Alphabeta Math
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A Bell pair has negative quantum conditional entropy

Example

Let HA=HB=C2 with computational bases e0,e1, let ϕ=(e0⊗e0+e1⊗e1)/2, and define the pure state ρABx:=⟨x,ϕ⟩ϕ. Then the conditional entropy is S(A∣B)ρ=−ln⁡2 nats.

Facts & Assumptions

Given: The two-qubit vector ϕ and the rank-one state ρAB in the Example.

[F1]

Finite quantum mutual, conditional, and conditional mutual information: S(A∣B)ρ=S(ρAB)−S(ρB).

[F2]

Finite-dimensional von Neumann entropy: S(ρ)=−∑λ>0λln⁡λ.

[F3]

Density operator: a state is non-negative with trace one.

[F4]

Partial trace over a finite tensor factor: Tr⁡B sums the matching second-factor basis indices.

[F5]

Partial trace over the first tensor factor: Tr⁡A sums the matching first-factor basis indices.

[F6]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: ln⁡ is strictly increasing, ln⁡1=0, and ln⁡(1/2)=ln⁡1−ln⁡2=−ln⁡2.

Verification

1.1givenF2F3construct

The vector ϕ is a unit vector by orthonormality of the product basis, so the rank-one projection ρAB has eigenvalues 1,0,0,0 and is a state by [F3]. Hence S(ρAB)=0 by [F2].

2.1givenF4F5step 1.1algebra

Expanding ρAB gives diagonal terms 12Pe0⊗e0+12Pe1⊗e1 and two off-diagonal cross terms. The partial traces [F4]–[F5] remove those cross terms because their traced basis indices differ, and leave ρA=ρB=12(Pe0+Pe1)=I2/2.

3.1givenF1F2F6step 1.1step 2.1algebra∎

The eigenvalues of I2/2 are 1/2,1/2, so [F2] and [F6] give S(ρB)=−2(12ln⁡12)=ln⁡2. Since 2>1, strict increase in [F6] gives ln⁡2>ln⁡1=0. Substitution into [F1] and step 1.1 yields S(A∣B)ρ=0−ln⁡2=−ln⁡2<0.

Depends on

Used by

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Sources