Quantum Teleportation

Protocol Statement

Quantum teleportation transfers the state of an input system from a sender, Alice, to a receiver, Bob, using:

1. one entangled pair shared in advance;
2. a joint Bell-basis measurement by Alice;
3. two classical bits sent from Alice to Bob;
4. a correction chosen from the Pauli operators.

For an ideal qubit protocol, the induced channel from Alice’s input register $Q$ to Bob’s output register $B$ is exactly the identity:

$$\mathcal T_{Q\to B} = \operatorname{id}.$$

The word state is essential. The protocol does not transport the input particle, matter, energy, or a classical description of its amplitudes. It transfers the operational quantum state to a different physical carrier. The input state at Alice is consumed, the shared entangled pair is consumed, and Bob cannot recover the state until the classical message arrives.

This page is the canonical home for the protocol identity, circuit, resource accounting, no-signaling argument, and fidelity benchmarks. Bell States owns the four states and the Bell basis. No-Cloning and No-Signaling owns the general impossibility theorems and the distinction between conditional and unconditioned remote states. Entanglement Swapping owns the link-extension protocol, while Entanglement Sharing owns its multipartite structural context.

Registers and Shared Resource

Let the unknown input qubit be

$$|\psi\rangle_Q = \alpha|0\rangle_Q + \beta|1\rangle_Q, \qquad |\alpha|^2+|\beta|^2=1.$$

Alice does not need to know $\alpha$ or $\beta$. She holds $Q$ and one qubit $A$ of a shared Bell pair. Bob holds the other qubit $B$:

$$|\Phi^+\rangle_{AB} = \frac{ |00\rangle_{AB} + |11\rangle_{AB} }{\sqrt2}.$$

The initial three-qubit state is

$$|\Omega\rangle_{QAB} = |\psi\rangle_Q \otimes |\Phi^+\rangle_{AB}.$$

The Bell pair must have been distributed before the protocol’s classical communication step. Creating that pair ordinarily required a quantum channel, an entangling interaction, or a larger network protocol. Teleportation does not create remote connectivity for free; it converts previously distributed entanglement plus classical communication into one use of a quantum state-transfer channel.

The Bell-Basis Identity

Use the Bell-state convention

$$\begin{aligned} |\Phi^\pm\rangle &= \frac{|00\rangle\pm|11\rangle}{\sqrt2}, \\ |\Psi^\pm\rangle &= \frac{|01\rangle\pm|10\rangle}{\sqrt2}. \end{aligned}$$

Expanding the input and regrouping the first two qubits in the Bell basis gives the central identity:

$$\begin{aligned} |\psi\rangle_Q|\Phi^+\rangle_{AB} = \frac12\Big( & |\Phi^+\rangle_{QA}|\psi\rangle_B \\ &+ |\Phi^-\rangle_{QA}Z|\psi\rangle_B \\ &+ |\Psi^+\rangle_{QA}X|\psi\rangle_B \\ &+ |\Psi^-\rangle_{QA}XZ|\psi\rangle_B \Big). \end{aligned}$$

This equation contains the whole protocol.

• Alice’s Bell measurement projects $QA$ onto one of four orthogonal outcomes.
• Each outcome occurs with probability $1/4$, independent of $|\psi\rangle$.
• Conditioned on that outcome, Bob holds $|\psi\rangle$ up to a known Pauli operator.
• Two bits distinguish the four possible correction classes.

Let $a$ be the Bell phase bit and $b$ the Bell parity bit:

Outcome $ab$	Bell result on $QA$	Bob before correction	Bob applies
$00$	$\lvert\Phi^+\rangle$	$\lvert\psi\rangle$	$I$
$01$	$\lvert\Psi^+\rangle$	$X\lvert\psi\rangle$	$X$
$10$	$\lvert\Phi^-\rangle$	$Z\lvert\psi\rangle$	$Z$
$11$	$\lvert\Psi^-\rangle$	$XZ\lvert\psi\rangle$	$ZX$

In compact form, Bob’s conditional state before correction is

$$X^bZ^a|\psi\rangle.$$

He applies

$$C_{ab} = Z^aX^b.$$

The corrected state is

$$\begin{aligned} C_{ab}X^bZ^a|\psi\rangle &= Z^aX^bX^bZ^a|\psi\rangle \\ &= |\psi\rangle. \end{aligned}$$

For $a=b=1$, equivalent correction conventions can differ by a minus sign because $XZ=-ZX$. That sign is a global phase on the corrected pure state and has no observable effect. A protocol description should nevertheless state its bit ordering and Pauli convention explicitly.

