University of Kurdistan logoNanoelectronics: A New Perspective on TransportWeek 11 · The second law and information
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WEEK 11 · THE SECOND LAW AND INFORMATION

Nanoelectronics:
A New Perspective on Transport

Entropy, the law of equilibrium, Fock space, the info-battery and Landauer's principle
Prof. Mohammad RazaghiUniversity of Kurdistan · Graduate course in Electronics Engineering
Based on S. Datta, Lessons from Nanoelectronics, Part A: Basic Concepts, Chapters 15–16
The last chapters of Part A: where the Fermi function, the Bose function and the e^(−ħω/kT) of Weeks 9–10 come from. Circular session C6 next week covers Weeks 11–12.

RecapWhere we are

We have used, without proof

  • f(E) = 1/(e(E−µ)/kT + 1) (Week 2) and n(ω) = 1/(eħω/kT − 1) (Week 10)
  • DAB/DBA = e−ħω/kT₀: absorbing is harder than emitting (Week 9)
  • "Heat goes into the contacts" (Week 1)

Today's questions

Why does everything go downhill if mechanics is reversible? Where do these distributions come from? And can information be used as fuel?

Week 11Today's goals

  1. State the second law for a channel with several reservoirs, and check the elastic resistor obeys it.
  2. Explain the asymmetry of absorption and emission from the reservoir's number of states.
  3. Use the law of equilibrium on Fock-space states: Fermi, Bose, interacting electrons.
  4. Derive S = −Nk Σ p ln p and the minimum of free energy.
  5. Explain the info-battery, NkT ln 2, Landauer's principle and Maxwell's demon.

Physics refresherRefresher: entropy is counting

  • A macrostate realised in W ways: S = k ln W (Eq. 15.11).
  • Independent systems: W multiplies, S adds.
  • Temperature measures how fast counting grows: 1/T = dS/dE (Eq. 15.13).
  • Equilibrium = the overwhelmingly most numerous macrostate.
N spins: the "half up" peak gets sharper as √N/N; at N = 10²³ nothing else ever happens.

Datta Sec. 15.1The second law for a channel

(E₁ − µ₁N₁)/T₁ + (E₂ − µ₂N₂)/T₂ + E₀/T₀ ≤ 0Eq. 15.3 · with N₁ + N₂ = 0, E₁ + E₂ + E₀ = 0
  • One temperature ⇒ µ₁N₁ + µ₂N₂ ≥ 0 (15.5): battery → heat, never back.
  • Elastic resistor: [(ε − µ₁)/T₁ − (ε − µ₂)/T₂]N₁ ≤ 0, always, because N₁ ∝ f₁ − f₂.
  • Different temperatures: heat engines (Weeks 9–10) are allowed.

Datta Sec. 15.2–15.3Absorbing is harder than emitting

P(+E, +N)/P(−E, −N) = e−(E − µN)/kTEq. 15.6 · any reservoir at (T, µ)
P(−ε)/P(+ε) = W(E₀ + ε)/W(E₀) = eε/kTEqs. 15.12–15.13
  • Emitting gives the reservoir energy, and eε/kT times more states.
  • Inelastic rates: D₂←₁/D₁←₂ = e(E₁−E₂)/kT₀ (15.9); I ∝ f₁ − f₂ only if elastic (15.10).
Any transport theory, semiclassical or quantum, must build in Eq. 15.9 or it can violate the second law.

Datta Sec. 15.3Why does an excited atom decay?

  • "Electrons go to the lowest energy": but mechanics is reversible.
  • The same coupling takes 2p → 1s and 1s → 2p.
  • The difference is the surroundings: at T → 0 they can only absorb.
  • An entropic, not mechanical, force. Transport is hard because it mixes both.

InteractiveExplore: a reservoir that counts states

Why systems go downhill: a reservoir that counts states  (Datta Ch. 15, Fig. 15.2)

A small system shares energy quanta ε with a reservoir of NR oscillators (an "Einstein solid"). Every move simply passes one quantum between two randomly chosen members: no force pushes the system down. Yet the system ends up mostly in its low states, because giving a quantum to the reservoir multiplies the reservoir's number of states W (Eq. 15.12). The histogram converges to the Boltzmann law pn ∝ e−En/kT with kT fixed by dS/dE of the reservoir (Eq. 15.13).

Reservoir temperature kT/ε = 1/ln(1 + NR/Q)–
Measured p(0)/p(1) · theory eε/kT–
Mean system energy ⟨n⟩ · theory–
Moves made–

Fock space: one law of equilibrium, three famous answers (Sec. 15.5)

Try this

  1. Restart: all energy starts in the system. Does it stay there?
  2. Compare p(0)/p(1) with eε/kT, kT/ε = 1/ln(1 + NR/Q).
  3. Is ⟨n⟩ the Bose function? Switch to a two-level system.
  4. Lower panel: raise U₀. Why does the plateau sit at n = 1?

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

The swap rule is symmetric; the asymmetry is entirely counting.

Datta Sec. 15.3, 15.6Entropy increases, free energy decreases

ΔS₁ + ΔS₂ + ΔS₀ ≥ 0Eq. 15.14 · ΔS = −(E − µN)/T for each reservoir
ΔF = ΔE − TΔS ≤ 0Eqs. 15.15–15.17 · system with one reservoir
F = Σ p̃i(Ei + kT ln p̃i), minimum at p̃i = e−Ei/kT/ZEqs. 15.26–15.27

Any other distribution has extra free energy ΔF = ΔE − TΔS (15.28): the fuel of Chapter 16.

