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Amplitude damping can decrease system entropy

Statement refuted

Every finite quantum channel Φ and every positive semidefinite trace-one matrix ρ satisfy S(Φ(ρ))≥S(ρ).

Facts & Assumptions

Given: The finite channel and entropy definitions, kB>0 with units J K−1, and the full Axiom of Choice as required by the KL supplier below.

[A1]

The Axiom of Choice: every family of nonempty sets has a choice function. It is an explicit hypothesis of the published KL nonnegativity supplier used for the strict entropy comparison.

[F1]

Finite completely positive trace-preserving maps: a finite channel is a complex-linear completely positive trace-preserving map.

[F2]

Complete finite Choi-Kraus equivalence: a finite Kraus map X↦∑aKaXKa∗ is completely positive, and it is trace-preserving when ∑aKa∗Ka=I.

[F3]

Quantum spectral entropy and relative-entropy domains: for a positive trace-one matrix with eigenvalues λi, S(ρ)=−kB∑iλiln⁡λi, with 0ln⁡0=0.

[F4]

Kullback–Leibler divergence: for probability laws P≪Q, D(P∥Q)=∫ln⁡(dP/dQ) dP, using the natural logarithm.

[F5]

Kullback–Leibler divergence is nonnegative: under full AC, D(P∥Q)≥0, with equality exactly when P=Q.

[F6]

The real exponential function and the number e by a power series: exp⁡(x) is the real exponential function defined by its power series, with exp⁡(0)=1.

[F8]

The exponential function is strictly increasing: x<y implies exp⁡(x)<exp⁡(y).

Counterexample

1.1givenF1F2F6F7F8algebra

Fix γ>0 in s−1 and t>0 in seconds, so 0<e−γt<1 by [F6]–[F8]. On C2, define K0=diag⁡(e−γt/2,1), K1=1−e−γt ∣1⟩⟨0∣, and Φt(X)=K0XK0∗+K1XK1∗. Since K0∗K0+K1∗K1=I2, [F2] and [F1] show that Φt is a finite channel.

2.1givenA1F3F4F5F6F7F8step 1.1algebra∎

The matrix ρ0=I2/2 is positive semidefinite with trace one. Direct multiplication gives Φt(ρ0)=diag⁡(p,1−p), where p=e−γt/2∈(0,1/2) by [F6]–[F8]. The entropy difference is S(ρ0)−S(Φt(ρ0))=kB(ln⁡2−h(p)), where h(p)=−pln⁡p−(1−p)ln⁡(1−p). For the binary laws P=(p,1−p) and Q=(1/2,1/2), [F4] gives D(P∥Q)=ln⁡2−h(p); since P≠Q, [F5] gives D(P∥Q)>0. With kB>0, this yields S(Φt(ρ0))<S(ρ0) and refutes the universal claim.

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