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A Bell-basis measurement is the four-outcome projective measurement whose projectors are onto the four Bell states. It jointly measures two qubits and is the sender's measurement in quantum teleportation.

Suppose independent resource pairs have density matrices $v_j|{\Phi }^{+}⟩⟨{\Phi }^{+}|+(1-v_j)I_4/4$. An ideal complete Bell-basis measurement, followed by outcome-dependent Pauli gates, produces a chosen Bell state with white-noise visibility $v_1{v}_2$: either white-noise input makes the remote pair maximally mixed. If the reported label is reliable with probability $q$ and otherwise uniformly random, averaging the erroneous corrections makes that branch maximally mixed too, so $v_{\mathrm{out}}=v_1{v}_{2}q$. A second independent detector of the same quality during quantum teleportation adds another factor $q$ to the teleported Bloch vector.

Distributing one half of $|{\Phi }^{+}⟩$ through a quantum bit-flip channel produces $p{\Pi }_{{\Phi }^{+}}+(1-p){\Pi }_{{\Psi }^{+}}$. Ideal quantum teleportation with error-free classical correction bits induces the same bit-flip map on the input: conjugating the resource error by any Pauli gate correction changes at most its sign. For a pure target the squared quantum fidelity is $p+(1-p)|⟨X⟩{|}^2$. This describes noise in resource distribution, not a perfect resource supplied by an independent channel.

Dik Bouwmeester and colleagues demonstrated selected-outcome photon-polarization quantum teleportation. An ultraviolet pump traversed a nonlinear crystal twice: one passage supplied an entangled pair, the other the input photon and its trigger partner. The input and one resource photon interfered at a beam splitter; output-detector coincidences selected their antisymmetric Bell state. Narrow filters made their coherence time exceed the pump duration, preserving temporal indistinguishability.

The reported pump pulses were $200\mathrm{fs}$ long at $76\mathrm{MHz}$; $788\mathrm{nm}$ photons filtered to $4\mathrm{nm}$ bandwidth had a quoted $520\mathrm{fs}$ coherence time. A polarization analyzer measured the remote output. Trigger-conditioned fourfold coincidences excluded a double-pair background without subtraction. The selected Bell result had ideal probability $1/4$; this demonstration did not distinguish and actively correct all four outcomes.^[ref]

Classical communication transmits a distinguishable classical record, such as measurement bits, between parties. In quantum teleportation it tells the receiver which local correction to apply. Quantum no-signalling prevents shared entanglement alone from transmitting the record. A classical message cannot propagate across a spacelike separation in a relativistic protocol.

Exact quantum teleportation that preserves an arbitrary reference requires at least two classical bits from sender to receiver, even with preshared quantum entanglement. For the essential reduction, encode two classical bits in the four Bell states by Pauli gates on the sender's half of a Bell pair. Teleportation as an identity channel on a reference transfers this half while preserving the encoding. The receiver then has both halves and recovers both bits by a Bell-basis measurement, as in superdense coding.

The classical bit bound itself has a short no-communication theorem proof. For $N$ possible messages and $K$ possible classical transmissions, let ${\sigma }_{m,c}$ be the receiver's subnormalized conditional density matrix before decoding. The no-communication theorem gives ${\sum }_c{\sigma }_{m,c}=\tau$, independent of $m$, so $0\le {\sigma }_{m,c}\le \tau$. For each transmission $c$, let $\{E_{m|c}{\}}_m$ be the decoding POVM. Averaging over equiprobable messages,Perfect decoding of four messages requires $K\ge 4$, hence at least two classical bits. This applies to the preshared resources of the proposed quantum teleportation protocol as well as the Bell pair used for the encoding.

A quantum teleportation protocol in which a further participant supplies a resource or classical measurement record needed for exact reconstruction of an arbitrary unknown input at the receiver. Without that cooperation, some classical information about the input may remain accessible; control need not mean complete information-theoretic concealment.

