advanced36 min studyLesson 517 in Physics

Compute conductance from transmission channels

Count modes and spin consistently, include thermal averaging, and interpret finite ballistic contact resistance.

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What you’ll understand

  • Apply the two-terminal Landauer formula with an explicit channel convention.
  • Compute finite-temperature conductance as an energy average.
  • Explain contact resistance in a perfectly transmitting conductor.

In a short conductor, the relation between current and voltage is often clearer when expressed through incoming quantum states and their transmission probabilities than through a local resistivity. Reservoirs supply occupied incident states, the intervening structure scatters them, and outgoing states enter the opposite reservoir. This viewpoint exposes conductance quantization, explains why a ballistic conductor has finite two-terminal resistance, and distinguishes the role of coherence from the simpler statement that no momentum-randomizing collisions occur inside a channel.

Specify what one channel includes

A transverse orbital mode supplies right-moving and left-moving states. The excess incident populations from reservoirs at slightly different electrochemical potentials generate net current. For an elastic scattering region, each transmission eigenvalue TnT_n lies between zero and one. A unit incoming state flux times the one-dimensional density of states cancels the group velocity, giving an electrical conductance quantum per spin-resolved channel. The transmission eigenvalues are probabilities, not wave amplitudes that should be squared a second time.

If each orbital mode has two identical unresolved spin channels, it is convenient to use G0=2e2/hG_0=2e^2/h and sum only orbital transmissions. If a field or spin-dependent scattering removes that duplication, use e2/he^2/h and enumerate spin-resolved channels explicitly. Valley multiplicity likewise must be either enumerated or included once as a prefactor under a justified degeneracy. The two propagation directions are already part of the reservoir imbalance argument; they do not provide an additional conductance factor two.

G=G0∑nTn(EF),G0=2e2h,G=e2h∑α spin-resolvedTα(EF)G=G_0\sum_nT_n(E_F),\qquad G_0=\frac{2e^2}{h},\qquad G=\frac{e^2}{h}\sum_{\alpha\,\text{spin-resolved}}T_\alpha(E_F)
These are alternative channel conventions for zero-temperature, linear-response, two-terminal elastic transport. The G₀ form assumes two equivalent spin channels for every enumerated orbital mode. e is positive elementary charge, h is Planck's constant, and G has units siemens. The formulas must not be combined to count spin twice.↔ Scroll wide equations horizontally.

Average over the reservoir thermal window

At finite temperature, a small voltage changes occupations over an energy window of order kBTk_BT around the chemical potential. Conductance is a weighted average of transmission over that window, with weight minus the energy derivative of the Fermi function. Evaluating transmission only at the Fermi energy is reliable when it varies little across the window. Near a subband threshold, a resonance, or a narrow tunneling feature, thermal averaging can substantially change the result even in linear response.

For an ideal step transmission above a threshold, the weighted integral reduces to the Fermi occupation evaluated at that threshold. The resulting conductance rises smoothly rather than jumping sharply at finite temperature. Larger bias requires an integral over the difference between two reservoir Fermi functions, and transmission can itself depend on the bias-dependent potential. Substituting a large terminal voltage into a small-signal conductance law without checking that potential response is not a general nonlinear transport calculation.

G(T)=G0∫−∞∞(−∂f∂E)∑nTn(E) dE,T1(E)=Θ(E−E1) ⟹ G=G0f(E1)G(T)=G_0\int_{-\infty}^{\infty}\left(-\frac{\partial f}{\partial E}\right)\sum_nT_n(E)\,dE,\qquad T_1(E)=\Theta(E-E_1)\ \Longrightarrow\ G=G_0f(E_1)
The reservoirs share temperature T and equilibrium chemical potential μ before the infinitesimal bias is applied. f(E) = [1 + exp((E − μ)/(k_BT))]⁻¹. The single-step result assumes one spin-degenerate orbital channel, no other transmissions, and an energy-independent threshold in linear response.↔ Scroll wide equations horizontally.

