University of Kurdistan logoNanoelectronics · Week 12 Study ToolkitProf. Mohammad Razaghi
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WEEK 12 · CONTACT-ING SCHRÖDINGER · C6

From the wave equation to a matrix: tight binding and graphene

Study toolkit for students · based on S. Datta, Lessons from Nanoelectronics, Part B, Chapter 17
Prof. Mohammad Razaghi · University of Kurdistan

1This week's plan

Read
Datta Part B, Chapter 17: Contact-ing Schrödinger (wave equation, matrix form, tight binding, graphene).
Run
Colab notebook week12_tight_binding_C6.ipynb: Part 1 follows the lecture, Part 2 is the C6 lab (B1–B4) and the problem clinic.
Practise
Quiz (aim for 8/10), practice problems, and prepare a two-minute explanation for Block D of C6.

By the end of the week you should be able to

  1. Obtain the Schrödinger equation from E(p) with E → iħ∂/∂t, p → −iħ∇ and add a potential U(r).
  2. Explain affinity vs ionization levels, the charging energy U₀ and when the self-consistent field works.
  3. Write EΨ = HΨ in a basis; know N = n × b and the overlap matrix S.
  4. Fit t = −ħ²/2ma², ε = Ec − 2t (1D) or Ec − 4t (2D); put U(x) on the diagonal and B into Peierls phases.
  5. Derive graphene's E = ε ± |h₀(k)|, its two valleys and the Dirac velocity.
Circular session C6 task (bring to class): on paper, write the 10 × 10 Hamiltonian of a chain with ε = 0, t = −1 eV, open ends. Predict its lowest and highest eigenvalues from ε + 2t cos(nπ/11) before you run notebook cell B1.

2Key ideas in five sentences

  1. Wave equation from E(p). Replacing E and p by operators turns E = p²/2m + U into iħ∂ψ/∂t = Hopψ; plane waves eik·r − iEt/ħ have p = ħk.
  2. Interactions. An electron added (affinity) or removed (ionization) sees different potentials, separated by U₀; the SCF approximation uses one average potential and fails when U₀ is large.
  3. Matrix form. In any basis the Schrödinger equation is E[S]{ψ} = [H]{ψ}; with localized orbitals (tight binding) H is sparse, of size atoms × orbitals per atom.
  4. Bands from lattices. Translational invariance gives E(k) = ΣmHnmeik(m−n)a: a cosine for a chain, which matches the parabola near the band bottom when t = −ħ²/2ma².
  5. Graphene. Two atoms per unit cell give a 2 × 2 h(k) and E = ε ± |h₀|; h₀ vanishes at two valleys where the bands touch in Dirac cones with v₀ ≈ 10⁶ m/s.

3Equation sheet

E → iħ ∂/∂t,   p → −iħ∇Eq. 17.1
From a dispersion relation to a wave equation.
Eψ = Hopψ,   Hop = −(ħ²/2m)∇² + U(r)Eq. 17.7
Time-independent Schrödinger equation.
En = −13.6 eV/n²,   a₀ = 0.0529 nmEq. 17.8
Hydrogen atom.
USCF = UHartree − (q²/4πε)(3n/8π)1/3Eq. 17.9
Self-consistent field with a simple DFT correction.
E[S]{ψ} = [H]{ψ};   Snm = ∫un*um, Hnm = ∫un*HopumEqs. 17.10–17.11
Matrix form in a basis; N = n × b.
E = (ε + t)/(1 + s), (ε − t)/(1 − s)Eq. 17.12
Hydrogen molecule (bonding, antibonding for t < 0).
E(k) = ε + 2t cos kaEqs. 17.14–17.15
1D chain; bandwidth 4|t|.
t = −ħ²/2ma²,   ε = Ec − 2tEq. 17.16
Matches Ec + ħ²k²/2m for ka ≪ 1.
E = ε + 2t cos kₓa + 2t cos kya,   ε = Ec − 4tEqs. 17.18–17.19
Square lattice.
t → t eiqA·a/ħ,   Φ₀ = h/qSec. 17.4.3
Peierls substitution for a magnetic field.
h₀ = t(1 + 2eikₓacos kyb),   E = ε ± |h₀|Eqs. 17.24–17.26
Graphene, a = 1.5acc, b = (√3/2)acc.
E = ε ± a|t|(kₓ² + βy²)1/2,   v₀ = a|t|/ħEqs. 17.27–17.29
Dirac cones near kyb = ±2π/3.

