Quantum teleportation

A qubit cannot be copied — the no-cloning theorem forbids any unitary that duplicates an unknown $∣ ψ ⟩$ . But a qubit's state can be moved from one place to another, by consuming a shared Bell pair and sending two classical bits over a classical channel. The state leaves the sender's qubit in the process — no copy of $∣ ψ ⟩$ remains at the sender afterwards — and the destination ends up holding a different physical qubit in that same quantum state. This is quantum teleportation ^{Bennett et al. 1993}.

The network's job is to deliver Bell pairs; teleportation is what users do with them. Each qubit teleported consumes one Bell pair and two classical bits. In the entanglement-mediated architecture covered by 1G and 2G repeaters, this is the only way to move an unknown qubit between nodes — no-cloning forbids copying it, and any single photon carrying the state would be lost to channel attenuation long before reaching the far end. 3G repeaters take a different route: they send QEC-encoded photonic qubits directly down the channel and correct loss/operation errors at each hop, with no Bell pair or classical sideband involved. The repeaters page covers that contrast in detail.

What teleportation does

Alice has a qubit Q in some unknown state $∣ ψ ⟩ = α ∣0 ⟩ + β ∣1 ⟩$ . She wants Bob to end up holding that same state. The setup: Alice and Bob already share a Bell pair — qubit A on Alice's side, qubit B on Bob's. Alice performs a Bell-state measurement (BSM) on her two qubits Q and A; this collapses them into one of the four Bell states and produces two classical bits $(a, b)$ . The same instant the BSM fires, B becomes $P ∣ ψ ⟩$ — call it $P (Q)$ — where the Pauli $P \in {I, X, Z, Z X}$ is fixed by the Bell-state outcome. Alice sends $(a, b)$ to Bob over a classical channel; Bob reads the bits, looks up which $P^{- 1}$ they correspond to, and applies it to $P (Q)$ — which gives back $Q$ , so B is now in $∣ ψ ⟩$ . Q and A are consumed in the measurement and the Bell-pair entanglement is gone — exactly one copy of $∣ ψ ⟩$ exists at any moment.

1. Alice has Q; A and B are a shared Bell pair
2. Alice applies BSM on Q and A to get 2 classical bits (a, b); B becomes P|ψ⟩
3. Bits travel to Bob
4. Bob applies P⁻¹
5. B holds |ψ⟩

Q (purple) is the qubit to teleport. A and B (teal) are halves of a shared Bell pair, drawn connected by the high-frequency wavy line. Q and A travel through Alice's BSM apparatus and arrive at two detectors; the path changes each cycle to match which pair of detectors fired. The two clouds turn grey because the original states are gone the moment the measurement begins. Two classical bits emerge on the channel below. At that same instant B turns purple, marked with the Pauli P the BSM outcome left on it — its state is now

P (Q)

. When the bits arrive at Bob, the matching

P^{- 1}

chip rises to meet that P badge and the two cancel. B now holds Q's state.

The theory

The starting resource is a Bell pair $∣ Φ^{+} ⟩_{A B} = (∣00 ⟩ + ∣11 ⟩) / 2$ shared between Alice and Bob — Alice holds qubit $A$ , Bob holds qubit $B$ . Alice additionally holds the qubit $Q$ in the state $∣ ψ ⟩$ she wants to teleport. The protocol then proceeds in four steps:

Initial joint state. The three qubits together are $∣ ψ ⟩_{Q} \otimes ∣ Φ^{+} ⟩_{A B}$ . Re-expanding on $Q, A$ in the Bell basis,
$∣ ψ ⟩_{Q} \otimes ∣ Φ^{+} ⟩_{A B} = \frac{1}{2} (∣ Φ^{+} ⟩_{Q A} ∣ ψ ⟩_{B} + ∣ Φ^{-} ⟩_{Q A} Z ∣ ψ ⟩_{B} + ∣ Ψ^{+} ⟩_{Q A} X ∣ ψ ⟩_{B} + ∣ Ψ^{-} ⟩_{Q A} Z X ∣ ψ ⟩_{B}) .$
The state of the three qubits hasn't changed — this is the same vector written in a different basis. But it now exposes that $B$ is correlated with $∣ ψ ⟩$ up to a Pauli operator, indexed by which Bell state $Q, A$ sit in.
Bell-state measurement at Alice. Alice measures $Q$ and $A$ jointly in the Bell basis. The measurement projects them onto one of the four Bell states with equal probability ¼ and yields two classical bits $(a, b)$ .
Classical communication. Alice sends $(a, b)$ to Bob over a classical channel. At this point Bob's qubit is in $P ∣ ψ ⟩$ — the right state up to a known Pauli operator $P \in {I, X, Z, Z X}$ .
Pauli correction at Bob. Bob applies $X^{b} Z^{a}$ , which equals the inverse of $P$ . His qubit is now in $∣ ψ ⟩$ .

