University of Kurdistan logoNanoelectronics: A New Perspective on TransportWeek 16 · Spin, quantum to classical, p-bits · C8
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WEEK 16 · FINAL · SPIN, QUANTUM TO CLASSICAL, p-BITS · C8

Nanoelectronics:
A New Perspective on Transport

Rotating an electron, from quantum to classical, and probabilistic spin logic; final circular session and presentations
Prof. Mohammad RazaghiUniversity of Kurdistan · Graduate course in Electronics Engineering
Based on S. Datta, Lessons from Nanoelectronics, Part B: Quantum Transport, Chapters 23–25
Part 1 is the fixed lecture (Ch. 23–25). Part 2 is circular session C8: the final lab and the preparation of the final presentations.

Week 16Where we are

Weeks 1–15

  • Part A: the elastic resistor, modes, Landauer
  • Part B: matrices, NEGF, broadening
  • Scattering, inelastic processes, Coulomb blockade
  • Spin so far: only up and down (Week 9)

Today: the last step

Let the spin point anywhere: every number becomes a 2×2 matrix. Then ask when a quantum object looks classical, and end with a new kind of logic: the p-bit, halfway between a bit and a q-bit.

The same NEGF equations hold. The only change is that each site carries a two-component spinor, so alpha, beta, Sigma, Gamma, G, Gn double in size.

Week 16Today's goals

  1. Move between a spinor and the arrow n̂ using the Pauli matrices.
  2. Write Γ for a contact pointing in any direction; get the spin-valve MR and the probe signal.
  3. Explain precession, Zeeman and Rashba, and spin–momentum locking.
  4. Read N and S from Gⁿ; relate pseudospin and entropy.
  5. Build a p-bit and a p-circuit AND gate that runs backwards.

Datta Sec. 23.1, 23.6A spin is an arrow, but it is stored as a spinor

ψ = ( cos(θ/2) e−iφ/2, sin(θ/2) e+iφ/2 )Eqs. 23.6–23.7
nk = ψ† σk ψ,   σx = [[0,1],[1,0]], σy = [[0,−i],[i,0]], σz = [[1,0],[0,−1]]Eq. 23.20
  • Two complex numbers (4 reals, minus norm and global phase) = two angles.
  • Half angles: a 360° rotation gives ψ → −ψ; only 720° restores it.
Pauli algebra: sigma_k^2 = I, sigma_x sigma_y = i sigma_z (cyclic), traces zero. Notebook Part 1 checks all of it.

Datta Sec. 23.4A magnet that points anywhere

Γ = (α+β)/2 [ I + P n̂·σ ]Eq. 23.14 · P = (α−β)/(α+β)
  • In its own axis the contact is diag(α, β). Transform with V whose columns are spinors along ±n̂: Γ = V diag(α, β) V† (23.16).
  • Special cases: θ = 0 gives diag(α, β); θ = 180° gives diag(β, α) (the antiparallel valve).
  • Same form for Σ = −i(α+β)/4 (I + σ·P).
Notebook: max |V diag V^dagger - Eq 23.14| = 6e-17.

Datta Sec. 23.2–23.3The probe: electrons versus photons

µP = µ + P µS cos θEq. 23.5 · reversing the magnet changes µP by 2P µS (23.4)
  • µ = (µup + µdn)/2, µS = (µup − µdn)/2 (23.1).
  • Polarizer + analyzer: electrons (1 + cos θ)/2 = cos²(θ/2); photons cos² θ.
  • Spin-½ rotates by θ/2 when the arrow rotates by θ.

Datta Sec. 23.3The spin valve

  • One level ε between two magnets: T̄ = Trace[Γ₁ GR Γ₂ GA] (23.10).
  • Parallel: Γ₂ = diag(α, β). Antiparallel: α ↔ β.
  • Far from resonance: GP ∝ α² + β², GAP ∝ 2αβ.
MR = GP/GAP − 1 = 2P²/(1 − P²)Eq. 23.9 (MTJ); resistor model: P²/(1 − P²)
T(theta) is linear in cos theta: (GP+GAP)/2 + (GP-GAP)/2 cos theta. Notebook: deviation 2e-5.