Locate the meaning of resistance and coherence

A perfectly transmitting single spin-degenerate orbital channel has two-terminal resistance h/(2e2)h/(2e^2), rather than zero. The finite count of available incident modes limits conductance even without internal backscattering. In the ideal reservoir picture, equilibration and dissipation occur in the reservoirs; the scattering region need not possess a local Ohmic voltage gradient. A four-terminal resistance depends on how voltage probes couple and cannot automatically be equated with the inverse of this two-terminal conductance.

Phase coherence allows amplitudes along alternative paths to interfere and makes resonant transmission strongly phase sensitive. Ballistic motion refers to the lack of momentum-randomizing scattering over the relevant length; an elastically disordered conductor may remain phase coherent while being diffusive. The simple coherent scattering description is therefore not restricted to perfectly clean wires. Strong inelastic processes or interactions may require additional reservoirs, kinetic equations, or a many-body treatment, rather than merely replacing every channel by an arbitrary classical transmission number.

Two partially transmitting orbital modes

A zero-temperature conductor has two spin-degenerate orbital channels with transmissions 0.90 and 0.40. Use G0=77.5 μSG_0=77.5\,\mu\mathrm S. Find its two-terminal conductance and resistance, then compare with perfect transmission through both channels.

  1. The orbital transmission sum is 0.90 + 0.40 = 1.30. Spin is already included in the supplied conductance quantum, so neither spin nor the left/right directions multiply this sum again.

  2. Conductance is G=(77.5)(1.30)=100.75 μSG=(77.5)(1.30)=100.75\,\mu\mathrm S. Its reciprocal is R=1/(100.75×10−6)=9.93 kΩR=1/(100.75\times10^{-6})=9.93\,\mathrm{k\Omega}. This resistance is for the complete two-terminal arrangement under the stated scattering assumptions.

  3. If both channels become perfectly transmitting, the sum is two, giving G=155 μSG=155\,\mu\mathrm S and R=6.45 kΩR=6.45\,\mathrm{k\Omega}. The resistance decreases, but remains finite because the two reservoirs communicate through only two orbital modes.

The result

The partially transmitting device has conductance 100.75 μS and resistance 9.93 kΩ. Its ideal two-mode limit is 155 μS and 6.45 kΩ, demonstrating finite contact resistance even without backscattering.

Explore the model

Interactive lab

Interactive lab · probability

Transmission as partition probability

Reset and set both controls to success probability p = 0.8 and trials n = 10; predict mean count 8. Reset and set both controls to p = 0.4 and n = 10; predict mean count 4. Exact binomial bars show independent partition trials and are not resampled by playback. This illustrates a transmission probability, but supplies no Fermi reservoirs, spin factors, conductance quantum, thermal energy integral, or current units.

Transmission as partition probabilityThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA transmission probability controls an average transmitted count; quantum conductance additionally requires reservoir state counting.

Check your understanding

What two-terminal resistance does one perfectly transmitting spin-degenerate orbital channel have in the ideal Landauer setting?

Try it yourself

At finite temperature, a single spin-degenerate channel has ideal step transmission T(E)=Θ(E−E1)T(E)=\Theta(E-E_1). If μ−E1=kBTln⁡3\mu-E_1=k_BT\ln3, find its linear conductance as a fraction of G0G_0 and numerically for G0=77.5 μSG_0=77.5\,\mu\mathrm S. Why does using only T(μ)T(\mu) give the wrong result?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Integrate minus the Fermi derivative from the threshold to infinity, rather than evaluating transmission at a single energy.

Compare with the solution

The integral is f(E1)=[1+e(E1−μ)/(kBT)]−1=1/(1+1/3)=3/4f(E_1)=[1+e^{(E_1-\mu)/(k_BT)}]^{-1}=1/(1+1/3)=3/4. Thus G=0.75G0=58.1 μSG=0.75G_0=58.1\,\mu\mathrm S. Since the chemical potential is above threshold, T(μ)=1T(\mu)=1, but some of the thermal response window lies below the threshold and cannot transmit. The reduction is thermal averaging of an ideal sharp channel, not an assertion that its transmission at the chemical potential has become 0.75.

Take it with you

  • Count spin once and keep orbital channels separate from propagation directions.
  • Finite-temperature conductance averages transmission over a Fermi window.
  • Perfect transmission gives finite two-terminal contact resistance.

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