• uk coefficients found from initial conditions
• ωk associated with the wavevector k for each superposing solution
3.2.3 First Brillouin Zone
Our solutions have ambiguity, as simply adding an extra 2π to ka we will have the same value of the exponential(as complex exponentials are periodic). To ensure we have independent solutions, we will restrict our wavevectorsto lie in the first Brillouin zone. This means they lie in the range
−πa< k ≤ π
a(3.2.10)
This first Brillouin zone is centred at k = 0, and has a width equal to the smallest non-zero reciprocal latticevector. It is a Wigner-Seitz unit cell of the reciprocal lattice, as previously mentioned.
3.2.4 Dispersion relation
Substituting trial solution (3.2.8) into equation of motion (3.2.7) gives
−mω2ukei(nka−ωt) = κuke
−iωt(ei(n+1)ka + ei(n−1)ka − 2einka
)= κuke
i(nka−ωt) (eika + e−ika − 2)
12
• The equations are now effectively uncoupled: the expression is only dependent on the particular mass,rather than both adjacent ones
• This means the equation is different for all solutions and thus simple harmonic waves do occur as solutions
• Equation must hold for all n and t and so we can divide by the ei(nka−ωt) term.
• Must also have non-zero uk for motion, so can divide by that.
• Now via Euler’s formula and trig identities we can write that
ω2 =2κ
m(1− cos ka) = 4ω2
D sin2
(ka
2
)(3.2.11)
• The dispersion relation can therefore be written as
ωk = 2ωD
∣∣∣∣sin(ka2)∣∣∣∣ (3.2.12)
for k in the first Brillouin zone.
• Taken only the positive root: negative one is already expressed with negative k and in fact just correspondsto the complex conjugate of the solution
• This complex conjugate is irrelevant as we take the real part, so the (k, -ωk) solution is identical to the(-k, ωk) one.
As the wavevectors are restricted in the first Brillouin zone, we cannot have infinitely large frequencies andinstead we will have a continuous frequency band from 0 to 2ωD. These modes are said to have a bandwidth of2ωD. This dispersion relation is seen in Figure 3
Figure 3: The dispersion relation ωk for longitudinal vibrations of a monatomic chain with lattice spacing a, asa function of k in the first Brillouin zone.
3.2.5 Different vibrational modes
Start with short wavelength modes:
• Short wavelength =⇒ large k ∴ at edge of Brillouin zone
• As a result ka = ±π
• Thus un = uk cos(±πn− ωkt) = (−1)nuk cos(ωkt)
Wavelength of oscillation is twice that of lattice spacing, so even-n atoms displacing in opposite direction to odd-natoms
Consider when wavelengths are long, so k → 0:
• As k → 0, ωk → 0
• Modes with 0 frequency are said to be gapless as there is no gap between 0 and the considered frequencyband
13
• This is because k = 0 corresponds to constant un and thus no relative displacement
Entire chain is being translated, therefore no restoring force on atoms and therefore no simple harmonic oscillations
Now consider small, non-zero k:
• Can approximate sinka
2as
ka
2
• Can then write dispersion relation as
ωk ' ωDa|k| for|k|a� 1, v = ωDa = a
√κ
m(3.2.13)
This is a sound wave with velocity v, a longitudinal compression wave
• Right-moving for k > 0, left-moving for k < 0
• When the sound wave wavelength is comparable to lattice spacing (edge of Brillouin zone), dispersionrelation does not hold and travelling waves disperse (wavepackets will spread out as they propagate)
3.3 Lattice vibrations in higher dimensions
For a Bravais lattice in three dimensions, we expect three possible branches of collective modes:
• Longitudinal oscillations in the same direction as k. These are sound waves at long wavelengths and thismode may be called the acoustic mode.
• 2 sets of transverse modes, oscillation perpendicular to k. These are known as shear waves in the longwavelength limit.
• Solids resist shear stresses but liquids do not, so we can see shear modes as a signature of solidity.
3.3.1 Dispersion relation: simple cubic lattice
Via a similar method as before we can show that each vibrational mode in a simple cubic lattice will have adispersion relation of the form
ω2k = 4ω2
D
(sin2 kxa
2+ sin2 kya
2+ sin2 kza
2
)(3.3.1)
For wavelengths that are large compared to the lattice spacing, we can make the same trigonometric approxi-mation as before in order to write that
ωk ' ωDa(k2x + k2
y + k2z)
12 = v|k| (3.3.2)
This is valid for |k|a� 1, where we say that v = ωDa.
3.3.2 Brillouin zones for higher dimensions
• As before we need to define a Brillouin zone in order to avoid redundancy in wavevectors.
• We will still expect plane-wave collective modes and thus write the displacement at a lattice site R as
uR =∑k
uk ei(k·R−ωkt) (3.3.3)
• Will restrict the first Brillouin such that it only contains points that cannot be joined by a reciprocallattice vector.
• This requirement is given as
−πa< kx,y,z ≤
π
a(3.3.4)
• As normal we will generally define the first Brillouin zone as the Wigner-Seitz unit cell of the reciprocallattice, centred at the origin.
14
3.4 Finite Systems and Density of States
3.4.1 Finite systems
We will now consider a finite monatomic chain, that we loop in order to connect the two ends together.
• Now have periodic boundary conditions
• Must therefore have wavelengths that fit in the length Na of the system
• Require that
k =2π
Nap (3.4.1)
for p ∈ Z and k in the first Brillouin zone so wavelengths are no greater than Na.
• Equations of motion and solutions are same as for infinite chain, but with discrete selections of wavevectors
• Number of allowed wavevectors is equal to the number of masses in the chain
3.4.2 Density of states in reciprocal space
• Consider the number of allowed wavevectors in a finite system as points in reciprocal space
• They have a density called the density of states
• For a one-dimensional chain of length L the separation between allowed wavevectors is2π
Na=
2π
Land
then the density of states is given as
gk =L
2π(3.4.2)
• In higher dimensions, different modes operate independently but still are subject to the same boundaryconditions applied separately
• Allowed wavevectors are therefore given as
k =
(2π
Lxpx,
2π
Lypy,
2π
Lzpz
)(3.4.3)
for px,y,z ∈ Z and kx,y,z in all in the first Brillouin zone.
• Density of states in 3D is therefore
gk =Lx2π· Ly
2π· Lz
2π=
V
(2π)3(3.4.4)
for the real space lattice volume V.
In order to determine the number of states in a given real-space volume, we can simply integrate the densityof states over the corresponding k-space volume (in this case, a cubic volume where k = 2π
a ). This integral iswritten as
Nstates = gk
ˆBZ
d3k =V
(2π)3
(2π)3
a3=V
a3= Nunit cells (3.4.5)
As expected the number of states equals the number of unit cells. It is important to note that this is only thecase for one of the three branches of vibrational modes, so we must apply the same method to each branchseparately. This gives a total number of states of 3N.
3.5 Thermodynamics
We wish to consider how lattice vibrations contribute to the thermodynamic properties of a solid.
• Expect equipartition to apply at high temperatures but will need quantum effects at low temperatures
• Need a quantized energy system to get an idea of the energy content of vibrational motion.
• We can quantize each simple harmonic collective mode in a way similar to a parabolic well (see QuantumTheory of Matter course in Year 4 for details).
15
• The Hamiltonian for this classical system is
H =∑
k∈BZ
~ωk(a†kak +
1
2
)(3.5.1)
• ak and a†k are the raising and lowering operators for each collective mode
• This Hamiltonian can be regarded as counting bosons labelled by the k.
• These bosons have energy given asεk = ~ωk (3.5.2)
• Transition from classical waves (frequency ωk )to particles (energy ~ωk) is called second quantization
• This ’boson’ from vibrational waves of a solid is known as a phonon
• A phonon is a discrete quantum of vibrational energy, just like a photon is for electromagnetic energy.
• We can say a phonon has momentum p = ~k, but this is not true momentum as it is periodic (can replacek with k + G and not change the result), so instead we call it crystal momentum
3.5.1 Energy and heat capacity
• Bose-Einstein statistics gives us an expression for the number of phonons of energy εk at a temperatureT, of
nB(εk, T ) =1
eβεk − 1, β =
1
kBT(3.5.3)
• The energy from thermally excited vibrations for a wavevector k is given as nB(εk, T ) εk.
• Internal energy at 0K (zero-point energy) is irrelevant for the thermodynamics of the system so definetotal energy as ∆U(T ) = U(T )− U(0).
• ∆U is the sum of the contribution from all of the phonons at all of the wavevectors in the first Brillouinzone, given as
∆U =∑
k∈BZ
nB(εk, T )εk =∑
k∈BZ
εkeβεk − 1
(3.5.4)
• This sum can be taken to an integral over the states, and thus we introduce a factor of the k-space densityof states outside the front. We thus have
∆U =V
(2π)3
ˆBZ
εkeβεk − 1
d3k (3.5.5)
The next step is to consider different regimes for this energy.
3.5.2 High-temperature regime
• High-temperature regime =⇒ βεk � 1
• Approximate eβεk ' 1 + βεk (using ex ' 1 + x for |x| � 1)
• Then we get
nB(εk, T ) ' 1
βεk + 1− 1=kBT
εk
• Require kBT � εk.
• Every wavevector mode contributes energy nBεk = kBT . This is the equipartition theory for vibrationalmodes.
16
• Total energy is then given as
U ' kBTˆBZ
V
(2π)3d3k (3.5.6)
• This integral is just kBT multiplied by the total number of states in the first Brillouin zone = number ofunit cells (see Section 3.4.2).
• At high temperatures, total energy isU ' NkBT (3.5.7)
in agreement with traditional thermodynamics, but only valid for the higher temperature limit when kBTis large compared to phonon energy.
• The corresponding heat capacity is given as
CV =
(∂U
∂T
)V
' NkB (3.5.8)
3.5.3 Low-temperature regime
• Any phonon band that has a non-zero lower bound frequency will not contribute energy for temperaturesbelow ~ωmin
kBas the Bose-Einstein factor is exponentially small
• In effect there is not enough thermal energy to excite these modes and so they are frozen out
• The dominant contribution to total energy comes from long wavelength phonon branches.
• These are those with wavevectors near the centre of the Brillouin zone whilst those at the edges are frozenout.
• As a result the integral is independent of the Brillouin zone edges, so we can discard the first Brillouinrestriction
• By considering an infinitesimal k-space volume as a shell of radius k and thickness dk (more detail inChapter 4), we can write that
d3k = 4πk2dk
• Can write total energy as
∆U ' Vˆ ∞
0
~vkeβ~vk − 1
4πk2 dk
(2π)3
using (3.2.13) to write εk = ~vk
• This is a standard integral in k to reach that
∆U ' π2(kBT )4N
30(~ωD)3, CV '
2π2
15
(kBT
~ωD
)3
NkB (3.5.9)
where this is only valid for kBT � ~ωD, and Va3 = N , the number of unit cells (or the number of states
in the band).
The heat capacity of this regime is a fraction of what is expected for an equipartition system. We also see thatthis fraction is the fraction of all modes that are excited at a temperature T.
We wish to count the number of these excitations (NOTE: this is a separate, heuristic argument for the resultswe just found):
• Density of states in energy is given asδNstates = g(ε)δε (3.5.10)
• This gives the number of states in the energy range ε to ε+ δε
17
• A long wavelength phonon uses the dispersion relation
εk = ~v|k| (3.5.11)
• Can relate energy and wavevector densities of state via g(ε)δε = g(k)δk and thus that
g(ε) = g(k)dk
dε(3.5.12)
• The DoS in energy goes to
g(ε) =V k2
2π2~v=
V ε2
2π2(~v)3=
Nε2
2π2(~ωD)3(3.5.13)
where this equation is valid where ε� ~ωD, N is the number of unit cells, and we have used that v = ωDafrom (3.2.13).
• The number of states up to the energy kBT as a fraction of all states in the band is
1
N
ˆ kBT
0
g(ε) dε ∝(kBT
~ωD
)3
• The contribution to internal energy of each of these states is
ˆ kBT
0
ε g(ε) dε ∝ kBT(kBT
~ωD
)3
This separate method of counting the states is just an heuristic argument, to roughly support the previousdiscussion of the heat capacity and internal energy. It is not a separate physical argument for the same result.
The internal energy can be written as a function of the energy density of states, as
∆U =
ˆ ∞0
εg(ε)
eβε − 1dε (3.5.14)
This power law scaling of internal energy is similar to that of a black body, as we have effectively treated thecrystal as a black body for phonons instead of photons.
3.6 Summary of Chapter 3: What you need to know/be able to do
• Write down potential energy between atoms in a monatomic chain
• Derive equation of motion and dispersion relation of a monatomic chain
• Understand effects in long and short wavelength limits for the monatomic chain
• Understand the three branches of vibrational modes in a three-dimensional lattice (sound vs shear waves)
• Identify first Brillouin zone in different types of lattice
• Derive both conditions on allowed wavevectors in finite systems (periodic boundary conditions and first BZ)
• Derive g(ε) for any spherically symmetric dispersion relation
• Derive heat capacity and internal energy in both high-temperature and low-temperature regimes, with eitherfull integration (first method shown) or the heuristic argument (second method).
18
4 Free Electron Model
We will now treat a metal as a lattice of positive ions with mobile electrons, in order to consider how electronsbehave in a metal. The major factors determining how they behave are as follows:
1. Electrons are very small and undergo zero-point motion (motion at T = 0K). We therefore require quantummechanics to describe their motion.
2. Electrons obey Fermi-Dirac statistics.
3. Electrons interact with the positively charged ions.
4. Electrons repel each other due to the Coulomb force.
When constructing this model, we will assume that electrons observe static ions and thus a periodic potential,and we will neglect the Coulomb interaction between electrons for the sake of simplicity.
4.1 Modelling electrons in a metal as a Fermi Gas
• Wish to model electrons in a metal as a non-interacting electron gas
• This model is known as the free electron model.
• Electrons obey the Pauli exclusion principle. As we add electrons to the system they fill up successivelyhigher energy estates up to the Fermi level EF , equal to the kinetic energy of the most energetic electronsin the solid
• Typically EF � kBT , which is a Fermi system in a quantum degenerate regime
• In these cases the Fermi energy is the dominant energy scale in a quantum degenerate system
• This somewhat justifies using the Fermi gas model to approximate electron behaviour in solids
4.2 Filling energy levels
We wish to define a ground state of a system with N electrons in a volume V. This has a few caveats:
• We add electrons level by level in levels of increasing energy
• By the Pauli exclusion principle, no two electrons can share the same state.
• This means there can be only 2 electrons in each energy level, a result of the two spin states ↑ and ↓.
• This means we will haveN
2occupied energy levels
• The highest filled energy level is called the Fermi level, denoted EF .