InteractiveExplore: Bloch sphere, probe and precession

Rotating an electron: Bloch sphere, probe and precession  (Datta Part B, Ch. 23)

A spin-½ is a rotatable arrow. A probe magnet with polarization P pointing along ˆm measures µP = µ + P(m·S)µS. Mode 1 rotates the probe (electron versus photon). Mode 2 lets the spin precess in a field along z.

probe signal (µP − µ)/µS–
Γ of the probe (α=1, β=(1−P)/(1+P))–
MTJ magnetoresistance 2P²/(1−P²)–
spin vector n–

Try this

  1. Mode 1, P = 1: read the signal at θ = 90°. Compare with the photon curve.
  2. Reduce P to 0.3: how does the probe signal and the valve MR change?
  3. Mode 2: set θ = 90° and increase ω. How many oscillations fit in the window?
  4. Turn on spin relaxation: what happens to the arrow and to the amplitude?

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

Mode 2 uses a probe in the x-y plane; signal P cos(omega t - theta_p) exp(-Gamma t).

Datta Sec. 23.6–23.7Spin precession

ψu,d(t) = ψu,d(0) e∓iωt/2,   ω = 2µelB/ħEqs. 23.29–23.30
  • The relative phase of the spinor components rotates the arrow: φ(t) = φ(0) + ωt.
  • dn/dt = ω ẑ × n (23.33): rotation matrix for an x-directed field is obtained by x → y → z.
  • B = 1 T, g = 2: f = 28 GHz. A probe sees P cos(ωt − θp).
Notebook Part 4 verifies n_x = cos(wt), n_y = sin(wt), n_z = 0 to 1e-16.

Datta Sec. 23.5Spin Hamiltonians: Zeeman and Rashba

Zeeman

HB = µel σ·B

Lattice: α = εI + µelσ·B, βx,y = tI (23.25). Only the on-site term changes.

Rashba

HR = η ẑ·(σ × k) = η(σxky − σykx)

On the lattice k → sin(ka)/a; βx = tI + (iη/2a)σy, βy = tI − (iη/2a)σx (23.28).

Rashba is believed to be the relativistic effect where the atomic-scale electric field is perceived as an effective magnetic field. Splitting 2 eta |sin ka|/a verified in the notebook.

Datta Sec. 23.8–23.9Spin–momentum locking

  • For each k the eigenspinor has its spin in the plane, perpendicular to k.
  • The two Rashba bands have opposite helicity; topological insulators have a single band with this locking.
  • Net current = net k = net spin: spin–charge coupling (Sec. 23.8), robust even at room temperature.
  • Superconducting contacts (Sec. 23.9): Nambu doubling of H, Andreev reflection.
Notebook example: k = (0.5, 0.3): spin (-0.525, 0.851, 0) for the lower band, spin.k = -0.007 (sin vs k).

Datta Ch. 24Chapter 24: from quantum to classical

Gⁿ/2π = ½ (N I + σ·S)Eq. 24.1
N = Trace[Gⁿ]/2π,   S = Trace[σ Gⁿ]/2πEq. 24.2 · σ traceless
  • If only Sz matters: N ± Sz = number of up/down electrons: two-component spin diffusion (Valet–Fert).
  • Four-component diffusion keeps Sx,y: Model C dephasing (24.8).
The elements of Gn at a point tell us the number of electrons N and the net spin vector S.

Datta Sec. 24.4Pseudospin: spins appear everywhere

  • Any two-component wavefunction can be viewed as a spinor: right/left movers e±ikz are pseudo-up and pseudo-down.
  • Left of a barrier: N = (1 + r²)/2, Sz = (1 − r²)/2, Sx = r cos 2kz, Sy = −r sin 2kz.
  • The probe Γ = [[1,1],[1,1]] is a pseudomagnet along x: it measures N + Sx, the standing-wave pattern of Fig. 19.7.
  • Graphene sublattices and valleys are other pseudospins.
The notebook Part 6 uses the unnormalised psi psi^dagger so N = 1 + r^2 (twice Datta convention).

Datta Sec. 24.5How much information can one spin carry?

S/k = −Σ pᵢ ln pᵢ = −Trace[ρ ln ρ]Eq. 24.11 · ρ = ½(I + σ·P)
  • A known pure spin (P = 1): S = 0. A random spin: S = k ln 2 = one bit.
  • P = 0.5: 0.562 nats = 0.811 bit.
  • Interactions with other spins can transfer entropy but the total of a closed system cannot decrease (Chapter 16).

