Week 11 Study Toolkit · Nanoelectronics

1This week's plan

Read
Datta Part A, Chapters 15–16 (pp. 195–226): the second law and the fuel value of information.

Run
Colab notebook week11_second_law.ipynb: both widgets, Einstein-solid counting, Fock-space averages, the info-battery and the exercises.

Practise
Quiz (aim for 8/10), the practice problems and one mini-investigation. These ideas return in C6 next week.

By the end of the week you should be able to

State the second law for a channel with several reservoirs and show the elastic resistor obeys it.
Explain P(−ε)/P(+ε) = e^ε/kT from the reservoir's number of states and 1/T = dS/dE.
Apply p_i ∝ e^{−(E_i−µN_i)/kT} to Fock-space states: Fermi, Bose and interacting levels.
Derive S = −Nk Σ p ln p and the Boltzmann distribution from minimum free energy.
Explain the info-battery, NkT ln 2, Landauer's principle and Maxwell's demon.

Optional video: Datta's nanoHUB-U course, Unit 4: L4.5–L4.9.

2Key ideas in five sentences

The second law for transport. The total entropy of all reservoirs increases: Σ (E_i − µ_iN_i)/T_i ≤ 0; at one temperature a battery can only lose energy.
Entropic asymmetry. Emitting ε into a reservoir multiplies its states by e^ε/kT, so emission beats absorption by that factor. Every model with energy exchange must build this in.
One law of equilibrium. States of the whole system (Fock space) are occupied with e^{−(E−µN)/kT}/Z; for one level it is the Fermi function, for one mode the Bose function, with interactions something new.
Entropy and free energy. S = −Nk Σ p ln p; the Boltzmann distribution minimises F = E − TS, and any other distribution carries extra free energy.
Information is fuel. A known state (S = 0) of N spins can drive an info-battery with up to NkT ln 2; erasing costs at least that (Landauer).

3Equation sheet

N₁ + N₂ = 0, E₁ + E₂ + E₀ = 0Eqs. 15.1–15.2

Particle and energy conservation for a channel.

(E₁ − µ₁N₁)/T₁ + (E₂ − µ₂N₂)/T₂ + E₀/T₀ ≤ 0Eq. 15.3

Second law; one temperature ⇒ µ₁N₁ + µ₂N₂ ≥ 0 (15.5).

P(+E, +N)/P(−E, −N) = e^{−(E−µN)/kT}Eqs. 15.6–15.7

Asymmetry of absorption and emission.

D₂←₁/D₁←₂ = e^{(E₁−E₂)/kT₀}Eq. 15.9

Inelastic rates; I ∼ D₂←₁f₁(1 − f₂) − D₁←₂f₂(1 − f₁) (15.10).

S = k ln W, 1/T = dS/dEEqs. 15.11, 15.13

Boltzmann entropy and temperature.

ΔS₁ + ΔS₂ + ΔS₀ ≥ 0; ΔF = ΔE − TΔS ≤ 0Eqs. 15.14–15.17

Entropy increases; free energy decreases.

p_i = e^{−(E_i−µN_i)/kT}/ZEq. 15.18

Law of equilibrium for Fock-space states.

n = 2/(1 + eˣ) (U₀ → 0); n = 1/(1 + eˣ/g) (U₀ → ∞)Eqs. 15.21–15.23

Interacting levels, x = (ε − µ)/kT.

S = −Nk Σ p̃_i ln p̃_iEqs. 15.24–15.25

From W = N!/ΠN_i! with Stirling.

F = Σ p̃_i(E_i + kT ln p̃_i)Eqs. 15.26–15.27

Minimum at the Boltzmann distribution.

E_available = F − F_eq = (E − E_eq) + T(S_eq − S)Eqs. 16.1, 16.6

Energy + info-energy.

Δµ = kT ln(P_U/P_D), E = TΔS = NkT ln 2Eqs. 16.10–16.12

Info-battery.

H = −Σ p_i ln p_i; E_erase ≥ NkT ln 2Eq. 16.13, Sec. 16.4

Shannon entropy; Landauer's principle.

