Quantum Teleportation

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:

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

Bennett et al. (1993) introduced this conversion of entanglement and classical communication into quantum state transfer. 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.

Required background. Bell states supplies the entangled basis, multi-qubit gates gives the controlled-NOT convention, and projective measurement supplies the outcome probabilities.

Input qubit and shared Bell pair

Let the unknown input qubit be

$$\begin{aligned} |\psi\rangle_Q&=\alpha|0\rangle_Q+\beta|1\rangle_Q,\\ |\alpha|^2+|\beta|^2&=1. \end{aligned}$$

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. Write the initial state as $|\Omega\rangle$ to keep its four branches visible:

$$|\Omega\rangle = \frac12\Big( |\Phi^+\rangle_{QA}|\psi\rangle_B + |\Phi^-\rangle_{QA}Z|\psi\rangle_B + |\Psi^+\rangle_{QA}X|\psi\rangle_B + |\Psi^-\rangle_{QA}XZ|\psi\rangle_B \Big).$$

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. Define the inverse Bell transform

$$W=(H_Q\otimes I_A)\operatorname{CNOT}_{Q\to A}.$$

It maps the four Bell states to the two-bit labels used above:

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

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

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. For a terminal Pauli measurement, he may measure first and relabel the result after the relevant bit arrives; exercise 8 shows this explicitly.

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 or which input she chose. Random measurement outcomes can still be correlated; the unchanged marginal prevents a controllable message.

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 Bell outcome changes which conditional output Bob has, while the prior entanglement makes that conditional output possible. Neither the Bell pair alone nor the random message alone specifies the input amplitudes.

Why Teleportation Is Not Cloning

Alice’s Bell measurement consumes the original input as an independently available state. A nondestructive ideal Bell measurement leaves $QA$ in its recorded Bell state; the circuit above instead leaves computational-basis records after the inverse Bell transform. In either implementation, those registers no longer retain an independent copy of the input, while Bob holds the Pauli-rotated state. 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.

Resources and timing

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.

The notation omits the cost of distributing and storing the pair, resolving Bell outcomes, waiting for the classical message, and recovering from loss. A photonic Bell analyzer that recognizes only some outcomes produces a heralded protocol with a success probability, even when successful outputs have high fidelity. Noise reduction can consume several initial pairs for one useful pair; entanglement distillation explains that tradeoff.

Teleportation can replace a difficult direct quantum transmission at the time of use by earlier entanglement distribution and later classical communication. Whether this helps depends on memory lifetime, link success probability and the application’s timing requirements.

Generalizations

Qudits

For a $d$-level system, use the shared state

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

Define $X|j\rangle=|j+1\pmod d\rangle$ and $Z|j\rangle=\omega^j|j\rangle$, with $\omega=e^{2\pi i/d}$. One generalized Bell convention is

$$\begin{aligned} U_{mn}&=X^nZ^m,\\ |\beta_{mn}\rangle&=(I\otimes U_{mn})|\Phi_d\rangle. \end{aligned}$$

Contracting Alice’s two registers with $\langle\beta_{mn}|$ leaves the unnormalized Bob vector

$$\frac1d U_{mn}^*|\psi\rangle =\frac1d X^nZ^{-m}|\psi\rangle.$$

The star is entrywise conjugation in the chosen computational basis. Thus each outcome has probability $1/d^2$, and the correction for this Bell convention is

$$C_{mn}=(U_{mn}^*)^\dagger=Z^mX^{-n}.$$

One must derive the conjugation from the chosen Bell convention rather than import the qubit formula: the qubit matrices $X$ and $Z$ happen to be real, which hides this distinction. Werner (2001) develops the general relationship between unitary bases, Bell bases, teleportation and dense coding.

The pair contains $\log_2d$ ebits. A message with $d^2$ possible symbols has entropy $2\log_2d$ bits, but a single fixed-length binary message requires $\lceil2\log_2d\rceil$ bits. These numbers agree for powers of two. The entropy value is approached per use by block coding; it is not a fractional length for one binary string.

Continuous variables

The protocol of Braunstein and Kimble (1998) for 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. Gottesman and Chuang (1999) used this idea for teleportation-based gates; it also underlies magic-state injection and parts of Measurement-Based Quantum Computation.

The same Pauli corrections also enter blind and delegated quantum computation and distributed quantum computing. Their privacy and gate costs depend on the specific protocol; the identity channel derived here is the elementary state-transfer step.

Noisy Resources and Teleportation Fidelity

Bell-diagonal resource

Suppose the shared pair is Bell diagonal:

$$\begin{aligned} \omega_{AB}&=\sum_{a,b}p_{ab}|\beta_{ab}\rangle\langle\beta_{ab}|,\\ \sum_{a,b}p_{ab}&=1, \end{aligned}$$

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 follows from the channel identity derived by Horodecki et al. (1999) and Nielsen (2002):

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

For a Bell-diagonal pair, a local Pauli can align the largest Bell weight with the target before using the protocol. More generally, distinguish a fully entangled fraction optimized only over local unitaries from the fraction attainable after deterministic local operations and classical communication. Their different allowed transformations can produce different optimized teleportation fidelities.

