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Pure entanglement and correlated mixtures

Physical scope

Exact two-qubit state, marginal, entropy and sharp-observable probability calculations under the adopted normal-state Born interpretation; no apparatus or experimental outcome is asserted.

Example

With $k_{B} > 0$ as fixed by Quantum spectral entropy and relative-entropy domains, for two qubits with basis $∣ 0 ⟩, ∣ 1 ⟩$ , compare the Bell pure state $∣ Φ ⟩ = (∣ 00 ⟩ + ∣ 11 ⟩) / \sqrt{2}$ , with density $ρ_{Φ} = ∣ Φ ⟩ ⟨ Φ ∣$ , and the classically correlated mixture $ρ_{m i x} = \frac{1}{2} ∣ 00 ⟩ ⟨ 00 ∣ + \frac{1}{2} ∣ 11 ⟩ ⟨ 11 ∣ .$ Both states have marginals $I_{2} / 2$ and hence marginal entropy $k_{B} \log 2$ . The global entropies differ: $S (ρ_{Φ}) = 0, S (ρ_{m i x}) = k_{B} \log 2.$ The Bell pure state is entangled for the declared two-qubit splitting; the mixture is separable. The dimensionless Hermitian observable $W = ∣ 00 ⟩ ⟨ 11 ∣ + ∣ 11 ⟩ ⟨ 00 ∣$ has outcome $+ 1$ with probability $1$ in $ρ_{Φ}$ , and probabilities $1 / 2, 1 / 2$ for $+ 1, - 1$ in $ρ_{m i x}$ . These are exact predictions of the adopted measurement model, not reports of an apparatus.

Facts & Assumptions

Given: The declared splitting $C^{2} \otimes C^{2}$ , the two density operators above, the finite partial trace, the positive entropy scale $k_{B} > 0$ in J/K, the entropy convention $0 \log 0 = 0$ , and the normal-state Born interpretation.

[A1]

The Axiom of Choice and The Axiom of Countable Choice ( ${A C}_{ω}$ ): retain the choice assumptions of the imported separability, spectral and normal-state supplier closure.

[F1]

Product vectors, separable states and entangled states of a two-factor composite: separable states are the trace-norm closed convex hull of pure product states, and entangled means outside that set for the declared splitting.

[F2]

Pure-state separability is equivalent to being a product: a Schmidt overlap witness: a pure state is separable exactly when its unit vector is a product vector.

[F3]

Partial trace over a finite tensor factor, The partial trace is characterized by the trace pairing and is basis-independent, and The partial trace preserves non-negativity and total trace: the finite partial trace is characterized by the trace pairing, sums the matching diagonal factor indices in a product basis, and preserves positivity and trace.

[F4]

Quantum spectral entropy and relative-entropy domains and Finite mixing, product entropy, and bipartite subadditivity: finite entropy is the spectral sum $- k_{B} \sum p \log p$ , and finite reduced-state entropies obey subadditivity.

[F5]

Normal-state quantum measurement interpretation: the adopted sharp-measurement probabilities are $Tr (ρ P)$ for spectral projections $P$ .

Proof

1.1givenA1F1F2algebra

Under the retained choice hypotheses [A1], the imported pure-state criterion applies. Suppose $∣ Φ ⟩ = a \otimes b$ , where $a = a_{0} ∣ 0 ⟩ + a_{1} ∣ 1 ⟩$ and $b = b_{0} ∣ 0 ⟩ + b_{1} ∣ 1 ⟩$ . Comparing coefficients with $(∣ 00 ⟩ + ∣ 11 ⟩) / \sqrt{2}$ gives $a_{0} b_{0} = a_{1} b_{1} = 1 / \sqrt{2}$ but $a_{0} b_{1} = a_{1} b_{0} = 0$ . The first pair makes all four factors nonzero, contradicting the second pair. Thus $∣ Φ ⟩$ is not a product vector, so [F2] makes $ρ_{Φ}$ entangled under the exact definition [F1].

1.2givenF3F4algebra

In the product basis, the partial-trace rule [F3] removes $∣ 00 ⟩ ⟨ 11 ∣$ and $∣ 11 ⟩ ⟨ 00 ∣$ , since the traced factor basis indices differ, and retains the two diagonal terms. Therefore both reductions of $ρ_{Φ}$ are $I_{2} / 2$ ; positivity and trace preservation in [F3] make them density matrices. The mixture already has only diagonal terms, and tracing either factor likewise gives $I_{2} / 2$ . The spectrum of $ρ_{Φ}$ is $(1, 0, 0, 0)$ , whereas that of $ρ_{m i x}$ is $(1 / 2, 1 / 2, 0, 0)$ ; each marginal has spectrum $(1 / 2, 1 / 2)$ . Substituting these eigenvalues into the entropy definition [F4] gives the displayed three entropy values. In particular both pairs obey the finite subadditivity inequality in [F4].

2.1givenF1F5algebra∎

The mixture is the convex combination with weights $1 / 2, 1 / 2$ of the two pure product densities $∣ 00 ⟩ ⟨ 00 ∣$ and $∣ 11 ⟩ ⟨ 11 ∣$ ; hence it lies in the finite convex hull in [F1] and is separable. Put $∣ Φ_{\pm} ⟩ = (∣ 00 ⟩ \pm ∣ 11 ⟩) / \sqrt{2}$ ; then $W ∣ Φ_{\pm} ⟩ = \pm ∣ Φ_{\pm} ⟩$ , and $W$ vanishes on the orthogonal span of $∣ 01 ⟩, ∣ 10 ⟩$ . The Bell state is $∣ Φ_{+} ⟩$ , so [F5] assigns probability one to $+ 1$ . Each of $∣ 00 ⟩, ∣ 11 ⟩$ has squared overlap $1 / 2$ with each of $∣ Φ_{+} ⟩, ∣ Φ_{-} ⟩$ , so the equal mixture gives probability $1 / 2$ to each of $+ 1, - 1$ and zero to $0$ .

Pure entanglement and correlated mixtures

Example

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources