 Solution

In quantum teleportation, Alice transfers an arbitrary unknown qubit $|\chi ⟩=\alpha |0⟩+\beta |1⟩$ using one shared Bell pair and two bits of classical communication. Define $|{\beta }_{mn}⟩=(I\otimes X^n{Z}^m)|{\Phi }^{+}⟩$, with $m,n\in \{0,1\}$. Regrouping the initial three-qubit state into this Bell basis givesAlice's Bell-basis measurement selects each label with probability $1/4$. She sends $(m,n)$ to Bob, who applies the inverse $Z^m{X}^n$, restoring $|\chi ⟩$. The correction depends on the measurement label, not on the unknown amplitudes. Before the message, Bob's average state is $I/2$, independent of the input; quantum no-signalling therefore prevents superluminal communication. The shared pair is consumed and Alice does not retain an independent copy, in agreement with the no-cloning theorem.

The Bouwmeester quantum teleportation experiment used photon polarization, a down-conversion entangled pair, a beam-splitter Bell-state analyzer and coincidence-conditioned output polarization measurements. Temporal indistinguishability of the two analyzer photons was essential. Its successful selected Bell outcome demonstrated the state transfer, rather than a deterministic implementation of all four outcomes.^[ref]

The general experimental tasks are to distribute entanglement, prepare and synchronize the input, perform a sufficiently discriminating joint measurement, communicate the label and apply or account for the correction. A demonstration must test nonorthogonal input states, not merely transfer a known classical basis label. Finite efficiency, accidental coincidences, distinguishability and phase drift reduce success probability or teleportation fidelity. Postselected verification, a heralded usable output and unconditional deterministic teleportation are distinct experimental claims.