Nanoelectronics: A New Perspective on TransportWeek 14 · Wires, scatterers and modes · C7

WEEK 14 · WIRES, SCATTERERS AND MODES · C7

Nanoelectronics:
A New Perspective on Transport

Quantum resistors in series, the potential drop and the quantum of conductance; NEGF lab

Prof. Mohammad RazaghiUniversity of Kurdistan · Graduate course in Electronics Engineering
Based on S. Datta, Lessons from Nanoelectronics, Part B: Quantum Transport, Chapters 19–20

Part 1 is the fixed lecture (Ch. 19–20). Part 2 is circular session C7: the NEGF lab on scatterers, series resistors and the potential drop.

Week 14Where we are

Last week

The four NEGF equations (18.1–18.4)
G(E) = (q²/h) Trace[Γ₁G^RΓ₂G^A]
Dephasing: Σ₀ = D × G^R

Today

Use them on the simplest real device, a wire: put in scatterers, put resistors in series, and see where the voltage drops. Then widen the wire into 2D and count the modes.

Week 14Today's goals

Write the contact self-energy of a wire, Σ = −t₀e^ika.
Find T for one scatterer: (2t₀ sin ka)²/[U² + (2t₀ sin ka)²].
Show why two quantum resistors in series are not additive.
Read the potential drop f(j) = Gⁿ/A and the role of dephasing.
Count modes in 2D wires, graphene and the quantum Hall regime.

Datta Sec. 19.1The contact of a wire: Σ = −t₀ e^ika

E = E_c + 2t₀(1 − cos ka)1D lattice, t₀ = ħ²/2ma²

Σ₁ = Σ₂ = −t₀e^ika, Γ = 2t₀ sin kaEq. 19.2 · end site only

A semi-infinite wire is the contact: the electron leaves as a travelling wave, never comes back.
Retarded: Im Σ ≤ 0. Outside the band, k is imaginary and Σ is real: no broadening.

Derive Σ from the surface Green function g = 1/(E − ε − t² g): the root with negative imaginary part. Check: Γ is largest at mid-band (ka = π/2).

Datta Sec. 19.1A perfect wire: T = 1, resistance at the contacts

T(E) = Trace[Γ₁G^RΓ₂G^A] = 1any length, any energy in the band · checked numerically in the notebook

G = q²/h (per mode): independent of length, so R = h/q² = 25.8 kΩ, not zero.
The resistance sits at the interfaces between the wide contacts and the one-mode wire (Week 3, R_B).

Check yourself

For N = 2, 5, 20 sites at E = 0.3, 1, 2, 3.5 t₀ the notebook gives T = 1.000000 every time. Why does the length drop out?

Because the wave is never reflected: Σ matches the lattice exactly.

Datta Sec. 19.2One scatterer

G^R(j,j) = 1/(−U + 2i t₀ sin ka)the two halves of the wire act as contacts on site j

T = (2t₀ sin ka)² / [U² + (2t₀ sin ka)²]Eq. 19.5

U = 0 gives T = 1; U ≫ Γ gives T ≈ (Γ/U)².
At E = t₀ and U = 2t₀: T = 3/7 = 0.43. Verified against NEGF to 10⁻¹⁵.

Datta Sec. 19.3Two resistors in series: the semiclassical answer

R(T) = (h/q²)·1/T = (h/q²)[1 + (1 − T)/T]contact term + scatterer

1/T_series = 1/T₁ + 1/T₂ − 1add the scattering parts

R₂ = R₁(2 − T)Eq. 19.7 · two identical scatterers, T₂ = T/(2−T)

Valid if the electron loses its phase between the scatterers, so probabilities (not amplitudes) multiply.
Example: T₁ = 0.39 gives T₂ = 0.24.
With a coherent electron, the two reflected waves interfere. Is the formula still true?

This is the Ohm's law result: N scatterers give R_N = R_B(1 + N(1−T)/T), linear in N.

Datta Sec. 19.3Two quantum resistors: not additive

Dashed: incoherent T₁/(2−T₁). Red: exact NEGF for two U = 2t₀ scatterers 6 sites apart.
The pair oscillates between almost 0 and almost 1: the cavity between them resonates when k d ≈ mπ.
At E = 0.81 t₀: T₁ = 0.39 but T(two) = 0.9996: two scatterers offer less resistance than one.
A Fabry–Pérot resonance for electrons.

Datta Fig. 19.4: N = 11, scatterers at sites 3 and 9; A (E = 0.81 t0) constructive, B (0.60 t0) destructive: T = 0.0665 against 0.34 for one scatterer.

