1This week's plan
Read
Datta Part B, Chapters 19 and 20: 1D wires, scatterers, quantum resistors in series; 2D wires, graphene, quantum Hall.
Datta Part B, Chapters 19 and 20: 1D wires, scatterers, quantum resistors in series; 2D wires, graphene, quantum Hall.
Run
Colab notebook week14_negf_lab_C7.ipynb:
Colab notebook week14_negf_lab_C7.ipynb:
wire_sigma, T_wire, wire_solve (with dephasing), subbands, modes_B; plus the circular-session lab.Practise
Quiz (aim for 8/10), the problems and one mini-investigation. Write the half-page for C7: why are two quantum resistors in series not additive?
Quiz (aim for 8/10), the problems and one mini-investigation. Write the half-page for C7: why are two quantum resistors in series not additive?
By the end of the week you should be able to
- Write Σ = −t₀eika, Γ = 2t₀ sin ka for a wire and show T = 1 for the clean wire.
- Derive and use T = Γ²/(U² + Γ²) for one scatterer (Eq. 19.5).
- State the incoherent series rule R₂ = R₁(2 − T) and explain why the coherent result differs (interference, Fig. 19.4).
- Describe the potential profile f(j) = Gⁿ/A for coherent, Model A and Model B conductors.
- Count modes in a 2D strip, graphene ribbon or nanotube, and relate the Hall plateaus to RH = h/(Mq²).
Optional video: Datta's nanoHUB-U course, Unit 3 (Quantum Transport).
2Key ideas in five sentences
- The wire is a contact-ed lattice. The semi-infinite 1D lattice gives Σ = −t₀eika on the end site, so Γ = 2t₀ sin ka, and a clean wire transmits perfectly: T = 1, R = h/q².
- One scatterer. T = (2t₀ sin ka)²/[U² + (2t₀ sin ka)²]: strong barriers (U ≫ Γ) give T ≈ (Γ/U)².
- Series resistors. If the electron loses its phase between scatterers, resistances add: R₂ = R₁(2 − T). If it does not, the reflected waves interfere and T of the pair can be anywhere between 0 and 1 (0.39 → 0.9996 in Fig. 19.4).
- The potential drop. f(j) = Gⁿ(j,j)/A(j,j): oscillatory and localised at scatterers when coherent, a straight line for Model B dephasing.
- 2D wires. Subbands En(kx) = εn − 2t₀ cos kxa, T(E) = M(E) in steps of q²/h; with B the edge states give RH = h/(Mq²).
3Equation sheet
E = Ec + 2t₀(1 − cos ka), t₀ = ħ²/2ma²Sec. 19.1
1D lattice dispersion; band 0 to 4t₀.
Σ = −t₀eika, Γ = i[Σ − Σ†] = 2t₀ sin kaEq. 19.2
Contact self-energy on the end site.
T = (2t₀ sin ka)²/[U² + (2t₀ sin ka)²]Eq. 19.5
One scatterer of strength U.
R = (h/q²)/T = (h/q²)[1 + (1 − T)/T]Sec. 19.3
Contact + scatterer.
1/Tseries = 1/T1 + 1/T2 − 1; R₂ = R₁(2 − T)Eq. 19.7
Incoherent series; identical scatterers.
f(j) = Gⁿ(j,j)/A(j,j)Eq. 19.8
Electrochemical potential profile.
En(kx) = εn − 2t₀ cos kxa, εn = 4t₀ − 2t₀ cos[nπ/(N+1)]Eqs. 20.3–20.4
Subbands of a strip with N sites across.
Eonset,n = Ec + 2t₀[1 − cos(nπ/(N+1))]Sec. 20.1
A new mode opens: a step of q²/h in T(E).
g−1 = (E + i0⁺)I − α − β†gβ, Σ = βgβ†Eq. 20.6
Surface Green function of a 2D contact.
G = M q²/h, M(E) = floor(2WE/hv₀) (graphene)Sec. 20.3
Ribbon with armchair-type boundary.
RH = h/(Mq²)Sec. 20.4
Quantum Hall plateaus: M edge channels.
