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In entanglement swapping, start with independent Bell pairs $|{\varphi }^{+}{⟩}_{12}|{\varphi }^{+}{⟩}_{34}$. Send qubits 2 and 3 to a central Bell state analyzer, retaining 1 and 4 at separate endpoints. Reordering the tensor product factors to group 23 and 14 gives the identityConsequently each ideal Bell-basis measurement outcome has probability $1/4$ and projects 14 onto the corresponding Bell state, even though those two photons have never interacted. Communicating the two-bit result permits an endpoint Pauli gate to convert every outcome into the same chosen Bell pair. Without that record, the endpoints have ${\rho }_{14}=I_4/4$, a product state, rather than a known entangled state.

The optical Bell state analyzer overlaps photons 2 and 3 at a balanced beam splitter, with matched spatial modes, spectra and arrival times, and resolves the output photon polarizations using polarizing beam splitters and single-photon detectors. Two-photon interference maps Bell states to different output patterns. The antisymmetric $|{\psi }^{-}⟩$ gives one photon in each output with opposite polarizations. The symmetric $|{\psi }^{+}⟩$ gives two opposite-polarization photons in the same output; the two polarization detectors there can resolve this pattern. The $|{\varphi }^{\pm }⟩$ states give same-polarization pairs and remain mutually ambiguous in this basic setup. The detected pattern therefore heralds a particular endpoint Bell state. With ideal detection this arrangement succeeds unambiguously on two of four equally likely inputs, with probability $1/2$; the successful events can be selected. A complete analyzer would accept all four outcomes. The endpoint quantum entanglement is conditional on the measurement record, so it enables no instantaneous signalling.^[ref]

Before the gates, write the two Bell states asThe CNOT gate sends the last bit to $b\oplus a$. Projection onto outcome 0 on qubit 4 therefore requires $b=a$. After the Hadamard gate on qubit 2, projection onto its outcome 0 contributes $1/\sqrt{2}$ for either remaining value of $a$. The unnormalized surviving state isThusThis is entanglement swapping: photons 1 and 3 become entangled even though the joint measurement acts only on their original partners.

Yes: deterministic entanglement swapping gives the required singlet. Denote Bob's two local qubits by $B_1,B_2$. Known local unitary operations first convert the two given spin singlet states into $|{\Phi }^{+}{⟩}_{A{B}_1}$ and $|{\Phi }^{+}{⟩}_{B_{2}C}$. For example applying $ZX$ to the second qubit of a singlet gives $|{\Phi }^{+}⟩$, where $X,Z$ are Pauli operators.

Use the binary-labelled Bell statesThey are ${\Phi }^{+},{\Phi }^{-},{\Psi }^{+},{\Psi }^{-}$ for labels $(0,0),(1,0),(0,1),(1,1)$. Reordering registers to group Bob's qubits, direct expansion givesAll four basis vectors are real in this convention; otherwise the paired expansion must account for complex conjugation. Bob performs the Bell-basis measurement on $B_1{B}_2$. This joint operation is local to him, so it is allowed by LOCC. Each outcome occurs with probability $1/4$ and leaves $AC$ in the corresponding Bell state.

Bob sends $(p,q)$ to Charlie. Charlie applies $Z^p{X}^{-q}$ to obtain $|{\Phi }^{+}{⟩}_{AC}$, and then $XZ$ to obtainThe correction works for every outcome, so no postselection or successful-outcome assumption is needed. This is entanglement swapping, equivalently teleportation of $B_1$ to $C$ while preserving its entanglement with $A$. If Bob's measurement result is ignored, the average $AC$ state is $I_4/4$; the protocol therefore does not supply entanglement usable without the classical outcome or violate the no-communication theorem.

Bob uses qudit teleportation to teleport his carrier $B_1$ to Clare through the pair $B_{2}C$. This transfers Bob's entanglement with Alice to Clare. Explicitly, Bob performs a generalized Bell basis measurement on $B_1{B}_2$. Contraction of the two initial pairs givesThe probability is $1/n^2$ for each outcome. Bob sends $(r,s)$ to Clare, who applies $Z_C^r{X}_C^{-s}$. Alice and Clare then share $|{\psi }_{00}{⟩}_{AC}$ with certainty. This is entanglement swapping; neither an additional entangled pair nor a quantum transmission during the protocol is required.

The two initial pairs are consumed, and Bob's measured carriers cease to be entangled with $AC$. Without Bob's classical record, averaging the possible generalized Bell states leaves $AC$ maximally mixed. Thus the conditional entanglement swapping does not supply faster-than-light communication.

In 1998, Jian-Wei Pan, Dik Bouwmeester, Harald Weinfurter and Anton Zeilinger created two photon pairs by two passages of a pulsed pump through a crystal using spontaneous parametric down-conversion. Interfering the inner photons at a nonpolarizing beam splitter implemented a partial Bell-basis measurement. Narrow optical filters and spatial filtering removed distinguishing information; changing the outer polarizer bases checked the conditional correlations. Fourfold detections selected the entanglement swapping events. The reported interference supported conditional quantum entanglement, but the paper reserved a statistically significant Bell inequality violation for improved visibility; it was not a demonstration closing the detection loophole.^[ref]

The ideal calculation makes the distinction between conditional and unconditional quantum states explicit. Start with spin singlet states $|{\psi }^{-}{⟩}_{12}|{\psi }^{-}{⟩}_{34}$, where $|{\psi }^{-}⟩=(|HV⟩-|VH⟩)/\sqrt{2}$ and $H,V$ denote orthogonal photon polarizations. Taking the inner product with $⟨{\psi }^{-}{|}_{23}$ gives $-|{\psi }^{-}{⟩}_{14}/2$: expanding the four terms leaves precisely the two antisymmetric outer terms. That Bell-basis measurement outcome has probability $1/4$ and leaves the outer pair in a spin singlet state. Ignoring the inner outcome instead leaves the outer density matrix $I_4/4$, a maximally mixed state. A conditional preparation flag is useful for later Bell tests only with the timing and setting conditions in setting-independent heralding preserves Bell locality.