For odd $n=2k+1$, a symmetric quantum teleportation resource can encode an unknown qubit into $E(\alpha |0⟩+\beta |1⟩)=\alpha |D_k^n⟩+\beta |D_{k+1}^n⟩$. Begin with $|D_{k+1}^{n+1}⟩$, apply a controlled-NOT gate from the input to the sender's resource qubit, apply a Pauli X gate to that resource qubit and a Hadamard gate to the input, and measure those two qubits. The four unnormalized remaining vectors are $E|\psi ⟩/2$, $EX|\psi ⟩/2$, $EZ|\psi ⟩/2$ and $EXZ|\psi ⟩/2$. Each outcome has probability $1/4$. Since $X^{\otimes n}$ interchanges $|D_k^n⟩$ and $|D_{k+1}^n⟩$, and $Z^{\otimes n}$ acts as logical $Z$ up to the common phase $(-1{)}^k$, local Pauli corrections make the output independent of the result. The one-qubit reduction of Dicke-state superpositions givesFor $n>1$ this distributes imperfect copies, as required by the no-cloning theorem. For $n=1$ it reduces to exact quantum teleportation.

A shared resource enables deterministic exact transmission of an arbitrary unknown qubit using only local operations and classical communication if and only if it can produce a shared Bell pair deterministically by such operations. Sufficiency follows from quantum teleportation. For necessity, prepare a Bell pair locally at the sender and apply the purported transmission protocol to one half: exact identity-channel action leaves a shared Bell pair between the retained reference and the receiver. A separable shared resource cannot satisfy this condition because local operations and classical communication preserve separability.

Nature 390, 575–579. Dik Bouwmeester and colleagues' original selected-outcome optical quantum teleportation report.

Author text hosted by ETH Zürich: qudev.phys.ethz.ch/static/website-2008-2015/content/courses/QSIT11/pdfs/Bouwmeester1997.pdf

Charlie's Hadamard basis measurement, Alice's Bell-basis measurement and communicated correction bits give deterministic GHZ-assisted quantum teleportation. Let the unknown input be $|\chi {⟩}_Q=\alpha |0⟩+\beta |1⟩$. Charlie measures $C$ in the Hadamard basis and labels $+$ by $c=0$, $-$ by $c=1$. The identityshows that $AB$ is conditionally a Bell pair with phase label $c$. Charlie sends that one classical bit to Bob. The procedure for localizing a Bell pair from a GHZ state supplies precisely the missing conditional information; forgetting $c$ leaves the mixed reduced state of the preceding part.

Alice performs the usual Bell-basis measurement on $QA$. Operationally, she applies ${\mathrm{CNOT}}_{Q\to A}$, applies $H_Q$, and measures $Q$ and $A$ in the computational basis, obtaining bits $s,t$. She sends both bits to Bob. To track the correction without a convention ambiguity, conditional on $s,t$ the normalized state on $BC$ isProjecting Charlie's factor onto his recorded Hadamard basis state gives Bob, up to the global phase $(-1{)}^{ct}$,Consequently the inverse Pauli correction isBob applies $X^t$ first and $Z^{s\oplus c}$ second. The measurements on Charlie and on Alice commute because they act on different systems, so deriving the conditional state with Alice's measurement first gives the same protocol as Charlie measuring first. Each triple $(s,t,c)$ has probability $1/8$, independent of the unknown input, and every triple succeeds after correction.

This uses one previously shared GHZ state, two classical bits from Alice and one from Charlie, with no quantum transmission during the transfer. The standard Bell-measurement/Pauli-correction mechanism was established in the original quantum teleportation proposal.^[ref] Here Charlie's record supplies the Bell-pair phase, giving a controlled quantum teleportation protocol. Alice's input is consumed by her measurement; Bob reconstructs its state without learning $\alpha ,\beta$ and without making a second surviving copy.

Combine the hypothetical protocol with superdense coding. Before a message is chosen, Alice and Bob share a further Bell pair $|{\varphi }_{+}{⟩}_{QR}$, with $Q$ at Alice and $R$ at Bob, in addition to any resources needed for the hypothetical quantum teleportation protocol. To encode two classical bits $u,v$, Alice applies the Pauli gates $Z_Q^u{X}_Q^v$, preparingThe four choices are the four mutually orthogonal Bell states. Alice now uses the hypothetical protocol to teleport $Q$ to Bob using only one classical bit. Its assumed preservation of the joint quantum state gives Bob both $B$ and $R$ inBob performs a Bell-basis measurement. Explicitly, a controlled-NOT gate $B\to R$, followed by a Hadamard gate on $B$, givesA computational-basis measurement therefore recovers both message bits with certainty after only one transmitted classical bit. The initial Bell pairs are shared before the message is chosen and carry no message themselves. This proves the classical communication lower bound for exact qubit teleportation: a perfect protocol preserving an external reference needs at least two transmitted classical bits.