4Physics refresher: waves, eigenvalues, matrices

Standing waves

Clamp a wave at x = 0 and L: only sin(nπx/L) survives, so k = nπ/L and En = ħ²π²n²/2mL². For L = 10 nm and m = m₀, E₁ = 3.76 meV; it scales as 1/(mL²).

Eigenvalues of a Hermitian matrix

H = H† has N real eigenvalues and orthonormal eigenvectors. In numpy: E, V = np.linalg.eigh(H); column V[:, n] is the n-th state, |V[j, n]|² its probability on site j.

Why the second derivative becomes (ψn+1 − 2ψn + ψn−1)/a²Taylor: ψ(x ± a) = ψ ± aψ′ + (a²/2)ψ″ ± …; adding the two removes ψ′ and gives ψ″ ≈ [ψ(x + a) − 2ψ(x) + ψ(x − a)]/a². Multiply by −ħ²/2m: diagonal −2t, neighbours t, with t = −ħ²/2ma².
Why an open chain has ε + 2t cos(nπ/(N + 1))Try ψj = sin(jka). Row j of HΨ gives εψj + t(ψj+1 + ψj−1) = (ε + 2t cos ka)ψj. The ends need ψ0 = ψN+1 = 0, so k(N + 1)a = nπ.
Ring vs open chainA ring adds H1N = HN1 = t; Bloch waves eijka with k = 2πm/Na; ±k have the same energy, so levels come in pairs.

5Glossary

TermMeaning
Hamiltonian HThe operator (or matrix) whose eigenvalues are the allowed energies.
Basis function umA known function (often an atomic orbital) used to expand ψ.
Overlap matrix SSnm = ∫un*um; the identity for an orthogonal basis.
Tight bindingMatrix H in a localized basis, with on-site ε and hopping t to neighbours.
Charging energy U₀The difference between affinity and ionization levels: the cost of one extra electron.
SCFSelf-consistent field: each electron moves in an average potential from all others.
Peierls phaset → t eiqA·a/ħ: how a magnetic field enters a tight-binding H.
Valley, Dirac conePoints in k where graphene's bands touch, with E linear in |k − K|.

6Calculators

Tight-binding parameters from m and a

t = −ħ²/2ma²
ε − Ec = −2t × dimension
Band width 4|t| × dimension
Parabola error at ka = 0.5

Eigenvalues of a chain

Eigenvalues (eV), lowest first
Lowest level spacing

Graphene near a valley

v₀ = 1.5acc|t|/ħ
E − ε = ħv₀|k − K|
de Broglie wavelength 2π/|k − K|

7Self-check quiz

Click an answer: you get instant feedback. Score: 0 / 0

8Practice problems

ħ²/2m₀ = 0.0381 eV·nm²; ħ = 6.582 × 10⁻¹⁶ eV·s; Φ₀ = h/q = 4.136 × 10⁻¹⁵ T·m².

P1. Matching the parabolaEASY

m = m₀ and a = 0.25 nm. Find t, ε − Ec and the band width of a 1D chain.

Solutiont = −0.0381/0.0625 = −0.610 eV; ε − Ec = −2t = +1.22 eV; band width 4|t| = 2.44 eV.
P2. Effective mass and spacingEASY

GaAs (m = 0.067m₀) on a grid with a = 1 nm: t and band width?

Solutiont = −0.0381/0.067 = −0.569 eV; 4|t| = 2.27 eV. Grids for low-mass materials can be coarser.
P3. A 6-site ringEASY

ε = 0, t = −1 eV, periodic boundary. List the eigenvalues.

Hintk = 2πm/6a.
Solution2t cos(πm/3): −2, −1, −1, +1, +1, +2 eV; the ±k pairs are degenerate.
P4. Hydrogen molecule with overlapMEDIUM

ε = −10 eV, t = −3 eV, s = 0.2. Bonding and antibonding energies?