A general energy level is defined in terms of the electron momentum, as
εk =~2k2
2m(4.2.1)
for an electron with momentum ~k, and m equal to the free electron mass. In general we will need to replacem with the effective mass of the electron in the relevant substance, but for a monovalent metal (metal withvalency one, such as sodium) this is very close to the free electron mass.
• We can write the Fermi level as
EF =~2k2
F
2m(4.2.2)
• kF is the Fermi wavevector.
• We represent all the occupied plane-wave states with a given momentum ~k as a point in reciprocal space
• This produces a Fermi surface, which is spherical for electrons in free space. This spherical surface hasradius equal to the Fermi wavevector
• The dispersion relation of electrons in the system defines the shape of the Fermi surface
• A Fermi surface is the locus of points in reciprocal space where electron energy εk = EF
19
• Surface is often spherical if particle wavelengths at the Fermi level are much larger than lattice spacing
• However if wavelength is comparable to lattice spacing electrons do not behave as if the environment isisotropic and so the dispersion relation (and by extension the Fermi surface) becomes more complicated
We can relate real-space number density of electrons to Fermi surface volume.
• Expect number of occupied states = Fermi surface volume × k-space density of states
• Therefore writeN
2=
4πk3F
3
V
(2π)3(4.2.3)
• Use N2 as we expect that many occupied states for N electrons
• Rearrange for definition of Fermi wavevector in terms of electron number density:
kF = (3π2n)13 (4.2.4)
for n = electron number density, and for a 3D Fermi gas
• We can use (4.2.2) and (4.2.4) to find an expression for the Fermi level in terms of electron number density,as
EF =~2
2m(3π2n)
23 (4.2.5)
• From the Fermi wavevector we can also define the Fermi wavelength, which is predictably the wavelengthof electrons at the Fermi level, as
λF =2π
kF=
2π
(3π2n)13
∝ n− 13 (4.2.6)
• We can then see that this wavelength is of the order of the size of the crystal unit cell
• Using the de Broglie relation we define the Fermi velocity as
vF =~kFm
(4.2.7)
4.3 Density of States
• Want to express number of states in the Fermi level via density of states
• Need to consider number of states in reciprocal space, in a particular range
• Begin with a spherical shell that is the range of k→ k + δk (illustrated in Figure 4)
Figure 4: Infinitesimal spherical shell in k-space.
• Volume of shell = shell surface area ×δk
20
• Number of states can be written down using k-space DoS, as
N = 4πk2δk × V
(2π)3(4.3.1)
• Can convert this to energy using dispersion relation, to find states in range ε→ ε+ δε
• Know that
k =(2mε)
12
~
• However, still need to relate δk to δε. Can say that
δk
δε=dk
dε=
(2m
ε
) 12 1
2~
• Now we can replace terms in our expression for number of states to get things in terms of ε:
N =V
(2π)34πk2δk =
mV
2π2~3(2mε)
12 δε = g(ε)δε (4.3.2)
• Therefore we can write
g(ε) =mV
2π2~3(2mε)
12 (4.3.3)
• This is the density of states in energy per spin state (as nowhere have we accounted for spin degeneracy).We will need to multiply by the spin degeneracy factor 2 to find the complete density of states.
4.4 Non-zero temperature
For a system with non-zero temperature we do not have all electrons in their lowest possible states. Instead,some will be thermally excited.
• The occupation number at a temperature T of a state with energy ε is given by the Fermi-Dirac distribu-tion, as
nF (ε, T ) =1
e
ε− EFkBT + 1
(4.4.1)
• No chemical potential as at & near T = 0 chemical potential ' EF .
• Integrate over entire range for states at a given level, i.e.
N = 2V
ˆ1
eβ(εk−EF ) + 1
d3k
(2π)3(4.4.2)
where the factor 2 accounts for spin degeneracy.
• Integrand is only dependent on k through the energy εk, so we can convert from 3D k integral to a 1Denergy integral:
N = 2
ˆ ∞0
g(ε)
eβ(ε−EF ) + 1dε (4.4.3)
• As temperature raises, Fermi-Dirac distribution only changes significantly if |ε − EF | is comparable tokBT such that the exponential is approximately unity
• ∴ only the occupation of single-particle states near the Fermi level are affected by thermal excitation
• Say that the affected single-particle states have energy within ±2kBT of the Fermi level (2 is nominallyarbitrary)
This can be summarised that in a quantum degenerate system, only states near the Fermi level play a role inthe system’s physical properties, whilst those further away are frozen out by the Pauli exclusion principle.
21
4.5 Electronic heat capacity
Note: this an heuristic argument and is examinable, whilst the mathematically rigorous argument is providedin Derek Lee’s notes but is not examinable.
• Expect states within δε ' kBT of the Fermi level to be thermally excited
• Electron of energy ε < EF can gain energy of kBT if the final state is unoccupied (typically if ε+kBT > EF )
• If we assume the density of states is approximately constant over the range of interest we can say thatthe number of states we expect to experience thermal excitation is given as
2 · g(ε)δε = 2g(EF )kBT (4.5.1)
• If density of states is not constant over range (e.g. having 0 occupancy at Fermi level) then we have tointegrate DoS over the range of interest, such that we calculate
N = 2
ˆ EF
EF−δεg(ε) dε
• If each electron that can be thermally excited gains energy kBT the energy change from raising thetemperature from 0 is expected to be
U(T )− U(0) ' 2g(EF )(kBT )2 (4.5.2)
for the constant DoS approximation.
• This means CV from electrons is given as
CV =
(∂U
∂T
)V
' 4kBg(EF )kBT (4.5.3)
4.6 Real metals
• Have contributions to heat capacity from lattice vibrations (Chapter 3) and from mobile electrons
• Vary with T 3 and T respectively
• Expect general expression for heat capacity to take the form
CV (T ) = γT +AT 3 (4.6.1)
where γ gives information on electronic properties of the system and A gives information on latticevibrations. Obviously the electronic factor dominates at low temperatures and the lattice vibration factorat high temperatures.
4.7 Summary of Chapter 4: Things you need to know/be able to do
• Derive energy density of states in 1, 2 and 3D
• Understand and be able to explain what the Fermi level and a Fermi surface are
• Apply the concept of a Fermi surface to a free Fermi gas to express occupation numbers of state ranges
• Understand only states near the Fermi level affect physical properties as others are frozen out by the Pauliexclusion principle
• Derive the electronic heat capacity using the heuristic argument
22
5 Electron states in a crystal
Wish to consider interaction of electrons with ions:
• Generally consider a periodic potential, first strong and then weak (tight-binding and nearly free electronmodels)
• Can write down Schrodinger equation for these systems as
Hψ =
[− ~2
2m∇2 + V (r)
]ψ = Eψ , V (r) = V (r + R) (5.0.1)
• The requirement on V is because we need it to be periodic, R is some lattice vector.
• Total periodic potential is sum of individual potential wells v from each of the ions in the lattice:
V (r) =∑R
v(r−R) (5.0.2)
• Only require limr→∞
v(r)→ 0
5.1 Tight binding model
• Model periodic potentials throughout a lattice as an array of potential wells
• For atoms far apart compared to atomic orbitals, each electron will have energy eigenstates like those ofisolated orbitals
• Will still have possible tunnelling from one well to another (allows for sea of mobile electrons)
5.1.1 Single well: quantum tunnelling
• Will refer to energy eigenstates of a single well as orbitals (by analogy to atomic orbitals)
• Denote normalized orbitals as uα(x) with energy εα for α = 0, 1, 2, 3, . . .
• These orbitals obey the Schrodinger equation
Hsingleuα =
[− ~2
2m
d2
dx2+ v(x)
]uα = εαuα (5.1.1)
We expect the orbital wavefunctions to oscillate in the classically allowed region (i.e. where εα > v(x), with αnodes in the oscillation. More nodes means faster spatial oscillations means more kinetic energy.
Particles can access the classically forbidden region where εα < v(x) via quantum tunnelling. The wavefunctionsthen take the form of a decaying exponential.
5.1.2 Double well: energy splitting due to quantum tunnelling
Consider two identical wells, separated by a large distance such that quantum tunnelling is weak.
• Low tunnelling probability as wells have large separation
• Amplitude of the single-well orbital in one well is small in the other
• Weak tunnelling approximation breaks down if well separation becomes smaller than decay length ofsingle-well orbital wavefunction
We construct total functions as superpositions of the wavefunctions from each well. We can have positive andnegative parity solutions at each value of α by adding or subtracting the two solutions e.g. the first two solutionsare given as
ψ0+(x) ≈ 1√2
[u0
(x− a
2
)+ u0
(x+
a
2
)]23
ψ0−(x) ≈ 1√2
[u0
(x− a
2
)− u0
(x+
a
2
)]for well separation a. Although for the α = 0 case the + and - signs in the wavefunction (not the subscript)correspond to the parity of the solution, this is not necessarily always the case. In general the odd-paritysolutions have higher energy than the even-parity as it has an additional node.
• Higher energy orbitals can tunnel further & have a greater probability of being found in the other well
• Quantum tunnelling is thus said to be stronger for higher energy orbitals
• As a result energy splitting is greater for higher orbitals (difference in energy between odd and even paritywavefunctions increases)
5.1.3 Double well: tight binding approximation (non examinable)
NOTE: Including this non-examinable derivation but some parts are important to know. These parts will behighlighted.
• Expect total potential to be a sum of the potential from both wells and thus total Schrodinger equationis
Hdoubleψ =
[− ~2
2m
d2
dx2+ vL(x) + vR(x)
]ψ = Eψ (5.1.2)
• vL(x) and vR(x) refer to the left and right single wells centred at x = ±a2 respectively.
• Expect pairs of states as solutions, each pair constructed by a single-well orbital level
• Approximate solution as having the form
ψ(x) ' cLuL(x) + cRuR(x) (5.1.3)
where uL(x) = uα(x+ a2 ) and uR(x) = uα(x− a
2 ).
• Now have reduced continuous quantum motion described by ψ(r) to a discrete problem in probabilities offinding an electron in each well (or indeed unit cell R)
This is the central simplification of the tight binding model.
Each shifted orbital will obey the single-well equations independently. We can apply the double-well Hamiltonianto the two separate solutions individually to say that
HdoubleuL =[H
(L)single + vR
]uL = [εα + vR(x)]uL
HdoubleuR =[H
(R)single + vL
]uR = [εα + vL(x)]uR
• Can apply Double-well Hamiltonian to the composite wavefunction to show
Hdoubleψ = Hdouble(cLuL(x) + cRuR(x)) = cLHdoubleuL + cRHdoubleuR
• Can then use previous result for applying double-well Hamiltonian to single well orbitals to write
cL[εα + vR(x)]uL(x) + cR[εα + vL(x)]uR(x) = E[cLuL(x) + cRuR(x)] (5.1.4)
• To extract cL and cR we need to do overlap integrals
• Multiply both sides by u∗L and integrate over all space, and repeat for u∗R.
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• Produces two simultaneous equations given as[εα +
ˆ|uL|2vR dx
]cL +
[εα
ˆu∗LuR dx+
ˆu∗LvLuR dx
]cR = E
[cL + cR
ˆu∗LuR dx
](5.1.5)
[εα +
ˆ|uR|2vL dx
]cR +
[εα
ˆu∗RuL dx+
ˆu∗RvRuL dx
]cL = E
[cR + cL
ˆu∗RuL dx
](5.1.6)
• Some integrals are removed from this result by the normalization condition on the uL and uR wavefunc-tions.
• The 6 integrals above are effectively only 3 distinct integrals along with reflection in the x = 0 (swappingthe L and R around).
We will consider each of these integrals in turn.
1. The most important of these is the hopping integral
tα = −ˆu∗LvLuR dx = −
ˆu∗RvRuL dx (5.1.7)
• This integral has dimensions of energy
• Minus sign is the conventional definition
• In general the sign of tα depends on chemistry of the atoms infolved
• Can interpret this as the scattering of an electron in the right well to the left (for the first integral)or from left to right (second integral).
• An electron sees the potential in the other well and hops across to that well
2. A second energy integral is
∆εα =
ˆ|uL|2vR dx =
ˆ|uR|2vL dx (5.1.8)
• Involves probability of finding an electron in the vicinity of one well when it is in an orbital of theother well
• Multiplies amplitudes in same way as orbital energy & thus can be seen as a shift to orbital energy
• Does not contribute to the energy splitting of the level & therefore is not that relevant to what wewant to know
• We will thus absorb it into εα and not worry about it
3. Final integral is ˆu∗RuL dx =
ˆu∗LuR dx ' 0 (5.1.9)
• Integral represents overlap of the two orbitals in different wells
• Weak tunnelling =⇒ this overlap is small
• We therefore ignore this term
With these simplifications, we can now turn our simultaneous equations into a sensible, solvable format.
• We can write the equations as
εαcL − tαcR = EcL
εαcR − tαcL = EcR(5.1.10)
• This is an eigenvalue problem for the energy E
25
• Can regard this as a discrete Schrodinger equation for this system
• Can express it in matrix form: (εα − tα−tα εα
)(cLcR
)= E
(cLcR
)(5.1.11)
• Characteristic equation to solve is ∣∣∣∣εα − E − t−t εα − E
∣∣∣∣ = 0 (5.1.12)
• The solutions are (cLcR
)=
1√2
(1±1
)for E = εα ∓ tα (5.1.13)
• The two levels therefore have an energy splitting of 2|tα|.
• Eigenstates are equal superpositions of the single-well orbitals - same as estimated in Section 5.1.3.
This method is an application of degenerate perturbation theory. We have considered a system with degenerateorbitals, and then introduced the hopping as a perturbation. This is valid so long as the splitting is smallcompared to the energy spacing between the single-well energy levels, i.e. |tα| � |εα − εα±1|.
We can improve the trial wavefunction but for our purposes it is unnecessarily complicated.
5.1.4 One-dimensional array: wave motion and energy bands
Consider an infinite one-dimensional array with lattice spacing a:
• Lattice sites at x = xn = na for integer n labelling a site
• There is a potential well centred at each lattice site, denoted vn
• Total periodic potential seen by an electron is
V (x) =
∞∑n=−∞
v(x− na) =
∞∑n=−∞
vn(x) (5.1.14)
• Want to solve Schrodinger equation with this potential
• Taking weak hopping again we can focus on electron states formed from the same single-well orbitals atdifferent sites
• In absence of hopping, for N lattice sites there are N degenerate orbitals
• When hopping is introduced degeneracy is lifted
For each orbital α, we have wave-like eigenstates when hopping exists.The energies of these eigenstates form a band centred near the isolatedorbital energy εα with a bandwidth of 4|tα|, where tα is a hoppingintegral.
There are no propagating states in the energy intervals between these energy bands. These intervals with nopropagating states are called band gaps.