4Physics refresher: waves, interference, cyclotron motion
Interference
Two waves of amplitude a, relative phase θ, give intensity 2a²(1 + cos θ): between 0 and 4a². A wave bouncing between two partial reflectors resonates when the round-trip phase 2kd is a multiple of 2π (Fabry–Pérot).
Magnetic field
A charge moves on a circle of radius r = ħk/(qB) with frequency ωc = qB/m. Orbits hitting a wall skip along it: edge states, one way on each edge.
Why does k become imaginary outside the band?
cos ka = 1 − E/2t₀ has no real solution for E < 0 or E > 4t₀. Then ka = iκ (or π + iκ): the wave decays as e−κ per site. Σ = −t₀eika is real, so Γ = 0: no broadening, no outflow.Why is Γ largest at mid-band?
Γ = 2t₀ sin ka: group velocity ∝ sin ka is maximum at ka = π/2, where E = 2t₀. A faster-moving wave leaves the contact more quickly.Why is ln T the natural variable?
Amplitudes multiply along a long disordered wire; a product turns into a sum of logarithms and the central limit theorem applies to ln T, not to T.5Glossary
| Term | Meaning |
|---|---|
| Scatterer | Local potential U on one site of the wire. |
| Mode | One transverse subband; each open mode is a 1D channel carrying q²/h. |
| Subband | En(kx) of the n-th transverse state. |
| Phase coherence | The electron keeps its phase between scatterers: amplitudes, not probabilities, add. |
| Localisation | ln T falls linearly with length in a coherent disordered wire. |
| Edge state | Quantum Hall mode propagating along one edge, in one direction. |
| Basis transformation | V: from site basis to the mode basis, in which Σ is diagonal. |
| Model A / B | Dephasing that randomises phase only / phase and momentum. |
6Calculators
One scatterer: T(E, U)
ka (rad)
Γ = 2t₀ sin ka (t₀)
T (Eq. 19.5)
R = (h/q²)/T (kΩ)
Two scatterers, incoherent series
T₂ = T/(2 − T)
R₂/R₁ = 2 − T
R₂ (kΩ) = (h/q²)(2−T)/T
Modes of a strip
M(E) = number of modes
G = M q²/h (µS)
Onset of mode n (t₀ / meV)
h/(Mq²) (kΩ)
Graphene ribbon
M = floor(2WE/hv₀)
R = h/(M q²) (kΩ)
7Self-check quiz
Click an answer: you get instant feedback. Score: 0 / 0
8Practice problems
q²/h = 38.74 µS; h/q² = 25.81 kΩ; hv₀ = 4.14 eV·nm (v₀ = 10⁶ m/s); t₀ = 91 meV for a = 2.5 nm, m = 0.067m₀.
P1. One scattererEASY
Wire, U = 2t₀, E = t₀ (ka = π/3). Find T.
Solution
2t₀ sin ka = √3 t₀; T = 3/(4 + 3) = 0.429.P2. Incoherent seriesEASY
A scatterer has T₁ = 0.39. Find T and R/R₁ for two in series with random phase.
Solution
T₂ = 0.39/1.61 = 0.242; R₂/R₁ = 2 − T = 1.61.P3. Outside the bandEASY
What is Σ for E = 4.5t₀, and what is Γ?
Hint
cos ka = 1 − E/2t₀ = −1.25.Solution
ka = π + iκ with cosh κ = 1.25, κ = 0.693; Σ = −t₀eika = +0.5t₀, real, so Γ = 0.P4. Modes of a stripEASY
N = 25 sites across. How many modes at E = t₀?
Solution
The n-th mode is open if 2t₀[1 − cos(nπ/26)] < t₀, i.e. cos(nπ/26) > 1/2, nπ/26 < π/3, n < 8.67: M = 8, G = 8 q²/h = 310 µS.P5. Onset energyMEDIUM
At what energy does the 5th mode of the 25-site strip open (t₀ = 91 meV)?
Solution
E = 2t₀[1 − cos(5π/26)] = 0.354 t₀ ≈ 32 meV.P6. Fig. 19.4MEDIUM
At E = 0.81t₀ one scatterer has T₁ = 0.3925 and the pair has T = 0.9996. By what factor does the quantum answer beat the incoherent one? At E = 0.6t₀: T₁ = 0.3377, T = 0.0665: factor?