Datta Sec. 19.3The numbers behind Fig. 19.4

Point	E	T (one scatterer)	T (two)	Incoherent T₁/(2−T₁)
A, constructive	0.81 t₀	0.3925	0.9996	0.244
B, destructive	0.60 t₀	0.3377	0.0665	0.204

N = 11 sites, U = 2t₀ at sites 3 and 9; the quantum answer is 4× larger or 3× smaller than the incoherent one.
Only after averaging over phase (energy, or random d) does Ohm's law reappear.
Mesoscopic samples: conductance fluctuations of order q²/h from sample to sample.

InteractiveExplore: scatterers in series and the potential drop

Can two scatterers offer less resistance than one? (Datta Part B, Ch. 19, Figs. 19.2–19.9)

A 1D wire (t₀ = 1) with 51 sites; every contact is a self-energy Σ = −t₀e^ika on the end site (Sec. 19.1). Left: transmission T(E) for the chosen scatterers (crimson), one scatterer alone (grey) and the semiclassical result (green dashed). Right: the occupation f(j) = Gⁿ(j,j)/A(j,j) at the marked energy, a picture of the potential drop (Eq. 19.8). Click the left plot to choose E.

Scatterers Energy E − E_c (units of t₀) Scatterer strength U (t₀) Spacing of the two scatterers (sites) Dephasing D₀ (t₀²) Dephasing model

T for one scatterer, T₁–

T of your structure (coherent)–

R/R₁ classical: N − (N−1)T₁–

R/R₁ quantum: T₁/T–

Try this

None → one scatterer: compare T(E) with Eq. 19.5. Click on the curve to move E.
Two scatterers: set E near 0.81 and vary the spacing d. Where is T > T₁?
Six scatterers: watch the quantum T collapse below the Ohm's-law curve (localisation).
Turn D₀ up in Model B: does f(j) become a straight line? Try Model A.

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

The semiclassical curve T1/(N−(N−1)T1) is what Ohm predicts for N identical scatterers; the quantum T averages lower in the localised regime.

Datta Sec. 19.3Many scatterers: localisation

n scatterers	mean T	exp⟨ln T⟩	Ohm: 1/(1 + nR₁)
1	0.419	0.410	0.399
2	0.261	0.161	0.249
3	0.132	0.065	0.181
4	0.101	0.027	0.142
5	0.083	0.012	0.117
6	0.056	0.005	0.100

Ohm: R grows linearly with n. Quantum: ln T falls linearly, so T decays exponentially: strong localisation.
Averages over 80 energies in 0.3–3.7 t₀, 51-site wire, U = 2t₀ at sites 4, 11, 19, 24, 34, 42.

Datta Sec. 19.4Where does the voltage drop?

f(j) = Gⁿ(j,j)/A(j,j)Eq. 19.8 · electrochemical potential at site j

Coherent: oscillations around the scatterer, no linear drop in the clean parts; the potential changes at the obstacle.
Model B: phase and momentum randomised at every site: straight line, the classical Ohmic profile.
Model A: in between, with interference ripples.

51 sites, U = 1 t0 at site 27, E = t0, D0 = 0.09 t0². Model B profile: 0.82 at the left end to 0.18 at the right.

Datta Sec. 20.1From 1D to 2D: subbands and modes

E_n(k_x) = ε_n − 2t₀ cos k_xaEq. 20.3

ε_n = 4t₀ − 2t₀ cos[nπ/(N+1)]Eq. 20.4 · N sites across

One 1D wire per subband; a new mode opens at E_c + 2t₀[1 − cos(nπ/(N+1))].
Ballistic T(E) = M(E): steps of q²/h (van Wees, Wharam 1988).

The number of crossings of the energy line with the subbands is 2M: one right-moving and one left-moving state per mode.

Datta Sec. 20.2Coupling to a 2D contact: basis transformation

g⁻¹ = (E + i0⁺)I − α − β^†gβ, Σ = βgβ^†Eq. 20.6 · surface Green's function by iteration

Σ̃ = V^†ΣV = diag(−t₀e^ik_na)transform to the mode basis V

In the mode basis Σ is diagonal: each mode has its own 1D value. The notebook checks off-diagonal elements < 10⁻¹⁵.
Closed modes (E below the subband) have real Σ: no broadening, no transmission.
Example N = 6, E = 1.3 t₀: 2 open modes.