Datta Ch. 25Chapter 25: spins, magnets and p-bits

  • Each electron is a magnet: µB = qħ/2m = 9.27×10⁻²⁴ A m² (25.1).
  • A stable magnet is a bit; a spin is a q-bit; a low-barrier (unstable) magnet fluctuates: a p-bit.
  • Three-terminal p-bit listens to an input ν:
Vi = sgn[ tanh(νi/V₀) + rand(−1,+1) ],   νi = Σj Jij VjEqs. 25.6a, 25.6b
p-bits can be built with a low-barrier MTJ and an ordinary transistor (Camsari 2017b).

Datta Sec. 25.4p-circuits: an AND gate that runs backwards

  • Choose JAB = −1, JAC = JBC = 2, h = (1, 1, −2): the four valid AND rows have the lowest energy.
  • Sequential updating samples p ∝ exp[I₀(½ mᵀJm + h·m)]: valid rows 98.6 % of the time at I₀ = 1.
  • Clamp C = 1: the circuit returns A = B = 1 (99.5 %). Clamp C = 0: A, B ∈ {00, 01, 10}, each ≈ 1/3.
  • The same weights perform inference, not just a function.
Numbers from the notebook: sampled 0.995/0.995 for clamp C=1, exact. At I0=0.25 valid fraction 0.776; at 2 almost 1.0 but slow mixing.

InteractiveExplore: a probabilistic AND gate

p-bits: a probabilistic AND gate that also runs backwards  (Datta Part B, Ch. 25, Eqs. 25.6)

Each node is a three-terminal p-bit: Vi = sgn[ tanh(I0νi) + rand(−1,+1) ] with input νi = Σj JijVj + hi. With JAB=−1, JAC=JBC=2 and h=(1,1,−2) the network spends its time in the four rows of the AND truth table (C = A AND B). Clamp C = 1 and the inputs are forced to A = B = 1: inversion.

samples–
time in a valid AND row–
exact (Boltzmann) valid fraction–
current state (A B C)–

Try this

  1. Leave everything free: are the four valid rows equally visited?
  2. Clamp C = 1: what do A and B do? Then clamp C = 0.
  3. Reduce I₀ to 0.25: how much time is spent in invalid rows (exact value 22 %)?
  4. Raise I₀ to 3 with C free: does the sampler still visit all four valid rows?

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

Counts are compared with the exact Boltzmann distribution (red ticks).

CheckConcept check

One minute with a neighbour, then click to reveal.

1. Why is the electron analyzer law (1 + P cos θ)/2? (click)
The arrow projects as cos θ onto the probe axis; P cos θ µS is added to µ. For P = 1 it equals cos²(θ/2), a half angle because spinors rotate by θ/2.
2. P = 0.6: MR of the MTJ spin valve? (click)
2P²/(1 − P²) = 2(0.36)/0.64 = 1.125 (NEGF: 1.124).
3. What does a Rashba channel do to a current? (click)
Spin–momentum locking: spin ⟂ k, so a net k means a net spin.
4. Entropy of a spin with P = 0.5? (click)
0.562 nats = 0.811 bits; ln 2 for P = 0, 0 for P = 1.
5. Why can a p-circuit run backwards? (click)
It samples a joint distribution; clamping any p-bits samples the others conditionally.

Wrap-upSummary of the lecture

Take-home

  • Spinor ↔ arrow: n = ψ†σψ
  • Γ = (α+β)/2 (I + σ·P)
  • MR = 2P²/(1−P²); signal (1 + P cos θ)/2
  • ω = 2µB/ħ; Rashba locks spin ⟂ k
  • Gⁿ = ½(N I + σ·S); S = −Σ p ln p
  • p-bit: tanh + random; p-circuits infer

Next: circular session C8

The final lab and the final presentations. Open week16_spin_pbit_lab_C8.ipynb and the Week 16 toolkit. Your own calculation will be part of your presentation.

PART 2

Circular session C8

Spin lab, p-bit lab and final presentation planning

A · 30 min
Spinors and Γ(n̂)
B · 40 min
Valve, probe, precession
C · 40 min
Rashba, Gⁿ, entropy
D · 70 min
p-bits; presentation plan
Groups of 3-4. At the end each group hands in a one-page plan for its final presentation.

Circular session C8Block A · Spinors and Γ(n̂)

A1. Verify the Pauli algebra numerically, and show that a 360° rotation gives ψ → −ψ.