26
5.1.5 The discrete Schrodinger equation for the one-dimensional array
NOTE: The derivation of these equation is effectively the same argument as for the double well and is alsonon-examinable, so will not be covered in these notes.
• The discrete Schrodinger equation is given as
εαcn − tαcn−1 − tαcn+1 = Ecn for all n (5.1.15)
• Equation shows each cn depends on the nearest neighbours and no others
• First term gives on-site energy of an electron
• Other two terms can be regarded as a lowering of kinetic energy from hopping from one site to itsneighbours
Like in the coupled equations in the lattice vibration analysis in Chapter 3, we should try a wave-like solution:
• Consider cn = Aeikxn
• As normal this wave-like solution has the same wavevector redundancy as normal so we restrict k to thefirst Brillouin zone
• Subbing this into the discrete Schrodinger equation gives
Aeinka[εα − E − tαe−ika − tαeika
]= 0 (5.1.16)
• A and the exponential must be non-zero therefore the expression in the bracket must be equal to 0, givingus an expression for E in terms of k
• Eigenstates and eigenenergies of the system are
ψαk(x) = A∑n
einkauα(x− na)
Eαk = εα − 2tα cos(ka)
(5.1.17)
for k in the first Brillouin zone.
• The energy bands for the α = 0 and α = 1 orbitals can be seen in Figure 5
Figure 5: Energy spectrum of the one-dimensional tight binding model. Top energy band is α = 1, bottom isα = 0. Dashed lines indicate the energies at zero tunnelling. t0 = 0.5, t1 = −0.75. Can observe the bandwidthof 4|tα| in terms of the range of energies in each band
• For positive tα the lowest energy state is at k = 0, but for negative tα it is at k = πa (and −πa but that’s
a redundant wavevector so is not included in the first Brillouin zone)
• Solutions are labelled both by orbital index and by k and are propagating waves.
27
These solutions are not eigenstates of the momentum operator. We therefore cannot say that ~k is the momen-tum of the electron. Although they are not the same everywhere in space like a pure wave, they are the sameunder a shift of any lattice vector R = ma i.e.
ψαk(x+R) = A∑
n=−∞einkauα(x− (n−m)a) = A
∞∑n′=n−m=−∞
ei(n′+m)kauα(x− n′a)
= Aeik(ma)∞∑
n′=−∞ein′kauα(x− n′a) = eikRψαk(x)
(5.1.18)
As before we call ~k the crystal momentum, associated with a discrete translation on the wavefunction
• If we have N wells we can arrange them in a ring much like in the finite monatomic chain
• Applying periodic boundary conditions will then suggest that
cm+N = cm (5.1.19)
• The probability amplitude is obviously the same if you go fully around the ring back to the same point(which is what this statement says).
• These should obey the same discrete Schrodinger equation as the infinite chain
• But allowed k no longer are continuous, and instead form discrete set defined as eiNka = 1.
• This gives a set of wavevectors with a spacing of 2πNa like in the infinite monatomic chain
• This means state counting in reciprocal space is exactly the same for this model as in the phonon case -just the number of allowed wavevectors.
5.1.6 Tight binding model in higher dimensions
Consider a lattice of potential wells centred at lattice positions R:
• Tight binding model gives approximate solution of
ψα(r) =∑R
uα(r−R) (5.1.20)
• uα(r−R) is the bound state in the well located at site R.
• From Equation 5.1.15 we can see that the equation involving the site m simply involves nearest neighboursm± 1 via the hopping integral tα.
• Can therefore easily extend discrete Schrodinger to higher dimensions: for 3D it is given as
εαcR − tα∑δ
cR+δ = EcR (5.1.21)
• δ denotes the set of vectors that connect R to all of its nearest neighbours (and so takes 6 values in 3D).
• Again we try a wave-like solution
cR = Aeik·R , k ∈ first Brillouin zone (5.1.22)
• Again this decouples equations and thus eigenstates and eigenenergies are given as
ψαk(r) = A∑R
eik·Ruα(r−R) , Eαk = εα − tα∑δ
eik·δ (5.1.23)
28
• In the long wavelength limit (|k|a � 1) we observe spherical (or circular) Fermi surfaces (contours ofconstant energy) as electrons behave as if they are in free space
• For a cubic lattice with spacing a the energy spectrum is given as
Ek = εα − 2tα [cos(kxa) + cos(kya) + cos(kza)] (5.1.24)
• When wavelength becomes comparable to lattice spacing the Fermi surface will take on more interestingshapes.
• The number of allowed wavevectors in the first Brillouin zone is equal to the number of unit cells, aspreviously discussed
• This means there are N distinct eigenstates in each tight-binding band α
• This means each tight-binding band is completely filled when there are 2 electrons per unit cell
• At low electron densities the Fermi surface is spherical (or circular) but diverges from this as electrondensity increases.
• In the same manner as for the one dimensional case but using vectors instead of discrete points m and n,we can show that the under a shift by a lattice vector R the wavefunction changes by a phase factor ofeik·R
5.2 Bloch’s Theorem
A typical conservation law that we have already encountered is the conservation of the crystal momentum ~kin the first Brillouin zone, even though electron momentum is not conserved. Bloch’s theorem is a more precisestatement of this.
• Consider electron in a periodic potential such that V (r + R) = V (r)
• Bloch’s theorem tells us that the electron eigenstates ψ(r) are also periodic, apart from a phase factorwhen we move from one lattice site to another
• Can label states by the wavevector k associated with the phase factor
Bloch’s theorem: For any eigenstate ψ(r) of a periodic system,
ψαk(r + R) = eik·Rψαk(r) for any lattice vector R (5.2.1)
for k ∈ first Brillouin Zone
For this general case the α is a band index, rather than specifically denoting orbital like it did in the tight-binding model.
• We can factor out the plane-wave part from ψαk to say that
ψαk(r) = eik·rφαk(r) (5.2.2)
with φαk(r) = φαk(r + R) and k ∈ 1st Brillouin zone
• The function φαk has the same periodicity as the potential
• We can find it by calculatingφαk = e−ik·rψαk
• A simple corollary of Bloch’s theorem is that
|ψαk(r + R)|2 = |ψαk(r)|2 (5.2.3)
• This means that the probability density of finding a particle at r is the same if we shift r by a latticevector
• Consistent with what is expected: physical environment around the points at r and r + R are the samefor an infinite periodic crystal
29
5.3 Nearly Free Electron Model
Want to consider free electrons moving in a weak periodic potential. We will expect it to act as a perturbation
to free electrons with energy εk = ~2k2
2m , becoming a relevant perturbation at the edge of the first Brillouin zone.
5.3.1 Phonon band gap in a diatomic chain
To introduce the idea of an energy band gap from a perturbation, we will look at vibrational modes in a diatomicchain
• Particles have alternating masses M & m along the chain
• All springs connecting particles are identical, spring constant κ
• Each unit cell contains one of each mass
• When in rest at equilibrium unit cell has length a, so particle separation = a2
NOTE: Full derivation of the dynamics of this system is found in Chapter 3 Section 6 of Derek Lee’s notes(non-examinable)
Consider the mass variation as a periodic perturbation of period a, on the monatomic chain with masses m0
and spacing a2 .
• The propagating modes of the monatomic chain can be considered as waves propagating through andundergoing Bragg scattering off of the perturbing ’chain’
• Experience Bragg scattering when kin = q2 for q = G = m 2π
a for integer m (Chapter 2, section 2.4 of thesenotes)
• Therefore expect Bragg scattering at k = ±πa - these modes are strongly backscattered into each other
• This combination of backscattered modes creates new normal modes
• These are standing waves of wavelength = 2πk = 2a
• If nodes of these waves are on light masses, effectively see large masses M oscillating with frequency√
2κM
• If nodes of these waves are on the heavy masses, effectively see light masses m oscillating with frequency√2κm
• This difference splits the single phonon band into two disjoint bands that have a discontinuity betweenthem
This band gap has no propagating states in it, and is larger the greater the difference between m and M. SeeFigure 6 for these bands.
To make the gap more visible, we can plot the diatomic curve in a reduced zone scheme, where we translateeverything outside of the first Brillouin zone a number of reciprocal lattice vectors such that it is in the firstBrillouin zone. See Figure 7.
Figure 6: The phonon energy spectrum of themonatomic and diatomic chains. The firstBrillouin zone for the monatomic chain is for− 2π
a < k ≤ 2πa , but the first Brillouin zone for
the diatomic chain is for −πa < k ≤ πa .
Figure 7: The reduced zone scheme represen-tation of the diatomic chain phonon band.
30
5.3.2 Electron band gap in a weak periodic potential
First consider electrons in one dimension with a periodic potential V(x) with a spatial period of a
• Expect strong Bragg scattering between states at k = πha for integer h
• This corresponds to a scattering wavevector q = G = 2πha (reciprocal lattice vectors)
• Energy band of free electrons now splits into disjoint bands like the phonon bands, with disjoints atk = ±mπ
a for integer m
• The eigenstates backscattered at these points will form new states that are standing waves, e.g. for k = ±πawe get simple cosine and sine waves (of πx
a ), nodes at x = (n+ 12 )a and x = na respectively
• Consider ions (source of periodic potential ) centred at x = na, so the potential seen by the electrons isgreatest at x = (n+ 1
2 )a and lowest at x = na.
• By looking at the probability amplitudes of the two eigenstates we can see that the sine eigenstate has thehighest amplitude in the high energy parts of the potential and the cosine eigenstate has lowest amplitudethere & thus avoids the high energy parts of the potential (seen Figure 8)
Figure 8: Probability amplitudes of the sine and cosine eigenstates formed at the edge of the first Brillouinzone. The dashed line is the potential observed by the electron. Sine is a max when energy is a max, cos is amax when energy is a min.
• This means the cosine eigenstate has lower energy and thus two non-degenerate eigenstates at k = πa
• This is the reason for the first band gap in the spectrum. This same effect is why each of the band gapsexist.
• Figures 9 and 10 show the electron spectrum and the reduced zone scheme for the same spectrum respec-tively
Figure 9: Electron energy band in the extendedzone scheme (i.e. displaying more than 1 Brillouinzone)
Figure 10: Reduced zone scheme for the electronbands. The colours correspond to those in Figure9.
31
We can label an energy spectrum with the crystal momentum, as this is conserved under translations of a latticevector, and then shift all k labels by a multiple of 2π
a until they fall in the first Brillouin zone. This createsthe reduced zone scheme, whilst plotting out the spectrum without doing this is called the extended zonescheme.
5.4 Summary of Chapter 5: What you need to know/be able to do
• Describe the physical assumptions behind the tight-binding model & its range of validity
• Explain the significance of the hopping integral tα
• Write down and solve the discrete Schrodinger equation for a given system (Equation 5.1.21)
• State both forms of Bloch’s theorem (Equation 5.2.1 and Equation 5.2.2)
• Describe effects of Bragg scattering on electron states and how it produces band gaps in energy
• Understand reduced and extended zone schemes
32
6 Electrons in Solids
6.1 Electron models
We have three main ways of describing electron states within a crystal:
1. Free electron wavefunctions (Sommerfeld (free electron) Model)
2. Strongly coupled free electron wavefunctions (Nearly Free Electron Model)
3. Weakly coupled bound atomic orbitals (Tight Binding Model)
We will review these three in turn.
6.1.1 Sommerfeld (free electron) model
• Assumes electrons are completely free with no binding potential
• Gives Hamiltonian as
H =p2
2m
• Has plane-wave like solutions with parabolic dispersion, given as
ψk(r) =ei(k·r−
Ekt
~ )
VEk =
~2k2
2m(6.1.1)
• Will describe some metals quite well
6.1.2 Nearly free electron model
• Improves theory by considering periodicity of crystal
• Uses free electron wavefunctions but allows Bragg scattering when the Laue condition is fulfilled
• Scattering of electrons creates forbidden energy bands as seen in Chapter 5.
• Insist that the dispersion respects crystal periodicity, i.e. that
Ek =~2(k−G)2
2mfor all G (6.1.2)
• Parabolic dispersion at every reciprocal lattice value
• Points where states intersect (where Laue condition is satisfied) are where we expect scattering and thenenergy bands
6.1.3 Tight binding model
• Starts from atomic wavefunctions and then add coupling between atoms
• Solutions dependent on electron state quantum number n and wavenumber k (n is equivalent to α inDerek Lee’s course)
• Solutions and energies given as
ψnk(x) = A∑l
eilkaun(x− la) Enk = En0 − 2tn cos(ka) (6.1.3)
for integer l used to identify a lattice site
• Labels n required to distinguish state of same k in different bands
33
6.2 Bloch’s Theorem
Must solve Schrodinger’s equation in a periodic potential to get a complete picture:
p2
2mψ(r) + V (r)ψ(r) = Eψ(r) (6.2.1)
When solving this rigorously, we get Bloch wavefunctions, labelled for a band n and wavevector k:
ψnk(r) = eik·runk(r) (6.2.2)
These wavefunctions are identical if we shift k by reciprocal lattice vectors:
THIS PROOF IS WRONG: Oulton gave it like this but as far as I can tell it doesn’t work nomatter how you spin it. There are some examples online that do it by assuming stuff abouttranslation operators so if you want you can look at those but this is what Oulton said. Hemight fix the explanation for later years but yeah this is garbage
‘Proof’: Wavefunctions shifted by G are identical
By definition
ψnk′ = unk′(r)ei(k+G)·r =[unk′(r)eiG·r
]eik·r
for k′ = k + G
Want to show the term in brackets is equal to unk.
