University of Kurdistan logoNanoelectronics: A New Perspective on TransportWeek 12 · Contact-ing Schrödinger · C6
University of Kurdistan logo
WEEK 12 · CONTACT-ING SCHRÖDINGER · C6

Nanoelectronics:
A New Perspective on Transport

From the wave equation to a matrix: tight-binding chains, 2D lattices, graphene; Weeks 11–12 lab
Prof. Mohammad RazaghiUniversity of Kurdistan · Graduate course in Electronics Engineering
Based on S. Datta, Lessons from Nanoelectronics, Part B: Quantum Transport, Chapter 17
Part B begins. Part 1 is the fixed lecture (Ch. 17). Part 2 is circular session C6 on Weeks 11–12, with the task: build a 10-site chain Hamiltonian by hand, then check it with code.

RecapWhere we are

Part A gave us

  • The elastic resistor: entropy in the contacts, mechanics in the channel
  • BTE = classical dynamics + contacts
  • D(E), M(E) from an assumed E(p)

Part B

NEGF = quantum dynamics + contacts. Inputs: a Hamiltonian H for the channel and self-energies Σ for the contacts. Today: how to write H.

Fig. 17.1: the channel is described by H, each contact (physical or conceptual, like the "0" contact for distributed scattering) by a Σ.

Week 12Today's goals

  1. Build the Schrödinger equation from E(p) and include a potential.
  2. Distinguish affinity and ionization levels; the self-consistent field.
  3. Turn the differential equation into a matrix EΨ = HΨ; size N = n × b.
  4. Fit tight-binding parameters: t = −ħ²/2ma², ε = Ec − 2t; add U(x) and magnetic phases.
  5. Derive graphene's bands ε ± |h₀(k)| and its Dirac cones.

Physics refresherRefresher: from E(p) to a wave equation

E → iħ ∂/∂t,   p → −iħ∇Eq. 17.1
iħ ∂ψ̃/∂t = −(ħ²/2m)∇²ψ̃ + U(r)ψ̃Eq. 17.6 · solutions eik·r − iEt/ħ, p = ħk
  • Time-independent: Eψ = Hopψ, Hop = −(ħ²/2m)∇² + U (17.7).
  • Group velocity (1/ħ)∇kE = particle velocity ∇pE.
  • |ψ|² is a probability density; summed over electrons, the density n(r).

Physics refresherRefresher: clamped waves have discrete levels

  • A guitar string: ω²u = −∂²u/∂z² with clamped ends ⇒ harmonics.
  • Particle in a box: En = ħ²π²n²/2mL².
  • Hydrogen: a tiny 3D box, En = −13.6 eV/n², a₀ = 0.0529 nm (17.8).
  • Matrices: Hψ = Eψ has N real eigenvalues for Hermitian H; numpy's eigh does it.

Datta Sec. 17.2Electrons interact: which levels?

  • Affinity levels (N → N+1 → N): the added electron feels N electrons.
  • Ionization levels (N → N−1 → N): it feels N − 1.
  • They differ by the charging energy U₀.
  • SCF: each electron in an average U(r) from n(r): Hartree minus a DFT correction ∝ n1/3 (17.9).
Helium: first ionization 24.6 eV, second 54.4 eV: one electron's repulsion matters.
Datta quotes U₀ ~ 25 eV for helium; the SCF picture fails when U₀ ≫ kT and coupling, the subject of Ch. 22.

Datta Sec. 17.3Differential equation → matrix

E[S]{ψ} = [H]{ψ}Eq. 17.10 · ψ(r) = Σ ψmum(r)
Snm = ∫un*um,   Hnm = ∫un*HopumEq. 17.11
  • Localized basis (atomic orbitals): tight binding.
  • Size N = n × b: atoms × orbitals per atom (b = 4 for sp³, 1 for graphene pz).

Datta Sec. 17.3.1The hydrogen molecule: two orbitals

H = [ε t; t ε],   S = [1 s; s 1]Eq. 17.12
E = (ε + t)/(1 + s), (ε − t)/(1 − s)bonding and antibonding (t < 0)
  • First principles: compute ε, t, s from u1,2.
  • Semi-empirical: fit them to experiment; s = 0 for an orthogonal basis.