Circuit Representation

A Bell-basis measurement can be implemented by a CNOT from $Q$ to $A$, followed by a Hadamard on $Q$ and computational-basis measurements. The inverse Bell transform obeys

$$\begin{aligned} (H_Q\otimes I_A) \operatorname{CNOT}_{Q\to A} |\Phi^+\rangle_{QA} &= |00\rangle_{QA}, \\ (H_Q\otimes I_A) \operatorname{CNOT}_{Q\to A} |\Psi^+\rangle_{QA} &= |01\rangle_{QA}, \\ (H_Q\otimes I_A) \operatorname{CNOT}_{Q\to A} |\Phi^-\rangle_{QA} &= |10\rangle_{QA}, \\ (H_Q\otimes I_A) \operatorname{CNOT}_{Q\to A} |\Psi^-\rangle_{QA} &= |11\rangle_{QA}. \end{aligned}$$

The first measurement bit $a$ records the Bell phase class, and the second bit $b$ records the parity class.

Canonical qubit teleportation circuit. Alice applies the inverse Bell transform and measures $Q$ and $A$, producing bits $a$ and $b$. Bob applies $X^b$ followed by $Z^a$, so the net correction operator is $Z^aX^b$. Double lines carry classical information; Bob’s quantum wire never crosses from Alice’s laboratory.

The circuit makes a timing fact visible. Bob may postpone the physical correction and track a Pauli frame, but he must receive $a$ and $b$ before he can interpret a noncommuting later measurement or deliver an unconditional output state. Classical feed-forward is part of the protocol, not optional bookkeeping.

Why There Is No Faster-Than-Light Signal

Bob without the message

Conditioned on $(a,b)$, Bob’s state is

$$\rho_B^{(a,b)} = X^bZ^a\rho_QZ^aX^b.$$

Before the message arrives, Bob must average over the four outcomes:

$$\rho_B = \frac14 \sum_{a,b\in\{0,1\}} X^bZ^a\rho_QZ^aX^b.$$

Write an arbitrary input density operator as

$$\rho_Q = \frac12 \left( I+r_xX+r_yY+r_zZ \right).$$

Conjugation by the four Pauli operators changes the signs of the Bloch components in all combinations. The components cancel in the average, leaving

$$\rho_B = \frac12 I.$$

Bob’s local state is therefore maximally mixed and independent of Alice’s input. No local measurement at Bob can reveal whether Alice has performed the Bell measurement, which input she held, or which outcome she obtained.

Only after a classical message traveling no faster than the allowed signaling speed can Bob select the correct conditional branch. Entanglement supplies correlations; it does not supply a controllable superluminal channel.

The two bits contain no state description

The outcome distribution is

$$p(a,b) = \frac14$$

for every input pure state. The bits do not encode estimates of $\alpha$ and $\beta$. Repeating the protocol on identically prepared inputs gives uniformly random Bell labels, not a tomography record of the input. Their role is to identify which Pauli frame Bob occupies.

This is why a continuum of possible qubit states can be teleported using only two classical bits. The continuous quantum information was already represented relationally in the input and the shared entanglement; the bits merely unlock the correct branch.

Why Teleportation Is Not Cloning

Alice’s Bell measurement consumes the original input as an independently available state. After the measurement, the $QA$ registers occupy a definite Bell-outcome branch, while Bob’s register contains the Pauli-rotated input. Once Bob corrects, there is one accessible copy at $B$, not one at $Q$ and another at $B$.

If a deterministic protocol left

$$|\psi\rangle_Q \longmapsto |\psi\rangle_Q|\psi\rangle_B$$

for every unknown $|\psi\rangle$, it would violate No-Cloning and No-Signaling. Teleportation avoids that conclusion by transferring rather than duplicating the state.

The word “unknown” does not mean no physical system has information about the input. The input may be entangled with an external reference. It means Alice need not possess a classical specification from which she could prepare another copy.