• From Derek Lee’s lectures we know the tight binding wavefunction, and can write down that
unk′(r) =∑R
eik′·(r−R)φn(r−R)
• φn(r) are the wavefunctions of each atom (uα in Derek Lee’s section)
• unk(r) is the lattice periodic function (φαk in Derek Lee’s section)
• Multiply both sides by eiG·r to get
unk′(r)eiG·r =∑R
eik′·(r−R)+iG·rφn(r−R)
• Can replace k′ with k′ −G if we add terms to compensate, such that we get
unk′(r)eiG·r =∑R
ei(k′−G)(r−R)−iG·Rφn(r−R)
• This is then equal tounk′(r)eiG·r = unk′−G(r)e−iG·R
• As by definition G ·R = n2π for n ∈ Z, we can write that
unk′(r)eiG·r = unk(r)
∴ ψnk′(r) = ψnk(r)
6.2.1 Zone schemes
• As a result of this property, wavefunctions outside the first Brillouin zone can just be translated byreciprocal lattice vectors into the first Brillouin zone
34
• This means we normally only need to use the reduced zone scheme i.e. the first Brillouin zone
• Extended zone scheme is relevant for the nearly-free electron model and the Sommerfeld model, whereperiodicity is less strict
• The periodic zone scheme lets us demonstrate the periodicity of a dispersion relation, by effectively justrepeating the reduced zone scheme, and reminds us that electrons can move across zone boundaries
6.3 Band theory, simplified
• In semiconductor physics, periodicity is less relevant as we restrict to first Brillouin zone
• We are interested in phenomena that occur within a few kBT of the boundaries of conduction and valencebands
• In these ranges bands are nearly parabolic & so we don’t need to know the full band structure
• Typically the energy gap Eg � kBT
• Typical range of energies in a band is also large ∆EC � Eg � kBT (see Figure 11)
Figure 11: Diagram of energy bands, illustrating the scales of Eg, ∆Ec and ∆Ev in comparison with kBT , inthe reduced zone scheme
• Number of valence electrons per unit cell = number of valence electrons per type of atom × number ofthese atoms per unit cell
• Electrons fill states up to the Fermi level
• HOWEVER: Fermi level could be in a band or in a band gap in band theory
35
Discussion of electrons per band
States per band in the First Brillouin zone:
• Can write number of electrons per energy band in the first Brillouin zone as the sum over all allowed wavevec-tors in FBZ, as
N = 2∑
k´FBZ
= 2V
(k)FBZ
( 2πL )3
= 2V
(k)FBZ
(2π)3V
• V is crystal volume, V(k)FBZ = (2π)3
Vprimitiveis k-space volume of FBZ, Vprimitive is primitive cell volume
• ∴ N = 2 VVprimitive
= 2 × Number of primitive cells in crystal
• Regardless of the size of the crystal we have at most 2 electrons per unit cell
6.4 Metals, Insulators and Conductors
Metals
• Highest occupied band is partially full
• This band is called the conduction band
• Conducts as electrons can ‘move’ into nearby enmpy states
• Fermi level in conduction band
Insulators
• Valence band is full, conduction band is empty
• Fermi level in the band gap
• No conduction as there are no empty states close for the electrons to ‘move’ into
• Band gap too large for thermal excitation
Semiconductors
• Valence band full, conduction band empty (like insulators)
• Fermi level lies in band gap
• Band gap small enough to allow thermal excitation from valence band to conduction band
• Conduction is possible, material can be doped to improve conductivity
There are a few exceptions to these rules:
• Group II elements conduct, even though they have a full s-shell: This is because in 2D a full s-shell doesnot mean the energy band is full (for reasons apparently). The Fermi wavevector will be outside the firstBrillouin zone and thus the Fermi level lies in the conduction band (and thus the conduction band is notfull & electrons can be excited to other states)
• Group IV elements form hybrid bands (due to stuff we don’t know), for diamond-like structures for C, Siand Ge.
6.5 Dynamics of Bloch Electrons and Holes
Want to simplify situation down to a mostly classical problem, with a few quantum aspects.
• Begin with defining a free electron velocity for electron wavefunctions in the Free Electron Model, to definean equation of motion
• These wavefunctions are just plane waves with a single wavevector
36
• Define vp, free electron velocity (phase velocity), in terms of electron energy E, wavevector k and momen-tum p
vp =E
~|k|=|p|
2me=
~2me|k| (6.5.1)
• Wavefunctions must also have a time-variation, as determined by TDSE
• Can define a group velocity as well, as
vg =1
~∇kE (6.5.2)
• However, Bloch electron wavefunctions are not just plane waves but instead have a lattice periodic com-ponent
• As a result the electron velocity is not the phase velocity but is instead the group velocity of the Blochwavefunction (see NON-EXAMINABLE derivation from lecture)
• As a result define Bloch electron velocity as
vg =1
~∇kEn =
1
~
(dEndkx
kx +dEndky
ky +dEndkz
kz
)(6.5.3)
6.5.1 Effective mass
Will discover that, in order to produce a mostly classical picture of how Bloch electrons move in a crystal, weneed to replace mass with an effective mass:
DERIVATION: Effective mass
• To describe motion of electron in the crystal, want to equate Newton II with electron equation of motion inthe crystal, as
F = mevgdt
• As vg is fixed by the wavevector and energy of the electron wavefunctions we can only change the mass me
• Thus replace mass m with effective mass m∗, so
F = m∗edvgdt
where this is Newton II
• Now wish to write out electron equation of motion in crystal
• Work done δW on an electron by an external force F in a time δt is
δW = F · δx = F · vgδt (6.5.4)
• From QM we write thatδW = ∇kEn · δk = ~vg · δk (6.5.5)
• Equate these two to get that~vg · δk = F · vgδt
37
• This then implies that
limδt→0
~∂k
∂t= ~
dk
dt= F
• Thus we find that
~dk
dt= F (6.5.6)
• Calculating vg is complicated in 3D so will restrict to 1D, so
d
dtvg =
1
~d2Endkdt
=1
~d2Endk2
· dkdt
• Subbing in expression for dkdt gives
dvgdt
=1
~d2Endk2
F
~
• Rearranging then gives that
F = ~2
[d2Endk2
]−1dvgdt
• We can then replace F with the Newton II we defined above, so we get that
m∗e = ~2
[d2Endk2
]−1
(6.5.7)
• This result is for the one dimensional case
• We may also have a multi-dimensional case, and we may even have an anisotropic crystal
• The general expression for effective mass is[1
m∗e
]ij
=1
~2
∂2En∂ki∂kj
(6.5.8)
where the effective mass m∗e is a 3 by 3 matrix (rank 2 tensor).
• The NON-EXAMINABLE derivation of this result is in the lecture scans
• This theory holds up well: for the free electron dispersion relation it pulls out the free electron mass asexpected
• For the parabolic band approximation we find precisely the coefficient of the energy dispersion parabolas(effective mass affects shape of bands) as
Ec(k) = Ec0 +~2k2
2m∗cEv(k) = Ev0 −
~2k2
2m∗v(6.5.9)
• We effectively see that we can use classical physics to describe electrons in solids, so long as we use theeffective mass of the electron
6.5.2 Consequence of Bloch’s theorem
• If electron wavefunction has same periodicity as underlying potential the electron has smooth & uniformmotion at constant velocity
• In a pure crystal (no defects) electrons move as if they are free electron wave packets
• Their velocity is given by the group velocity of the Bloch electron wavefunction
• Limited conductivity of true materials is a result of defects, impurities phonons and other electrons, notfrom periodicity
38
6.5.3 Holes
Consider a full valence band, with energy dispersion given as
Ev(k) = Ev0 −~2k2
2m∗v(6.5.10)
• For the full band, we must have∑k kei = 0 i.e. the sum over all electron wavevectors in the band is 0
• Now remove an electron with wavevector kej . Can consider this in two ways:
1. Total crystal momentum goes to ∑i 6=j
kei = −kej
and deal with all the electrons minus one
2. Have added a hole to the system, which has momentum
kh = −kej
and then look at how the single hole behaves
• The lower down the band we take the electron from, the system has higher energy (hole will move up theband)
• This means thatEh(kh) = −E(kej ) (6.5.11)
• This is a hypothetical hole dispersion
• This means holes must have an effective mass of opposite sign to electrons - negative in conduction band,positive in valence band
• Hole velocity:
vh =1
~∇kEh =
1
~∇−ke (−E(ke)) = ve
vh = ve (6.5.12)
In general we say it is far easier to consider a hole rather than an entire band of electrons.
6.6 Counting ‘free’ electrons and holes
‘Free’ electrons refers to electrons that are available to conduct
6.6.1 Counting electrons in a solid
• Electrons obey Fermi-Dirac statistics (and the Pauli Exclusion principle) so Statistical Physics tells ushow to count electrons in solids
• For a solid of carrier density n(E) at temperature T, the number of electrons between energies E1 and E2
is given as
NE1−E2=
ˆ E2
E1
g(E) f(E) dE (6.6.1)
where f(E) is the Fermi-Dirac probability function
f(E) =1
eE−EFkBT + 1
(6.6.2)
and g(E) is the density of states
g(E) =dn(E)
dE(6.6.3)
39
• At T = 0 f(E) accounts for electrons filling up to the Fermi energy EF
• For T > 0 f(E) accounts for thermal excitation of carriers (in this specific case electrons) beyond EF by∼ kBT
• Some metals can be well described by the Sommerfeld (Free Electron) model and this gives an expectedenergy dispersion of
E =~2k2
2m0(6.6.4)
DERIVATION: Density of states in a crystal
• Consider a crystal of volume V = L3
1. With physical boundaries the crystal reflects Bloch electrons: expect the lowest possible state to be a half-wavelength sinusoid, but this is a sum of forward and backward states & so would double count the states
2. Want to be able to treat forward and backward states separately so will introduce periodic boundaryconditions, i.e. don’t want the wavefunction to have to change at the boundaries. Require that for
V (x) = V (x+ L) (6.6.5)
and then the lowest possible state is a full sinusoid with wavelength equal to the crystal side length L.
This then only sums over electrons moving in one direction, rather than double counting
• Thus |kmin| = 2πL
• Therefore
kx,y,z =2π
Lnx,y,z =⇒ Vk1 =
(2π
L
)3
for nx = ny = nz = 1
• For free electron model a constant energy surface is a sphere so the associated k-space volume is given as
Vk =4
3πk3 (6.6.6)
• Using the energy dispersion relation E = ~2k2
2m0we can write the wavevector k as
k =
(2moE
~2
) 12
• Therefore k-space volume is
Vk =4
3π
(2m0E
~2
) 32
(6.6.7)
40
• The number of states up to energy E is given as
n(E) = 2× VkVk1
=8
3π
(2m0E
~2
) 32(L
2π
)3
=V
3π2
(2m0E
~2
) 32
• Density of states is then given as
g(E) =dn(E)
dE=
V
2π2
(2m0
~2
) 32 √
E (6.6.8)
• This DoS holds for any parabolic band
• Figure 12 shows the Fermi-Dirac curve, the density of states curve, and the product curve that we integrateover in order to find NE1−E2
Figure 12: Left: Fermi-Dirac distribution f(E). Middle: Density of states g(E). Right: Product of f(E) and g(E),what we wish to integrate to find number of states in energy range
• For Bloch electrons we encounter different densities of states in the conduction and valence bands
• For the conduction band we replace E with E(k)− Ec0 and m with m∗c
• In valence band replace E with Ev0 − E(k) and m with m∗v
• In the band gap, g(E) = 0. These are illustrated in Figure 13
Figure 13: Comparison of the density of states for the conduction and valence bands
41
6.6.2 Number of ‘free’ electrons in a metal
• For metals, EF is in the conduction band
• For sufficiently large gap (Eg � kBT ), can neglect Fermi-Dirac
• Can then use free electron theory to write that
N =
ˆ ∞Ec0
g(E)dE (6.6.9)
• As Eg � kBT , difference between band is negligible (more than enough energy for thermal excitation) so
can use normal density of states g(E) ∝√E
• Know that we have no occupied states above EF so integral goes up to EF , not ∞
• Evaluating this integral gives
N =V
2π2
(2m∗c~2
) 32ˆ EF
0
√E dE =
V
3π2
(2m∗cEF
~2
) 32
(6.6.10)
in agreement with free electron theory in that we have states up to the Fermi energy
6.6.3 ‘Free’ electrons in other solids
• For semiconductors and insulators we must account for electron transfers between bands
• Must use DoS in each band AND the Fermi-Dirac function
• We find the number of electrons in the conduction band as
N =V
2π2
(2m∗c~2
) 32ˆ ∞Ec0
√E − Ec0
eE−EFkBT + 1
dE (6.6.11)
• Could count electrons in valence band but this is impractical
• Instead count ‘free’ holes
• Fermi-Dirac function for holes must account for missing electrons i.e. require that
fh(E) = 1− f(E) (6.6.12)
• Number of holes is thus
P =
ˆ Ev0
0
g(E)(1− f(E)) dE
P =V
2π2
(2m∗v~2
) 32ˆ Ev0
0
√Ev0 − E
e−E−EF
kBT + 1dE (6.6.13)
• If we substitute through the definition of the hole Fermi-Dirac distribution we get that fh(E) = f(−E),so long as we define a hole Fermi energy EFh
= −EF
• To solve the integrals that make up the definitions of N and P we need to make some approximationsnormally, such as (Ec0 − EF ), (EF − Ev0)� kBT
• After this we can reach closed form solutions by a standard integral (or substitution, see PS5)
Once these integrals have been evaluated, we can write the number of electrons and holes in an insulator orsemiconductor (including thermal exchange of electrons between bands) as
N = 2V
(m∗ckBT
2π~2
) 32
e−Ec−EF
kBT (6.6.14)
P = 2V
(m∗vkBT
2π~2
) 32
e−EF−Ev
kBT (6.6.15)
42
• Typically write carrier densities rather than number of carriers, to avoid using volume:
n =N
V= 2
(m∗ckBT
2π~2
) 32
e−Ec−EF
kBT = NCe−Ec−EF
kBT (6.6.16)
p =P
V= 2
(m∗vkBT
2π~2
) 32
e−EF−Ev
kBT = NV e−EF−Ev
kBT (6.6.17)
• For insulators, ‘free’ carrier densities are negligible due to the large gap energy
6.6.4 Law of mass action
• Charge carrier concentration only depend on conduction/valence band dispersion and Fermi energy
• Product np is independent of EF and defines intrinsic concentration at a temperature T
• Write that
np = n2i = NCNV e
− EgkBT (6.6.18)
• If impurities change the balance of one charge carrier the other must change to compensate
• This is called the law of mass action
6.7 Intrinsic and Extrinsic Semiconductors
• A pure semiconductor material (with no impurities) is called an intrinsic semiconductor
• Carrier densities are matched i.e. there is 1 hole for every 1 electron =⇒ n = p = ni
• Intrinsic carrier concentration is defined as
ni =√np =
√NVNCe
− Eg2kBT (6.7.1)
• Intrinsic semiconductors are very poor conductors; we must dope them to make them useful
• Doping is the process of replacing individual atoms in a lattice with different atoms, to add or removefree electrons.