The classical benchmark

For one unknown pure qubit drawn uniformly from the Bloch sphere, any deterministic measure-and-prepare strategy has (Massar and Popescu, 1995)

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

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 ensemble matters. If the only promised inputs are the orthogonal states $|0\rangle$ and $|1\rangle$, measuring $Z$ and preparing the result achieves fidelity one with no entanglement. Multiple copies, side information and continuous-variable ensembles also change the comparison.

Postselection requires its own calculation, but it does not automatically defeat the $2/3$ bound. For one exactly Haar-uniform unknown pure qubit, a physical measure-and-prepare strategy cannot raise its success-weighted conditional fidelity above $2/3$ merely by accepting selected measurement outcomes. Exercise 7 proves this statement. Biased test-state sampling, input-label-dependent acceptance or extra input copies change its assumptions.

Conditional fidelity and useful output rate answer different questions. An analyzer with high fidelity on rare heralds may preserve a state well when it succeeds while providing very few usable outputs. Compare protocols using their success probability, input ensemble, treatment of no-output trials and statistical uncertainty as well as fidelity. If verification destroys the output, that trial demonstrates correlations but does not also deliver a surviving state for another computation.

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

$$|\Omega\rangle={}\frac{\alpha}{\sqrt2}(|000\rangle+|011\rangle) +\frac{\beta}{\sqrt2}(|100\rangle+|111\rangle).$$

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

$$\frac12\Big[ |\Phi^+\rangle \left( \alpha|0\rangle+\beta|1\rangle \right) + |\Phi^-\rangle \left( \alpha|0\rangle-\beta|1\rangle \right) + |\Psi^+\rangle \left( \beta|0\rangle+\alpha|1\rangle \right) + |\Psi^-\rangle \left( -\beta|0\rangle+\alpha|1\rangle \right) \Big],$$

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

2. 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.

3. 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

$$\tfrac14\bigl(\rho+X\rho X +Z\rho Z+XZ\rho ZX\bigr) =\tfrac12I.$$

Bob’s unconditioned state is independent of the input.

4. 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.

5. 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. The entropy of the uniform outcome label is

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

bits of outcome entropy. A single fixed-length binary message instead requires $\lceil2\log_2d\rceil$ bits. For a qutrit, nine outcomes need four binary bits in one use, while their entropy is $\log_2 9\approx3.17$ bits. Bob applies the correction $Z^mX^{-n}$ for the Bell convention above.

6. 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.

7. Can accepting fewer outcomes beat the classical bound?

For a Haar-uniform pure qubit $\psi=|\psi\rangle\langle\psi|$, a classical strategy accepts an outcome with POVM effect $E\geq0$ and prepares a pure guess $\phi$. Show that its conditional fidelity is at most $2/3$, even when $E$ depends on the selected outcome. You may use

$$\begin{aligned} \int d\psi\,\psi&=I/2,\\ \int d\psi\,\psi^{\otimes2}&=(I+S)/6, \end{aligned}$$

where $S$ swaps the two qubits and the Haar measure is normalized to one.

Solution

For $\operatorname{Tr}E>0$, the mean success probability and fidelity numerator are

$$\begin{aligned} p_E&=\operatorname{Tr}E/2,\\ n_E&=\frac{\operatorname{Tr}E+\operatorname{Tr}(E\phi)}6. \end{aligned}$$

Positivity gives $\operatorname{Tr}(E\phi)\leq\operatorname{Tr}E$, so

$$F_E=\frac{n_E}{p_E}\leq\frac23.$$

Several accepted outcomes form a success-probability-weighted average of these ratios, still at most $2/3$. Mixed guesses are convex combinations and cannot improve the bound. A zero-probability branch does not define a conditional fidelity. This argument requires physical outcome selection on the single unknown input, not selection using an externally supplied input label or a distorted test ensemble.

8. 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 sign $z_{\mathrm{raw}}\in\{+1,-1\}$ 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.

References

• C. H. Bennett, G. Brassard, C. Crépeau, R. Jozsa, A. Peres, and W. K. Wootters, “Teleporting an unknown quantum state via dual classical and Einstein–Podolsky–Rosen channels,” Physical Review Letters 70, 1895–1899, 1993, doi:10.1103/PhysRevLett.70.1895.
• S. L. Braunstein and H. J. Kimble, “Teleportation of continuous quantum variables,” Physical Review Letters 80, 869–872 (1998), doi:10.1103/PhysRevLett.80.869.
• D. Gottesman and I. L. Chuang, “Demonstrating the viability of universal quantum computation using teleportation and single-qubit operations,” Nature 402, 390–393, 1999, doi:10.1038/46503.
• M. Horodecki, P. Horodecki, and R. Horodecki, “General teleportation channel, singlet fraction, and quasidistillation,” Physical Review A 60, 1888–1898, 1999, doi:10.1103/PhysRevA.60.1888.
• S. Massar and S. Popescu, “Optimal extraction of information from finite quantum ensembles,” Physical Review Letters 74, 1259–1263, 1995, doi:10.1103/PhysRevLett.74.1259.
• M. A. Nielsen, “A simple formula for the average gate fidelity of a quantum dynamical operation,” Physics Letters A 303, 249–252 (2002), doi:10.1016/S0375-9601(02)01272-0.
• R. F. Werner, “All teleportation and dense coding schemes,” Journal of Physics A: Mathematical and General 34, 7081–7094, 2001, doi:10.1088/0305-4470/34/35/332.