Solution
Incoherent: 0.3925/1.6075 = 0.244 → ratio 0.9996/0.244 = 4.1. At 0.6t₀: 0.3377/1.6623 = 0.203 → 0.0665/0.203 = 0.33 (three times smaller).P7. Resistance of the pairMEDIUM
Convert the two transmissions of P6 to resistances (units of h/q²).
Solution
1/T: A, one: 2.55; pair quantum 1.0004; incoherent 4.1. B: pair quantum 15.0 against 4.9 incoherent. The pair at A is almost a perfect conductor.P8. Graphene ribbonMEDIUM
W = 30 nm, E = 0.3 eV: number of modes and resistance?
Solution
M = floor(2 × 30 × 0.3/4.14) = floor(4.35) = 4; R = 25.81/4 = 6.45 kΩ.P9. Hall plateauMEDIUM
A sample shows RH = 3.2 kΩ. How many edge channels?
Solution
M = 25.81/3.2 = 8.07: 8 channels (plateau at 3.227 kΩ).P10. LocalisationHARDER
If ln T falls by 0.5 per scatterer on average, what is T for 6 scatterers? How does this compare with Ohm (T = 1/(1 + nR₁) with R₁ = (1 − T₁)/T₁ for a single scatterer of T₁ = e−0.5)?
Hint
Ohm: R₁ = (1 − T₁)/T₁ with T₁ = e−0.5.Solution
T₆ = e−3 = 0.050; Ohm: R₁ = 0.65, T = 1/(1 + 6×0.65) = 0.20: four times larger. Quantum transmission decays exponentially, Ohm only as 1/n.P11. Why a linear drop?HARDER
Explain in three sentences why Model B dephasing gives a straight-line f(j) while a coherent wire does not.
Solution
Model B randomises phase and momentum at every site, so electrons reach local equilibrium and diffuse: the drop is spread uniformly (Ohmic). The coherent wire has no such averaging, so the drop is concentrated at the scatterer and Friedel-like ripples remain.9Common mistakes
MISTAKE"A clean wire has zero resistance."
CORRECTR = h/(Mq²) with M the number of modes of the wire: the resistance sits at the contacts.
MISTAKE"Two scatterers in series always have a larger resistance than one."
CORRECTIn a coherent wire the pair can transmit better (0.39 → 0.9996): resonance of the cavity between them.
MISTAKE"R₂ = 2R₁."
CORRECTThe correct classical rule is R₂ = R₁(2 − T) for the total (contact + scatterer) resistance, and it holds only if the phase is lost between the scatterers.
MISTAKE"ln T and T average the same way."
CORRECT⟨ln T⟩ falls linearly with the number of scatterers; ⟨T⟩ is dominated by rare high-T resonances.
MISTAKE"A 2D wire has a continuous conductance."
CORRECTT(E) = M(E) is a staircase of q²/h; the steps are rounded only by temperature or disorder.
MISTAKE"Edge states scatter back like other modes."
CORRECTIn a strong B the two directions live on opposite edges, so backscattering needs a transfer across the whole sample: RH stays quantised.
10Explore
Files are in the course folder (course_materials).
Mini-investigations (15 min each)
- In the series widget, find the spacing d at which two scatterers have the highest T at E = 0.81t₀. Is it periodic in d? What is the period in lattice sites, and how does it relate to k?
- Switch on Model B dephasing: at what D₀ does f(j) look linear? How does T change?
- In the modes widget, vary N from 10 to 40 at E = t₀ and compare the width of the steps with 2t₀[1 − cos(nπ/(N+1))].
11Can I…?
Tick what you can do without notes. Your ticks are saved in this browser.
12Looking ahead: Week 15
Beyond elastic, non-interacting transport (Datta Part B, Ch. 21–22).
- What changes when electrons exchange energy with phonons?
- What is Coulomb blockade, and why does the one-level picture of Week 13 need correcting for charging?
- How do the Fock-space ideas of Week 11 reappear?