Coherent discrimination bias is not a channel metric

Abstract

Coherent discrimination bias fails the triangle inequality for three qubit replacement channels.

The input A, output B and flag F have basis Fin(2); the four Kraus operators are indexed by Fin(4). Fin(n) consists of the integers from 0 to n − 1. Matrices and vectors have complex entries. Matrix.unitaryGroup consists of unitary matrices, and a dagger denotes conjugate transpose. A finite reference R is retained throughout the experiment. The finite ancilla E, representing E′, has a specified zero state e₀, denoted e in the formulas. Products of types index tensor-product bases. Prod.fst and Prod.snd select the first and second coordinates of a pair; tuples are nested to the right. The anonymous brackets supply finite enumerations and decidable equality. All finite sums include every index.

Definition 1.1 (The canonical isometric extension).

$$\forall R\in Type,\forall K\in \mathrm{Fin}\left(4\right)\to \mathrm{Matrix}\left(\mathrm{Fin}\left(2\right),\mathrm{Fin}\left(2\right),Complex\right),\forall psi\in R\times \mathrm{Fin}\left(2\right)\to Complex,\forall r\in R,\forall b\in \mathrm{Fin}\left(2\right),\forall k\in \mathrm{Fin}\left(4\right),\mathrm{extend}\left(K,psi\right)\left((r,(b,k))\right)=\sum _{a:\mathrm{Fin}\left(2\right)}(K\left(k,b,a\right)\cdot psi\left((r,a)\right))$$

Formalization. D5/S3/Quantum/QuantumChannels/CoherentDiscriminationTriangleRefutation.extend (✓ std3).

Citation. Mark M. Wilde (2020). Coherent Quantum Channel Discrimination. DOI: 10.1109/ISIT44484.2020.9174425. URL: https://arxiv.org/abs/2001.02668v1.

Commentary.

For a Kraus family K, the vector extend(K, ψ) has coordinate (r,b,k) equal to the sum over a of K_k(b,a) ψ(r,a). This is the canonical extension of Eq. (10), with the reference coordinate unchanged. The formula is defined for every K and ψ. For a channel, the Kraus normalization is the matrix identity Σ_k K_k† K_k = 1; under this condition the extension is an isometry. The identity matrix acts on the input qubit.

Definition 1.2 (Success of a coherent discrimination strategy).

$$\forall R\in Type,\forall E\in Type,[\mathrm{Fintype}\left(R\right)],[\mathrm{Fintype}\left(E\right)],[\mathrm{DecidableEq}\left(R\right)],[\mathrm{DecidableEq}\left(E\right)],\forall K\in \mathrm{Fin}\left(4\right)\to \mathrm{Matrix}\left(\mathrm{Fin}\left(2\right),\mathrm{Fin}\left(2\right),Complex\right),\forall L\in \mathrm{Fin}\left(4\right)\to \mathrm{Matrix}\left(\mathrm{Fin}\left(2\right),\mathrm{Fin}\left(2\right),Complex\right),\forall psi\in R\times \mathrm{Fin}\left(2\right)\to Complex,\forall V\in \mathrm{Matrix}.\mathrm{unitaryGroup}\left(R\times \mathrm{Fin}\left(2\right)\times E\times \mathrm{Fin}\left(2\right),Complex\right),\forall e\in E,\mathrm{successProbability}\left(K,L,psi,V,e\right)=\frac{1}{4}\cdot \sum _{x:R\times \mathrm{Fin}\left(2\right)\times E}({\Vert \sum _{i:\mathrm{Fin}\left(2\right)}(\sum _{b:\mathrm{Fin}\left(2\right)}(\sum _{k:\mathrm{Fin}\left(4\right)}(\mathrm{conj}\left((\mathrm{if}(i=0)\mathrm{then}K\mathrm{else}L)\left(k,b,\mathrm{Prod}.\mathrm{fst}\left(\mathrm{Prod}.\mathrm{snd}\left(x\right)\right)\right)\right)\cdot \sum _{y:R\times \mathrm{Fin}\left(2\right)\times E\times \mathrm{Fin}\left(2\right)}(\mathrm{Subtype}.\mathrm{val}\left(V\right)\left((\mathrm{Prod}.\mathrm{fst}\left(x\right),(b,(\mathrm{Prod}.\mathrm{snd}\left(\mathrm{Prod}.\mathrm{snd}\left(x\right)\right),i))),y\right)\cdot (\mathrm{if}((\mathrm{Prod}.\mathrm{fst}\left(\mathrm{Prod}.\mathrm{snd}\left(\mathrm{Prod}.\mathrm{snd}\left(y\right)\right)\right)=e)\wedge (\mathrm{Prod}.\mathrm{snd}\left(\mathrm{Prod}.\mathrm{snd}\left(\mathrm{Prod}.\mathrm{snd}\left(y\right)\right)\right)=0))\mathrm{then}\mathrm{extend}\left((\mathrm{if}(i=0)\mathrm{then}K\mathrm{else}L),psi\right)\left((\mathrm{Prod}.\mathrm{fst}\left(y\right),(\mathrm{Prod}.\mathrm{fst}\left(\mathrm{Prod}.\mathrm{snd}\left(y\right)\right),k))\right)\mathrm{else}0)))))\Vert }^2)$$

Formalization. D5/S3/Quantum/QuantumChannels/CoherentDiscriminationTriangleRefutation.successProbability (✓ std3).