• These behave like adding a hydrogen atom to the system
• For example for a Si lattice we can add an As atom to add an extra electron, or a B atom to add an extrahole (remove an electron)
• For Si we take n-type dopants (donors, add an electron to the lattice) from group V, and p-type dopants(acceptors, add a hole to the lattice) from group III
6.7.1 Hydrogen model of dopants
Can model dopants as hydrogen-like atoms, as a single ion with a circulating electron (or in some cases acirculating hole)
• We can describe the dopant electrons via the Bohr model for hydrogen (see Atomic Physics course)
• Binding energies to the parent ion (i.e. the dopant ion like the As or B ion) are given as
En = − m∗ee4
2(4π~εrε0n)2= −13.6[eV ]
ε2rn2
m∗eme
(6.7.2)
for relative permittivity εr
43
• Orbital radii are given as
rn = n2aB =4π~2εrε0e2m∗e
= n2εrme
m∗e× 0.053[nm] (6.7.3)
• These are different from the traditional Bohr model results as we replace mass with effective mass andneed to consider the permittivity of the material we are working in
• We can do some simple calculations for the binding energy and atomic radius of the ground state in Si toshow that the electron is only weakly bound to the parent ion (E1 ∼ kBT ) and it will experience a lot ofthe crystal (r1 ∼ 7 atoms)
• This means the majority of the electron’s behaviour is not determined by the parent ion but is insteaddetermined by the crystal
• Good semiconductor materials will have dopant electron bound levels close to band edges so electrons arethermally excited into crystal at room temperature
6.7.2 Donor ionization probability: NON-EXAMINABLE
This section is not an examinable derivation but is somewhat useful for understandingUnderstanding the number of ionized dopants allows us to understand how adding dopants changes carrierdensities
• We use the grand canonical ensemble (see Statistical Physics course) to examine donor ionization proba-bility
• By this model the probability of a state having N particles and an energy E is
p(E, N) = Ae−E−NEF
kBT (6.7.4)
• Ed is the energy of the donor level
• Probabilities associated with the donor level (ignoring degeneracy) are:
1. Ionized electron:p(E = 0, N = 0) = A
2. Bound electron:
p(E = Ec − Ed, N = 1) = Ae−Ec−Ed−EF
kBT
3. Coulomb repulsion N < 2:p(E = Ec − Ed, N > 1) = 0
• Require that sum of probabilities is 1, i.e. that
A(1 + e−Ec−Ed−EF
kBT ) = 1 (6.7.5)
• Can therefore write that the probability of the donor level being ionized (i.e. it having no electrons) as
p(E = 0, N = 0) = A =1
1 + e−Ec−Ed−EF
kBT
(6.7.6)
• For a given donor doping density ND the dopant ion density is then given as
N+D = ND p(E = 0, N = 0) =
ND
1 + e−Ec−Ed−EF
kBT
44
Can use definition of electron carrier density n (Equation 6.6.16) to remove Fermi energy from this expression,to write that
N+D =
NDNC
NC + neEd
kBT
(6.7.7)
Important to note that presence of dopant carriers shifts the Fermi level as
n = NCe−Ec−EF
kBT
we can see that
Ec − EF = kBT ln
(Ncn
)As n increases, EF moves closer to conduction band.
• By similar arguments, we can find acceptor ionization probabilities and thus the ionized acceptor density
N−A =NANV
NV + peEa
kBT
(6.7.8)
where Ea is the acceptor level
• Again presence of these carriers shifts the Fermi level: as p increases, the Fermi level moves closer to thevalence band
45
6.7.3 Extrinsic carrier density
Can calculate new carrier concentrations using the numbers of ionized donors and acceptors, using the chargeneutrality condition and the law of mass action np = n2
i :
• For donors: n = N+D + p
=⇒ n =n2i
n+
NDNC
NC + neEd
kBT
(6.7.9)
• For acceptors: p = N−A + n
=⇒ p =n2i
p+
NANV
NV + peEa
kBT
(6.7.10)
• Assume that doping concentration ND � ni at room temperature (i.e. doping concentration is much greaterthan intrinsic concentration, otherwise doping is pointless)
• Can then make a simplification that
n ≈ NDNC
NC + neEd
kBT
• We then reach a quadratic in n:
n2eEd
kBT + nNC −NDNC = 0 (6.7.11)
• Take the positive solution of this (via quadratic equation), to get that
n =1
2
(NC
√1 + 4
NDNC
eEd
KBT −NC
)e− Ed
kBT (6.7.12)
• At room temperature, eEd
KBT ∼ 1 and ND < NC
∴ 4NDNC
eEd
kBT � 1
• Can then use identity that√
1 + x ≈ 1 + 12x for |x| � 1 to get that
n ≈ 1
2
(NC
(1 + 2
NDNC
eEd
kBT
)−NC
)e− Ed
kBT ≈ ND (6.7.13)
• Therefore under these assumptions, all dopants are ionized (same arguments apply to p)
• If dopant levels are within 3kBT of the band edges they are described as shallow dopants
• These completely ionize at room temperature such that N+D ≈ ND and N−A ≈ NA
• Only valid for doping concentrations in the range of ni(T = 300K) < ND < NC
• Only doping above the intrinsic carrier concentration ni increases conductivity
• For ND > NC the semiconductor is degenerately doped and becomes metallic (must use integral approx-imation from Section 6.6 to solve)
6.7.3.1 n-type semiconductor
• An n-type semiconductor is doped with donor atoms
• New carrier density is n = N+D + p
• Electrons are majority carriers as n > p, holes are minority carriers
46
• Usually dope so N+D � ni so n ≈ N+
D (p is small)
• Fermi energy moves closer to conduction band
6.7.3.2 p-type semiconductor
• A p-type semiconductor is doped with acceptor atoms
• New carrier density is p = N−A + n
• Holes are majority carriers as p > n, electrons are minority carriers
• Dope so N−A � ni such that p ≈ N−A (n is small)
• Fermi energy moves closer to valence band
6.7.3.3 Extrinsic semiconductor temperature dependence
Want to revisit how the majority carrier concentration of an n-type semiconductor varies with temperature:Know that
n =1
2
(NC
√1 + 4
NDNC
eEd
KBT −NC
)e− Ed
kBT
1. For kBT � Ed and ND < NC donors are ionized and n ≈ ND
2. For kBT � Ed, donor electrons are bound and n =√NDNCe
− Ed2kBT
3. For kBT > Eg, intrinsic carriers dominate and we see that n = ni =√NCNV e
− Eg2kBT
At low temperatures, the dopant carriers become frozen-out and do not contribute (point 2)
6.8 Summary of Chapter 6: What you need to know/be able to do
• Prove wavefunctions shifted by a reciprocal lattice vector are identical Difficult given Oulton can’t do it, itseems
• Describe the different zone schemes
• Describe band structure of semiconductors and insulators and explain how energy gap and Fermi level positiondetermine whether a solid is a metal, semiconductor or insulator
• State and justify the parabolic band approximation
• Explain the effective mass concept (including why and how it is used)
• Derive isotropic effective mass and use both isotropic and anisotropic effective mass
• Describe holes, their properties & their uses
• Know & derive carrier densities and density of states
• State & use law of mass action
• Describe intrinsic semiconductors via intrinsic carrier concentration
• Understand doping to create extrinsic semiconductors, the difference between n-type and p-type semiconduc-tors, and the temperature dependence of extrinsic semiconductors
47
7 Charge Transport
7.1 The Drude model and carrier mobility
The Drude model is a model of conduction, considering electrons as a gas around solid ions within a solid. Ithas the following assumptions:
– Collisions occur between electrons and ion cores only
– Electrons do not interact between collisions
– Collisions instantaneously redistribute electron velocities
– Electrons collide with ions at a rate of τ−1
– Electrons achieve thermal equilibrium through collisions alone
We will compare this model with the physical picture we have been building up from Bloch’s theorem and thenmove forward in understanding how carriers move.
DERIVATION: The Drude model of conduction
• Define current density J
J = nq〈v〉 = nq
m〈P〉
for n = carrier dnesity, q = charge, m = mass of charge, 〈v〉 = average velocity
• The average momentum is given as
〈P(t)〉 =1
N
N∑n
pn(t)
where we are summing over all of the momenta of every electron
• No applied field i.e. E = 0 implies that 〈P(t)〉 = 0, but pn(t) 6= 0
• Apply a force F [= −eE] (e.g. Lorentz force) over a time δt
• This changes average momentum:
δP =1
N
N∑n
(pn − Fδt)− 1
N
N∑n
pn = Fδt
• If carriers scatter, they return to equilibrium and no longer contribute to this change in total momentum (asin equilibrium every carrier momentum is cancelled out by the momentum of other carriers)
• δP ∴ only includes the fraction of unscattered electrons in time δt
• The probability an electron is scattered is equal to δtτ (if δt� τ)
• Probability of unscattered electron is therefore(1− δt
τ
)• Can then write the average total momentum at a time t+ δt as
〈P(t+ δt)〉 = (〈P(t)〉+ δP)
(1− δt
τ
)(7.1.1)
• The multiplying factor on the right is such that this momentum is only of unscattered electrons, as required
48
• Sub in the definition of δP = Fδt to get that
〈P(t+ δt)〉 − 〈P(t)〉 = (〈P(t)〉+ δP)
(1− δt
τ
)− 〈P(t)〉
= Fδt− 〈P(t)〉δtτ
+O(δt2)
• Divide by δt and take the limit of δt→ 0, so we can produce a differential equation, i.e.
limδt→0
〈P(t+ δt)〉 − 〈P(t)〉δt
=d〈P〉dt
= F− 〈P〉τ
• Then, as 〈P〉 = m〈v〉, we can write that
d〈v〉dt
+〈v〉τ
=F
m=−eEm
(7.1.2)
where the force we are considering is the Lorentz force for a zero magnetic field
• The equation of motion for individual electrons, mass m, in electric field E, is given in Equation 7.1.2.
• For a zero field∑i vi = 0 for summing over all electrons
• For non-zero field, average net motion 〈v〉 opposing field direction
• In steady state d〈v〉dt = 0
• 〈v〉 is given as
〈v〉 = −eτm
E (7.1.3)
• Electron density n gives current density
J = −ne〈v〉 =e2nτ
mE = σE (7.1.4)
Although the Drude model gives a decent prediction on scattering time of metals, we know that the physicalmechanism is not correct: we know from Bloch’s theorem that electrons do not scatter from ion cores.
Instead, we will develop a conduction model first from the Sommerfeld free electron model and then adapt itfor Bloch wavefunctions.
• We know typical wavefunctions and energies for the free electron model:
ψk(r, t) =1
Veik·r−
Et~ Ek =
~2k2
2me
• States fill up to the Fermi level, creating a Fermi sphere in k-space
• Can define Fermi level and Fermi wavevector as
EF =~2
2me(3π2n)
23 kF = (3π2n)
13 (7.1.5)
• The Fermi surface is also a surface of constant velocity, as we can relate the velocity of an electron to itswavevector (in 1D) by
vx =~kxme
(7.1.6)
The same follows in higher dimensions
49
• It follows that as we increase in energy & go up our band we will have higher velocities
• This is the first correction to the Drude model: electrons higher in the energy band can have much highervelocities than can be achieved thermally
• As net velocity is 0 for an electron moving in one direction the electron on the other side of the band mustmove in the opposite direction (illustrated in Figure 14)
Figure 14: An energy band with no applied force. Electrons on either side of the band move in oppositedirections so the total velocity is 0. Electrons further up the band have higher velocities.
• When we apply an electric field, electrons must accelerate subject to
medvxdt
= −eEx
• We know that kx must also change, subject to
~dkxdt
= −eEx
• In a time δt all electrons change momentum by
δkx = −eEx~δt (7.1.7)
• This causes a net current flow as there are more electrons with +kx than −kx (see Figure 15)
• In theory, electrons should continuously accelerate with time - but this does not happen
• As electrons move up the band on one side states below the Fermi level become available
• Electrons can then scatter from phonons back to the other side of the band and occupy these availablestates
• This prevents the system from running away with itself and instead it reaches a steady state
• Within the scattering time τ momentum can only increase by
δkx = −eEx~τ (7.1.8)
• The entire Fermi surface drifts in 3D, by δkx
50
Figure 15: Energy band showing electrons with a change in kx as a result of the application of a field Ex in thenegative x direction. There is a net current flowing in the positive x direction.
• This gives a net drift velocity of
vd =~me
δkx = −eExme
τ (7.1.9)
• Thus define electron mobility µ as
µ =eτ
me(7.1.10)
This adaptation of the Drude model, known as the Sommerfeld-Drude model, correctly attributes scattering tophonons rather than to ions. However, it only applies when our Sommerfeld model applies, which is not in allcases. Now need to think about how electrons move in semiconductors, insulators and more complex metals
7.1.1 Charge transport in semiconductors
Must account for band-structure and dispersion, through:
1. Effective mass of charge carriers
2. Dynamics of electrons and holes
3. Brillouin zone boundaries
Mathematics is similar to before, with parabolic band and Bloch electron wavefunctions
• In semiconductors, conduction band electrons of effective mass m∗n are shifted by
δkx = −eEx~τ (7.1.11)
• Electrons scatter into empty states in conduction band (phonons they scatter from have low energy buthigh momentum)
• Drift velocity of conduction band electrons is therefore
vdn = − eτ
m∗nE = −µnE (7.1.12)
• Valence band holes move in the opposite direction to electrons as they have opposite charge - have effectivemass m∗p
51
• Shifted by
δkx =eEx~τ (7.1.13)
• Hole scattered into empty valence band states to prevent system running away with itself
• Valence band hole drift velocity is
vdp =eτ
m∗pE = µpE (7.1.14)
• Can also scatter across Brillouin zones, but will not deal with this in this course
7.1.2 Charge transport in real metals
• For real metals must consider displacement of real Fermi surface (including its shape) by δkx
• Same rules apply but must be mindful how parts of the Fermi surface could end up in other Brillouinzones (again not considered in this course)
7.1.3 Transport Equations
Much like the Drude equations of motion, we can write equations of motion for drifting electrons and holes inthis situation
Electron transport equation is
m∗ndvdndt
+m∗nvdnτ
= −eE (7.1.15)
Hole transport equation is
m∗pdvdpdt
+m∗pvdpτ
= eE (7.1.16)
7.1.4 Bloch Oscillations
With no scattering processes, we just have electrons moving through a band (can increase k-vector over time).This leads to Bloch oscillations (very difficult to achieve in reality).
• Start with an initial stationary condition
• Apply an electric field Ex
• Electrons accelerate as time increases, in the direction opposite to the electric field
• This fills some states in the second Brillouin zone, which we can shift back to the first Brillouin zone
• Eventually produces average velocity of zero
• Then observe a velocity of electrons in the same direction as the field (unexpected)
• We overall observe an oscillation, i.e. an AC effect from a DC applied bias
The period of Bloch oscillation is related to effective mass, electric field and lattice parameter.
• Period of oscillation will be related to time taken for an electron to traverse FBZ
• Therefore our ∆kx is given as
−∆kx =2π
a=eEx~τB (7.1.17)
for Bloch oscillation time period τB
• The frequency of Bloch oscillation is therefore
ωB =2π
τB=eaEx~
(7.1.18)
52
• Can observe Bloch oscillation if τB < τ the scattering time
• Various approaches to produce this:
– High quality materials at low temperatures removes phonon scattering
– Change lattice parameter (build structure out of different materials, called a superlattice)
As without scattering we only observe Bloch oscillation, we cannot have conduction if we do not have scattering.