The Reference-System Test

The pure-state derivation is useful, but the strongest statement treats the input as part of an arbitrary joint state $\rho_{RQ}$ with an inaccessible reference $R$. An ideal teleportation protocol satisfies

$$\begin{aligned} \rho_{RB}^{\mathrm{out}} &= \left( \operatorname{id}_R \otimes \mathcal T_{Q\to B} \right) \left( \rho_{RQ} \right) \\ &= \rho_{RQ}, \end{aligned}$$

where the last expression relabels subsystem $Q$ as $B$.

Thus teleportation preserves:

• mixed input states;
• entanglement between the input and a reference;
• coherence relative to degrees of freedom outside Alice’s control;
• the action of the identity channel on every operator, not only selected test states.

This test distinguishes genuine quantum state transfer from “measure the input and prepare a guessed state.” A measure-and-prepare channel is entanglement breaking: when applied to half of an entangled state, it cannot preserve the original entanglement with $R$.

The same viewpoint connects teleportation to the channel–state correspondence. A maximally entangled resource is the Choi state of an identity channel, and the Bell measurement plus feed-forward turns that static bipartite resource into an operational channel.

Resource Accounting

For one deterministic ideal qubit teleportation:

Resource	Consumed or required
shared entanglement	one Bell pair, or one ebit
classical communication	two bits from Alice to Bob
local quantum operations	inverse Bell transform, two measurements, and a Pauli correction or frame update
input	one qubit state, consumed at Alice
output	one qubit at Bob in the transferred state

A standard resource inequality is

$$[qq] + 2[c\to c] \geq [q\to q].$$

Here $[qq]$ denotes a shared ebit, $[c\to c]$ one classical bit from Alice to Bob, and $[q\to q]$ one ideal use of a qubit channel. The inequality describes convertibility, not a conservation law.

Several costs lie outside the compact notation:

• distributing and verifying the entangled pair;
• storing it until the input and classical controller are ready;
• synchronizing independent sources;
• implementing a complete Bell measurement;
• losses, heralding probability, detector dead time, and reset;
• classical latency and feed-forward electronics;
• purification or error correction for noisy network links.

Teleportation can trade a difficult direct quantum transmission at the time of use for earlier entanglement distribution and later classical communication. Whether that trade is advantageous depends on the platform and network.

Generalizations

Qudits

For a $d$-level system, define

$$|\Phi_d\rangle = \frac1{\sqrt d} \sum_{j=0}^{d-1}|j\rangle_A|j\rangle_B,$$

and generalized Pauli operators

$$X|j\rangle = |j+1\!\!\!\pmod d\rangle, \qquad Z|j\rangle = \omega^j|j\rangle, \qquad \omega=e^{2\pi i/d}.$$

The generalized Bell basis has $d^2$ outcomes:

$$|\beta_{mn}\rangle = \left( I\otimes X^nZ^m \right) |\Phi_d\rangle.$$

Alice sends the pair $(m,n)$, requiring

$$\log_2(d^2) = 2\log_2d$$

classical bits in a fixed-length encoding. The shared maximally entangled state has entanglement entropy $\log_2d$ ebits. Bob applies the inverse generalized Pauli correction.

Continuous variables

Continuous-variable teleportation replaces the finite Bell basis by joint quadrature measurements and uses an EPR-like two-mode squeezed state. The two real-valued outcomes determine phase-space displacement feed-forward. Finite squeezing adds noise, so the ideal identity channel is approached only in a limiting resource regime. The finite-dimensional qubit protocol should not be transferred to continuous variables without changing both the resource and benchmark conventions.

Gate teleportation and measurement-based computation

If a resource state has a gate $U$ embedded in it, Bell measurement and feed-forward can produce $U|\psi\rangle$ at the output. For Clifford $U$, Pauli byproducts remain Pauli under propagation; for non-Clifford gates, correction structure is more demanding. This idea underlies teleportation-based gates, magic-state injection, and parts of Measurement-Based Quantum Computation.

The same correction structure appears in Blind and Delegated Quantum Computation, where a client masks adaptive measurement angles and tracks encrypted outcomes while logical information propagates through a graph state. Distributed Quantum Computing owns remote-gate constructions, teledata–telegate choices, partitioning, scheduling, and program-level resource accounting. The state-transfer identity remains the core primitive here. Protocol-specific privacy, verification, gate synthesis, and fault-tolerance costs belong to the corresponding computation pages.

