Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 15E (Quantum Information and Computation)

After transmission Alice and Bob announce bases, retain matching-basis bits, reveal a random sample, and abort if its error rate is too large; privacy amplification and error correction produce the key. A Breidbart intercept-resend causes error $1/4$ conditional on matching Alice and Bob bases. Since matching occurs with probability $1/2$, one transmitted qubit detects Eve with probability $1/8$.

Conditional on Alice and Bob using the same basis, Eve chooses the wrong basis with probability $2/3$, then Bob disagrees with probability $1/2$, so the conditional error is $1/3$. Alice and Bob match with probability $1/3$, hence the per-transmission detection probability is $1/9$.

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.