Week 15 Study Toolkit · Nanoelectronics

1This week's plan

Read
Datta Part B, Chapters 21 and 22: inelastic scattering; does NEGF include "everything?" (Coulomb blockade, Fock space, entanglement).

Run
Colab notebook week15_inelastic_coulomb.ipynb: golden-rule lifetime, Bose factors, ladder relaxation, inelastic() NEGF on an energy grid, double-dot rectification, Fock-space master equation, entangled two-dot model.

Practise
Quiz (aim for 8/10), the problems and one mini-investigation. No circular session this week; next week is the final one (C8).

By the end of the week you should be able to

Derive ħ/lifetime = 2πΣ|τ|²δ(E − E′) from Γ = τ^†aτ.
State the absorption (∝ N_ω) and emission (∝ N_ω + 1) rates and their ratio e^ħω/kT.
Write Σ₀ⁱⁿ(E) = D × Gⁿ(E+ħω) and Γ₀(E) = D × [Gⁿ(E+ħω) + G^p(E−ħω)].
Describe Coulomb blockade in the SCF picture (U_i = U₀(N − N_i)) and in Fock space.
Derive the plateau I = 2qν₁ν₂/(2ν₁ + ν₂) and say why it signals correlation.

Optional video: Datta's nanoHUB-U course, Unit 3: L3.7 (golden rule), L3.8 (inelastic), L3.9 (Does NEGF include everything?).

2Key ideas in five sentences

Scattering is a partition problem. Take the initial state as the channel and all other states as the contact; Γ = τ^†aτ gives the golden rule ħ/lifetime = 2πΣ|τ|²δ(E − E′).
Emission beats absorption. With N_ω phonons, absorption ∝ N_ω and emission ∝ N_ω + 1; their ratio e^ħω/kT is detailed balance and is what thermalises electrons.
NEGF with inelastic scattering. Σ₀ⁱⁿ(E) = D × Gⁿ(E + ħω) and Γ₀ = D × [Gⁿ(E + ħω) + G^p(E − ħω)] couple energies; with the right N and N + 1 factors an equilibrium device stays in equilibrium.
Interactions can break the one-electron picture. Coulomb blockade (U₀ ≫ kT, γ) needs Fock space: n(µ) has steps at ε and ε + U₀ and the high-bias current has a 2/3 plateau with p₀₀ = p₀₁ = p₁₀ = 1/3, p₁₁ = 0.
Entanglement. Two coupled dots have 16 Fock states; the lowest two-electron state is a singlet, a superposition not writable as two one-electron states.

3Equation sheet

Γ = i[Σ − Σ^†] = τ^† a τ, Σ = τ^†gτEqs. 21.3–21.5

Contact broadening from the contact spectral function a.

ħ/Lifetime(m) = Γ_mm = 2π Σ_m′|τ_m′m|² δ(E_m − E_m′)Eq. 21.12

Inverse lifetime = sum of golden-rule rates.

S(m′←m) = (2π/ħ)|τ_m′m|² δ(E_m − E_m′)Eq. 21.14

Elastic golden rule.

S_abs ∝ N_ω δ(E_m − E_m′ + ħω); S_em ∝ (N_ω + 1) δ(E_m − E_m′ − ħω)Eq. 21.16

Inelastic golden rule.

Σ₀ⁱⁿ(E) = D(ħω) × Gⁿ(E + ħω)Eq. 21.17a

In-scattering from E + ħω (emission, ħω > 0).

Γ₀(E) = D(ħω) × [Gⁿ(E + ħω) + G^p(E − ħω)]Eq. 21.17b

G^p = A − Gⁿ.

D(ħω) = D(N_ω + 1) [ħω > 0]; D N_ω [ħω < 0]; D_ij = ⟨U_s(i)U_s(j)⟩Eqs. 21.18–21.19

Phonon-number factors.

U_i = U₀(N − N_i)Eq. 22.1

SCF with self-interaction correction.

p_i = e^{−(E_i − µN_i)/kT}/Z, n = Σ N_ip_iEq. 15.18

Fock-space equilibrium.

I = qν₂p₁, p₁ = ν₁/(ν₁ + ν₂)Eq. 22.2

One-level resistor in Fock space.

I = 2qν₁ν₂/(2ν₁ + ν₂)Sec. 22.2.2

High-bias plateau of the two-level channel; = 2/3 of the maximum for ν₁ = ν₂.

