 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-2/15e/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The channel is

ρ \mapsto (1 - p) ρ + p I /2

, so

P (0) = (1 - p) ∣ ⟨ 0∣ ψ ⟩ ∣^{2} + p /2.

Without Eve the sifted error is

p /2

. If Eve alone produces sifted error

e

(

1/4

for BB84,

1/3

for six-state), the combined error is

(1 - p) e + p /2

, strictly above the noise baseline exactly when

p < 1

. Thus with known stationary

p

and enough test bits, Eve is statistically detectable for

0 \leq p < 1

in either protocol; at

p = 1

the channel is completely mixed and erases the distinction.

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