Let

∣ θ^{⊥} ⟩ = sin θ ∣0 ⟩ - cos θ ∣1 ⟩, c = \frac{1}{1 + c o s θ} .

The three effects

E_{1} = c ∣ θ^{⊥} ⟩ ⟨ θ^{⊥} ∣, E_{2} = c ∣1 ⟩ ⟨ 1∣, E_{?} = I - E_{1} - E_{2}

form a POVM, because the largest eigenvalue of the sum of the two rank-one projectors is

1 + cos θ

. Outcome 1 never occurs on

∣ θ ⟩

, while outcome 2 never occurs on

∣0 ⟩

. Thus conclusive outcomes are never wrong;

E_{?}

ii

Solution

Arrange the amplitudes of

∣ ϕ_{1} ⟩

C_{1} = (a c b d) .

The second state has

C_{2} = C_{1} X

, where

X

swaps the

A_{2}

basis states. If Bob receives

A_{1}

ρ_{A_{1}}^{(1)} = C_{1} C_{1}^{†}, ρ_{A_{1}}^{(2)} = C_{2} C_{2}^{†} = C_{1} X X^{†} C_{1}^{†},

and are identical. No measurement on

A_{1}

contains any information about the shared state.

If Bob instead receives

A_{2}

ρ_{A_{2}}^{(1)} = (a^{2} + c^{2} ab + c d ab + c d b^{2} + d^{2}), ρ_{A_{2}}^{(2)} = X ρ_{A_{2}}^{(1)} X .

They differ because

a^{2} + c^{2} < b^{2} + d^{2}

. Bob can therefore distinguish them with better-than-random success. For equal priors,

D (ρ_{A_{2}}^{(1)}, ρ_{A_{2}}^{(2)}) = b^{2} + d^{2} - a^{2} - c^{2},

so the optimal success probability is

\frac{1}{2} (1 + D)

. It is generally below one, so a single copy does not permit certain identification.

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