University of Kurdistan logoNanoelectronics: A New Perspective on TransportWeek 15 · Inelastic scattering and Coulomb blockade
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WEEK 15 · INELASTIC SCATTERING AND COULOMB BLOCKADE

Nanoelectronics:
A New Perspective on Transport

Beyond elastic, non-interacting transport: golden rule, phonon emission and absorption, Fock space
Prof. Mohammad RazaghiUniversity of Kurdistan · Graduate course in Electronics Engineering
Based on S. Datta, Lessons from Nanoelectronics, Part B: Quantum Transport, Chapters 21–22
Fixed lecture only. Week 16 (final) has circular session C8. The two themes: (1) electrons exchange energy with the surroundings, (2) electrons interact with each other and a one-electron picture fails.

Week 15Where we are

So far (Weeks 13–14)

  • NEGF: GR, Gⁿ, A, Ĩ
  • Contacts 1 and 2: Σ₁, Σ₂
  • Interactions: only elastic dephasing (Σ₀ = D × GR)
  • Electrons independent, mean-field U

Today

Two ways to go beyond: let electrons exchange energy with phonons (Ch. 21), and let electrons repel each other so strongly that a one-electron picture fails (Ch. 22).

Week 15Today's goals

  1. Obtain the golden rule from Γ = τ†aτ.
  2. Explain why emission ∝ Nω + 1 beats absorption ∝ Nω.
  3. Write the inelastic Σ₀in, Γ₀ (Eq. 21.17) and see it obey the second law.
  4. See Coulomb blockade fail in a self-consistent-field picture.
  5. Use Fock space: n(µ), the 2/3 plateau, entangled states.

Datta Sec. 21.0–21.1Scattering = an abstract contact

Σ = τ† g τ,   Γ = τ† a τEqs. 21.3, 21.5 · a = i[g − g†]
  • The channel H exchanges electrons with contacts 1, 2 and with the surroundings (contact 0): Σ₀, Σ₀in.
  • Elastic dephasing (Week 13) = no energy exchange, but momentum and phase are lost.
  • Inelastic = the electron changes energy. Emission is easier than absorption (Part A, Ch. 15).

Big idea

A scattering problem is just a channel–contact partitioning problem: the initial state is the channel, all the other states are the contact.

Datta Sec. 21.1.1The golden rule from broadening

ħ/Lifetime(m) = Γmm(Em)
= 2π Σm′ |τm′m|² δ(Em − Em′)Eq. 21.12 · a = 2πδ(E − Em′) for eigenstates
  • The inverse lifetime = the sum of golden-rule rates S(m′←m) = (2π/ħ)|τ|²δ(Em − Em′) (21.14).
  • Numerical test: one state + 2001 bath levels (Δ = 1 meV, τ = 3 meV): Γ = 56.5 meV, lifetime 11.6 fs; the exact dynamics gives 11.4 fs.
Matrix elements: tau(m',m) = integral phi*(m') Us phi(m) dr (21.15), with the exact eigenfunctions of the channel.

Datta Sec. 21.1.2Energy conservation: elastic and inelastic

  • Static potential Us(r): the delta function conserves energy; elastic.
  • Time-dependent potential 2Us cos ωt: the electron leaves with energy Ei ± ħω: absorption or emission.
  • The classical argument says the two rates are equal. They are not:
absorption ∝ Nω,   emission ∝ Nω + 1Eq. 21.16

Datta Sec. 21.1.2Why N + 1 and N: the composite picture

  • Treat electron + phonon system as one: states (m, Nω).
  • Absorption (m, N) → (m′, N − 1): rate ∝ N.
  • Emission (m′, N) → (m, N + 1): rate ∝ N + 1.
  • Ratio (N+1)/N = eħω/kT: detailed balance.
  • 300 K, ħω = 30 meV: N = 0.456, ratio 3.2; 77 K: ratio 92.
Elastic scatterers are usually rigid with no internal degrees of freedom. A spin bath out of equilibrium behaves like Maxwell's demon (Part A Ch. 16): not considered here.

InteractiveExplore: emission is easier than absorption

Emission is easier than absorption  (Datta Part B, Ch. 21, Eqs. 21.16–21.18)

A scatterer that exchanges one quantum ℏω with the electron has rates ∝ Nω for absorption and ∝ Nω+1 for emission. Left: the rate arrows from one level. Middle: Nω(T) and the ratio (Nω+1)/Nω = eℏω/kT. Right: a ladder of levels spaced by ℏω, started in a non-equilibrium distribution, relaxes to the Boltzmann distribution only because the two rates differ.

