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Selective-conditioning counterexample

Statement refuted

False claim: for every density operator $ρ$ on a tensor product $H \otimes K$ of finite-dimensional complex inner product spaces and every finite local family $(M_{a})$ on $K$ with $\sum_{a} M_{a}^{*} M_{a} = I_{K}$ , the conditional reduced state on $H$ after the outcome $a$ is again ${Tr}_{K} ρ$ whenever its probability is positive.

The claim confuses the nonselective identity ${Tr}_{K} (\sum_{a} (I \otimes M_{a}) ρ (I \otimes M_{a}^{*})) = {Tr}_{K} ρ$ of Local trace-preserving Kraus maps do not change the opposite reduced state with the corresponding statement for one selected outcome. The displayed unit vector in $C^{2} \otimes C^{2}$ and one outcome of the explicitly defined projection instrument below refute the selective claim.

Facts & Assumptions

Given: $H = K = C^{2}$ with orthonormal bases $(e_{+}, e_{-})$ ; the unit vector $Ψ = \frac{1}{\sqrt{2}} (e_{+} \otimes e_{-} - e_{-} \otimes e_{+})$ ; the density operator $ρ = T_{Ψ, Ψ}$ with $T_{u, v} x = ⟨ x, v ⟩ u$ ; the rank-one projections $P_{0} : = T_{e_{-}, e_{-}}$ and $P_{1} : = T_{e_{+}, e_{+}} = I_{K} - P_{0}$ on $K$ .

[L1]

On $H \otimes K$ the pairing is $⟨ u \otimes v, u^{'} \otimes v^{'} ⟩ = ⟨ u, u^{'} ⟩_{H} ⟨ v, v^{'} ⟩_{K}$ ; the four tensors $e_{\pm} \otimes e_{\pm}$ form an orthonormal basis, $∥ Ψ ∥^{2} = 1$ , and $ρ = T_{Ψ, Ψ}$ is a density operator with $tr ρ = 1$ (Hilbert tensor construction: inner product, completion and product bases, Density operators and scalar spectral probability(a)); a partial trace ${Tr}_{K}$ is the unique linear map with $tr (({Tr}_{K} T) A) = tr (T (A \otimes I_{K}))$ , it is linear, and ${Tr}_{W} T_{Φ, Φ} = C C^{*}$ for the coefficient map $C$ of $Φ$ (Finite partial trace, Schmidt decomposition and finite no-signalling(a),(b)).

[L2]

${Tr}_{W}$ is linear, and for endomorphisms $M_{a}$ of $W$ with $\sum_{a} M_{a}^{*} M_{a} = I_{W}$ the nonselective sum satisfies ${Tr}_{W} (\sum_{a} (I \otimes M_{a}) R (I \otimes M_{a}^{*})) = {Tr}_{W} R$ for every endomorphism $R$ ; the lemma covers only the nonselective sum and says nothing about one selected summand (Finite partial trace, Schmidt decomposition and finite no-signalling(d), Local trace-preserving Kraus maps do not change the opposite reduced state).

[L3]

The refuted claim: for every unit $Ψ$ and every outcome $a$ of a local family $(M_{a})$ with $\sum_{a} M_{a}^{*} M_{a} = I$ , the conditional reduced state ${Tr}_{K} ((I \otimes M_{a}) ρ (I \otimes M_{a}^{*})) / tr ((I \otimes M_{a}) ρ (I \otimes M_{a}^{*}))$ equals ${Tr}_{K} ρ$ whenever the probability is positive.

Counterexample

1.1L1algebra

[The instrument] On $K$ take the self-adjoint projections $M_{0} : = P_{0} = T_{e_{-}, e_{-}}$ and $M_{1} : = P_{1} = T_{e_{+}, e_{+}}$ , so that $M_{a}^{*} M_{a} = P_{a}$ for $a = 0, 1$ and $\sum_{a} M_{a}^{*} M_{a} = P_{0} + P_{1} = I_{K}$ : this is the stipulated Lüders instrument for the sharp observable $P_{0}$ . Its outcome- $0$ non-normalized conditional operator on $H \otimes K$ is $(I \otimes P_{0}) ρ (I \otimes P_{0})$ .

2.1step 1.1L1algebra

[The selected outcome] Let $Ψ_{0} : = (I \otimes P_{0}) Ψ = \frac{1}{\sqrt{2}} e_{+} \otimes e_{-}$ . Its coefficient map in the chosen bases has matrix $C_{0} = \frac{1}{\sqrt{2}} (\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array})$ . The outcome probability is $tr ((I \otimes P_{0}) ρ (I \otimes P_{0})) = ∥ Ψ_{0} ∥^{2} = \frac{1}{2} > 0$ , and the non-normalized conditional operator is $T_{Ψ_{0}, Ψ_{0}} = \frac{1}{2} T_{e_{+} \otimes e_{-}, e_{+} \otimes e_{-}}$ . After normalization the conditional state is $T_{e_{+} \otimes e_{-}, e_{+} \otimes e_{-}}$ .

3.1step 2.1L1algebra

[The conditional reduced state differs] The coefficient map of the unit vector $e_{+} \otimes e_{-}$ has one nonzero matrix entry, equal to $1$ in the $e_{+}$ row and $e_{-}$ column, so its product with its adjoint is $T_{e_{+}, e_{+}}$ . By [L1], the conditional reduced state on $H$ is therefore ${Tr}_{K} T_{e_{+} \otimes e_{-}, e_{+} \otimes e_{-}} = T_{e_{+}, e_{+}}$ . The unconditional reduced state is ${Tr}_{K} ρ = C C^{*}$ for the coefficient matrix $C = \frac{1}{\sqrt{2}} (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ of $Ψ$ , namely $\frac{1}{2} I_{H}$ . These differ, since $⟨ T_{e_{+}, e_{+}} e_{+}, e_{+} ⟩ = 1 \neq \frac{1}{2} = ⟨ \frac{1}{2} I_{H} e_{+}, e_{+} ⟩$ . Hence the selective conditional reduced state is not the unconditional one, and the claim of [L3] fails for this unit vector and instrument.

4.1step 3.1L1L2algebra∎

[The nonselective identity still holds] The outcome- $1$ summand is $(I \otimes P_{1}) ρ (I \otimes P_{1})$ , and $(I \otimes P_{1}) Ψ = - \frac{1}{\sqrt{2}} e_{-} \otimes e_{+}$ , so this summand equals $\frac{1}{2} T_{e_{-} \otimes e_{+}, e_{-} \otimes e_{+}}$ . The coefficient map of the unit vector $e_{-} \otimes e_{+}$ has one nonzero entry in the $e_{-}$ row, so its reduced state on $H$ is $T_{e_{-}, e_{-}}$ by [L1]. Adding both summands and using the linearity of ${Tr}_{K}$ [L2] gives ${Tr}_{K} (\sum_{a} (I \otimes M_{a}) ρ (I \otimes M_{a}^{*})) = \frac{1}{2} T_{e_{+}, e_{+}} + \frac{1}{2} T_{e_{-}, e_{-}} = \frac{1}{2} I_{H} = {Tr}_{K} ρ$ , in agreement with the nonselective identity of [L2]. The refuted claim is thus exactly the illegitimate replacement of that nonselective identity by its selective counterpart.

Selective-conditioning counterexample

Statement refuted

Facts & Assumptions

Counterexample

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