Thermodynamic Uncertainty Relations Lab

How precise can a heat engine be? Thermodynamic uncertainty relations (TURs) say that a steady current cannot be both precise and cheap: low noise costs entropy. Build a coherent thermoelectric heat engine from a hot and a cold reservoir and an energy filter, and see where it sits with respect to the classical TUR, the quantum TUR for coherent transport, and the heat-engine bound that forces the Fano factor of any fermionic engine above 1/2.

Energy diagram · hot reservoir (red, TH, μH) and cold reservoir (blue, TC, μC) shaded by their occupation n(E) · the scattering region transmits carriers only through the energy window 𝒯(E)
Carriers
Working mode
Current I · power P
Efficiency η/ηC
Dissipation X = σ/2kB|I|
Fano factor F = SI/|I|
Classical TUR: F·X ≥ 1
Precision vs dissipation · white dot: your engine
Classical TUR, F = 1/X Quantum TUR (Brandner–Saito), F = 1/sinh X Fermionic heat engine, Eq. (3) Heat engine with TH ≤ 2TC, Eq. (4) Bosons & classical carriers, F = coth X Quantum advantage for a fermionic engine
Maximum efficiency allowed by each bound vs Fano factor, at the current γ = kBTC/|δμ| · white dot: your engine

What you are seeing

  • The engine. Carriers leave the hot reservoir, cross the energy filter and enter the cold one against the chemical-potential difference δμ = μC − μH, producing power P = δμ·I. It works as a heat engine when P > 0 and heat flows out of the hot reservoir.
  • Precision and cost. The Fano factor F = SI/|I| measures the relative noise of the current, and X = σ/2kB|I| the entropy produced per transferred carrier. The classical TUR, derived for Markov processes, demands F ≥ 1/X: precise currents are expensive.
  • Quantum coherence relaxes it, but not for engines. Coherent fermionic transport can beat the classical TUR, down to the quantum bound F ≥ 1/sinh X. Once the device is required to be a heat engine, the Fano factor is bounded by Eq. (3), F ≥ (1 + cosh²X)/sinh 2X, so F > 1/2 always. Violations of the classical TUR are confined to 0 < X < 1.606 (green region) and never exceed 6.78%. When TH ≤ 2TC, the tighter Eq. (4) gives F > 1 and no quantum advantage.
  • Bosons and classical carriers. They obey F ≥ coth X, which is stricter than the classical TUR: no quantum advantage and F > 1.
  • Efficiency. Every bound on F becomes a bound on efficiency, η/ηC ≤ 1/(1 + 2γXmin(F)), with γ = kBTC/|δμ|. A fermionic engine needs F > 1/2 to deliver any efficiency at all.
  • Try it. Narrow unit-height filters placed where nH ≈ 1/2 and nC ≈ 0 approach the bound. Lowering 𝒯0 adds partition (shot) noise and moves the dot up. The random sample fills the region above the bound, never below it.

Model: two-terminal coherent conductor described by Landauer–Büttiker scattering theory, with a boxcar transmission of height 𝒯0 between E0 ± w/2. Charge and heat currents, entropy production σ = −Σα Qα/Tα and the zero-frequency noise SI (thermal plus shot) are computed exactly; units e = kB = h = 1, energies in kBTC, μH = 0. Bounds from N. Sobrino, R. Fazio & M. Acciai, arXiv:2607.11704 (2026); quantum TUR from K. Brandner & K. Saito, Phys. Rev. Lett. 135, 046302 (2025); classical TUR from A. C. Barato & U. Seifert, Phys. Rev. Lett. 114, 158101 (2015) and T. R. Gingrich et al., Phys. Rev. Lett. 116, 120601 (2016).