Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 10D (Quantum Information and Computation)

For each position $i$, Alice uses $y_i$ to choose a basis and $x_i$ to choose a vector in that basis. Explicitly, she sends the qubitThus $y_i=0$ uses the computational basis $\{|0⟩,|1⟩\}$, while $y_i=1$ uses the Hadamard basis $\{|+⟩,|-⟩\}$.

Alice sends the $m$ conjugate-coded qubits through the noiseless quantum channel. Bob independently chooses a uniformly random basis bit $y_i^{\prime }$ for each qubit and performs a projective measurement in the computational basis when $y_i^{\prime }=0$ or the Hadamard basis when $y_i^{\prime }=1$.

After all measurements, Alice and Bob publicly reveal $Y$ and $Y^{\prime }$ but keep the data bits secret. They discard every position with $y_i\ne y_i^{\prime }$ and retain the measured bits at positions where the bases agree. In the ideal channel those retained bits equal Alice's $x_i$ with certainty. Each pair of bases agrees with probability $1/2$, so the shared secret key has expected length $m/2$.

Condition on a position retained after basis reconciliation, so Bob used Alice's basis. If Eve does not intercept, no error occurs. If she intercepts, her uniformly chosen basis agrees with Alice's with probability $1/2$, again causing no error. With probability $1/2$ she uses the conjugate basis; her measurement and resent state then make Bob's bit uniformly random, giving an error with probability $1/2$.

Therefore the bit error rate in the retained key is