Datta Sec. 15.4–15.5The law of equilibrium in Fock space

pi = e−(Ei − µNi)/kT/ZEq. 15.18 · Feynman's "summit"
  • One level (empty, full): p₁ = Fermi function.
  • One mode (0, 1, 2, … quanta, no µ): Bose function.
  • Two levels with U₀: n = 2/(1 + eˣ) (U₀ → 0), n = 1/(1 + eˣ/2) (U₀ → ∞) (15.21–15.22).

States of the whole system, not one-electron levels. Interactions enter naturally.

Eq. 15.23 with g levels is the familiar donor-occupancy formula of semiconductor physics.

Datta Sec. 15.6Two expressions for entropy

S = k ln W = Nk ln 2Eq. 15.24 · N random spins
S = −Nk Σi p̃i ln p̃iEq. 15.25 · N identical independent systems
W = N!/(N₁!N₂!…),  ln n! ≈ n ln n − nStirling gives 15.25 from 15.24

Shannon's entropy

H = −Σ pi ln pi (Eq. 16.13) measures the information in a message: the same form. Knowing the exact state lowers S by NkH and raises F by NkTH.

p = ¼, ¾: S/Nk = 0.562; p = ½, ½: ln 2 = 0.693.

Datta Sec. 16.1The fuel value of information

(µ₁ − µ₂)N₁ ≥ ΔE − TΔS = ΔFEq. 16.4 · ΔF < 0 lets electrons flow uphill
Eavailable = (E − Eeq) + T(Seq − S)Eq. 16.6 · energy + info-energy
  • Lithium-ion battery: mostly ΔE.
  • All-up spins: ΔE = 0, ΔS = Nk ln 2 ⇒ ΔF = −NkT ln 2 (16.5). The energy comes from the surroundings.

Datta Sec. 16.2–16.3The info-battery

u + D ⇔ U + dEq. 16.7 · perfect anti-parallel spin valve
Δµ = kT ln(PU/PD)Eqs. 16.9–16.10
E = −∫Δµ dNU = TΔS = NkT ln 2Eqs. 16.11–16.12

NkT ln 2 = 17.9 meV per spin, 1.73 kJ/mol at 300 K: about 200× below burning carbon. The point is conceptual.

Datta quotes about 2.5 kJ/mol; N_A kT ln 2 at 300 K is 1.73 kJ/mol (RT = 2.49 kJ/mol).

InteractiveExplore: the info-battery and erasure

The info-battery: energy from information  (Datta Ch. 16, Fig. 16.3; Landauer's principle)

N localized spins sit in a perfect anti-parallel spin valve: the source only exchanges up-spin electrons (u), the drain only down-spin electrons (d). The exchange "reaction" u + D ⇔ U + d flips spins. Starting from the known all-up state (S = 0), the spins randomize while pushing electrons through an external load: the open-circuit voltage is Δμ = kT ln(PU/PD) (Eq. 16.10). The spins' energy never changes: the work comes from the surroundings, paid for by the entropy increase, at most TΔS = NkT ln 2 (Eqs. 16.5, 16.11).

Fraction of up spins PU–
Open circuit Δμ/kT = ln(PU/PD) · terminal qV/kT–
Entropy S/Nk (max ln 2 = 0.693)–
Energy to the load E/NkT (max ln 2)–

Try this

  1. Discharge with a large load: what fraction of NkT ln 2 reaches it?
  2. Repeat with a small load: faster, but how much energy?
  3. Switch to Erase: is the work always above the orange area?
  4. Make η small: does the erasure cost approach NkT ln 2?

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

Dimensionless load: r = R_L·Nγq²/kT. Large r is slow and efficient.

Datta Sec. 16.4–16.5Landauer and Maxwell's demon

Erase + info-battery is a cycle: Eerase − (F − Feq) ≥ 0 ⇒ Eerase ≥ NkT ln 2. Not necessarily dissipated, only usually.

Demon ↔ localized spins, hot/cold molecules ↔ u/d electrons. Both run on information and stop when their entropy saturates.

Boltzmann machines in machine learning design interactions so that the most probable Fock-space state encodes the answer (Datta, end of Ch. 16; Part B Ch. 25 on p-bits).

CheckConcept check

One minute with a neighbour, then click to reveal.

1. Why does an excited atom decay if the coupling is symmetric? (click)
Emitting ε multiplies the surroundings' states by eε/kT; at T → 0 they can only absorb.
2. Which Fock-space states give the Bose function, and why no µ? (click)
N = 0, 1, 2, … quanta; photons and phonons are not conserved.
3. Two levels, U₀ → ∞, ε = µ: average n? (click)
1/(1 + ½) = 2/3 (Eq. 15.22).
4. A random-looking but known spin pattern: can it power a battery? (click)
Yes, with contacts arranged to match (Fig. 16.4): S = 0 for any known state.

Wrap-upSummary and the road to Part B

Take-home

  • Second law: total entropy of the reservoirs increases (15.3, 15.14)
  • P(−ε)/P(+ε) = eε/kT comes from counting the reservoir's states
  • pi ∝ e−(Ei−µNi)/kT in Fock space ⇒ Fermi, Bose, interacting levels
  • Known states carry free energy NkT·H; erasure costs ≥ NkT ln 2

Next: Part B, Week 12 and C6

Contact-ing Schrödinger (Datta Part B, Ch. 17): the wave equation becomes a matrix, bands come from tight-binding chains, and contacts become open boundaries. Open week11_second_law.ipynb and the Week 11 study toolkit.

Datta: the elastic resistor separates entropy-driven processes (contacts) from force-driven ones (channel). Part B keeps that separation with a quantum channel.
Prof. Mohammad Razaghi · University of Kurdistan · after S. Datta, Lessons from Nanoelectronics