1This week's plan
Datta Part A, Chapters 4–5 (pp. 43–56) and Appendix B (angular averaging).
Colab notebook week04_ballistic_diffusive_C2.ipynb: Part 1 before class, Part 2 (lab) in circular session C2.
Quiz (aim for 8/10), practice problems, and the C2 preparation list.
By the end of the week you should be able to
- Explain t ∝ L (ballistic) and t ∝ L² (diffusive), and use t = L/u + L²/2D = tB(1 + L/λ).
- Derive G = GBλ/(L + λ) and the conductivity in 1D, 2D and 3D.
- Explain λ = vτ{2, π/2, 4/3} from angular averaging.
- Explain modes and conductance quantization, GB = (q²/h)M.
- Use the fluctuation–dissipation theorem G₀ = CI/2kT and the Johnson–Nyquist formula.
Optional video: Datta's nanoHUB-U course, Unit 1: L1.5–L1.8 (ballistic, diffusive, angular averaging).
2Key ideas in five sentences
- Two time scales. A bullet crosses in tB = L/u; a random walker needs ≈ L²/2D; Datta joins them as t = tB(1 + L/λ).
- The new Ohm's law. With G = q²D/2t and D ∝ L, G = GBλ/(L + λ): length-independent when L ≪ λ, Ohm's law when L ≫ λ.
- λ includes an angular factor. λ = 2D/u = vτ × {2, π/2, 4/3} in 1D, 2D, 3D, because collisions in 2D/3D usually deflect rather than reverse.
- Modes. GB = (q²/h)M: M parallel channels, which makes small conductors' conductance rise in steps.
- Noise knows the conductance. At equilibrium, random current pulses give CI = 2kTG₀ (fluctuation–dissipation); in engineering form irms = √(4kTGΔf).
3Equation sheet
4Physics refresher: random walks and averages
Each step in a random direction: the mean displacement stays zero but ⟨x²⟩ = 2Dt grows linearly in time, so the typical distance grows as √t. Crossing L takes ≈ L²/2D.
| 1D | 2D | 3D | |
|---|---|---|---|
| ⟨|cosθ|⟩ | 1 | 2/π | 1/2 |
| ⟨cos²θ⟩ | 1 | 1/2 | 1/3 |
| λ/vτ | 2 | π/2 | 4/3 |
Why 2 in 1D?
With isotropic scattering in 1D, after a collision the electron goes forward or backward with equal chance, so only half the collisions reverse it: the backscattering time is 2τ and λ = 2vτ.Spectral density vs correlation
Datta's CI is the zero-frequency current correlation. The familiar one-sided noise spectral density is SI = 2CI = 4kTG, so irms² = 4kTGΔf.5Glossary
| Term | Meaning |
|---|---|
| Ballistic | L ≪ λ: electrons cross without backscattering; t ∝ L, G independent of L. |
| Diffusive | L ≫ λ: random walk; t ∝ L², G ∝ 1/L (Ohm's law). |
| Diffusion constant D | ⟨vz²τ⟩; ⟨x²⟩ = 2Dt. |
| Mode (channel) | An independent transverse state (like a waveguide mode) that carries q²/h of conductance. |
| Sharvin resistance | Ballistic resistance of a wide contact, inversely proportional to its area. |
| Conductance quantization | Steps of 2q²/h in the conductance of narrow point contacts (two spins per spatial mode). |
| Fluctuation–dissipation theorem | Links the low-bias conductance to equilibrium current fluctuations: G₀ = CI/2kT (Kubo formula). |
| Johnson–Nyquist noise | Thermal current noise of a resistor, SI = 4kTG. |
6Calculators
Transit time and the new Ohm's law
From vτ to Datta's λ
Johnson–Nyquist noise
7Self-check quiz
Click an answer: you get instant feedback. Score: 0 / 0
8Practice problems
q²/h = 38.74 µS, k = 1.381 × 10⁻²³ J/K.
L = 100 nm, λ = 50 nm, u = 10⁵ m/s. Find tB, t and G/GB.
Solution
tB = 1 ps; t = 1 × (1 + 2) = 3 ps; G/GB = 50/150 = 1/3.M = 20, λ = 200 nm, L = 1 µm. Find G and R.
Solution
G = 38.74 µS × 20 × 200/1200 = 129 µS; R = 7.74 kΩ.For vτ = 30 nm, find Datta's λ in 2D and in 3D.
Solution
2D: (π/2) × 30 = 47.1 nm; 3D: (4/3) × 30 = 40 nm.3D conductor: v = 10⁵ m/s, τ = 0.1 ps. Find D, u, λ and the transit time through L = 1 µm.
Hint
D = v²τ/3, u = v/2.Solution
D = 3.33 × 10⁻⁴ m²/s (3.33 cm²/s), u = 5 × 10⁴ m/s, λ = 2D/u = 13.3 nm = (4/3)vτ. t = L/u + L²/2D = 20 ps + 1.50 ns = 1.52 ns: deep in the diffusive regime.A narrow constriction in a 2D electron gas with Fermi wavelength 40 nm is 100 nm wide. Estimate the number of spatial modes and the conductance, counting two spins.
Hint
Modes ≈ floor(2W/λF).Solution
floor(200/40) = 5 spatial modes; G = 5 × 2q²/h = 387 µS, R = 2.58 kΩ.A one-level resistor with ε = µ₀, transit time t = 1 ps, at 300 K. Find the one-way current I₀ and G₀ from the noise.
Hint
f₀(1 − f₀) = 1/4 at ε = µ₀.Solution
I₀ = (q/t)/4 = 40 nA. G₀ = (q²/t)f₀(1 − f₀)/kT = 1.55 µS (as in the noise widget).Find CI and irms in 1 MHz at 300 K for R = 1 kΩ, 25.8 kΩ and 1 MΩ.
Solution
1 kΩ: CI = 8.3 × 10⁻²⁴ A²s, irms = 4.07 nA; 25.8 kΩ: 0.80 nA; 1 MΩ: 0.13 nA. Better conductors are noisier in current.9Common mistakes
10Prepare for circular session C2
- Flux = population ÷ residence time, with a non-physics example
- Why diffusion time grows as L²
- What a mode is, and why GB comes in steps
- Degenerate vs non-degenerate: sketch both
Run Part 2 of the Week 4 notebook. Bring answers: Is λ closer to vτ or (4/3)vτ? Which part of R can better contacts reduce? What changes when vτ = 200 nm?
Problems like P1, P2 and P7 above.
Derive σ = q²(D/AL)Ddiff from G = q²D/2t and t = L²/2Ddiff. Be ready to explain any Week 3–4 idea for two minutes without equations.
11Explore
Files are in the course folder (course_materials).
Mini-investigations (15 min each)
- In the transit-time widget, measure t for λ = 30 nm and λ = 120 nm. Plot t/tB against L/λ for both: do they fall on one line?
- In the noise widget, find the level position ε at which G₀ falls to 10% of its peak (300 K). Compare with the f₀(1 − f₀) prediction.
- In the 3D nanowire widget, use the quick measurement at L = 100, 200, 400 nm and extract λ from 1/T = 1 + L/λ.
12Can I…?
Tick what you can do without notes. Your ticks are saved in this browser.
13Looking ahead: Week 5
Energy band model (Datta Ch. 6): where D(E) and M(E) come from.
- In a box of length L, which wave vectors k are allowed for an electron wave?
- If allowed k values are evenly spaced, are allowed energies E = ħ²k²/2m evenly spaced too?
- How does the number of states with energy below E grow with E in 1D, 2D and 3D?