7.1.5 Charge carrier mobility and electrical conductivity
• The charge carrier mobility µ is the ratio of the drift velocity vd to the applied electric field E
• See electron and hole mobilities with opposite signs
• Mobility is determined by scattering time and effective mass i.e.
µn =eτ
m∗nµp =
eτ
m∗p(7.1.19)
• We know that current density J = σE for conductivity σ
• We can therefore find σ for a semiconductor from the current density as a result of electrons and holesmoving
• Total current density isJ = −envdm + epvdp = e(nµn + pµp)E
• Therefore we can see conductivity is equal to
σ = e(nµn + pµp) (7.1.20)
• Might not be a scalar: could be anisotropic & therefore a tensor
7.2 The Hall Effect and Magnetoresistance
7.2.1 Crystals in a static magnetic field
When a crystal is in a magnetic field, the charge carriers will have cyclotron orbits:
DERIVATION: Cyclotron orbits have constant energy
• Equation of motion for electrons in this magnetic field (no electric field) is
~dk
dt= −ev ×B (7.2.1)
• Cross product =⇒ dkdt ⊥ to both v and B
• Define v in terms of group velocity & therefore have v = 1~∇kE(k)
• This means thatdk
dt· v =
dk
dt· ∇kE(k) = 0
as dkdt ⊥ v
• Expanding out the dot product gives that
∂kx∂t· ∂E∂kx
+∂ky∂t· ∂E∂ky
+∂kz∂t· ∂E∂kz
= 0
=⇒ dE(k)
dt= 0 (7.2.2)
∴ cyclotron orbits have constant energy
53
Note: the next derivation is non-examinable, but important-to-know steps will be in blue.
NON-EXAMINABLE DERIVATION: Cyclotron frequency of Bloch electrons
• Electrons take orbits of constant energy (and therefore on constant energy contours of the dispersion curve)
• The time between k1 → k2 of an orbit is
t2 − t1 =
ˆ t2
t1
dt
• As dk = dkdt · dt we can say that
|dk| = |dkdt|dt
for dt > 0
• Thus can write that
t2 − t1 =
ˆ k2
k1
|dk||dkdt |
• Using the equation of motion we can write that
dk
dt= − e
~v ×B = −e|B|
~2∇kEn × B = −e|B|
~2[∇kEn]⊥
where ⊥ denotes perpendicular to B
• Change in energy in changing wavevector by δk is given as
δE = ∇kEn · δk
as illustrated in Figure 16
Figure 16: Illustration of moving from one constant energy contour to another, by shifting by a wavevector δk
• By carefully considering directions of our vectors (δk ⊥ k, k ⊥ B, etc.) we can see that
[∇kEn]⊥ ‖ δk
• Can then rewrite a couple earlier expressions, for δE and |dkdt |, as
δE = |∇kEn|⊥|δk|∣∣∣∣dkdt
∣∣∣∣ =e|B|~2|∇kEn|⊥
• Can then use these expressions (second first, then the δE expression) to rewrite the expression for t2 − t1, as
t2 − t1 =~2
e|B|
ˆ k2
k1
|dk||∇kEn|⊥
=~2
e|B|
ˆ k2
k1
|δk||dk|δE
54
• Can define an infinitesimal area between the two constant-energy contours, of sides δk and dk (see Figure16), such that
|δk||dk| = δ(dA)
• Using this we can sub into the integral to get that
ˆ k2
k1
|δk||dk|δE
=δA12
δE
• Then can take the limit of δE → 0 to get that
t2 − t1 = limδE→0
~2
e|B|δA12
δE=
~2
e|B|∂A12
∂E(7.2.3)
• For a closed orbit (k1 = k2), A12 is the area of k-space enclosing the orbit (i.e. the circle around the Fermisurface for k-values parallel to the magnetic field)
• Say A12 = A(En,k‖)
• The time period τc of this orbit is given as
τc =~2
e|B|∂EA(E,k‖) (7.2.4)
• Frequency of orbit is
ωc =2π
τc=eB
m∗cm∗c =
~2
2π∂EA(E,k‖) (7.2.5)
Can see effective mass is different for cyclotron motion, where cyclotron effective mass is written as m∗c .
Can determine type of carrier from sign of ∂EA(E,k‖).
For an isotropic parabolic band A(E,k‖) = πk2, then using the dispersion relation we can show that m∗c = m∗e
7.2.2 Transport equations and the Hall effect
• General form of transport equations are
m∗ndvdndt
+m∗nvdnτ
= −e(E + vdn ×B) (7.2.6)
m∗pdvdpdt
+m∗pvdpτ
= e(E + vdp ×B) (7.2.7)
• From here on we will simplify notation by saying vd ≡ v
55
DERIVATION: Anisotropic conductivity and resistivity (steady state solution v = 0)
• Consider orthogonal electric and magnetic fields applied to a crystalline solid, fields given as
B = (0, 0, Bz) E = (Ex, Ey, 0)
• The equation of motion is given as, for a carrier (density c and charge q)
m∗v
τ= q(E + v ×B) (7.2.8)
• Evaluating the cross product givesv ×B = x [vyBz]− y [vxBz]
• Thus equation of motion goes to
m∗v
τ= q(Ex + vyBz)x + q(Ey − vxBz)y (7.2.9)
• We can write this as two separate equations, one in each of the x-y directions:
x :m∗vxτ
= qEx + qvyBz
y :m∗vyτ
= qEy − qvxBz
• Can now write this as a matrix equation if we collect the v components on the left, as(m∗
τ −qBzqBz
m∗
τ
)(vxvy
)= q
(ExEy
)(7.2.10)
• Want to be able to define this in terms of mobility and cyclotron frequency, so multiply by e, divide by q, andtake m∗
τ to the other side, so the equation goes to(eq − eBzτ
m∗eBzτm∗
eq
)(vxvy
)=
eτ
m∗
(ExEy
)(7.2.11)
• We thus reach the full matrix equation of( eq −ωcτωcτ
eq
)(vxvy
)= µ
(ExEy
)(7.2.12)
• We have used definitions of µ and ωcτ of
µ =eτ
m∗ωcτ = µBz =
eBzτ
m∗
• Know that current density is given asJ = cqv
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• Can therefore write a matrix equation for J, of( eq −ωcτωcτ
eq
)(JxJy
)= cqµ
(ExEy
)(7.2.13)
• We also know that ρJ = E and so in this case we have ρ = ρ i.e. have anisotropic resistivity ρ that is instead
a matrix, rather than a scalar
• The conductivity is similarly anisotropic and can be found by taking the inverse of the resistivity matrix
• See that the conductivity is given as
σ =cqµ
1 + (ωcτ)2
( eq ωcτ
−ωcτ eq
)(7.2.14)
Note that as ωcτ = µBz, the anisotropic conductivity and resistivity are both dependent on the applied magneticfield.
• When we apply an electric field in one direction to a slab of material (e.g. x direction) and a magnetic fieldin the z direction, charge carriers follow circular orbits and charge builds up on top & bottom surfaces (inthe y-direction) and creates a vertical (in the y-direction) electric field
• This is described by the anisotropic resistivity
• Electric field in x-direction =⇒ Jy = 0 in steady state, magnetic field Bz in z-direction
• Can therefore write that ( eq −µBz
µBzeq
)(Jx0
)= cqµ
(ExEy
)(7.2.15)
• As a result can see that
Ey =BzJxcq
= RHBzJx (7.2.16)
by definition of the Hall coefficient
• ∴ RH = 1cq
• For electrons RH = − 1ne and for holes RH = 1
pe
• As RH is independent of mobility it is independent of effective mass and scattering - only dependent onconcentration c, so this allows us to measure c
Can investigate a material using the van der Pauw method, where you connect the four corners of a slab ofmaterial. Then, applying various voltages across a pair of opposite corners and measuring the current betweenthe other two, can find expressions for resistivity and the Hall coefficient, and then from there find the Carrierdensity and type and then the mobility.
• Now consider a system with electrons and holes
• When we apply an electron field electrons move opposite to electric field & associated current whilst holesmove in the direction of the electric field & current
• Magnetic field deflects the trajectory of the carriers towards one of the other edges, opposite directionsfor opposite signs of charge on the carriers
• This accumulation of charge & the resultant field from them is illustrated in Figure 17.
57
Figure 17: Illustrations of accumulation of carriers on one surface or another. The two diagrams separatelyshow electrons (above) and holes (below). The total effect will be a summation of these two. Ey is the resultantfield from the accumulation of charge on either surface.
• Must sum anisotropic conductivities for the net effect, from each carrier type
• Can write that (JxJy
)=
(σD −σAσA σD
)(ExEy
)(7.2.17)
σD =
(neµn
1 + µ2nB
2z
+peµp
1 + µ2pB
2z
)σA =
(neµ2
n
1 + µ2nB
2z
−peµ2
p
1 + µ2pB
2z
)Bz (7.2.18)
• σD is just the sum of the diagonal components of the two conductivities (one for electrons, one for holes)and σA is just the off-diagonal sum
• Typically consider low magnetic field so carriers do not make full orbits (µBz � 1) so can write that(JxJy
)=
(e(nµn + pµp) −e(nµ2
n − pµ2p)Bz
e(nµ2n − pµ2
p)Bz e(nµn + pµp)
)(ExEy
)(7.2.19)
• Can again solve for the Hall field by assuming Jy = 0 (electric field is only along x-direction)
• This leads to an expression for Ey of
Ey = −(nµ2
n − pµ2p)
(nµn + pµp)BzEx (7.2.20)
• For weak fields we can say thatJx
neµn + peµp≈ Ex
• Can therefore write Hall coefficient for both carriers as
RH = −(nµ2
n − pµ2p)
e(nµn + pµp)2(7.2.21)
More work on this can be found in PS7
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7.2.3 Magnetoresistance
Consider the single-carrier case again, in the Hall configuration:
• Find that Ex = Jxcqµ
• This means the resistivity is independent of the magnetic field
• To observe magnetic field-dependent resistance (known as magnetoresistance) we require more thanone carrier type
DERIVATION: Magnetoresistance
• Work in the weak field regime µBZ � 1
• Have σD = e(nµn + pµp) = σ0
• Can write thatσAσD
=(nµ2
n − pµ2p)
(nµn + pµp)Bz
• From previous definition of RH for system with electrons and holes, know that
σ0RH = −(nµ2
n − pµ2p)
(nµn + pµp)
• Can therefore say thatσAσD
= −σ0RHBz (7.2.22)
• Writing out the equations from the matrix equation gives us that
Jx = σ0Ex − σAEy = σ0Ex + σ20RHBzEy (7.2.23)
0 = σAEx + σDEy = −σ20RHBzEx + σ0Ey (7.2.24)
• Can rearrange the second of these to get that
Ey = σ0RHBzEx (7.2.25)
• Substituting this back in gives thatJx = σ0(1 + σ2
0R2HB
2z )Ex
• Coefficient of Ex ≡ σ therefore find ρ = σ−1 & binomially expand to see that the resistivity ρ(Bz) is given as
ρ(Bz) =1− σ2
0R2HB
2z
σ0(7.2.26)
These is a particular case when we can observe magnetoresistance for a single carrier, known as a Corbino disc:
• Electrons around a disc of semi-conductor material used to produce a radial electric field
• Radial symmetry means no charge accumulates like for the typical Hall configuration
• As a result, only have Er, whilst Eθ is 0
• Then observe magnetoresistance (see derivation in PS7)
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7.2.4 Scattering mechanisms
Consider three scattering mechanisms:
1. Phonons: phonons have low energy and high momentum, scatter electrons across Fermi sea. Possiblemomenta span first Brillouin zone.
2. Impurities: doping introduces impurities to a lattice that can significantly reduce mobility. Ion core leftby ionized dopants scatters carriers. Temperature dependence related to ionization probability
3. Defects: defects in the crystal such as grain boundaries and dislocations of particles in the lattice (couldinclude deformations of the lattice). Temperature independent ∴ dominates at low temperatures wherephonon & impurity scattering is minimal
Scattering time is dependent on number of scatterers Nσ, average carrier velocity v and the cross section σ ofthe scattering process, as
τ =1
σvNσ(7.2.27)
Average carrier velocity is linked to thermal energy via equipartition such that v ∝ T 12
7.2.4.1 Phonon scattering
• At or above the Debye temperature (Section 3.2) large phonon population (High temperature regime,Section 3.5.2) dominates other scattering processes
• Below Debye temperature phonon number roughly scales as T 3, above scaling is with T (on top of thephonons up to the Debye temperature) - see Section 4.6
• Reason for this is not important (and not covered in either half of the course? Correct me if I’m wrong)but what does matter is how it affects scattering times
• Taking this information and assuming that carrier-phonon scattering cross section is temperature inde-pendent tells us that
τphonon =1
σvNσ∝ 1
T12 × T
∝ T− 32 (7.2.28)
7.2.4.2 Impurity scattering
• Impurity scattering cross-section takes a similar form to the Rutherford scattering cross section & scaleswith v−4 and thus with T−2
• Can say that
τimpurity =1
σvNσ∝ 1
R−2 × T 12 × (N+
D (T ) +N−A (T ))∝ T
32
(N+D (T )N−A (T ))
(7.2.29)
• This scattering decreases with increasing temperature
7.2.4.3 Matthiessen’s Rule
Probability that a carrier is scattered in time δt is δtτi
where τi is the scattering time for the process we areconsidering
Total scattering probability is a sum over these and thus can say that
δt
τ=∑i
δt
τi
=⇒ 1
τ=
1
τphonon+
1
τdefect+
1
τimpurity+ . . .
This is obviously not valid if any of these scattering rates are dependent on each other
60
7.3 Drift, Diffusion and Einstein’s relation
Now going to look at the physics at semiconductor junctions i.e. between two pieces of semiconductor. Willrestrict ourselves to spatially varying doping.
