Entanglement Swapping

A single Bell pair only reaches as far as one optical link. To connect nodes that are too far apart to share a direct pair, we swap entanglement through an intermediate node. The remarkable part: the two end qubits become entangled even though they have never interacted and were created independently.

The setup

Place a repeater station between Alice and Bob. Each end shares a Bell pair with the repeater:

|\Phi^+\rangle_{A R_1} \otimes |\Phi^+\rangle_{R_2 B},

where $R_1, R_2$ are the repeater's two qubits. In our register, $A = q_0$ , $R_1 = q_1$ , $R_2 = q_2$ , $B = q_3$ . Initially Alice is entangled only with $R_1$ , and Bob only with $R_2$ ; Alice and Bob share nothing.

Bell-state measurement at the repeater

The repeater performs a Bell-state measurement (BSM) jointly on its two qubits $R_1, R_2$ . A BSM projects a two-qubit system onto one of the four Bell states. Operationally it is the inverse of the entangler that creates a Bell pair: apply a CNOT and then a Hadamard, then measure in the computational basis.

\text{BSM} = \big(\text{measure}\big)\circ \big(H\otimes I\big)\circ \mathrm{CNOT}_{R_1\to R_2}.

This measurement consumes both repeater qubits and, by entanglement monogamy, transfers the correlations: Alice's $q_0$ and Bob's $q_3$ are now entangled, with which Bell state they share determined by the two measurement bits.

Feed-forward corrections

The four BSM outcomes leave $q_0,q_3$ in one of the four Bell states $|\Phi^\pm\rangle$ , $|\Psi^\pm\rangle$ . To standardise on $|\Phi^+\rangle$ , Bob applies a Pauli correction based on the classically communicated bits $(m_{R_1}, m_{R_2})$ :

\text{correction on } B = Z^{\,m_{R_1}}\, X^{\,m_{R_2}}.

This is the same Pauli-frame bookkeeping that appears in teleportation — entanglement swapping is literally teleportation of one half of a Bell pair. In a circuit we model the classically conditioned corrections with the deferred-measurement trick: a CNOT from $q_2$ onto $q_3$ implements $X^{m_{R_2}}$ , and a CZ from $q_1$ onto $q_3$ implements $Z^{m_{R_1}}$ .

The swap signature

After correction, the defining observable is that Alice's and Bob's outcomes are perfectly correlated: a computational-basis readout never finds $q_0 \neq q_3$ . The repeater qubits, having been measured, are uniformly random and carry no remaining link to the data — exactly as they should after being consumed.

Try it

Run the swap. The grader verifies the swap signature directly: it checks that all probability lies on outcomes with $q_0 = q_3$ , and that the probability of any $q_0 \neq q_3$ outcome is zero.

Run your code to see the quantum state.

Inspect the Probabilities tab: every populated bar has matching first and last bits, confirming that two never-interacting qubits are now entangled across the repeater.