Datta Sec. 20.3Graphene and carbon nanotubes

Ribbon of width W

E = ±ħv₀|k| gives
M(E) = floor(2WE/hv₀)
(armchair-type edges); a zigzag edge adds one mode at E = 0.

hv₀ = 4.14 eV·nm for v₀ = 10⁶ m/s. W = 30 nm at 0.3 eV: M = 4.

Carbon nanotube

Close the width with periodic boundary conditions: allowed k_y are multiples of 2π/circumference; metallic tubes keep k_y = 0 and have M = 2 at the Dirac point (the K and K′ valleys).

With spin: 4 modes, G = 4q²/h = 155 µS, R = 6.5 kΩ.

Datta counts M per spin unless said otherwise. A metallic nanotube has two modes at E = 0 (one per valley); with spin there are four, so R = h/4q² = 6.5 kOhm.

Datta Sec. 20.4The quantum Hall effect as mode counting

Landau gauge: x-hopping gets a phase e^iφy, φ = qBa²/ħ (Peierls).
Subbands flatten into Landau levels in the bulk; at the edges they bend up and cross E: edge states.
Each edge state is a mode carrying q²/h, moving one way on one edge and the other way on the opposite edge: no backscattering.
R_H = h/(Mq²): plateaus (von Klitzing 1980).

In the widget the Hall plateaus appear as M(B): M = 8 at 4 T, 7 at 6 T, 6 at 10 T, 5 at 12 T for N = 25, E = t0.

InteractiveExplore: the quantum of conductance

The quantum of conductance: modes of a 2D wire (Datta Part B, Ch. 20, Figs. 20.1–20.4)

A ballistic 2D conductor with N sites across its width is a set of 1D wires in parallel, one per subband E_n(k_x) = ε_n − 2t₀cos k_xa with ε_n = 4t₀ − 2t₀cos[nπ/(N+1)] (Eq. 20.4). Transmission is the number of modes M(E): half the number of times the energy line crosses a subband. Add a magnetic field (Peierls phases) to see the subbands bend into Landau levels and edge states.

Width: N sites across (a = 2.5 nm) Energy E − E_c (units of t₀) Magnetic field B (T), m = 0.067 m₀ Right panel

Modes M(E) = T(E)–

Conductance M q²/h (×2 with spin)–

Resistance h/(M q²)–

t₀, width W = (N+1)a–

Try this

B = 0: slide N and read M at E = t₀. Does M match floor of (N+1)/π · arccos(1 − E/2t₀)?
Sweep E: where does each new step appear? Compare with the dotted analytic steps.
Switch the right panel to "M versus width N": the van Wees staircase.
Turn on B = 6 T and switch to "M versus B": which plateau of R_H do you land on?

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

The strip units are a = 2.5 nm, m = 0.067 m0, t0 = 91 meV, giving phi = 9.5e-3 rad per tesla.

CheckConcept check

One minute with a neighbour, then click to reveal.

1. Why is the resistance of a clean wire h/q² and not 0? (click)

The wire has one mode only; the other modes of the wide contacts are reflected at the interface.

2. T₁ = 0.5: incoherent T for two in series? (click)

0.5/1.5 = 1/3; a coherent pair can be anything from near 0 to 1.

3. Which dephasing model gives a linear potential drop? (click)

Model B (phase and momentum randomised).

4. Where does the 5th mode of a 25-site strip open? (click)

2t₀[1 − cos(5π/26)] = 0.354 t₀.

Wrap-upPart 1 summary

Take-home

Wire: Σ = −t₀e^ika, Γ = 2t₀ sin ka, T = 1
Scatterer: T = Γ²/(U² + Γ²)
Series: classical R₂ = R₁(2 − T); quantum can be anything
f(j) = Gⁿ/A: linear only with Model B
2D wire: T(E) = M(E); R_H = h/(Mq²)

Next: circular session C7

An NEGF lab in four blocks. Open week14_negf_lab_C7.ipynb and the Week 14 toolkit. Write-up task: why are two quantum resistors in series not additive?

PART 2

Circular session C7

NEGF lab: scatterers, series resistors, potential drop

A · 15 min
One scatterer

B · 25 min
Two scatterers

C · 25 min
Potential drop

D · 25 min
Modes; write-up

Groups of 3-4. Each group must produce the write-up on why two quantum resistors in series are not additive, with its own numerical evidence.

Circular session C7Block A · One scatterer

A1. Build the wire (N = 21, U on site 11). Compute T(E) with T_wire and overlay Eq. 19.5.

A2. At E = t₀ vary U: T = 1, 0.75, 0.43, 0.16 for U = 0, 1, 2, 4 t₀.

A3. Why is T largest at mid-band for a given U? (Γ = 2t₀ sin ka.)

