Week 13 Study Toolkit · Nanoelectronics

1This week's plan

Read
Datta Part B, Chapter 18: NEGF Method (Secs. 18.1–18.4).

Run
Colab notebook week13_negf.ipynb: the rate equation, the Lorentzian, incoherent sources, the NEGF function negf(), current and conductance, self-consistent dephasing.

Practise
Quiz (aim for 8/10), the practice problems and one mini-investigation. Next week these equations are applied to wires (C7 follows).

By the end of the week you should be able to

Solve the one-level resistor with the rate equation and with E ψ = (ε − iγ/2)ψ + s, and show the two currents agree.
Explain broadening: D(E) is a Lorentzian of width γ = γ₁ + γ₂ and area one; τ = ħ/γ.
Explain why sources add incoherently and why Gⁿ ~ ψψ^† is the right object.
Write G^R, Gⁿ, A, the current Ĩ_m, and G(E) = (q²/h) Trace[Γ₁G^RΓ₂G^A].
State Σ₀ = D × G^R and what Models A and B destroy.

Optional video: Datta's nanoHUB-U course, Unit 2: L2.2–L2.5, L2.9.

2Key ideas in five sentences

Contacts in the Schrödinger equation. E ψ = [H + Σ₁ + Σ₂] ψ + s₁: outflow (non-Hermitian Σ) and inflow (s₁) from a contact; Γ = i[Σ − Σ^†] plays the role of γ in the one-level case.
Broadening. Coupling spreads a sharp level into D(E) = (γ/2π)/[(E − ε)² + (γ/2)²]; the current agrees with the rate equation, I = (q/ħ)γ₁γ₂/γ, but now we know the energy resolution.
Incoherent sources. Different contacts do not interfere; we insert one source at a time and add the bilinear quantities Gⁿ = G^RΣⁱⁿG^A.
Current and conductance. Ĩ_m = (q/h) Trace[Σⁱⁿ_mA − Γ_mGⁿ], and for coherent transport G(E) = (q²/h) Trace[Γ₁G^RΓ₂G^A].
Dephasing. Fluctuating potentials add Σ₀ = D × G^R, Σ₀ⁱⁿ = D × Gⁿ; Model A (D_ij = D₀) destroys phase only, Model B (D_ij = D₀δ_ij) phase and momentum.

3Equation sheet

dN/dt = −(ν₁ + ν₂)N + S₁ + S₂, S_m = ν_mf_mEqs. 18.5–18.7

Semiclassical one-level rate equation.

N = (ν₁f₁ + ν₂f₂)/(ν₁ + ν₂), I = q ν₁ν₂/(ν₁ + ν₂) (f₁ − f₂)Eq. 18.10

Steady state.

E ψ = (ε − iγ/2)ψ + s₁, ψ = s₁/(E − ε + iγ/2), γ = ħ(ν₁ + ν₂)Eqs. 18.13–18.16

Contact-ed Schrödinger equation.

2π s₁s₁* = γ₁, D(E) = (γ/2π)/[(E − ε)² + (γ/2)²]Eqs. 18.20, 18.23

Source strength; broadened level.

I = (q/h) ∫ dE γ₁γ₂/[(E − ε)² + (γ/2)²] (f₁ − f₂)Eq. 18.22

One-level current; peak G = q²/h for γ₁ = γ₂.

E{ψ} = [H + Σ₁ + Σ₂]{ψ} + {s₁}Eq. 18.24

N-level channel.

G^R = [E I − H − Σ]⁻¹, Gⁿ = G^RΣⁱⁿG^AEqs. 18.1–18.2

Σⁱⁿ = Γ₁f₁ + Γ₂f₂ (18.26).

A = G^RΓG^A = i[G^R − G^A], Γ = i[Σ − Σ^†]Eq. 18.3

Spectral function (2π × DOS).

Ĩ_m = (q/h) Trace[Σⁱⁿ_mA − Γ_mGⁿ]Eq. 18.4

Current per unit energy at contact m.

G(E) = (q²/h) Trace[Γ₁G^RΓ₂G^A]Eq. 18.30

Coherent conductance function.

Ĩ_m = (q/h) Σ_n T̄_mn(f_m − f_n)Eqs. 18.31–18.32

Multi-terminal (Büttiker).

Σ₀ = D × G^R, Σ₀ⁱⁿ = D × Gⁿ, Γ₀ = D × AEq. 18.35

Elastic dephasing; D_ij = ⟨U_R(i)U_R(j)⟩.

4Physics refresher: rates, Lorentzians, matrix inverses

Rate equation

dN/dt = −νN + S gives N(t) = (S/ν)(1 − e^−νt): the leaky bucket. The steady state S/ν is where inflow equals outflow.