Noisy Resources and Teleportation Fidelity

Bell-diagonal resource

Suppose the shared pair is Bell diagonal:

$$\omega_{AB} = \sum_{a,b} p_{ab} |\beta_{ab}\rangle \langle\beta_{ab}|, \qquad \sum_{a,b}p_{ab}=1,$$

with $|\beta_{00}\rangle=|\Phi^+\rangle$. Running the standard Bell measurement and correction produces a Pauli channel:

$$\mathcal T_\omega(\rho) = \sum_{a,b} p_{ab} X^bZ^a \rho Z^aX^b.$$

The ideal resource has $p_{00}=1$. The other Bell components appear as residual Pauli errors at Bob. This is a precise sense in which imperfect entanglement becomes channel noise.

For this qubit Pauli channel, the entanglement fidelity relative to the identity is

$$F_{\mathrm e} = p_{00},$$

and the Haar-averaged pure-state fidelity is

$$F_{\mathrm{avg}} = \frac{2F_{\mathrm e}+1}{3} = \frac{2p_{00}+1}{3}.$$

Local unitary optimization can relabel which Bell component is treated as the target. More general resources and protocols are characterized by their maximal singlet fraction and the allowed local operations.

The classical benchmark

For one unknown pure qubit drawn uniformly from the Bloch sphere, any deterministic measure-and-prepare strategy has

$$F_{\mathrm{avg}} \leq \frac23.$$

An ideal teleporter reaches $F_{\mathrm{avg}}=1$. Exceeding $2/3$ under the stated ensemble and deterministic accounting rules rules out that classical strategy class.

The number $2/3$ is not a universal teleportation benchmark. It changes or becomes inapplicable when:

• the input ensemble is not Haar uniform;
• the protocol postselects successful events;
• loss and no-output events are omitted from the denominator;
• multiple input copies or side information are available;
• the figure of merit is worst-case, entanglement, or process fidelity;
• continuous-variable states use an energy-constrained ensemble.

A trustworthy experiment states the ensemble, heralding rule, success probability, correction policy, loss treatment, and confidence interval alongside fidelity.

What an implementation should report

A complete teleportation demonstration separates:

1. resource quality: fidelity or entanglement of the shared pair;
2. Bell analyzer: which outcomes are resolved and with what success probability;
3. input independence: whether the input source is independent of the entanglement source;
4. feed-forward: whether Bob receives and uses the classical result in real time;
5. output availability: whether a freely usable output remains after heralding and verification;
6. state-transfer quality: conditional and unconditional fidelities with uncertainties;
7. rate: attempts, heralds, accepted outputs, and latency;
8. channel test: whether selected-state tomography or a reference-entanglement/process test is used.

Postselected teleportation can be valuable, especially in photonic systems, but success probability and output conditioning are part of the protocol claim. A high conditional fidelity at vanishing success rate is not equivalent to a deterministic high-rate channel.

What Is and Is Not Transmitted

Teleportation transfers:

• the complete quantum state supported by the input Hilbert space;
• coherence and entanglement with external systems;
• the ability to reproduce every later measurement statistic at Bob;
• an unknown state without Alice learning its amplitudes.

Teleportation does not transfer:

• the original material carrier;
• a classical list of amplitudes;
• a second perfect copy;
• usable information before the classical message arrives;
• energy or matter instantaneously;
• entanglement without consuming or transforming network resources;
• immunity to loss, decoherence, detector error, or finite-rate constraints.

The safest operational statement is: the protocol simulates an identity quantum channel from Alice’s input register to Bob’s output register, using pre-shared entanglement and classical communication.

Common Mistakes

• Saying the particle itself is teleported.
• Saying Alice measures or learns $\alpha$ and $\beta$.
• Omitting the two classical bits from the protocol.
• Treating Bob’s conditional state as usable before he knows the condition.
• Forgetting that the input and shared Bell pair are consumed.
• Describing teleportation as cloning.
• Treating entanglement distribution as free or instantaneous.
• Assuming every physical Bell analyzer is deterministic and complete.
• Quoting $2/3$ without defining the input ensemble and postselection rule.
• Verifying only a few convenient states and calling that an unconditional identity channel.
• Confusing state teleportation with entanglement swapping, where no independent input state is transferred.

Exercises

1. Derive the Bell decomposition

Starting from

$$|\psi\rangle_Q = \alpha|0\rangle+\beta|1\rangle,$$

expand $|\psi\rangle_Q|\Phi^+\rangle_{AB}$ and recover the four-term Bell-basis identity.