4Physics refresher: Bose factor, rate equations, Fock counting

Bose–Einstein

N_ω = 1/(e^ħω/kT − 1). At ħω = 30 meV: N = 0.456 at 300 K and 0.011 at 77 K. For ħω ≪ kT, N ≈ kT/ħω.

Master equation

dp_i/dt = Σ_j(W_ijp_j − W_jip_i). Steady state: the vector p with W p = 0 and Σp = 1; the current counts electrons crossing a contact per unit time.

Why Fock space has 2^N states

Each of N one-electron levels is empty or full: 2 choices each, so 2^N many-electron states; those with n electrons number C(N, n). For N = 4: 1, 4, 6, 4, 1.

Why (N + 1)/N = e^ħω/kT

(N + 1)/N = 1 + 1/N = 1 + (e^x − 1) = e^x with x = ħω/kT.

Singlet and triplet

For two spin-½ electrons the singlet (antisymmetric in spin, S = 0) lies below the three triplets (S = 1) when the orbital part is symmetric; they differ by the exchange energy ≈ 4t²/U₀ for a weakly coupled pair.

5Glossary

Term	Meaning
Golden rule	Transition rate (2π/ħ)\|τ\|²δ(E_i − E_f) (elastic) or with ± ħω (inelastic).
N_ω	Number of phonons of energy ħω: Bose–Einstein factor.
G^p	A − Gⁿ: hole density (matrix) per unit energy (×2π).
Charging energy U₀	q²/C: the energy to add one electron to a small conductor.
Coulomb blockade	Suppression of current when the next electron costs U₀ ≫ kT, γ.
Fock space	Space of many-electron occupation states 00, 01, 10, 11, …
Entangled state	A many-electron state that cannot be written as a product of one-electron states.
Singlet / triplet	Two-electron spin states with S = 0 / S = 1.

6Calculators

Phonon number and emission/absorption ratio

ħω (meV)T (K)

N_ω

(N+1)/N = e^ħω/kT

ħω/kT

Golden-rule lifetime (state + quasi-continuum)

coupling τ (meV)level spacing Δ (meV)

Γ = 2πτ²/Δ (meV)

lifetime ħ/Γ (fs)

Two-level Coulomb blockade plateau

ν₁/ν₂

plateau/maximum = (ν₁+ν₂)/(2ν₁+ν₂)

I/(qν₂) plateau = 2ν₁/(2ν₁+ν₂)

Charging energy U₀ = q²/C

C (aF)

U₀ (meV)

U₀/k_B (K)

7Self-check quiz

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

8Practice problems

ħ = 0.6582 meV·ps; kT = 25.85 meV at 300 K; k_B = 0.08617 meV/K; q²/h = 38.74 µS.

P1. Golden-rule lifetimeEASY

A state couples with τ = 3 meV to a bath of levels spaced by Δ = 1 meV. Find Γ and the lifetime.

Solution

Γ = 2πτ²/Δ = 2π × 9/1 = 56.5 meV; lifetime = ħ/Γ = 0.6582/56.5 ps = 11.6 fs.

P2. Phonon numberEASY

ħω = 30 meV, T = 300 K: N_ω and the emission/absorption ratio?

Solution

x = 30/25.85 = 1.16; N = 1/(e^1.16 − 1) = 0.456; ratio = e^1.16 = 3.19.

P3. Low temperatureEASY

Same phonon at 77 K.

Solution

kT = 6.64 meV, x = 4.52: N = 0.011; ratio = e^4.52 = 92. Absorption almost disappears.

P4. PlateauEASY

ν₁/ν₂ = 2. Plateau value relative to the maximum?

Solution

(ν₁+ν₂)/(2ν₁+ν₂) = 3/5 = 0.600.

P5. Charging energyMEDIUM

A dot has C = 10 aF. Find U₀ and the temperature below which blockade is visible (U₀ > 5kT).

Hint

U₀ = q²/C.

Solution

U₀ = 160.2 meV/10 = 16 meV; 5kT = U₀ ⇒ T = 16/(5 × 0.0862) = 37 K.

P6. CorrelationMEDIUM

In the plateau each spin level is occupied with probability 1/3. What would independent electrons give for double occupancy? What does Fock space give?

Solution

Independent: 1/3 × 1/3 = 1/9. Fock space: 0 (p₁₁ = 0): strong correlation.