Solution(ε + t)/(1 + s) = −13/1.2 = −10.83 eV; (ε − t)/(1 − s) = −7/0.8 = −8.75 eV. With s ≠ 0 the antibonding level rises more than the bonding level falls.
P5. A 10-site open chainMEDIUM

ε = 0, t = −1 eV. Lowest and highest eigenvalues? Is the band width 4|t|?

Solution2t cos(π/11) = −1.919 eV and +1.919 eV; the finite chain spans 3.84 eV, approaching 4|t| as N → ∞.
P6. Matrix sizeMEDIUM

A silicon nanowire of 2000 atoms in an sp³s* basis (5 orbitals per atom), with spin. Size of H?

SolutionN = n × b = 2000 × 5 × 2 = 20 000; a dense matrix would have 4 × 10⁸ entries, so sparsity matters.
P7. Graphene Dirac velocityMEDIUM

|t| = 2.7 eV, acc = 0.142 nm. Find v₀ and the energy 0.1 nm⁻¹ from a valley.

Solutionv₀ = 1.5 × 0.142 nm × 2.7 eV/ħ = 8.74 × 10⁵ m/s; ħv₀ × 0.1 nm⁻¹ = 57.5 meV.
P8. One flux quantumHARDER

A ring of diameter 100 nm. Which B threads one flux quantum? What happens to its levels as B goes from 0 to twice that?

SolutionB = Φ₀/(π × (50 nm)²) = 0.53 T. The spectrum is periodic in Φ with period Φ₀: at 2Φ₀ it is back to the B = 0 spectrum.
P9. Square lattice bandHARDER

Show the 2D square lattice band runs from Ec to Ec + 8|t| and near the bottom E ≈ Ec + ħ²k²/2m.

Hintcos x ≈ 1 − x²/2.
SolutionWith ε = Ec − 4t and t < 0: k = 0 gives Ec, (π/a, π/a) gives Ec − 8t = Ec + 8|t|. Expanding: E ≈ ε + 4t − t(kₓ² + ky²)a² = Ec + ħ²k²/2m.

9Common mistakes

MISTAKE"Tight binding only works for crystals."
CORRECTThe matrix H works for any structure; translational invariance is only needed to label states by k.
MISTAKE"t is positive because it is a hopping energy."
CORRECTFor s-like orbitals and the parabola fit, t = −ħ²/2ma² is negative; the band bottom is at k = 0.
MISTAKE"The cosine band is wrong because it is not a parabola."
CORRECTIt is the exact band of the discrete model; it agrees with the parabola for ka ≪ 1, which is where a fine grid puts the electrons.
MISTAKE"A potential changes the band structure E(k)."
CORRECTA non-uniform U(x) breaks translation symmetry: there is no E(k), but the matrix with U on the diagonal still gives the levels.
MISTAKE"Graphene has one atom per cell."
CORRECTThe A and B sites have different neighbours, so the cell has two atoms and h(k) is 2 × 2.
MISTAKE"Affinity and ionization levels are the same orbital."
CORRECTThey differ by U₀; ignoring this is fine for large conductors but not for small molecules or quantum dots.

10Explore

Files are in the course folder (course_materials).

Mini-investigations (15 min each)

  1. In the chain widget, make N = 40 and add the linear field. Which states are localized near the low-potential end, and why does the spacing become uneven?
  2. In the lattice widget, find the energy at which the square-lattice contours become a square (the "van Hove" energy). What is special about the density of states there?
  3. In the notebook, compute armchair ribbon gaps for N = 3p, 3p + 1, 3p + 2 up to N = 30 and fit gap ∝ 1/width for the two gapped families.

11Can I…?

Tick what you can do without notes. Your ticks are saved in this browser.

12Looking ahead: Week 13

The NEGF method (Datta Part B, Ch. 18): the H of this week meets the contacts through self-energies Σ.

  1. How does connecting a level to a contact broaden it?
  2. What are GR = [EI − H − Σ]⁻¹, the spectral function and the transmission?
  3. How do we include scattering ("dephasing") as one more contact?