 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2015/s3/3/3/solution

ρ = \frac{1}{2} ∣ HH ⟩ ⟨ HH ∣ + \frac{1}{2} ∣ VV ⟩ ⟨ VV ∣

. Average the probabilities of its two alternatives:

p (A, B) = \frac{1}{2} [cos^{2} θ_{A} cos^{2} θ_{B} + sin^{2} θ_{A} sin^{2} θ_{B}] .

Alice's single-detector transmission probability is

1/2

at every angle. Dividing by it gives

p (B ∣ A, θ_{A} = 4 5^{\circ}) = \frac{1}{2}, p (B ∣ A, θ_{A} = 0) = cos^{2} θ_{B} .

4 5^{\circ}

, Alice's click does not distinguish the two classical alternatives. At zero, it selects the horizontally polarized alternative.

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