Use partition noise to learn beyond conductance
Calculate a Fano factor from transmission eigenvalues and distinguish shot noise from thermal noise and interaction effects.
On this page
The question for today
What you’ll understand
- Compute the conductance and partition-noise transmission sums.
- Apply a declared noise-spectrum convention with correct units.
- Explain how equal conductances can conceal different channel distributions.
Average conductance measures how much charge is transmitted, but not how regularly it arrives. A partially transmitting channel partitions occupied incoming states into transmitted and reflected outcomes, creating current fluctuations even at zero temperature under bias. The resulting shot noise depends on a different combination of transmission probabilities than conductance does. Measuring both can therefore distinguish conductors with the same average current while still leaving some ambiguity about their complete microscopic channel structure.
Separate incoming occupation from partition randomness
For noninteracting elastic transport at zero temperature, occupied incoming states inside the bias window provide the available attempts. Fermi statistics suppress the occupation fluctuations that an unconstrained classical Poisson arrival model would assign to these filled states. A channel with transmission T still randomly partitions each attempt, giving a variance proportional to . A closed channel contributes neither current nor partition noise. A perfectly transmitting channel carries current but has no zero-temperature partition noise in this model.
For spin-degenerate orbital channels, conductance is with . Noise weights the same channels by , so spin must follow the same counting convention in both sums. If transmissions mix modes, the relevant numbers are the eigenvalues of the transmission-probability matrix, rather than arbitrary diagonal amplitudes in a chosen basis. The ratio of these two sums is the Fano factor, a dimensionless measure of noise relative to the Poisson reference.
Check the temperature and spectral convention
A noise spectral density is not a variance in amperes squared until integrated over a measurement bandwidth with its specified convention. Here a one-sided low-frequency spectrum is used, so a Poisson stream of elementary charges has . Using a two-sided convention changes factors of two. A quoted numerical noise value without its bandwidth or spectral units can therefore be misleading even when its current and Fano factor are correct. This convention should also match the thermal-noise formula used for comparison.
At finite temperature, a perfectly transmitting conductor still has equilibrium Johnson–Nyquist noise in the classical low-frequency one-sided limit. The simple zero-temperature partition expression is most useful when the bias energy greatly exceeds thermal energy, with negligible heating and otherwise valid scattering assumptions. In the crossover region, the full Fermi-occupation integrals combine thermal and partition contributions. Subtracting an equilibrium background and calling the remainder excess noise does not make every finite-temperature expression equal to its zero-temperature limit.
Infer channel information without overclaiming uniqueness
Two channels with transmissions 0.8 and 0.2 have the same conductance sum as channels with transmissions one and zero. The former partition both streams and make noise; the latter transmit one stream deterministically and reject the other. Thus conductance alone cannot distinguish their channel composition. Noise supplies a second moment-like constraint, but many larger channel sets can still share both sums. Recovering every transmission eigenvalue generally requires additional information, assumptions, or higher current cumulants.
When all transmissions are small, is approximately T, giving nearly Poissonian F near one for weak independent tunneling. Strongly transmitting channels suppress noise. Interactions and correlated transfer can produce super-Poissonian noise, so an observed F above one is evidence against this simple partition model rather than a violation of charge quantization. Coulomb blockade, dynamical trapping, and avalanche processes each require their own transport description; the elementary formula cannot identify one uniquely from a single excess-noise ratio.
Work it through
Equal conductance does not mean equal noise
At zero temperature take two spin-degenerate orbital transmissions and , , and bias . Use . Find current, Fano factor, and one-sided shot-noise density. Compare with channels 1 and 0 at the same bias.
The transmission sum is one, so and . The supplied G₀ already includes the two equivalent spins of each orbital channel.
The partition sum is . Dividing by the transmission sum gives , below the independent Poisson limit because these are partly open fermionic channels.
The noise is . Channels one and zero have the same transmission sum and current, but their partition sum is zero, hence no shot noise in this zero-temperature model.
The device carries 77.5 nA with F = 0.32 and noise density 7.94 × 10⁻²⁷ A²/Hz. A fully open channel paired with a closed one carries the same current with zero partition noise, demonstrating the extra information in a fluctuation measurement.
The living notebook
Explore the model
Interactive lab · probability
Compare partition variance with deterministic transmission
Reset and set both controls to success probability p = 0.5 and trials n = 20; predict mean 10 and variance 5, so variance divided by mean is 0.5. Reset and set both controls to p = 1 and n = 20; predict mean 20 and variance 0. These exact binomial bars illustrate partition statistics, not random resampling, a Fermi reservoir, or a calibrated current-noise spectrum in A²/Hz.
- Mean · np
- 5
- Variance · np(1−p)
- 2.5
- Total probability
- 1
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 forPerfect transmission can carry a nonzero mean count with no partition variance; thermal and interaction noise require additional physics.
Pause & reason
Check your understanding
Your turn
Try it yourself
Three spin-degenerate orbital channels have transmissions 1, 0.5, and 0.5 and carry in the shot-dominated low-frequency regime. Find their conductance in units of , Fano factor, and one-sided partition noise. Compare with transmissions 1, 1, and 0 at the same current, neglecting thermal noise.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Compute the transmission sum and partition sum separately. A fully open channel contributes to only the first sum.
Compare with the solution
The transmission sum is two, giving . The partition sum is , hence . The noise is . Transmissions (1,1,0) have the same conductance but zero partition sum, giving zero shot noise in this approximation. Neither comparison predicts zero finite-temperature equilibrium noise, and the noninteracting Fano bound does not exclude super-Poissonian noise in other transport regimes.
Keep the idea
Take it with you
- Partition noise weights channels by T(1 − T), while conductance weights them by T.
- Spectral convention and temperature determine the correct numerical noise formula.
- Noise adds channel information but does not generally identify a unique microscopic transmission set.
Read further
Original sources and deeper explanations to follow at your own pace.
Optional progress, saved in this browser when available. Revisit any lesson at any time.
Lesson not marked complete.