Lorentzian

L(E) = (γ/2π)/[(E − ε)² + (γ/2)²] has half-maximum width γ and area 1. Its Fourier transform in time is e^−iεt/ħe^−γ|t|/2ħ: exponential decay of the level.

Why ∫dE γ/[(E−ε)² + (γ/2)²] = 2π

Substitute E − ε = (γ/2) tan θ; dE = (γ/2) sec²θ dθ and the denominator is (γ/2)² sec²θ, so the integrand becomes (γ/2)·(2/γ)·... = 2 dθ over (−π/2, π/2): total 2π.

Why the current is a trace

Trace[ψψ^†] = ψ^†ψ is the number of electrons. Its time derivative is a sum of terms, one for each contact; each is a trace of a matrix product (the trace of AB equals the trace of BA).

Matrix inverse numerically

Never invert explicitly if you only need G^Rs: np.linalg.solve(E*I - H - Sigma, s). For a tridiagonal channel the cost is linear in N.

5Glossary

Term	Meaning
Self-energy Σ	Matrix added to H describing the coupling to a contact; its anti-Hermitian part is Γ.
Broadening Γ	Γ = i[Σ − Σ^†]: how easily electrons enter and leave the channel (γ for one level).
Green's function G^R	[E I − H − Σ]⁻¹: the response of the channel to a source.
Spectral function A	2π × the (matrix) density of states.
Electron density Gⁿ	2π × (matrix) number of electrons per unit energy.
In-scattering Σⁱⁿ	Rate of injection of electrons: Γ₁f₁ + Γ₂f₂ (+ Σ₀ⁱⁿ).
Transmission T̄(E)	Trace[Γ₁G^RΓ₂G^A]: conductance function in units of q²/h.
Dephasing Σ₀	Self-energy of a conceptual contact that randomises the phase (Model A) or phase and momentum (Model B).

6Calculators

One-level resistor

ε − µ₀ (meV)γ₁ (meV)γ₂ (meV)

γ = γ₁ + γ₂

Dwell time τ = ħ/γ

Peak G = 4γ₁γ₂/γ² (q²/h)

G at µ₀ for kT ≪ γ (q²/h)

Rate-equation I, f₁ = 1, f₂ = 0

N, f₁ = 1, f₂ = 0

One-level current at T = 0

ε − µ₀ (meV)Bias qV (meV)γ₁ (meV)γ₂ (meV)

I = (q/h)∫_windowT dE

Rate-equation limit (q/ħ)γ₁γ₂/γ

Ratio

Two-site molecule

ε₁ (eV)ε₂ (eV)t (eV)γ₁ on site 1 (eV)γ₂ on site 2 (eV)E (eV)

T(E) = γ₁γ₂t²/|det|²

Levels of isolated H (eV)

7Self-check quiz

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

8Practice problems

ħ = 0.6582 meV·ps; q²/h = 38.74 µS; h/q² = 25.81 kΩ; kT = 25.85 meV at 300 K.

P1. Dwell timeEASY

A level is coupled with γ₁ = γ₂ = 20 meV. Find its dwell time and the width of its D(E).

Solution

γ = 40 meV ⇒ D(E) has full width at half maximum 40 meV; τ = ħ/γ = 0.6582/40 ps = 16.5 fs.

P2. Rate-equation currentEASY

Same device, f₁ = 1, f₂ = 0. Find N and I.

Solution

N = ν₁/(ν₁ + ν₂) = 0.5; ν = γ/ħ = 3.04 × 10¹³ s⁻¹; I = qν₁ν₂/(ν₁ + ν₂) = 2.43 µA.

P3. Unequal contactsEASY

γ₁ = 10 meV, γ₂ = 40 meV at very low temperature. Peak conductance and resistance?

Solution

4γ₁γ₂/γ² = 4 × 400/2500 = 0.64 q²/h = 24.8 µS ⇒ R = 25.81/0.64 = 40.3 kΩ.

P4. Incoherent sourcesMEDIUM

Two sources s₁, s₂ with |s₁|² = |s₂|² have a random relative phase. What is ⟨ψψ*⟩ at E = ε compared with the coherent in-phase and opposite-phase cases?

Hint

Average the cross term.

Solution

Coherent in phase: 4|s₁|²/(γ/2)²; opposite phase: 0; random phase: the cross terms vanish on average, leaving 2|s₁|²/(γ/2)², half the in-phase value.

P5. Window much larger than γMEDIUM

ε = 0, γ₁ = γ₂ = 20 meV, qV = 200 meV, T = 0. Find I from Eq. 18.22 and compare with the rate-equation value 2.43 µA.

Solution

I = (q/h)(2γ₁γ₂/γ)[atan(2(qV/2 − ε)/γ) − atan(2(−qV/2 − ε)/γ)] = (q/h)(0.02 eV)(2 atan 5) = 38.74 µS × 0.0549 V = 2.13 µA, 87 % of 2.43 µA: the Lorentzian tails outside the window are lost.

