Checkpoint: Run BB84

This checkpoint ties the protocol together: you will simulate two sifted BB84 rounds — rounds where Alice and Bob already chose the same basis — and watch Bob deterministically recover Alice's bits. It is the operational core of BB84: matched bases ⇒ perfect agreement.

Setup

You build a 2-qubit circuit, one qubit per round:

Round A (qubit 0). Alice encodes the bit $1$ in the diagonal ( $X$ ) basis, i.e. she prepares $|-\rangle = \tfrac{1}{\sqrt2}(|0\rangle - |1\rangle)$ . Bob, having sifted to the same basis, measures in $X$ .
Round B (qubit 1). Alice encodes the bit $0$ in the rectilinear ( $Z$ ) basis, i.e. she prepares $|0\rangle$ . Bob measures in $Z$ .

Measuring in a rotated basis

The simulator measures in the computational ( $Z$ ) basis. To realize an $X$ -basis measurement, you first rotate the diagonal basis onto the computational one with a Hadamard, then measure:

H|+\rangle = |0\rangle, \qquad H|-\rangle = |1\rangle.

So Bob's $X$ -basis measurement of qubit 0 is "apply $H$ , then read $Z$ ." On round A the qubit's life is $|0\rangle \xrightarrow{X} |1\rangle \xrightarrow{H} |-\rangle \xrightarrow{H} |1\rangle$ , so Bob reads $1$ — exactly Alice's bit. On round B nothing rotates: $|0\rangle$ measured in $Z$ reads $0$ .

What the grader checks

Both rounds are deterministic when the bases match, so the joint outcome is the single basis state 10 (qubit 0 $\to 1$ , qubit 1 $\to 0$ ) with probability $1$ :

p(\texttt{10}) = 1, \qquad p(\texttt{00}) = p(\texttt{01}) = p(\texttt{11}) = 0.

Bob's sifted key 1,0 equals Alice's transmitted bits 1,0. That perfect agreement on sifted rounds is what makes the key usable — and any deviation from it (the QBER) is exactly the eavesdropping signal you studied earlier. You have now run, end to end, the quantum heart of BB84.

Run your code to see the quantum state.