The bit pattern $(a, b)$ tells Bob which Bell state Alice's BSM collapsed onto, and therefore which Pauli to apply:

(a, b)	BSM outcome	State at Bob ( $P ∣ ψ ⟩$ )	Bob applies $P^{- 1}$
(0, 0)	$∣ Φ^{+} ⟩$	$∣ ψ ⟩$	$I^{- 1} = I$
(0, 1)	$∣ Ψ^{+} ⟩$	$X ∣ ψ ⟩$	$X^{- 1} = X$
(1, 0)	$∣ Φ^{-} ⟩$	$Z ∣ ψ ⟩$	$Z^{- 1} = Z$
(1, 1)	$∣ Ψ^{-} ⟩$	$Z X ∣ ψ ⟩$	$(Z X)^{- 1} = X Z$

$I$ , $X$ , $Z$ are each their own inverse ( $X^{2} = Z^{2} = I$ ); $(Z X)^{- 1} = X^{- 1} Z^{- 1} = X Z$ .

Circuit diagram

As a quantum circuit, the protocol runs on three wires — Alice's input qubit $Q$ , Alice's half of the Bell pair $A$ , and Bob's half $B$ — with time flowing left to right. The three conceptual stages are Bell-pair creation, the Bell-state measurement, and the Pauli correction ^{Cacciapuoti et al. 2020}.

Bell-pair creation: a Hadamard on Alice's

∣0 ⟩

followed by a CNOT onto Bob's

∣0 ⟩

produces the Bell pair

∣ Φ^{+} ⟩_{A B}

. Bell-state measurement: a CNOT, a Hadamard, and two single-qubit measurements project Alice's

Q

and

A

jointly onto a Bell state, yielding two classical bits — a from the

∣ ψ ⟩

measurement, b from the Bell-half measurement. Double lines are the classical wires carrying those bits to Bob. Pauli correction: Bob applies

X^{b}

followed by

Z^{a}

to undo the Pauli operator the BSM outcome left on his qubit, completing the transfer of

∣ ψ ⟩

onto

B

No faster-than-light shortcut

The instant non-locality of entanglement was exactly what Einstein, Podolsky and Rosen objected to in their 1935 paper — Einstein later dismissed it as spukhafte Fernwirkung, spooky action at a distance. Teleportation looks like a natural test case for that worry: Alice's BSM instantaneously fixes Bob's qubit into $P ∣ ψ ⟩$ . So why doesn't it let Alice signal superluminally by choosing what to measure?

Before Alice's classical message arrives, Bob's qubit $B$ is in a uniform mixture of $∣ ψ ⟩$ , $X ∣ ψ ⟩$ , $Z ∣ ψ ⟩$ , and $Z X ∣ ψ ⟩$ — its local density matrix is $\frac{1}{2} I$ , the maximally mixed state, regardless of $∣ ψ ⟩$ . Bob's local statistics are independent of Alice's outcome, so he learns nothing about $∣ ψ ⟩$ from his half alone. Only when he receives the two bits and applies the matching correction does the qubit become useful. Teleportation is therefore compatible with relativity — no information crosses faster than the classical bits do, and Einstein's worry survives as a piece of intuition about a phenomenon that turns out to obey the speed limit after all.

Resources and limits

Each teleported qubit consumes a fixed budget:

1 Bell pair, consumed in Alice's BSM. The Bell pair is the quantum resource — getting it to both ends is the network's job (covered in the distribution and links subjects).
2 classical bits, sent over a classical channel from Alice to Bob. The classical message is what makes teleportation no faster than light.
The original $∣ ψ ⟩$ at Alice, gone. Alice's BSM projects $Q$ and $A$ jointly onto a Bell state, leaving no copy of $∣ ψ ⟩$ at Alice — consistent with no-cloning. In photonic BSM the $Q$ and $A$ photons are absorbed at the detectors; in matter implementations (trapped ions, NV centres) the qubits remain physically but are now in a known Bell state with each other. Either way, exactly one copy of the state exists in the protocol at any time ^{Wootters et al. 1982}.

Two consequences follow from this resource ledger:

Teleportation inherits the Bell pair's fidelity. A noisy pair gives Bob a noisy copy of Q's state; the cleaner the pair, the cleaner the teleported state. This is why purification exists — to take noisy pairs and produce cleaner ones.
One Bell pair per qubit, no reuse. The protocol is one-shot: each teleport consumes its Bell pair entirely. Teleporting many qubits requires many Bell pairs, which is why entanglement-distribution rate (Bell pairs per second) is the standard network rate metric.

Network role

Teleportation sits at the boundary between the network and the application. The network promises one thing: deliver Bell pairs at adequate rate, fidelity, and reach. The application's "send a qubit" operation is implemented by teleportation on top of that promise. Other applications — entanglement-based QKD, blind quantum computing, distributed quantum computing — also consume delivered Bell pairs; teleportation is the simplest case of using one ^{Kumar et al. 2025}.

Two related primitives, covered in their own subjects, share teleportation's BSM machinery:

Entanglement swapping uses a BSM at a Bell-state analyser (BSA) to extend a Bell pair across an extra hop, without any of the four end-to-end qubits ever interacting. Memory-based repeaters chain swaps to span long distances. Covered in the swapping subject.
Purification applies local CNOTs and measurements at both ends of a noisy pair to trade two low-fidelity pairs for one higher-fidelity pair. Covered in the purification subject.