Datta Sec. 17.4.1A chain: matching the parabola

E(k) = ΣmHnmeik(m−n)a = ε + 2t cos kaEqs. 17.14–17.15 · translational invariance
t = −ħ²/2ma²,   ε = Ec − 2tEq. 17.16
  • m = m₀, a = 0.25 nm: t = −0.61 eV, band width 2.44 eV.
  • A potential U(x): add it to the diagonal (Fig. 17.6); no k any more, the matrix still works.

InteractiveExplore: the tight-binding chain

Schrödinger's equation as a matrix: a tight-binding chain  (Datta Part B, Ch. 17, Figs. 17.5–17.6)

N atoms spaced by a, one orbital each: the Hamiltonian is an N×N matrix with ε + Un on the diagonal and t between neighbours (Eq. 17.10). Choosing t = −ℏ²/2ma² and ε = Ec − 2t makes the band ε + 2t cos ka match the parabola near k = 0 (Eqs. 17.15–17.16). Diagonalise it live, pick an eigenstate, add a potential.

t = −ħ²/2ma² (Eq. 17.16a)–
Band: ε + 2t cos ka spans–
Selected level E (matrix) · box/ring formula–
Continuum particle in a box, L = (N+1)a–

Try this

  1. Open chain, N = 20: compare level 1 with the continuum box.
  2. Look at the highest levels: why do they deviate from the parabola?
  3. Switch to a ring: why are the levels in pairs?
  4. Add the square well, then the field: where do the low states live?

Scroll inside the panel for all controls. Swipe or use the arrow buttons outside it to change slide.

Eigenvalues by Jacobi rotation in the browser; t = −ħ²/2m₀a².

Datta Sec. 17.4.2–17.4.32D lattices and magnetic fields

E = ε + 2t cos kₓa + 2t cos kyaEq. 17.18b · ε = Ec − 4t (17.19)
t → t eiqAxa/ħSec. 17.4.3 · Peierls phase, B = ∇ × A
  • Phases around a loop: 2πΦ/Φ₀, Φ₀ = h/q.
  • Levels of a ring repeat with period h/q in flux (notebook).
Choose A not varying along the transport direction, e.g. A = (−By, 0) for current along x.

Datta Sec. 17.4.4Graphene: a lattice with a basis

h(k) = [ε h₀*; h₀ ε],   h₀ = t(1 + 2eikₓacos kyb)Eqs. 17.24–17.25 · a = 1.5acc, b = (√3/2)acc
E = ε ± |h₀(k)|Eq. 17.26
h₀ = 0 at kₓa = 0, kyb = ±2π/3
⇒ E = ε ± a|t|(kₓ² + βy²)1/2Eqs. 17.27–17.29 · two valleys, Dirac cones
v₀ = a|t|/ħ = 0.87 × 10⁶ m/s for |t| = 2.7 eV, a_cc = 0.142 nm.

InteractiveExplore: square lattice and graphene

Bands of 2D lattices: square lattice and graphene  (Datta Part B, Secs. 17.4.2, 17.4.4)

Axes are kₓa and k_yb, so circles in k look like ellipses for graphene. A solution ψn = ψ₀eik·rn turns the matrix equation into E(k) = Σm Hnmeik·(rm−rn) (Eq. 17.18). For graphene, a lattice with a two-atom basis, it becomes a 2×2 matrix h(k) with eigenvalues ε ± |h₀(k)| (Eqs. 17.22–17.26). Click on the k-map to pick a k-point.

Selected k–
Energies at that k–
Parabola at that k–
––

Try this

  1. Square lattice: at low E are the contours circles? At 4|t|?
  2. Graphene: find the dark valleys of |h₀|. Where is E = ε?
  3. Click near a valley: is the linear (Dirac) estimate good?
  4. Change |t|: how does the Dirac velocity change?

Scroll inside the panel for all controls. Swipe or use the arrow buttons outside it to change slide.

Axes are kₓa and k_yb: circles in k look like ellipses for graphene.