Citation. Mark M. Wilde (2020). Coherent Quantum Channel Discrimination. DOI: 10.1109/ISIT44484.2020.9174425. URL: https://arxiv.org/abs/2001.02668v1.

Commentary.

Equation (17) adjoins the basis states e₀ and 0 of E′ and F, applies the prover’s unitary V on R, B, E′ and F while retaining the Kraus index, and applies the adjoint extension. For the first channel the final flag is 0; for the second it is 1. The sum over i adds the two resulting vectors on R, A and E′ coherently, using K when i = 0 and L when i = 1. Their squared Euclidean norm is the sum of squared complex moduli displayed here. The factor 1/4 is the initial factor 1/2 in Eq. (17) multiplied by the factor 1/2 from contraction with the normalized Bell bra. The conditional inserts the two initial zero states. For a physical strategy ψ is a unit vector and the Kraus families satisfy the normalization Σ_k K_k† K_k = 1. The expression itself is defined for arbitrary ψ and Kraus matrices.

Definition 1.3 (The optimal coherent success probability).

$$\forall K\in \mathrm{Fin}\left(4\right)\to \mathrm{Matrix}\left(\mathrm{Fin}\left(2\right),\mathrm{Fin}\left(2\right),Complex\right),\forall L\in \mathrm{Fin}\left(4\right)\to \mathrm{Matrix}\left(\mathrm{Fin}\left(2\right),\mathrm{Fin}\left(2\right),Complex\right),\mathrm{coherentValue}\left(K,L\right)=\mathrm{sSup}\left(\{p:Real|\exists R\in Type,\exists E\in Type,[\mathrm{Fintype}\left(R\right)],[\mathrm{Fintype}\left(E\right)],[\mathrm{DecidableEq}\left(R\right)],[\mathrm{DecidableEq}\left(E\right)],\exists e\in E,\exists psi\in R\times \mathrm{Fin}\left(2\right)\to Complex,\exists V\in \mathrm{Matrix}.\mathrm{unitaryGroup}\left(R\times \mathrm{Fin}\left(2\right)\times E\times \mathrm{Fin}\left(2\right),Complex\right),(\sum _{x:R\times \mathrm{Fin}\left(2\right)}({\Vert psi\left(x\right)\Vert }^2)=1)\wedge (p=\mathrm{successProbability}\left(K,L,psi,V,e\right))\}\right)$$

Formalization. D5/S3/Quantum/QuantumChannels/CoherentDiscriminationTriangleRefutation.coherentValue (✓ std3).

Citation. Mark M. Wilde (2020). Coherent Quantum Channel Discrimination. DOI: 10.1109/ISIT44484.2020.9174425. URL: https://arxiv.org/abs/2001.02668v1.

Commentary.

Equation (18) takes the real supremum over every finite reference R, finite pointed ancilla E′, unit input ψ on R and A, and prover unitary V on R, B, E′ and F. The set explicitly includes every such choice of finite types in Type and its instances. For normalized Kraus families this is p_s^coh. The source states after Eq. (17) that its value is independent of the chosen isometric extensions. No normalization condition on the Kraus matrices is imposed by this definition alone.

Definition 1.4 (The proposed triangle inequality).

$$(claim)\iff (\forall K\in \mathrm{Fin}\left(4\right)\to \mathrm{Matrix}\left(\mathrm{Fin}\left(2\right),\mathrm{Fin}\left(2\right),Complex\right),\forall L\in \mathrm{Fin}\left(4\right)\to \mathrm{Matrix}\left(\mathrm{Fin}\left(2\right),\mathrm{Fin}\left(2\right),Complex\right),\forall M\in \mathrm{Fin}\left(4\right)\to \mathrm{Matrix}\left(\mathrm{Fin}\left(2\right),\mathrm{Fin}\left(2\right),Complex\right),(\sum _{k:\mathrm{Fin}\left(4\right)}(\mathrm{Matrix}.\mathrm{conjTranspose}\left(K\left(k\right)\right)\cdot K\left(k\right))=1)\Rightarrow ((\sum _{k:\mathrm{Fin}\left(4\right)}(\mathrm{Matrix}.\mathrm{conjTranspose}\left(L\left(k\right)\right)\cdot L\left(k\right))=1)\Rightarrow ((\sum _{k:\mathrm{Fin}\left(4\right)}(\mathrm{Matrix}.\mathrm{conjTranspose}\left(M\left(k\right)\right)\cdot M\left(k\right))=1)\Rightarrow (\mathrm{coherentValue}\left(K,M\right)-\frac{1}{2}\le (\mathrm{coherentValue}\left(K,L\right)-\frac{1}{2})+(\mathrm{coherentValue}\left(L,M\right)-\frac{1}{2})))))$$

Formalization. D5/S3/Quantum/QuantumChannels/CoherentDiscriminationTriangleRefutation.claim (✓ std3).