Nω (Bose–Einstein)–
emission / absorption–
ℏω / kT–
mean level ⟨k⟩ (now / equilibrium)–

Try this

  1. Keep "quantum" rates: start the ladder in the hot state. Does it relax to the red ticks?
  2. Switch to "classical" equal rates: where does the distribution end up?
  3. Lower T to 50 K: what happens to the ratio (N+1)/N and to the time to relax?
  4. Raise ħω: does a larger quantum make the asymmetry larger or smaller?

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

Time unit s0 = 1/(rate scale). The final distribution with quantum rates is exactly p(k) proportional to exp(-k hbar omega / kT), on a ladder truncated at 12 levels.

Datta Sec. 21.2Self-energy functions for inelastic scattering

Σ₀in(E) = D(ħω) × Gⁿ(E + ħω)Eq. 21.17a · ħω > 0 emission
Γ₀(E) = D(ħω) × [Gⁿ(E+ħω) + Gp(E−ħω)]Eq. 21.17b · Gp = A − Gⁿ
  • D(ħω) = D(N + 1) for emission, D·N for absorption (21.18); Dij = ⟨Us(i)Us(j)⟩ (21.19).
  • Reduces to elastic dephasing (18.35) when ħω → 0.
The equations couple energies E-hw, E, E+hw: solve on a grid with dE = hw/m. The notebook implements it with one phonon mode.

Datta Sec. 21.2Does it obey the second law?

Test (notebook)Result
Equilibrium: max |Gⁿ/A − f(E)|5 × 10⁻⁸
Net current into phonon "contact"≈ 10⁻¹⁹
Equal weights instead of N, N+1fails: no Fermi function
  • With emission ∝ N+1 and absorption ∝ N the device in equilibrium stays in equilibrium.
  • Many devices are forgiving: levels equally connected to both contacts do not test the physics.
  • A good test: current driven entirely by inelastic processes (Fig. 12.6a).

Datta Sec. 21.2A phonon-driven current: the double dot

  • Levels ε₁ = 50 meV, ε₂ = 0; t = 5 meV; γ = 10 meV; ħω = 50 meV.
  • Forward bias, 100 K, D₀ = 4×10⁻⁴ eV²: 27.4 nA against 7.1 nA coherent (×3.8).
  • Reverse bias needs absorption (N = 0.003 at 100 K for 50 meV): 7.5 nA, so the device rectifies (ratio 3.6).
  • At 300 K the reverse current rises (11.8 nA): N grows.
Numbers from the notebook. Stronger scattering (D0 = 1.6e-3) reduces the forward current again: the electron is scattered back and forth.

Datta Chapter 22Chapter 22: does NEGF include "everything"?

MBPT and the (1 − x)⁻¹ series

Many-body perturbation theory evaluates (1 − x)⁻¹ as 1 + x + x² + …: fine for x ≪ 1, useless for x > 1. NEGF equations (18.1–18.4) and the self-energies Σ, Σin are separable: the equations stay, the approximation of Σ may change.

A failure case

Two levels (spin up/down) at the same ε, µ at ε, strong charging energy U₀. The SCF picture predicts high conductance; the real device shows Coulomb blockade.

Datta Sec. 22.1Coulomb blockade: the one-electron picture

Ui = U₀(N − Ni)Eq. 22.1 · self-interaction correction
  • If the down level is full, the up level floats up by U₀ and empties; a down level feels no potential from itself.
  • Self-consistent solution: Ndn = 1, Nup = 0 (or vice versa: random switching).
  • Observable only if U₀ ≫ kT and U₀ ≫ broadening ħ/t: for molecules U₀ ~ 100s of meV.

I–V curves (Fig. 22.3–22.4)

  • U = 0: one step to Imax
  • U = U₀N: step stretched over U₀/q
  • Ui = U₀(N − Ni): plateau at 1/2
  • Experiment / exact: plateau at 2/3

Datta Sec. 22.2Fock space: many-electron states

  • N one-electron levels give 2N Fock states: for two levels 00, 01, 10, 11.
  • Energies 0, ε, ε, 2ε + U₀ (the interaction sits in E11).
  • Equilibrium: pi ∝ e−(Ei − µNi)/kT (Eq. 15.18).
  • n(µ) = Σ Nipi: a step at µ = ε and another at ε + U₀; G ∝ dn/dµ shows two peaks.
Week 11: Fock space and the probability p_i = e^{-(E_i - mu N_i)/kT}/Z. The middle plateau of n(mu) has n = 1 exactly (1.0000 in the notebook for eps = 10 kT, U0 = 20 kT).