Write the arbitrary initial pure state asThe reference vectors need not be normalized or orthogonal. By linearity, the same quantum teleportation calculation, with the identity operator on $R$, gives the post-gate quantum stateThus each outcome still has probability $1/4$. Bob's Pauli gate correction leaves the normalized joint quantum stateAll operations act trivially on $R$, so the result is independent of its holder or location. This establishes teleportation as an identity channel on a reference, preserving quantum entanglement with the untouched reference. Since every corrected branch is the same, discarding the classical outcomes also leaves the required joint density matrix.

Alice holds the input qubit $Q$ in $|\psi {⟩}_Q=a|0⟩+b|1⟩$ and her half $A$ of a shared Bell pair $|{\varphi }_{+}{⟩}_{AB}$; Bob holds $B$. Alice applies a controlled-NOT gate $Q\to A$, then a Hadamard gate on $Q$. In the system order $Q,A,B$, the initial quantum state isAfter these two quantum gates, direct expansion yields the quantum teleportation identityAlice performs computational-basis measurements on $Q,A$, and sends the two classical bits $m,n$ to Bob. Each outcome has probability $1/4$, independently of the input pure state. Bob applies the indicated Pauli gates: first $X^n$, then $Z^m$, so his total correction is $Z^m{X}^n$. In every branch,This proves exact quantum teleportation using one Bell pair and two classical bits. Alice's original qubit has been measured; the protocol transfers the quantum state without leaving her another copy.

Let the unknown input qubit be $|\psi {⟩}_S=\alpha |0⟩+\beta |1⟩$, with $|\alpha {|}^2+|\beta {|}^2=1$. Alice holds $S,A$, Bob holds $B$, and the shared Bell pair is $|{\varphi }^{+}{⟩}_{AB}$. All three-qubit strings below use the order $S,A,B$. Initially,The Bell-basis teleportation identity givesThus a perfect Bell state analyzer on Alice's two qubits identifies the Pauli operator affecting Bob's quantum state.

To display the intermediate three-qubit states explicitly, realize that Bell-basis measurement by a controlled-NOT gate $S\to A$, a Hadamard gate on $S$, and two computational-basis measurements. After the controlled-NOT gate,After the Hadamard gate,Measuring $S,A$ gives $(m,n)$ with probability $1/4$ for each outcome, independently of $\alpha ,\beta$. Conditional on that result, the normalized three-qubit post-measurement state isIf the Bell state analyzer is instead regarded as an ideal direct projective measurement preserving the measured qubits, replace $|mn{⟩}_{SA}$ here by the corresponding Bell state; Bob's factor is identical. A destructive optical analyzer records the same label but absorbs Alice's photons.

Alice sends the two classical bits $(m,n)$ to Bob. During transmission the above conditional three-qubit state is unchanged. Bob then applies the inverse Pauli operator $Z^m{X}^n$:The final conditional three-qubit state is therefore $|mn{⟩}_{SA}\otimes |\psi {⟩}_B$. The protocol transmits two classical bits per teleported qubit, consuming one shared Bell pair. Without Alice's record Bob averages over four Pauli operators and has density matrix $I/2$, so quantum teleportation respects the no-communication theorem. Alice's input has been measured and is no longer an independent copy, consistent with the no-cloning theorem.

Label the unknown input $1$, Alice's resource photon $2$, and Bob's resource photon $3$. First distribute photon $3$ through the channel while keeping photon $2$ with Alice; at $p=1$ this leaves the perfect shared Bell state $|{\Phi }^{+}{⟩}_{23}=(|00⟩+|11⟩)/\sqrt{2}$. When the input arrives, the three-qubit state isAlice applies ${\mathrm{CNOT}}_{1\to 2}$, givingShe next applies $H_1$, producingThese two gates followed by computational measurement implement a perfect Bell state analyzer. For outcome $(m,n)$ the normalized conditional state is $|mn{⟩}_{12}{X}^n{Z}^m|\psi {⟩}_3$, with probability $1/4$. Alice sends the two classical bits $m,n$; Bob applies $Z^m{X}^n$, recovering $|\psi ⟩$ exactly. Averaging before reception of the bits gives Bob $I/2$, independent of the input, so this protocol provides no faster-than-light communication.