Citation. Mark M. Wilde (2020). Coherent Quantum Channel Discrimination. DOI: 10.1109/ISIT44484.2020.9174425. URL: https://arxiv.org/abs/2001.02668v1.

Commentary.

Put d(N,M) = p_s^coh(N,M) − 1/2. The assertion requires the triangle inequality for every three qubit channels represented by four Kraus operators each. Each displayed normalization Σ_k K_k† K_k = 1 is exactly the channel condition. Section VI asks whether p_s^coh − 1/2 is a metric on quantum channels. That metric property implies this triangle inequality, because the optimal value is independent of the choice of isometric extensions.

Theorem 1.5 (Three replacement channels violate the triangle inequality).

$$\neg (claim)$$

Proof. Machine-checked in Lean as D5/S3/Quantum/QuantumChannels/CoherentDiscriminationTriangleRefutation.result (✓ std3). ∎

Resolves. Problems/wilde-2020-coherent-discrimination-triangle-refutation (refuted) by D5/S3/Quantum/QuantumChannels/CoherentDiscriminationTriangleRefutation.result.

Source. Repository-derived.

Acknowledgement. Mark M. Wilde (2020). Coherent Quantum Channel Discrimination. DOI: 10.1109/ISIT44484.2020.9174425. URL: https://arxiv.org/abs/2001.02668v1.

Commentary.

For a strictly positive probability vector t = (t₀,t₁), consider the qubit channel R_t(X) = Tr(X) diag(t₀,t₁). Its four Kraus operators are √t_j |j⟩⟨k|, with j,k in Fin(2), indexed by the equivalence Fin(2) × Fin(2) ≃ Fin(4). Their conjugate-transpose products sum to the identity.

For two such vectors t,u, set s_j = √(t_j² + u_j²). Then p_s^coh(R_t,R_u) = (s₀ + s₁)²/4. To obtain the upper bound, fix any strategy and write Q_ij for the part of V from (B = j, E′ = e₀, F = 0) to (B = j, F = i): explicitly, (Q_ij ψ)(r,a,e′) = Σ_q V((r,j,e′,i),(q,j,e₀,0)) ψ(q,a). Unitarity gives ‖Q_0j ψ‖² + ‖Q_1j ψ‖² ≤ ‖ψ‖². The vector in the success expression is Σ_j (t_j Q_0j + u_j Q_1j) ψ. Cauchy–Schwarz bounds each summand’s norm by s_j ‖ψ‖, and the triangle inequality gives the probability bound (s₀ + s₁)²/4.

Equality is attained with trivial reference and ancilla, input |0⟩, and the unitary that, conditioned on B = j, acts on F by (1/s_j)[[t_j,−u_j],[u_j,t_j]]. Its two relevant entries are t_j/s_j and u_j/s_j, so their weighted sum is s_j. Thus the upper bound belongs to the set of strategy values and is its supremum.

Take t = (3/4,1/4), u = (1/2,1/2), and v = (1/4,3/4). The closed form gives p_s^coh(R_t,R_v) = 5/8 and p_s^coh(R_t,R_u) = p_s^coh(R_u,R_v) = (9 + √65)/32. Consequently d(R_t,R_v) = 1/8, whereas d(R_t,R_u) = d(R_u,R_v) = (√65 − 7)/32. Their sum (√65 − 7)/16 is strictly less than 1/8 because √65 < 9. These normalized Kraus families contradict claim and hence give a negative answer to the metric question in Section VI.

References

• Truth anchor: D5/S3/Quantum/QuantumChannels/CoherentDiscriminationTriangleRefutation.claim
• Truth anchor: D5/S3/Quantum/QuantumChannels/CoherentDiscriminationTriangleRefutation.coherentValue
• Truth anchor: D5/S3/Quantum/QuantumChannels/CoherentDiscriminationTriangleRefutation.extend
• Truth anchor: D5/S3/Quantum/QuantumChannels/CoherentDiscriminationTriangleRefutation.result
• Truth anchor: D5/S3/Quantum/QuantumChannels/CoherentDiscriminationTriangleRefutation.successProbability
• Dependency: D5/S3/Quantum/Dynamics/PositivePauliClockOrder
• Dependency: D5/S3/Quantum/QuantumChannels/OutputMaximalCoherenceTensorizationRefutation