Datta Sec. 22.2.2The 2/3 plateau

2ν₁p₀₀ = ν₂(p₀₁ + p₁₀)source level between ε and ε + U₀; drain empty
I = 2qν₁ν₂/(2ν₁ + ν₂)plateau; for ν₁ = ν₂: 2/3 of 2qν₁ν₂/(ν₁+ν₂)
  • General plateau (ν₁+ν₂)/(2ν₁+ν₂): ν₁/ν₂ = 2 gives 0.600, 4 gives 0.556.
  • p₀₀ = p₀₁ = p₁₀ = 1/3, p₁₁ = 0: strongly correlated.
The same current formula follows from the master equation solved in the notebook (fock_current): 0.6667, 0.6000.

InteractiveExplore: Coulomb blockade in Fock space

Coulomb blockade in Fock space  (Datta Part B, Ch. 22, Figs. 22.1–22.9)

A two-level channel (spin up and down, both at ε = 0) with charging energy U0. Left: the four Fock states 00, 01, 10, 11 and their steady-state probabilities for the chosen source level μ1 (drain empty). Middle: the equilibrium n(μ), for one dot or for two coupled dots with entangled states. Right: current versus μ1 for three models; the correct plateau is at 2/3, not 1/2.

P(00)   P(01)+P(10)   P(11)–
I / Imax, Fock-space (exact)–
I / Imax, SCF U = U0N–
I / Imax, non-interacting–

Try this

  1. Set µ₁ − ε = 0.25 eV (between ε and ε + U₀): read P(00), P(01)+P(10), P(11). What is the plateau value?
  2. Change ν₁/ν₂ to 2: does the plateau move to 0.60?
  3. Raise kT to 60 meV: what happens to the steps in n(µ) and to the plateau?
  4. Switch the middle panel to two coupled dots, t = 0.1 eV: where is the n = 2 plateau and how wide is it?

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

Drain empty: electrons only leave to the drain. SCF curve: U = U0 N solved by bisection; non-interacting: levels at eps.

Datta Sec. 22.2Why the 2/3 plateau needs correlations

  • Look at the up level alone: occupied 1/3 of the time. Same for down.
  • Independent electrons: both occupied with probability 1/3 × 1/3 = 1/9.
  • Fock space: both occupied with probability 0.
  • No one-electron occupation factor f, nor a matrix Gⁿ (a sum of ψψ†), can describe this: Gⁿ says nothing about two-electron correlations.

Moral for NEGF

NEGF + SCF is excellent for weak interactions. For strong correlations one needs Σ and Σin chosen with care, or a Fock-space / quantum rate-equation approach.

Datta Sec. 22.3Entangled states: two coupled dots

  • Four one-electron states (u₁, d₁, u₂, d₂) give 2⁴ = 16 Fock states: 1, 4, 6, 4, 1 with 0–4 electrons.
  • The 6 two-electron states are mixed by t (Eq. 22.4): lowest = singlet, then three triplets.
  • For t = 0.1, U₀ = 0.5 eV: singlet −0.0702 eV, triplets 0, 0, 0, then 0.5 and 0.570 eV.
  • Entanglement is the key entity of quantum information.
The singlet is (u1 d2 + u2 d1) + s1 u1 d1 + s2 u2 d2 (22.5); the notebook prints its coefficients: -0.667, +0.667 on the mixed terms and -0.234 on the double-occupied terms. Equilibrium works as before (energies from the Fock Hamiltonian). High bias needs quantum rate equations (NDR, Muralidharan 2007).

CheckConcept check

One minute with a neighbour, then click to reveal.

1. Where does the golden rule come from in NEGF? (click)
Γ = τ†aτ with a = 2πδ(E − Em′) gives ħ/lifetime = Σ S(m′ ← m).
2. 300 K, ħω = 30 meV: Nω and emission/absorption? (click)
N = 0.456; ratio = eħω/kT = 3.19.
3. Why does the double dot rectify? (click)
Forward: emission ∝ N+1; reverse: absorption ∝ N, tiny at low T.
4. ν₁ = ν₂: plateau of the two-level Coulomb blockade? (click)
2/3 of the maximum (p₀₀ = p₀₁ = p₁₀ = 1/3, p₁₁ = 0).

Wrap-upSummary and what comes next

Take-home

  • Γ = τ†aτ ⇒ golden rule
  • Emission ∝ N+1, absorption ∝ N
  • Σ₀in = D × Gⁿ(E+ħω), Γ₀ = D × [Gⁿ + Gp]
  • Coulomb blockade: Fock space, 2/3 plateau
  • Entangled singlet/triplets

Week 16 (final)

Part B, Ch. 23–25: rotating an electron (spin, Bloch sphere), quantum to classical, and probabilistic spin logic. Circular session C8 and the final presentations.

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