Start with contact between intrinsic & extrinsic (doped) semiconductor
• We know that adding donor dopants raises Fermi level and adding acceptor dopants lowers the Fermi level
• When placed in contact electrons from extrinsic region migrate to regions of lower Fermi energy
• Energy bands change to produce a constant Fermi level - see Figure 18
Figure 18: LEFT: Junction showing the change in Fermi level moving from intrinsic to extrinsic region (extrinsicin the middle). RIGHT: Junction after EC & EV have changed to achieve a constant Fermi level. EFi
denotesthe initial intrinsic Fermi level
• Now have many spatially-varying parameters: EC(x), EV (x), n(x), p(x)
• Drives new carrier currents, such as diffusion current
7.3.1 Carrier drift & diffusion
• Fermi level adjusts by diffusion of charges
• Diffusion initially establishes an in-built field that adjusts band-edge energies
• Drift and diffusion will eventually balance each other i.e. can write electron and hole current densities as
Jn(x) = Jdriftn (x) + Jdiff
n = 0 Jp(x) = Jdriftp (x) + Jdiff
p = 0 (7.3.1)
• Previously defined drift velocity in an electric field E as
vdn = −µnE vdp = µpE
• Drift current density = drift velocity × carrier charge × carrier density, i.e. can write that
Jdriftn (x) = [−en(x)]× [−µnE ] = en(x)µnE (7.3.2)
Jdriftp (x) = [ep(x)]× [µpE ] = ep(x)µpE (7.3.3)
• Define total drift current as the sum of these, as
Jdrift(x) = Jdriftn (x) + Jdrift
p (x) = (en(x)µn + ep(x)µp)E (7.3.4)
• Inhomogeneous populations of carriers near junctions will diffuse & create a diffusion current
• This current opposes carrier density gradient i.e. carriers go from an area of high concentration to one oflow concentration
61
• Define these currents as equal to some constants multiplied by the rate of change of carrier number density
Jdiffn (x) = [−eDn]
[−dn(x)
dx
]= eDn
dn(x)
dx(7.3.5)
Jdiffp (x) = [eDp]
[−dp(x)
dx
]= −eDp
dp(x)
dx(7.3.6)
• Dn and Dp are diffusion coefficients for electrons and holes
• Note that Jdiffn is in the same direction as the electron carrier density gradient, but the corresponding
current density for holes opposes the hole carrier density gradient
• Total current includes all of these factors i.e. is written as
J = Jdriftn (x) + Jdiff
n (x) + Jdriftp (x) + Jdiff
p (x) = constant (7.3.7)
• In steady-state, Kirchoff’s law (current continuity) means that J(x) must be constant in all positions,otherwise would increase carrier density
7.3.2 Einstein’s Relation
DERIVATION: Einstein’s relation
• Know electron carrier density is given as
n(x) = Nce−(Ec(x)−EF )
kBT
• In equilibrium electron balanced current is given as
Jn(x) = en(x)µnE(x) + eDndn(x)
dx= 0
• Sub the first expression into the second, to get that
eµnE(x)Nce−(Ec(x)−EF )
kBT − e Dn
kBT
dEc(x)
dxn(x) = 0
• Divide by en(x) to get that
µnE(x)− Dn
kBT
dEc(x)
dx= 0
• By definition E(x) = −dV (x)dx , and Ec(x) = −eV (x) + constant. Have the −eV (x) term to represent the extra
energy each electron has from this electrostatic potential, added to the base band energy (the constant)
• This then tells us that
E(x) =1
e
dEc(x)
dx
• Can use this expression to rewrite our balanced current expression as
µnE(x) =Dn
kBTeE(x)
• Therefore reach Einstein’s relation, ofDn
µn=kBT
e(7.3.8)
62
A similar argument can be made for holes i.e.Dp
µp= kBT
e .
7.3.3 Generation and recombination
• Semiconductor devices strives for equilibrium
• Can have excess carriers but will have generation of electron-hole pairs, or recombination (effectivelyannihilation) of these pairs
• Define generation rate g and recombination rate r
• In equilibrium g = r
7.3.3.1 Generation processes
Figure 19: Three generation processes
Figure 19 shows several generation processes, from left to right:
1. Thermal energy (such as a phonon) or light (single photon) can excite an electron-hole pair
2. Dopant energy level can be between conduction and valence bands and act as a ‘trap state’ which effectivelyprovides a ‘jumping-off’ point so rather than needing 1 high energy phonon/photon, 2 lower energy oneswill work
3. Electric field can move an electron sufficiently far quickly enough that the energy band drops away beneathit, so it must release that energy as it moves to the band. This energy can generate an electron-hole pair.
7.3.3.2 Recombination processes
Figure 20: A pair of possible recombination processes
Figure 20 shows a pair of recombination processes, analogous to the generation processes (left to right):
63
1. Electron-hole pair can combine and produce a photon
2. Trap state produces a point at which electrons and holes can meet and annihilate
There are other, higher-order effects, but specifics are unimportant for this course.
7.3.4 Excess carrier density
• Semiconductor devices generally operate under external fields that cause a net current to flow
• Produces excess carriers within various regions of a device
• Define these excess carriers as
∆n = n− n0 ∆p = p− p0 (7.3.9)
• Require for charge neutrality that ∆n = ∆p, and in equilibrium that n0p0 = n2i
• Under typical operation current densities are small enough such that majority carrier densities are notaffected, i.e.
n-type: |∆n| � n0 ≈ ND =⇒ n ≈ n0
p-type: |∆p| � p0 ≈ NA =⇒ p ≈ p0
• Minority carriers are affected
• Typically excess minority carriers are consumed by generation and recombination processes, but maypersist under fields where there is a net current
• Minority carrier lifetime is the average time excess minority carriers survive in sea of majority carriers
• Time variation of excess minority carriers is given as
n-type:dpndt
= −∆pnτp
=⇒ d∆pndt
= −∆pnτp
(7.3.10)
p-type:dnpdt
= −∆npτn
=⇒ d∆npdt
= −∆npτn
(7.3.11)
• To track how diffusion, generation and recombination processes affect current we use a continuity equation:
DERIVATION: Current continuity equation
• Consider current density Jn(x) normal a slab of area A and thickness δx (see Figure 21)
Figure 21: Geometry for current continuity derivation
• Write out equation for rate of change of number of carriers in the slab (from ‘death’ of minority carriers andnet current from drift and diffusion), given as
Aδx∂n
∂t= −1
e[Jn(x)− Jn(x+ δx)]A− ∆n
τnAδx (7.3.12)
n(x) is generic carrier density, not specifically electrons
64
• Divide by Aδx & take limit δx→ 0 to get
∂n(x)
∂t=
1
elimδx→0
(Jn(x+ δx)− Jn(x)
δx
)− ∆n(x)
τn
• Can subtract n0 off from n as it is constant, to then finally get that
∂∆n(x)
∂t=
1
e
∂Jn(x)
∂x− ∆n(x)
τn(7.3.13)
DERIVATION: Minority carrier diffusion equation
• Can use continuity equation and electron current Jn:
Jn(x) = en(x)µn(x)E(x) + eDn∂n(x)
∂x
∂n(x)
∂t=
1
e
∂Jn(x)
∂x= −∆n(x)
τn
• Make two key assumptions:
1. Weak applied electric field E2. Minority carrier density is low i.e. n(x) < ni
• This means we ignore the first term in the current density expression
• Consider minority of electrons in a p-type semiconductor i.e. n(x)→ np(x)
• Substitute the current density into the continuity equation to get
∂∆np(x)
dt= Dn
∂2∆np(x)
∂x2− ∆np(x)
τn(7.3.14)
• Equation valid only for minority carriers
• Can solve this equation for a steady state current i.e. such that
∂∆np∂t
= 0 =⇒ ∂2∆np∂x2
=∆npDnτn
=∆npL2n
• We define diffusion length Ln asLn =
√Dnτn (7.3.15)
• Solutions will be exponential functions
See example in scans for how we can apply these equations for electron current in a p-type semiconductor.
65
7.4 Summary of Chapter 7: What you need to know/be able to do
• Describe conduction in some metals using the Sommerfeld-Drude model (including Fermi surface drift &electron mobility
• Explain electrical conduction in semiconductors through effective mass
• Understand the concept of Bloch oscillations and the conditions in which they arise
• Write down & use equations for charge carrier mobility and electric conductivity/resistivity
• Describe cyclotron orbits of electrons in a crystal within a static magnetic field
• Derive and use anisotropic conductivity/resistivity
• Describe Hall effect & how it is described by anisotropic resistivity
• Describe & derive magnetoresistance & describe the Corbino disc magnetoresistance
• Describe what factors dominate scattering processes and how they depend on temperature
• Explain and know equations for electron & hole drift, diffusion currents, and total current
• Write down & derive Einstein’s relation
• Explain generation and recombination processes & how they help maintain an equilibrium in the semicon-ductor
• Write down & use equations for minority carrier lifetime and diffusion length
• Understand & derive minority carrier diffusion equation and semiconductor current continuity equation
66
8 The pn Junction
8.1 Thermodynamics and Electrostatics of a pn Junction
• A pn junction is a semiconductor junction between a piece of p-type semiconductor and a piece of n-typesemiconductor
• Easiest method of construction is to add donor atoms to a p-type semiconductor (acceptors to an n-typesemiconductor), resulting in a carrier gradient of the new n-type region to the original p-type region
• Typically pn junctions have non-uniform doping profiles (i.e. moving between the p and n regions doesnot instantly change concentrations of carriers)
• Will assume a step function for doping profile in this course - see Figure 22
Figure 22: An illustration of a doping profile for a n-type semiconductor with acceptor atoms added to createa pn junction. The real profile is compared with the idealized profile we consider for this course.
• p-type and n-type regions have different Fermi levels EFp and EFn respectively
• Electrons & holes diffuse across junction to balance Fermi energies
• May undergo recombination processes
Figure 23: pn junction indicating the initial diffusion of electrons and holes
• This initial diffusion is indicated in Figure 23
• V0 is the built-in potential and defines the difference between EFnand EFp
(EFn− EFp
= eV0)
• After a period of time, the diffusion current creates an electric field
• Electric field causes a drift current which opposes diffusion current
67
• Band shape changes as a result - see Figure 24
Figure 24: The pn junction when carrier drift is occurring. Recombination processes remove carriers from thecentral region illustrated, whilst as the band deforms the two Fermi levels from the p-type and n-type regionscome closer together.
• Eventually the drift and diffusion currents are equal and the system is in a steady state.
• This is illustrated in Figure 25
Figure 25: The pn junction in a steady state. In the central region, electrons and holes recombine. This isknown as the depletion region and has few carriers.
• In this steady state, the Fermi levels EFn = EFp = EF
• This means there is no energetic advantage for carriers to change positions.
• Ecp − Ecn = eV0 here
8.1.1 Depletion region and built-in potential
• Net charge only in depletion region as no carriers but there are ionized dopants
• In other regions majority carriers and dopant ions cancel charge
68
Figure 26: LEFT: Illustration of charges from different sources at the pn junction. The range for −xp < x < xnis the depletion region. RIGHT: Illustration of the net charge, along with the approximate distribution we willassume.
• Use depletion region approximation, that we have uniform charge in depletion region of{ρn = eND for 0 < x < xn
ρp = −eNA for − xp < x < 0
• This is illustrated in Figure 26
• Define depletion region width W = xn + xp
DERIVATION: Built-in field
• Use the previously described depletion region approximation
• Gauss’ law tells us that∇ · E =
ρ
ε
• This means thatdEx(x)
dx=ρ(x)
ε
in the 1D case.
• For this 1D analysis, let Ex = E
• In the n-type region, have that
dE(x)
dx=eNDε
=⇒ Ex =eNxεx+ c
• Similarly for the p-type region
dE(x)
dx= −eNA
ε=⇒ E(x) = −eNA
εx+ c′
• n-type and p-type materials are conducting =⇒ E(x) is 0 outside of the depletion region & must thereforego to 0 at the edges
69
• This gives boundary conditions, from which we can find c and c′, as
E(x = xn) = 0 =⇒ c = −eNDxnε
E(x = −xp) = 0 =⇒ c′ = −eNAxpε
• Thus get an expression for this built-in field of
E(x) =
{eND
ε (x− xn) 0 < x < xn
− eNA
ε (x+ xp) − xp < x < 0(8.1.1)
• At x = 0, for constant permittivity across the junction we expect E(x = 0+) = E(x = 0−) for approaching 0from above and below respectively
• This gives thateNDε
xn =eNAεxp = −E0 (8.1.2)
• We can also say thatNDxn = NAxp (8.1.3)
DERIVATION: Built-in potential V0
• By definition know that the electrostatic potential V (x) is given as
V (x) = −ˆE(x) dx
• Can now look to find potentials associated with the fields in each region:{E(x) = eND
ε (x− xn) =⇒ V (x) = − eND
2ε (x− x2n) + Vn0 0 < x < xn
E(x) = − eNA
ε (x+ xp) =⇒ V (x) = eNA
2ε (x+ xp)2 + Vp0 − xp < x < 0
• Will define Vp0 = 0 and Vn0 = V0
• Then require that the potential is continuous at x = 0, so
V (x = 0) = −eND2ε
x2n + V0 =
eNA2ε
x2p
• Therefore haveV0 =
e
2ε
(NAx
2p +NDx
2n
)(8.1.4)
• Can write expressions for spatial variation of conduction & valence band energies across the depletionregion in terms of electrostatic potential and the constant conduction and valence bands in the p-typeregion:
Ev(x) = −eV (x) + Evp (8.1.5)
Ec(x) = −eV (x) + Ecp (8.1.6)
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• It follows that the conduction & valence band energies in the n-type region , Ecn and Evn, are eV0 lessthan the respective quantities in the p-type region
DERIVATION: Depletion region width
• Have two expressions from previous derivations, from the continuity results:
NDxn = NAxp
V0 =e
2ε
(NAx
2p +NDx
2n
)• Add NDxp to the first expression to get that
ND(xn + xp) = (NA +ND)xp =⇒ xp =ND(xn + xp)
NA +ND=
NDW
NA +ND
• Similarly now add NAxn to the same expression to find an expression for xn:
NAW = (ND +NA)xn =⇒ xn =NAW
NA +ND
• Can then use these expressions to eliminate xn and xp from the expression for V0, to get that
V0 =e
2ε
(NA
N2DW
2
(NA +ND)2+ND
N2AW
2
(NA +ND)2
)=
e
2ε
W 2
(NA +ND)2(NAND(NA +ND)) =
e
2εW 2 NAND
NA +ND
• This then gives that
W =
√2ε
eV0
(NA +ND)
NAND(8.1.7)
8.1.2 Applied bias
• We can also apply a bias (potential difference) Va to reduce potential drop across depletion region
• Creates imbalance between Fermi levels in the p-type and n-type regions, resulting in two quasi-Fermilevels EFn
and EFp
• Decreases built-in field & ∴ decreases xn and xp (see Equation 8.1.2), decreasing depletion region width
• 4 possible situations:
1. Equilibrium i.e. Va = 0 gives an unchanged W
2. Forward bias (positive terminal connected to p-type end) Va < V0 gives a reduction in the differencebetween Ecp and Ecn and decreases W as electrons and holes from the n and p-type regions arepushed towards the depletion zone and neutralize it
3. Va = V0 creates a flat band & W = 0. pn junction becomes a resistor (current flow is limited bydiffusion of minority carriers)
4. Reverse bias (negative terminal connected to p-type end) Va < V0 increases gap between Ecp andEnp and increases W as holes and electrons are ‘pulled’ away from the depletion region and chargedions are left with no carriers screening (effectively neutralizing) them
• Can write the equations out as
W (Va) =
√2εe V0
(NA+ND)NAND
Va = 0√2εe (V0 − Va) (NA+ND)
NANDVa < V0 forward bias
0 Va = V0√2εe (V0 + |Va|) (NA+ND)
NANDVa < V0 reverse bias
(8.1.8)
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These situations are illustrated in Figure 27.