Circular session C7Block B · Two scatterers

B1. Spacing d = 2…14 sites at E = 0.81 t₀: plot T(d).

B2. Compare with the incoherent T₁/(2 − T₁) = 0.244 (T₁ = 0.3925).

B3. What fraction of energies has T(two) > T(one)? (Notebook, d = 6: 20 %.) Largest ratio: 5.6 at E = 0.21 t₀.

B4. Count resonances (T > 0.9) for d = 4, 6, 10: 3, 5, 8.

Circular session C7Block C · The potential drop

C1. Wire of 51 sites, one scatterer U = t₀, E = t₀: plot f(j) for D₀ = 0, Model A, Model B (D₀ = 0.09 t₀²).

C2. Where does the potential change in each case? What is the effective T (Trace[Γ₂Gⁿ])?

C3. Check the zero-current condition at the dephasing "contact".

Circular session C7Block D · Problem clinic

1. T₁ = 0.39: incoherent T for two scatterers, and the largest quantum value? (click)

0.39/1.61 = 0.242; the quantum maximum is 1 (full resonance).

2. At which energy does a 15-site strip open its 3rd mode? (click)

2t₀[1 − cos(3π/16)] = 0.337 t₀; M(2t₀) = 7.

3. A Hall plateau at R_H = 3.2 kΩ: how many edge channels? (click)

25.813/3.2 ≈ 8.

4. Why is the E-averaged T of six scatterers (0.056) below Ohm (0.100)? (click)

Localisation: ln T, not T, adds, and the average is dominated by the dips.

Circular session C7Model answer: why two quantum resistors are not additive

The argument

Resistances add only if the electron arrives at the second resistor with a random phase: then probabilities multiply, giving T₁T₂-type series rules.
In a coherent wire the waves reflected from both scatterers interfere: the phase 2kd between them decides.
The transmission amplitudes, not the resistances, are what combine: T ranges from 0 to 1.

The evidence

E = 0.81 t₀: T₁ = 0.39, two scatterers T = 0.9996 (incoherent 0.244).
E = 0.60 t₀: T = 0.0665 (incoherent 0.204).
With Model B dephasing the classical result returns, with a linear drop.

A good write-up cites its own numbers, distinguishes phase coherence length from scattering length, and mentions the experimental consequence (conductance fluctuations).

Wrap-upSummary and what comes next

Take-home

Coherent series resistors: T interferes (0.39 → 0.9996)
Dephasing restores Ohm's law
2D wire: T = M(E), R_H = h/(Mq²)

Week 15

Inelastic scattering (Part B, Ch. 21) and Coulomb blockade (Ch. 22): what changes when electrons exchange energy and repel each other.

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

Nanoelectronics:A New Perspective on Transport

Week 14Where we are

Last week

Today

Week 14Today's goals

Datta Sec. 19.1The contact of a wire: Σ = −t₀ eika

Datta Sec. 19.1A perfect wire: T = 1, resistance at the contacts

Check yourself

Datta Sec. 19.2One scatterer

Datta Sec. 19.3Two resistors in series: the semiclassical answer

Datta Sec. 19.3Two quantum resistors: not additive

Datta Sec. 19.3The numbers behind Fig. 19.4

InteractiveExplore: scatterers in series and the potential drop

Can two scatterers offer less resistance than one? (Datta Part B, Ch. 19, Figs. 19.2–19.9)

Try this

Datta Sec. 19.3Many scatterers: localisation

Datta Sec. 19.4Where does the voltage drop?

Datta Sec. 20.1From 1D to 2D: subbands and modes

Datta Sec. 20.2Coupling to a 2D contact: basis transformation

Datta Sec. 20.3Graphene and carbon nanotubes

Ribbon of width W

Carbon nanotube

Datta Sec. 20.4The quantum Hall effect as mode counting

InteractiveExplore: the quantum of conductance

The quantum of conductance: modes of a 2D wire (Datta Part B, Ch. 20, Figs. 20.1–20.4)

Try this

CheckConcept check

Wrap-upPart 1 summary

Take-home

Next: circular session C7

Circular session C7

Circular session C7Block A · One scatterer

Circular session C7Block B · Two scatterers

Circular session C7Block C · The potential drop

Circular session C7Block D · Problem clinic

Circular session C7Model answer: why two quantum resistors are not additive

The argument

The evidence

Wrap-upSummary and what comes next

Take-home

Week 15

Slides

Nanoelectronics:
A New Perspective on Transport

Datta Sec. 19.1The contact of a wire: Σ = −t₀ e^ika