P6. N = 1 from the general formulaMEDIUM

Show that Trace[Γ₁G^RΓ₂G^A] = γ₁γ₂/[(E − ε)² + (γ/2)²] when all matrices are numbers.

Solution

G^R = 1/(E − ε + iγ/2), so G^RG^A = 1/[(E − ε)² + γ²/4], and Γ₁G^RΓ₂G^A = γ₁γ₂ × that.

P7. Two-site moleculeMEDIUM

ε₁ = ε₂ = 0, t = −1 eV, γ₁ = γ₂ = 0.1 eV on sites 1 and 2. T at E = ±1 eV and at E = 0?

Solution

det = (E + 0.05i)² − 1. At E = 1: det = −0.0025 + 0.1i, |det|² = 0.0100 ⇒ T = γ²t²/|det|² = 0.01/0.0100 = 0.999. At E = 0: det = −1.0025 ⇒ T = 0.01/1.005 = 0.0100.

P8. Spectral sum ruleHARDER

Why does ∫ Trace[A] dE/2π equal N? Estimate the value if the integral is cut at ±4 eV for a five-site chain with γ = 0.1 eV.

Solution

Each level contributes a Lorentzian of area 1. The tails missing beyond ±4 eV amount to about γ/(π × distance) ≈ 0.1/(π × 2.3) per level: total ≈ 4.98, as in the notebook.

P9. Zero current at the dephasing contactHARDER

Show that Trace[Σ₀ⁱⁿA − Γ₀Gⁿ] = 0 for Model A with D_ij = D₀.

Hint

Write the trace as a double sum.

Solution

Σ_ij D_ij[Gⁿ_ijA_ji − A_ijGⁿ_ji]. D_ij is symmetric, and swapping i ↔ j changes the sign of the bracket ⇒ the sum is 0.

9Common mistakes

MISTAKE"The one-level device has conductance q²/h whatever the contacts."

CORRECTOnly for γ₁ = γ₂ and kT ≪ γ; otherwise the peak is 4γ₁γ₂/γ² of q²/h (and thermal smearing lowers it when kT is not small).

MISTAKE"Σ is Hermitian like H."

CORRECTΣ is non-Hermitian: its anti-Hermitian part Γ is the broadening, which makes the equation describe leakage.

MISTAKE"Gⁿ = ψψ^† for the whole device."

CORRECTGn adds the contributions ψψ† from each contact, weighted by its Fermi function; ψ alone belongs to one source.

MISTAKE"Dephasing is just averaging over random potentials in H."

CORRECTIt is not exactly equivalent: Σ₀ also averages over different sections of the same sample (Sec. 18.4).

MISTAKE"A and the density of states are the same thing."

CORRECTA = 2π × the matrix density of states: ∫ Trace[A] dE/2π = N.

MISTAKE"Model A and Model B give the same result."

CORRECTA keeps momentum, B randomises it; B also needs only the diagonal of G, A the full matrix.

10Explore

Files are in the course folder (course_materials).

Week 13 slides Widget: one-level resistor Widget: NEGF on a small molecule Colab notebook Week 12 toolkit

Mini-investigations (15 min each)

In the one-level widget, set γ₁ = γ₂ = 20 meV and find the temperature above which the peak of G(µ₀) falls below 0.8 q²/h. How does it compare with γ/kB?
In the NEGF widget choose the benzene ring and compare T(E) for para and meta contacts at E = 0 and E = 1 eV. Explain the zero.
In the notebook, change the dephasing strength D₀ from 0 to 0.5 eV² for Models A and B and plot the peak transmission of the middle level against D₀.

11Can I…?

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

solve the one-level resistor semiclassicallysolve it quantum mechanically and get the Lorentzianexplain why γτ ≥ hexplain why sources are added incoherentlywrite G^R, Gⁿ, A and the current equationcompute T̄(E) for a small H with numpystate what Σ₀ = D × G^R doesdistinguish Models A and B

12Looking ahead: Week 14 and circular session C7

Using the equations (Datta Part B, Ch. 19–20), then C7: an NEGF lab.

How does a wire get its contact self-energy t e^ika?
Can two scatterers in series offer less resistance than one?
Why does the conductance of a narrow 2D wire go in steps of q²/h?

The NEGF method: contacts for the Schrödinger equation

1This week's plan

By the end of the week you should be able to

2Key ideas in five sentences

3Equation sheet

4Physics refresher: rates, Lorentzians, matrix inverses

5Glossary

6Calculators

One-level resistor

One-level current at T = 0

Two-site molecule

7Self-check quiz

8Practice problems

9Common mistakes

10Explore

Mini-investigations (15 min each)

11Can I…?

12Looking ahead: Week 14 and circular session C7