4Physics refresher: counting, temperature, Boltzmann

Counting

N spins with n up: W = N!/(n!(N − n)!). The half-up macrostate has W ≈ 2^N; for N = 10²³ every other macrostate is negligible. S = k ln W adds for independent systems.

Boltzmann factor

A small system in contact with a big reservoir: p(E) ∝ W_R(E_tot − E) ∝ e^−E/kT, because ln W_R decreases by E/kT (1/T = dS/dE).

Stirling in one line

ln n! = Σ ln k ≈ ∫₁ⁿ ln x dx = n ln n − n + 1 ≈ n ln n − n for large n.

Why the Fermi function has E − µ but the Bose function only ħω

Electrons are conserved: an electron entering a contact at E dissipates E − µ, the rest is the energy µ of the added particle. Photons and phonons are created and absorbed, so the full ħω is exchanged.

5Glossary

Term	Meaning
Reservoir	A system so large that exchanges do not change its T and µ.
Entropy S	k times the log of the number of accessible states.
Free energy F	E − TS; decreases in any spontaneous process at fixed T.
Fock space	States of the whole system labelled by occupation numbers (00, 01, 10, 11, …).
Law of equilibrium	p_i ∝ e^{−(E_i−µN_i)/kT} for Fock-space states.
Info-energy	T(S_eq − S): extractable because the state is known.
Landauer's principle	Erasing N bits costs at least NkT ln 2.
Maxwell's demon	A sorter that uses information to create a temperature difference.

6Calculators

Boltzmann, Fermi and Bose

Energy difference or ε − µ (meV)Temperature T (K)

Boltzmann e^−E/kT

Fermi 1/(e^E/kT + 1)

Bose 1/(e^E/kT − 1)

Entropy and information

Probabilities (comma separated)Number of systems N

S/Nk = −Σ p ln p

H in bits

S for N systems

T(S_eq − S) per system at 300 K

Info-battery and Landauer

Fraction of up spins P_UT (K)Bits to erase

Δµ/q = (kT/q) ln(P_U/P_D)

Energy still available per spin

Landauer minimum for those bits

kT ln 2

7Self-check quiz

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

8Practice problems

k = 1.381 × 10⁻²³ J/K = 86.17 µeV/K; kT = 25.85 meV and kT ln 2 = 17.92 meV at 300 K.

P1. A small reservoirEASY

An Einstein solid with N = 100 oscillators holds Q = 100 quanta. Find W(Q)/W(Q − 1) and kT/ε. Compare with the large-N formula.

Hint

W = C(Q + N − 1, Q).

Solution

W(Q)/W(Q − 1) = (Q + N − 1)/Q = 1.99 ⇒ kT/ε = 1/ln 1.99 = 1.453; large-N 1/ln 2 = 1.443.

P2. No free lunchEASY

Show from Eqs. 15.2–15.3 with T₁ = T₂ = T₀ that µ₁N₁ + µ₂N₂ ≥ 0, and interpret.

Solution

Multiply 15.3 by T: E₁ + E₂ + E₀ − (µ₁N₁ + µ₂N₂) ≤ 0, and E₁ + E₂ + E₀ = 0 ⇒ µ₁N₁ + µ₂N₂ ≥ 0: the battery can only deliver energy, never be charged by surroundings at the same temperature.

P3. Why atoms stay excited only brieflyEASY

Hydrogen 2p–1s: ε = 10.2 eV. Ratio of upward to downward rates at 300 K?

Solution

e^{−10.2/0.02585} = e^−394.6 ≈ 10⁻¹⁷¹: absorption from the thermal field essentially never happens.

P4. Fermi and Bose from Fock spaceMEDIUM

(a) Level 0.1 eV above µ at 300 K: occupation? (b) Phonon mode ħω = 10 meV at 300 K: average number?

Solution

(a) 1/(e^3.868 + 1) = 0.0205. (b) 1/(e^0.3868 − 1) = 2.12.

P5. Strongly interacting levelsMEDIUM

g levels, U₀ → ∞, ε = µ. Average n for g = 2 and g = 6? What would non-interacting levels give?

Solution

n = 1/(1 + 1/g): 2/3 and 6/7 = 0.857. Non-interacting: g/2 = 1 and 3.