CheckConcept check

One minute with a neighbour, then click to reveal.

1. Size of H for 1000 silicon atoms, sp³ basis? (click)
N = n × b = 4000.
2. m = 0.067m₀, a = 0.5 nm: t and band width? (click)
t = −0.61 × (0.25/0.5)²/0.067 = −2.28 eV; 4|t| = 9.1 eV.
3. Why are the ring levels doubly degenerate? (click)
+k and −k (left and right movers) have the same ε + 2t cos ka.
4. Why is h(k) 2×2 for graphene? (click)
The A and B atoms see different environments: only the pair is translationally invariant.

Wrap-upFixed session summary

Take-home

  • E(p) → wave equation (17.1); Hop = −(ħ²/2m)∇² + U
  • Matrix form EΨ = HΨ, N = n × b
  • Chain: ε + 2t cos ka, t = −ħ²/2ma²; U on the diagonal; phases for B
  • Graphene: ε ± |h₀|, two valleys, v₀ ≈ 10⁶ m/s

Next: circular session C6

Weeks 11–12 lab: build a 10-site chain by hand, spectra with potentials, graphene nanoribbon gaps, and Fock-space statistics. Open week12_tight_binding_C6.ipynb (Part 2) and the Week 12 study toolkit.

PART 2

Circular session C6

Weeks 11–12 · tight-binding lab and problem clinic

A · 15 min
Recap
B · 40 min
Lab: H by hand, bands, ribbons
C · 25 min
Problem clinic
D · 10 min
Two-minute explanations
Pairs for B1 (one writes the matrix on paper, the other types it), groups of 3–4 for the rest.

Circular session C6Block A · Weeks 11–12 in one table

WeekIdeaKey result
11Entropy is counting; reservoirs prefer to receiveP(−ε)/P(+ε) = eε/kT; pi ∝ e−(Ei−µNi)/kT in Fock space
11Information is fuelinfo-battery NkT ln 2; Landauer Eerase ≥ NkT ln 2
12Schrödinger as a matrixEΨ = HΨ, t = −ħ²/2ma², ε = Ec − 2t × dimension
12Bands from latticesε + 2t cos ka; graphene ε ± |h₀(k)|, Dirac cones

In pairs: where in a transport calculation does each row enter? (Hint: H for the channel, Σ and Fermi functions for the contacts.)

Circular session C6Block B · Lab

B1 (student task). Build a 10-site chain H by hand (ε = 0, t = −1 eV), diagonalise, compare with ε + 2t cos(iπ/11).

B2. Ring, well, field: how does the spectrum change, where is the ground state?

B3. Armchair graphene nanoribbons: gap vs width; the 3p + 2 family is metallic.

B4. Week 11: a two-level molecule with U₀ in equilibrium: a staircase in n(µ).

Real armchair ribbons all have small gaps because of edge relaxation and next-neighbour hopping; the 3p+2 rule is a property of the simplest model.

Circular session C6Block C · Problem clinic

1. A reservoir at 300 K receives 0.1 eV. By what factor does its number of states grow? (click)
e0.1/0.02585 ≈ 48.
2. Info-battery with PU = 0.75 at 300 K: open-circuit voltage? (click)
25.85 mV × ln 3 = 28.4 mV.
3. 4-site ring, ε = 0, t = −1: eigenvalues? (click)
−2cos(πm/2): −2, 0, 0, +2.
4. GaAs (0.067m₀): what a gives 4|t| = 1 eV? (click)
a = ħ/√(2m·0.25 eV) = 1.51 nm.

Circular session C6Block D · Two-minute explanations, and what comes next

Pick one, explain in two minutes

  1. Why an excited atom decays
  2. Why erasing a bit costs kT ln 2
  3. Why a chain's E(k) is a cosine
  4. Why graphene has Dirac cones

Peers score clarity and correctness (1–5 each).

Week 13

The NEGF method (Part B, Ch. 18): a one-level resistor, broadening by contacts, GR = [E − H − Σ]⁻¹, transmission and density of states, and dephasing.

Prof. Mohammad Razaghi · University of Kurdistan · after S. Datta, Lessons from Nanoelectronics