The entropy photon-number inequality

The explanation

Send two beams of light into a beam splitter, a partly silvered mirror. With transmissivity one half, each output mode is an equal blend of the two inputs. Let beam A be thermal light averaging one photon per mode. Let beam B be the vacuum, with no photons at all. Their mixture is thermal light averaging half a photon.

Entropy measures the disorder of a quantum state, here in natural units. A thermal mode with mean photon number t has entropy g(t) = (t + 1) log(t + 1) − t log t. So beam A has entropy g(1) = 2 log 2, about 1.386. Beam B has entropy 0. The output has entropy g(0.5), about 0.955.

Can a cleverer, non-thermal input with the same entropies give a less disordered output? Earlier bounds allow it. König and Smith proved a quantum entropy power inequalitySource 1. One form says output entropy is at least the weighted average, here 0.693. De Palma, Mari and Giovannetti proved an exponential version. Here it gives at least log 2.5, about 0.916. Neither reaches the thermal value 0.955. For unequal entropy levels, the exponential bound misses the exact thermal answerSource 2.

Guha, Erkmen and Shapiro conjectured a sharper rule in 2007Source 3. Convert each entropy back into an equivalent photon number using the inverse of g, per mode. Call this the entropy photon number. The conjecture says the output's entropy photon number is at least the beam-splitter blend of the two input values. In the example the blend is half of 1 plus half of 0, which is 0.5. That gives the thermal value 0.955 exactly. Thermal inputs give equality, even when the two ports have different meansSource 4. The case of Gaussian inputs, a well-studied family of light states, was already knownSource 5.

The paper proves the inequality for two independent inputs with finite mean energy, in any finite number of modesSource 6. Modes inside one input may be entangled. The hard part is comparing two freely varying inputs with thermal states at their own entropy levels. The proof assumes a violating pair and studies a penalized minimum. It then shows the minimizing inputs must be thermal, which gives a contradiction.

Two consequences follow. The paper finds the exact minimum output entropy, at fixed input entropy, for repeated identical thermal attenuatorsSource 7. A thermal attenuator is a lossy optical line that also adds thermal noise. It also finds the exact classical capacity region of a degraded pure-loss broadcast channel with a mean photon limitSource 8. A broadcast channel sends light to two receivers at once. The attenuator result claims that a minimum is reached, not that it is uniqueSource 9.

How sure is this? Lean, a computer proof checker, has a formal proof of the inequality as stated. That covers finite energy, any finite number of modes, and entanglement within each input. The attenuator and broadcast consequences are outside the Lean statement. They rest on the paper's arguments, which have no published specialist review.

What it doesn't show. This does not cover inputs with infinite mean energy, amplifiers or additive-noise channels, or attenuators that differ from mode to mode.