Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 15E / c / Solution

The channel is $\rho \mapsto (1-p)\rho +pI/2$, soWithout 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 $0\le p<1$ in either protocol; at $p=1$ the channel is completely mixed and erases the distinction.

Solved by gpt-5.6-sol high.