P6. Entropy of a partially polarised sampleMEDIUM

N = 10²⁰ spins with p = ¼, ¾. Find S.

Solution

S/Nk = 0.5623 ⇒ S = 10²⁰ × 1.381 × 10⁻²³ × 0.5623 = 7.76 × 10⁻⁴ J/K.

P7. Info-battery voltageMEDIUM

P_U = 0.9 at 300 K: open-circuit voltage, and energy still available per spin?

Solution

Δµ/q = 25.85 mV × ln 9 = 56.8 mV. Available T(S_eq − S) = (0.693 − 0.325)kT = 9.5 meV per spin.

P8. Erasing a gigabyteHARDER

Landauer minimum to erase 1 GB (8 × 10⁹ bits) at 300 K?

Solution

8 × 10⁹ × 2.871 × 10⁻²¹ J = 2.3 × 10⁻¹¹ J, many orders of magnitude below what real memories spend.

P9. Minimum free energyHARDER

Show that F = Σ p̃_i(E_i + kT ln p̃_i) with Σ p̃_i = 1 is minimised by the Boltzmann distribution, and that F_min = −kT ln Z.

Hint

Lagrange multiplier.

Solution

∂/∂p̃_i: E_i + kT ln p̃_i + kT − λ = 0 ⇒ p̃_i ∝ e^−E_i/kT. Then E_i + kT ln p̃_i = −kT ln Z for all i ⇒ F = −kT ln Z.

9Common mistakes

MISTAKE"Electrons go down in energy because of a force."

CORRECTThe mechanical coupling is symmetric; the surroundings make emission e^ε/kT times more likely (entropic).

MISTAKE"The Fermi function describes each electron."

CORRECTIt is the probability that a one-electron level is full, from Fock-space states of that level; with interactions it fails (Eq. 15.22).

MISTAKE"Ordered states store energy."

CORRECTThe all-up spins have the same energy as random spins; they store free energy through low entropy, and any known state works.

MISTAKE"Erasing a bit must dissipate kT ln 2."

CORRECTIt must cost at least kT ln 2 of work, which an info-battery could in principle recover.

MISTAKE"The second law is about a single system."

CORRECTIt concerns the total entropy of all reservoirs involved (Eq. 15.14).

MISTAKE"Temperature is defined for any system."

CORRECTOnly in equilibrium; a metastable spin collection has entropy but no temperature.

10Explore

Files are in the course folder (course_materials).

Week 11 slides Widget: reservoir and Fock space Widget: info-battery Colab notebook Week 10 toolkit

Mini-investigations (15 min each)

In the reservoir widget, use N_R = 20 and Q/N_R = 0.2. Is the histogram still Boltzmann? Why not?
In the info-battery, find the load at which half of NkT ln 2 is delivered. How long did it take compared with the largest load?
Erase with η = 0.1 and η = 2: how much extra work did the fast erasure cost?

11Can I…?

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

state the second law for a channel (Eq. 15.3)show the elastic resistor obeys itexplain P(−ε)/P(+ε) = e^ε/kT from countingobtain the Fermi and Bose functions from Fock spacecompute n for two interacting levelsderive S = −Nk Σ p ln pexplain the info-battery and NkT ln 2state and justify Landauer's principle

12Looking ahead: Week 12 and circular session C6

Part B begins: Contact-ing Schrödinger (Datta Part B, Ch. 17), then C6: a Colab lab on tight-binding chains and on this week's statistics.

How can a differential equation become a matrix?
Why does a chain of atoms give a band E(k) = ε + 2t cos ka?
How do we connect a matrix Hamiltonian to two contacts?

The second law, Fock space and the fuel value of information

1This week's plan

By the end of the week you should be able to

2Key ideas in five sentences

3Equation sheet

4Physics refresher: counting, temperature, Boltzmann

5Glossary

6Calculators

Boltzmann, Fermi and Bose

Entropy and information

Info-battery and Landauer

7Self-check quiz

8Practice problems

9Common mistakes

10Explore

Mini-investigations (15 min each)

11Can I…?

12Looking ahead: Week 12 and circular session C6