For $p=1$, directly sending the original qubit is already perfect. Quantum teleportation therefore has no fidelity advantage in this ideal case and consumes a shared entangled pair plus classical communication. Its operational advantage is that entanglement can be distributed beforehand: once the unknown input appears, only classical information need travel, and the original input photon need not itself propagate to Bob. The Bell measurement consumes the input; it does not create a second copy.

After the first Bell-basis measurement and the announced correction, the shared Werner state has visibility $v=p^{2}q$. Ideal quantum teleportation through that resource is the depolarizing channelOne way to see this is to separate the singlet branch, which teleports perfectly with the appropriate known correction convention, from $I_4/4$, which has no correlations and yields $I_2/2$.

The additional Bell-basis measurement at Alice's station has its own independent reliable branch of probability $q$. In its random-label branch the correction averages the four Pauli gates, again giving $I_2/2$. Hence the actual teleported density matrix isOn sifted BB84 rounds, Bob measures in Alice's preparation basis. The preserved component gives the correct bit with certainty; the maximally mixed component gives an error with probability $1/2$. ThereforeFor $p=q=1$ it vanishes; if either parameter is zero it is $1/2$. The result presumes successful full Bell-basis measurement outcomes and independent stated imperfections, without extra loss, polarization drift or dark counts.

The no-cloning theorem states that no fixed deterministic quantum operation can perfectly duplicate every unknown pure state. For a unitary copier with a fixed blank, suppose it cloned two normalized vectors $|a⟩,|b⟩$. Preservation of inner products would give $⟨a|b⟩=⟨a|b{⟩}^2$, impossible for a nonorthogonal distinct pair with overlap strictly between zero and one in modulus. Including an environment does not help: its final overlap has modulus at most one, so one would require $|⟨a|b⟩|\le |⟨a|b⟩{|}^2$. Thus a universal perfect copier is impossible. Orthogonal known alternatives do not contradict this restriction.

Label the three qubits as input $S$, Alice's entangled qubit $A$ and Bob's qubit $B$. Initially their joint state isThe controlled-NOT gate from $S$ to $A$ changes this toApply $H|0⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $H|1⟩=(|0⟩-|1⟩)/\sqrt{2}$ to the first qubit and group by Alice's two computational-basis outcomes. The result isIn expanded form the last two Bob factors are ${\alpha }_0|0⟩-{\alpha }_1|1⟩$ and ${\alpha }_0|1⟩-{\alpha }_1|0⟩$. Their Alice labels must be $10$ and $11$; the repeated $00$ labels in the source are printing errors.

Alice measures her two qubits and sends the two classical bits $(a,b)$ to Bob. Each outcome has probability $1/4$ and Bob's conditional state is $X^b{Z}^a|\psi ⟩$. He applies $\boxed{Z^a{X}^b}$ to recover the input, up to physically irrelevant global phase. Explicitly the corrections for $00,01,10,11$ are $I,X,Z,ZX$. This is quantum teleportation: Alice's input is consumed by measurement, so no second surviving copy is created. Before her classical message, Bob's reduced density operator is $I/2$, independent of the input, so the protocol also supplies no faster-than-light communication.

A measurement in quantum mechanics with outcomes $m$ is specified, in the single-operator-per-outcome description, by linear operators $M_m$ satisfying ${\sum }_m{M}_m^{†}{M}_m=I$. Its effects $F_m=M_m^{†}{M}_m$ form a positive operator-valued measure. The two measurement postulates are the Born rule for outcomes and the conditional state update:For a density operator the corresponding formulas are $p_m=\mathrm{Tr}(M_m\rho M_m^{†})$ and ${\rho }_m=M_m\rho M_m^{†}/p_m$. More general quantum instruments have multiple operators for each outcome and sum these terms. For a projective measurement of an observable $A={\sum }_m{a}_m{P}_m$, take $M_m=P_m$; the outcome value is $a_m$ and the postmeasurement state lies in its eigenspace. A POVM alone specifies probabilities, whereas the measurement operators or instrument also specify the backaction.