Solution

The computational-basis expansion is

$$\frac1{\sqrt2} \left( \alpha|000\rangle + \alpha|011\rangle + \beta|100\rangle + \beta|111\rangle \right).$$

Use

$$\begin{aligned} |00\rangle &= \frac{|\Phi^+\rangle+|\Phi^-\rangle}{\sqrt2}, & |11\rangle &= \frac{|\Phi^+\rangle-|\Phi^-\rangle}{\sqrt2}, \\ |01\rangle &= \frac{|\Psi^+\rangle+|\Psi^-\rangle}{\sqrt2}, & |10\rangle &= \frac{|\Psi^+\rangle-|\Psi^-\rangle}{\sqrt2}. \end{aligned}$$

Collecting Bob’s qubit in each Bell sector gives

$$\begin{aligned} \frac12\Big[ & |\Phi^+\rangle \left( \alpha|0\rangle+\beta|1\rangle \right) \\ &+ |\Phi^-\rangle \left( \alpha|0\rangle-\beta|1\rangle \right) \\ &+ |\Psi^+\rangle \left( \beta|0\rangle+\alpha|1\rangle \right) \\ &+ |\Psi^-\rangle \left( -\beta|0\rangle+\alpha|1\rangle \right) \Big], \end{aligned}$$

which is the stated identity because the four Bob states are $|\psi\rangle$, $Z|\psi\rangle$, $X|\psi\rangle$, and $XZ|\psi\rangle$.

2. Uniform outcomes

Why does every Bell outcome have probability $1/4$, independent of the input state?

Solution

Each Bell branch in the decomposition has amplitude factor $1/2$. The corresponding Bob state is a unitary Pauli transform of a normalized input, so its norm is one. Therefore each branch has squared norm

$$\left|\frac12\right|^2 = \frac14.$$

The result does not depend on $\alpha$ or $\beta$. Consequently, Alice’s two-bit record is uniformly random and carries no classical state description.

3. Correction for outcome 11

For outcome $a=b=1$, Bob holds $XZ|\psi\rangle$. Show that applying $ZX$ recovers the input.

Solution

Using $X^2=Z^2=I$,

$$(ZX)(XZ) = Z(XX)Z = I.$$

Hence

$$ZX\,XZ|\psi\rangle = |\psi\rangle.$$

Applying $XZ$ instead would give $-|\psi\rangle$ because $(XZ)^2=-I$. The minus sign is a global phase, so both conventions describe the same output ray.

4. No-signaling Pauli twirl

Let

$$\rho = \frac12 \left( I+r_xX+r_yY+r_zZ \right).$$

Evaluate the average of $\rho$, $X\rho X$, $Z\rho Z$, and $XZ\rho ZX$.

Solution

Pauli conjugation changes Bloch-vector signs:

$$\begin{aligned} X\rho X &= \frac12 \left( I+r_xX-r_yY-r_zZ \right), \\ Z\rho Z &= \frac12 \left( I-r_xX-r_yY+r_zZ \right), \\ XZ\rho ZX &= \frac12 \left( I-r_xX+r_yY-r_zZ \right). \end{aligned}$$

Adding these to $\rho$ cancels every Bloch component. Dividing by four yields

$$\frac14 \left( \rho+X\rho X+Z\rho Z+XZ\rho ZX \right) = \frac12I.$$

Bob’s unconditioned state is independent of the input.

5. Reference entanglement

Suppose $RQ$ begins in $|\Phi^+\rangle_{RQ}$. What must an ideal teleportation channel produce, and why can a measure-and-prepare channel not do the same?

Solution

Ideal teleportation applies the identity to $Q$ while leaving $R$ untouched, so the final state is

$$|\Phi^+\rangle_{RB}.$$

The entanglement has moved from the $RQ$ partition to $RB$. A measure-and-prepare channel first converts $Q$ into a classical outcome and then prepares $B$ conditionally. Such a channel is entanglement breaking, so its output across $R|B$ is separable and cannot equal a Bell state.

6. Why the original is gone

Explain why teleportation does not violate no-cloning even though Bob obtains the exact unknown state.