A2. Check Eq. 23.14 against V diag(α,β) V† for three random directions.

A3. For P = 0.6, plot Γ₁₁, Γ₂₂ and Γ₁₂ versus θ. At which θ is Γ₁₂ largest?

Expected: max error ~10⁻¹⁶; Γ₁₂ largest at θ = 90°.

Circular session C8Block B · Valve, probe, precession

B1. NEGF spin valve: compute GP, GAP for P = 0.6: MR = 1.124 versus 2P²/(1−P²) = 1.125.

B2. Show G(θ) is linear in cos θ (deviation 2×10⁻⁵).

B3. Integrate the Schrödinger equation of the spinor and compare with the arrow rotation; probe signal for θp = 60°.

B4. Compute f(B = 0.1 T) = 2.80 GHz.

Circular session C8Block C · Rashba, Gⁿ and entropy

C1. At k = (0.5, 0.3) find the two Rashba energies and spins. Check splitting = 2η|sin ka|/a = 0.2253 (η = 0.2 t).

C2. Plot the spin texture on a ring: arrows ⟂ k, opposite for the two bands.

C3. Build Gⁿ from 1000 random spins; recover N and S with Eq. 24.2.

C4. S/k for P = 0, 0.5, 0.9, 1: 0.693, 0.562, 0.199, 0.

Circular session C8Block D · p-bits and the AND gate

SettingResult
Free, I₀ = 14 valid rows ≈ 0.25 each; 98.6 % valid
Clamp C = 1A = B = 1: 99.5 %
Clamp C = 000, 01, 10 ≈ 1/3 each
I₀ = 0.25valid fraction 0.776
Family tree⟨F₁F₂⟩ = 0.5, ⟨C₁C₂⟩ = 0.125

D3 (design). Search for J and h of an OR gate.

Numbers from the notebook Part 7.

Final presentationsFinal presentation: structure

Format (15 min)

  1. The physical question (1 min)
  2. The model: which equation, which approximation (3 min)
  3. Your own calculation from the notebooks (4 min)
  4. What the model cannot do (1 min)
  5. Questions (5 min)

Assessment

  • Physics correctness: 40 %
  • Clarity of the visual: 25 %
  • Own calculation: 25 %
  • Answers to questions: 10 %
The rubric is repeated in the toolkit with a self-check list.

Final presentationsFinal presentation: choosing a topic

Part A topics

  • Elastic resistor and where the heat goes
  • Number of modes and the quantum of conductance
  • Nanotransistor: quantum capacitance, 60 mV/dec
  • Thermoelectrics: Seebeck, ZT
  • Information battery and Landauer erasure

Part B topics

  • NEGF for a molecule: broadening, Gⁿ
  • Two scatterers: coherent versus incoherent
  • Phonon-assisted current, double dot rectifier
  • Coulomb blockade: the 2/3 plateau
  • Spin valve / Rashba / p-bits
Each student states the topic and the numerical experiment in the plan handed in at the end of C8.

ClosingThe whole course on one page

WeeksThemeKey result
1–4Elastic resistor, Fermi functions, transit timeG = (q²/h) ∫ T M (−∂f/∂E) dE
5–8Modes, nanotransistor, Boltzmann, Landauer, HallG = (q²/h)M T, 60 mV/dec
9–11Spin valve, thermoelectrics, Fock space, informationZT, Landauer erasure kT ln 2
12–14Schrödinger matrix, NEGF, scatterers, modesGⁿ = GRΣinGA, T = Trace[Γ₁GRΓ₂GA]
15Inelastic scattering, Coulomb blockadeemission ∝ N+1, plateau 2/3
16Spin, quantum to classical, p-bitsΓ = (α+β)/2 (I + σ·P), p-circuits

One idea throughout: a channel connected to contacts, described by a few numbers (Γ, Σ, M, T) and a Fermi function.

ClosingThank you

Where next?

  • Datta, Quantum Transport: Atom to Transistor
  • nanoHUB-U: Fundamentals of Nanoelectronics, Parts A and B
  • Camsari, Datta et al., probabilistic computing with p-bits

Last tasks

  • Hand in the one-page presentation plan
  • Run all notebooks once on Colab
  • Prepare questions for the final presentations
Prof. Mohammad Razaghi, University of Kurdistan.
Prof. Mohammad Razaghi · University of Kurdistan · after S. Datta, Lessons from Nanoelectronics