P7. CountingMEDIUM

How many Fock states for 10 levels? How many with 3 electrons among 6 levels?

Solution

2¹⁰ = 1024; C(6,3) = 20.

P8. Derive the plateauMEDIUM

With the source adding electrons to 01 and 10 only, derive I.

Solution

2ν₁p₀₀ = ν₂(p₀₁ + p₁₀), p₀₀ + (p₀₁ + p₁₀) = 1 ⇒ p₀₁ + p₁₀ = 2ν₁/(2ν₁ + ν₂); I = qν₂(p₀₁ + p₁₀) = 2qν₁ν₂/(2ν₁ + ν₂).

P9. Double dotMEDIUM

At 100 K the double dot carries 27.4 nA forward and 7.5 nA reverse (coherent: 7.1 nA). Rectification ratio and enhancement?

Solution

Ratio 27.4/7.5 = 3.6; enhancement over coherent 27.4/7.1 = 3.8.

P10. Detailed balanceHARDER

Show that (N_ω + 1)/N_ω = e^ħω/kT and explain why equal emission and absorption rates cannot thermalise electrons.

Solution

1/N = e^x − 1, so (N+1)/N = 1 + 1/N = e^x. Thermal equilibrium requires p(k+1)/p(k) = absorption/emission = e^−x; equal rates would make p flat (diffusion) and the Fermi function could not be reached (notebook: ⟨k⟩ = 5.6 instead of 0.456).

P11. Singlet energyHARDER

For two coupled dots the lowest two-electron energy at t = 0.1, U₀ = 0.5 eV is −0.0702 eV. Compare with 4t²/U₀.

Hint

Superexchange: −4t²/U₀.

Solution

4t²/U₀ = 0.080 eV; the exact −0.0702 is smaller in magnitude because t/U₀ = 0.2 is not very small. At t = 0.2 eV the estimate −0.32 eV overshoots the exact −0.222 eV.

9Common mistakes

MISTAKE"Absorption and emission rates are equal."

CORRECTEmission ∝ N_ω + 1 exceeds absorption ∝ N_ω; the ratio is e^ħω/kT.

MISTAKE"Elastic dephasing exchanges energy with the lattice."

CORRECTElastic dephasing (Σ₀ = D × G^R) randomises phase and momentum but leaves each electron energy unchanged.

MISTAKE"SCF with U = U₀N shows Coulomb blockade."

CORRECTIt stretches the step; the plateau needs self-interaction correction, and the correct 2/3 needs Fock space.

MISTAKE"Coulomb blockade is seen in any small conductor."

CORRECTNeeds U₀ ≫ kT and U₀ ≫ ħ/t; with good contacts the broadening washes it out.

MISTAKE"Fock-space probabilities are one-electron occupations."

CORRECTThe joint probability of both levels full is not the product of single-level probabilities.

MISTAKE"NEGF fails whenever there are interactions."

CORRECTNEGF equations (18.1–18.4) hold; the difficulty lies in evaluating Σ and Σⁱⁿ.

10Explore

Files are in the course folder (course_materials).

Week 15 slides Widget: emission versus absorption Widget: Coulomb blockade in Fock space Colab notebook

Mini-investigations (15 min each)

In the phonon widget find the temperature at which the ratio emission/absorption drops below 2 for ħω = 30 meV. Check with the calculator.
In the Coulomb widget, increase kT until the 2/3 plateau disappears. How does kT compare with U₀?
In the notebook, vary D₀ in the double-dot model and plot forward and reverse currents. Where does the forward current peak, and why does it decrease for strong scattering?

12Looking ahead: Week 16 (final)

Spin, quantum to classical, probabilistic logic (Datta Part B, Ch. 23–25) and circular session C8.

How does a spin-½ rotate? (Pauli matrices, Bloch sphere)
When does a quantum system behave classically?
What is a p-bit?

Beyond elastic, non-interacting transport

1This week's plan

By the end of the week you should be able to

2Key ideas in five sentences

3Equation sheet

4Physics refresher: Bose factor, rate equations, Fock counting

5Glossary

6Calculators

Phonon number and emission/absorption ratio

Golden-rule lifetime (state + quasi-continuum)

Two-level Coulomb blockade plateau

Charging energy U₀ = q²/C

7Self-check quiz

8Practice problems

9Common mistakes

10Explore

Mini-investigations (15 min each)

11Can I…?

12Looking ahead: Week 16 (final)