Figure 27: Illustration of effects on the energy bands when under bias
1. TOP LEFT: Bands in equilibrium. States within ∼ 3kBT of the conduction band account for about 5%of carriers, but cannot conduct due to the potential barrier (band slope).
2. TOP RIGHT: System in forward bias. Electrons are introduced into the n-type region and holes into thep-type region, narrowing the depletion region and decreasing the potential barrier. This allows conduction.
3. BOTTOM LEFT: System in reverse bias. Depletion region grows & so does potential barrier. Circuittries to remove electrons from n-type and holes from p-type but this is thermally limited.
4. BOTTOM RIGHT: Flat band - depletion region disappears and diode acts like a resistor, with currentlimited by minority carrier diffusion. This is because as electrons are added to the n-type region (andholes to the p-type region) and they cross the junction such that they are now minority carriers in theopposite type region, they have a finite lifetime and so only a limited possible current will continue.
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8.2 The Shockley Equation and Diode Electronics
• All effects at a pn junction are driven by minority carrier diffusion (electrons injected into n-type regionforced into p-type region as minority carriers to conduct)
• Must therefore evaluate excess carriers at depletion region boundaries xn and xp to find diffusion currents
DERIVATION: Minority carrier density and current
• Excess minority holes can be found by comparing carriers at x = xn with equilibrium value (see Section 7.3.4),i.e.
∆pn(xn) = pn(xn)− pn0
for pn0 as the equilibrium number of holes in the n-type region
• In depletion region EFnand EFp
are constant (carriers would move to compensate otherwise)
• Can use carrier density equation (Section 6.6.3) to say that the equilibrium hole number density is
pn0 = pn(x� xn) = NV e−
EFn−EVn
kBT (8.2.1)
i.e. number of holes far into the n-type region, where NV is the carrier density in the valence band and EVn
is the valence band edge energy in the n-type region
• Also get that the number of holes at xn is equal to
pn(xn) = NV e−
EFp−EV (xn)
kBT = pn0e−
EFp−EFn
kBT (8.2.2)
NOTE: this result effectively comes from changing the limits on the Fermi-Dirac integrals in Section 6.6.3
• We know the difference between the two quasi-Fermi levels is equal to
EFn− EFp
= eVa
• Can substitute the expressions for pn0 and pn(xn) into the expression for excess minority carriers to get that
∆pn(xn) = pn0
(e
eVakBT − 1
)(8.2.3)
• From the minority carrier diffusion equation (see Section 7.3.4) our excess holes takes on a negative exponentialform, of
∆pn(x ≥ xn) = ∆pn(xn)e− x−xn
Lp
for diffusion length Lp. Expontential decay as we expect less excess holes the further into the n-type regionwe go (boundary conditions on solving minority carrier diffusion equation)
• Hole diffusion current is seen in Equation 7.3.6 & is equal to
Jdiffp (x) = −eDp
d∆pn(x)
dx
as p(x) = p0 + ∆p(x) and p0 is constant
• Can therefore say that the minority carrier diffusion current at xn is given as
Jdiffp (xn) = −eDp ×
d
dx(∆pn(xn)) = −eDp ×
d
dx
(pn0
(e
eVakBT −1
))
Jdiffp (xn) = e
Dp
Lppn0
(e
eVakBT − 1
)(8.2.4)
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• Can equally show that the electron minority diffusion current at −xp, the p-type depletion region boundary,is given as
Jdiffn (−xp) = e
Dn
Lnnp0
(e
eVakBT − 1
)(8.2.5)
• Combine these to get total current
J = e
(Dp
Lppn0 +
Dn
Lnnp0
)(e
eVakBT − 1
)(8.2.6)
• Define the Shockley Ideal Diode equation as
J = J0
(e
eVakBT − 1
)J0 = e
(Dn
Lnnp0 +
Dp
Lppn0
)(8.2.7)
Note: current through diode must be constant. Any loss of minority diffusion current must be compensated forby an increase in majority carrier current
Figure 28 shows the I-V curve of a pn junction. If Va is increased beyond V0 then the diode equation breaksdown and the diode essentially behaves as a resistor. In reverse bias have I = I0 as current is limited by thermalgeneration of carriers in the depletion region.
Figure 28: Current vs voltage for a pn junction
8.2.1 pn-junction devices
1. Diode: one way gate for current, or a variable capacitor
2. Light Emitting Diode (LED): Forward bias, achieves efficient electron injection into a semiconductor.Recombination causes light emission
3. Laser diodes: same as LED but produces fully coherent light
4. Photodiode: operating in equilibrium or reverse bias, absorbs light to generate electron-hole pairs &conduct (generating a current)
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8.3 Summary of Chapter 8: What you need to know/be able to do
• Describe structure of a pn junction and how they might be made
• Solve equations for electric field, potential and band structure across junction
• Describe the variation of these quantities in real space
• Derive & use equation for depletion region width & understand how it applies in different bias situations
• Describe four operating modes of a pn junction & how and why the depletion width is affected by thesedifferent modes
• Derive Shockley Ideal Diode equation
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9 Solids in Other External Fields
NOTE: Any questions on this section will only be qualitative, no calculations - mostly a preparation for Light& Matter in Year 3 . As a result there will only be fairly brief detail in this section, with reference to the slidesfrom lectures for any extra mathematics.
9.1 Optical properties of solids and Optoelectronics
• Form Drude-Lorentz model, saying electrons behave as simple harmonic oscillators & can then solve fora polarizability when an electric field is applied from EM waves propagating in the dielectric (see lecturederivation and EM Section 20 Waves in Dielectrics for mathematical detail)
• Polarizability of a solid depends on internal resonances at frequencies ωi (including electronic or vibrationalresonances)
• ‘Resonances’ refer to processes that occur depending on the applied EM wave frequency
• Permittivity becomes complex valued - imaginary part tells us about absorption (decaying exponentialterm) whilst real part tells us about dispersion (oscillatory term) (see EM)
• This permittivity is equal to
εr = 1 +∑i
e2Niε0m(ω2
i + ω2 − iωγi)(9.1.1)
where we are summing over all of the possible resonances i.e over the different frequencies from thedifferent sources of oscillation. ω is applied frequency, ωi is frequency of the relevant effect (e.g. electronictransitions), γi is from equation of motion (see derivation)
• E.g. in water we have three ‘resonances’ that cause these changes in permittivity:
1. Electronic transitions when (applied waves in visible range)
2. Vibrational states of atoms (infrared range)
3. Intermolecular vibrations (lower frequencies beyond infrared e.g. towards microwaves)
• Total permittivity is then determined by the summing over all of these processes
• Can relate the absorption and dispersion via Kramer-Kronig relations (which rely on complex analysis,seen in Maths Methods)
• Allows us to see that a materials refractive index will arise from internal absorption processes
• See slides/EM course for information about dielectric properties of metals
9.1.1 Radiative and non-radiative absorption and emission
• Generation and recombination processes in solids can be driven by absorption & emission of photons &phonons
• Obviously these processes must satisfy energy & momentum conservation (typically photons have highenergy & low momentum whilst phonons have low energy & high momentum)
• A radiative transition is one that occurs with only a photon being absorbed or emitted i.e. E changes butk is (mostly) conserved (transitions are nearly vertical)
• A direct gap has the conduction band directly above the valence band, whilst an indirect gap will havethe two bands offset by some wavevector
• Direct gaps may require absorption or emission of a phonon as well as a photon for electrons to transitionbetween conduction and valence bands (phonons shift the wavevector k but have very little change in E=⇒ mostly horizontal transition)
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• Direct and indirect gaps are shown in Figure 29.
Figure 29: An illustration of direct and indirect band gaps. To transfer from the valence band to the conductionband, we require at least a photon to be either emitted or absorbed (to go down or up respectively) but for theindirect gap we also require a phonon to be absorbed or emitted to provide the shift in wavevector.
• Direct band gaps are much more efficient than indirect band gaps
9.1.2 Illuminated p-n junction
• Illuminating a p-n junction allows for generation and recombination via absorption of photons
• In the p and n-type regions these generated electron-hole pairs recombine very quickly
• Those generated in the depletion region and then separated by the in-built field of the depletion region
• Photocurrent flows in opposite direction to incident radiation
• Only light absorbed in the depletion region contributes to this current
• This is illustrated in Figure 30
Figure 30: An illuminated p-n junction with air to the left. Incident radiation with power P0 comes in from theleft, an amount is reflected (RP0) and an amount is transmitted (TP0). Electron-hole pairs generated in thedepletion region separate because of the built-in field and thus cause a current flow
9.1.3 Optical diode operation modes
• Consider 3 main operating modes:
1. Under forward bias i.e. injection of majority carriers - Laser/LED
2. No bias gives photovoltaic mode
3. Reverse bias gives photoconduction & fast photodiodes
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9.1.3.1 Emission under forward bias
• Under forward bias electrons and holes can recombine in the depletion zone
• This allows for radiative emission of photons
• LEDs are diodes of direct band gap semiconductors, so light emission occurs near the band-edge underforward bias
• Several different direct band gap semiconductors can be used to produce light efficiently
• Insulators can emit light too but typically only via optical excitation & decay of electrons (exciting tohigher energy levels & dropping down again), known as photoluminescence
• As semiconductors can be doped they can be electrically excited (this electron-hole recombination processof emission), which is known as electroluminescence
• LEDs have issues with getting emitted photons out of the semiconductor as a result of the relatively largerefractive index the materials have (so a lot of emission doesn’t escape the semiconductor)
9.1.3.2 Photovoltaic mode
• No applied bias
• Typically connect diode in circuit with a resistor (called a load circuit)
• Incident light on the diode generates electrons & holes and they create a current
• The resistance of the resistor defines the potential difference you can get out of the circuit
• Typically choose resistance to get maximum power (balancing potential difference and current as neces-sary)
• Basis of all solar cells
9.1.3.3 Photoconduction mode
• Effectively have the same circuit as the photovoltaic mode but with a cell added to apply a reverse bias
• Output collected through load circuit as before
• Get a fairly linear response with power
• Also produces a fast response as carriers are separated by a strong built-in field (like in a non-illuminateddiode, but the reverse photocurrent increases this separating effect)
• See lecture for information about avalanche photodiodes (effectively very efficient photodetectors thatoperate using impact ionization, a recombination process that is not something we need to know)
9.2 Origin of Magnetism in Solids
9.2.1 Magnetization
• We derive magnetization the same way as in EM Lecture 16, by treating atoms as small current loops
• Whilst this is technically incorrect it is close enough to the quantum result for our understanding
• As well has having magnetization arising from orbit of electrons, also get a spin-magnetization
• Therefore must consider interaction of both these spin and orbit terms with the applied magnetic field
• By Pauli exclusion principle for atoms with paired electrons spin up and down electrons must move inopposing orbits (so no angular momentum states have more than 1 electron in them)
• As a result filled bands in solids have no orbital or spin magnetization, if the solid has only paired electrons
• Need a solid with unpaired electrons to have spin-orbit magnetization
• See magnetic potential µB |B| ∼ 58µeV for electrons in a 1T field, which is tiny compared to kBT = 26meVat room temperature
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• Means at room temperature there is plenty of energy available to swap spin of an electron & changemagnetization
• Can then link magnetization to thermodynamics, via
M = −(∂F
∂B
)V, T
(9.2.1)
See slides/lecture 11 for more details
9.2.2 Change to Hamiltonian for electrons in a solid with external magnetic field
• As magnetization results in a change in energy we will observe a different Hamiltonian for electrons inthese solids than the normal Hamiltonian
• From Lorentz force we can define the electron canonical momentum as an effective momentum thatsubstitutes into the Hamiltonian instead of traditional momentum, as
p = mv − eA (9.2.2)
for the vector potential A associated with magnetic field B
• Subbing this into the traditional Hamiltonian for Z electrons, and adding our terms for spin-orbit magne-tization, gives
H =
Z∑i=1
(p2i
2m+ V (ri)
)+ µB(L + gsS) ·B +
e2
2m
Z∑i=1
(B × r)2 (9.2.3)
where the second term sums only over unpaired electrons
NOTE: You do not need to be able to derive this Hamiltonian but knowing the rough form it takes isuseful to understand the sources of diamagnetism and paramagnetism
9.2.2.1 Diamagnetism
• Diamagnetism arises whether there are unpaired electrons or not
• Source of magnetization comes from the third term in our Hamiltonian
• From classical view, applied magnetic field attempts to modify current from electron orbit & thus createsa magnetic field opposing the applied field
• The magnetization acts in opposition to the applied magnetic field and thus results in a reduced net field
• Formulae related to this can be seen in the lecture derivation
• Diamagnetism occurs in all materials but is a weak effect
• Superconductors are perfect diamagnets and thus reject all applied magnetic fields
9.2.2.2 Paramagnetism
• Paramagnetism arises when there are unpaired electrons
• Second term in Hamiltonian is relevant (assuming diamagnetic term is small)
• Resultant magnetization is aligned with applied field & thus increases it
• Formulae in lecture derivation
• Can observe Pauli paramagnetism in conductors as spin up and down electrons are not equally likely -energy for spin up & spin down is different in presence of magnetic field & therefore part of band structureshifts vertically (see slides)
• Find Pauli paramagnetism is very weak in comparison to regular paramagnetism
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9.2.2.3 Ferromagnetism
• Permanent magnetism arises from the interaction between electrons in neighbouring atoms
• If there are different spins in different atomic sites there is some energy configuration that can be minimisedi.e. it may be energetically advantageous for two electrons in neighbouring atoms to spin in the samedirection or in the opposite direction
• Gives rise to exchange energy i.e. the energy if the spins are swapped between being aligned and anti-aligned
• In most materials it is lower energy to be anti-aligned (i.e. ground state will have anti-aligned electrons),but for ferromagnetic materials the aligned configuration is of lower energy
• This greatly increases applied magnetic field as all spins align under an applied field
• See slides for some related mathematics (that is completely non-examinable)
• External fields can permanently change magnetization of a ferromagnetic material (so can store informa-tion in these materials)
9.3 Chapter 9: What you need to know/be able to do
• Describe radiative and non-radiative transitions & how they involve photons and phonons
• Describe direct and indirect gap semiconductors
• Explain three operation modes of photo-diodes
• Describe origins of diamagnetism, paramagnetism and ferromagnetism
• Explain how orbital angular momentum, spin angular momentum and spin exchange interactions betweenelectrons in neighbouring atoms lead to these forms of magnetism
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