Solution

Alice performs a joint measurement on the input $Q$ and her entangled qubit $A$. After that measurement, $Q$ is not still available in the state $|\psi\rangle$; it belongs to a recorded Bell-outcome branch. Bob’s corrected register is the only output copy. The map is state transfer,

$$|\psi\rangle_Q \longrightarrow |\psi\rangle_B,$$

with other registers changed, not the forbidden cloning map

$$|\psi\rangle_Q \longrightarrow |\psi\rangle_Q|\psi\rangle_B.$$

7. Qudit resources

How much entanglement and classical communication does ideal teleportation of a $d$-level state use?

Solution

The shared state $|\Phi_d\rangle$ has $d$ equal Schmidt coefficients, so its entanglement entropy is

$$S = \log_2d$$

ebits. The generalized Bell measurement has $d^2$ outcomes. A fixed-length message identifying one outcome requires

$$\log_2(d^2) = 2\log_2d$$

classical bits. Bob then applies one of $d^2$ generalized Pauli corrections.

8. Bell-diagonal noise

A shared resource has $p_{00}=0.85$ and the remaining probability distributed among the other Bell states. Find the entanglement fidelity and average qubit teleportation fidelity of the standard protocol.

Solution

For the Bell-diagonal resource,

$$F_{\mathrm e} = p_{00} = 0.85.$$

The qubit average fidelity is

$$\begin{aligned} F_{\mathrm{avg}} &= \frac{2F_{\mathrm e}+1}{3} \\ &= \frac{2(0.85)+1}{3} \\ &= 0.90. \end{aligned}$$

This calculation assumes the target Bell component has already been aligned with $|\Phi^+\rangle$ and uses the deterministic standard protocol.

9. Benchmark audit

An experiment reports a conditional average fidelity of $0.75$ and says it beats “the classical limit.” What additional information is needed?

Solution

At minimum, the report must state:

• the input-state ensemble and sampling weights;
• whether inputs are independently prepared;
• the accepted-event or heralding rule;
• the success probability and treatment of no-output events;
• whether Pauli feed-forward is applied or inferred afterward;
• the classical comparison strategy and allowed side information;
• statistical uncertainty and state-estimation bias.

The familiar $2/3$ bound applies to a deterministic measure-and-prepare channel acting on one Haar-uniform unknown pure qubit. A conditional $0.75$ does not establish that comparison unless the experiment’s contract matches it.

10. Deferred correction

Bob plans to measure $Z$ immediately after teleportation. How can he use a Pauli frame instead of physically applying $Z^aX^b$?

Solution

A preceding $Z$ correction commutes with a $Z$ measurement and does not change its outcome. A preceding $X$ correction anticommutes with $Z$ and flips the outcome label. Bob can therefore measure directly and reinterpret the raw bit as

$$z_{\mathrm{corrected}} = (-1)^b z_{\mathrm{raw}}.$$

The phase bit $a$ is irrelevant for this particular final measurement. For a different basis or a later non-Clifford operation, both frame bits may affect the required interpretation or control.

Further Connections

• No-Cloning and No-Signaling proves the general operational restrictions that teleportation obeys.
• Communication with Quantum Systems places teleportation in the common task, channel, assistance, error, rate, and constraint framework and states its resource inequality.
• Superdense Coding uses the same Bell-pair and Bell-basis machinery for the complementary conversion of one transmitted qubit into two classical bits.
• Entanglement Swapping uses a Bell measurement to convert two neighboring entangled links into one conditional remote link, with no independent input state.
• Entanglement Distillation explains how several noisy shared pairs can be converted into fewer higher-fidelity teleportation resources, including the acceptance and yield cost.
• Bell States
• Conditional States
• Entanglement Sharing
• Multi-Qubit Gates
• Circuit Model
• Quantum Instruments
• No-Cloning Theorem
• Entanglement Measures
• Quantum Information Roadmap

References

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Summary

Quantum teleportation turns one shared Bell pair and two classical bits into one use of an identity quantum channel. Alice’s Bell measurement places Bob in one of four Pauli frames; the two-bit message identifies the frame, and Bob corrects it. Before that message arrives, Bob’s state is maximally mixed, so the protocol cannot signal faster than light.

The input is transferred rather than copied. The strongest test is preservation of entanglement with an arbitrary reference, which establishes channel identity rather than agreement on a few test states. In realistic implementations, entanglement quality, Bell-measurement success, feed-forward, loss, heralding, rate, and a declared fidelity benchmark all belong to the protocol claim.