For quantum teleportation, Alice and Bob initially share the Bell pair $|{\Phi }^{+}{⟩}_{AB}=(|00⟩+|11⟩)/\sqrt{2}$. Alice also holds the input $|\psi {⟩}_C=\alpha |0⟩+\beta |1⟩$, with $|\alpha {|}^2+|\beta {|}^2=1$. The shared entanglement is a resource; the two transmitted classical bits alone could not transfer an arbitrary unknown quantum state.

Alice applies a controlled-NOT gate with $C$ as control and $A$ as target, then a Hadamard gate to $C$. Starting from $|\psi {⟩}_C|{\Phi }^{+}{⟩}_{AB}$, direct expansion givesFor example, the four receiver vectors are $\alpha |0⟩+\beta |1⟩$, $\beta |0⟩+\alpha |1⟩$, $\alpha |0⟩-\beta |1⟩$, and $-\beta |0⟩+\alpha |1⟩$. Thus the gate calculation fixes the ordering of the Pauli corrections, including the last branch's sign.

Alice measures $C,A$ in the computational basis, obtaining $(a,b)$. The Born rule gives each outcome probability $1/4$, independent of the input amplitudes. Bob's normalized conditional state is $X^b{Z}^a|\psi ⟩$. Alice sends the two outcome bits. Bob applies the inverse, $Z^a{X}^b$, and henceThe same circuit is a Bell-basis measurement on Alice's two qubits, expressed using ordinary gates followed by computational-basis measurements. It restores the state without Alice or Bob knowing $\alpha ,\beta$.

By linearity, the calculation also holds when the input is entangled with a reference: each outcome has branch map $|\psi ⟩\mapsto X^b{Z}^a|\psi ⟩/2$, and after correction every branch is one-half of the identity map. Summing the four corrected density operators gives exactly the original input density operator, with all reference correlations preserved. This is teleportation as an identity channel on a reference. Without the two classical bits Bob must average the four uncorrected states, giving $\frac{1}{4}{\sum }_{a,b}{X}^b{Z}^a\rho Z^a{X}^b=I/2$, so the process cannot signal before the message arrives. The Bell pair is consumed and Alice retains no copy of the unknown state, in accordance with the no-cloning theorem.

Let $Q$ be Alice's unknown qubit, with pure state $|\varphi {⟩}_Q=\alpha |0⟩+\beta |1⟩$. Alice and Bob initially share the Bell state $|{\Phi }^{+}{⟩}_{AB}=(|00⟩+|11⟩)/\sqrt{2}$. Alice holds $Q,A$ and Bob holds $B$. Define the Bell basisExpanding in this orthonormal basis gives the Bell-basis teleportation identityFor example, the last branch on $B$ is $\alpha |1⟩-\beta |0⟩$, which is $XZ|\varphi ⟩$; the ordering of the Pauli operators therefore matters.

Alice performs a projective measurement in the Bell basis on her two local qubits. Each outcome has probability $1/4$, independent of the unknown amplitudes. She encodes the four possible outcomes as two classical bits and sends them to Bob. For outcomes ${\Phi }^{+},{\Phi }^{-},{\Psi }^{+},{\Psi }^{-}$, respectively, Bob applies the local unitary operationsThese are inverses of the corresponding branch Pauli operators, so Bob obtains exactly the unknown input qubit in every branch. Using $XZ$ instead of $ZX$ as the last correction also succeeds up to an irrelevant overall phase. Alice need not know $\alpha$ or $\beta$. Every quantum operation is local to one of the parties, and the only transmitted information is the two-bit classical outcome. The shared Bell pair is consumed.

Before the message arrives, Bob's density operator isso no unknown-state information is available without the classical communication. Alice's original qubit has been included in the destructive Bell-basis measurement; the protocol creates no extra copy. Since the corrected transformation is the identity on every input vector and is linear, it also preserves entanglement with an external reference and transfers mixed inputs. Thus this is deterministic quantum teleportation, using precisely one shared maximally entangled Bell pair, two classical bits, and local operations.