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The $m$-qubit bit-flip repetition code stores one logical qubit in $|0_L⟩=|0{⟩}^{\otimes m}$ and $|1_L⟩=|1{⟩}^{\otimes m}$. Its stabilizer generators are $Z_j{Z}_{j+1}$ for $j=1,\dots ,m-1$, and its logical operators are $\overline{X}=X_1\cdots X_m$ and $\overline{Z}=Z_1$. Interchanging the Pauli X gate and Pauli Z gate gives the phase-flip repetition code. For odd $m$, majority-vote decoding of a repetition code corrects up to $(m-1)/2$ physical Pauli X gate errors, assuming ideal operations and a noise family containing only these bit flips.

For $m=3$, prepare two quantum ancillas in $|00⟩$ and apply controlled-NOT gates from the input qubit to each of them. By linearity, encoding sendsThis does not produce $(\alpha |0⟩+\beta |1⟩{)}^{\otimes 3}$, which would contain cross terms. For example, encoding $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ produces $(|000⟩+|111⟩)/\sqrt{2}$, whereas three independent copies would have eight equal computational basis amplitudes. The encoding creates quantum entanglement when both coefficients are nonzero and respects the no-cloning theorem.

The three-qubit bit-flip syndrome table corrects the family $\{I,X_1,X_2,X_3\}$ and its entire linear span, as specified by the Knill--Laflamme condition. It cannot correct arbitrary errors on one physical qubit: the phase flipcommutes with both checks and acts as logical $\overline{Z}$. In particular, it takes the encoded $|+⟩$ to the orthogonal encoded $|-⟩$ with error syndrome $00$. Thus its full quantum code distance is one, despite correcting one error of the restricted Pauli X gate type. The Shor code, Steane code and five-qubit error correcting code provide protection against arbitrary one-qubit errors.

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.

Assume perfect cloning of both specified pure states. Allowing irrelevant output global phases, this would requireA unitary operator preserves inner products, and therefore their magnitudes. But the input overlap and required output overlap have different magnitudes:This contradiction proves that no such two-qubit unitary exists, without assuming the no-cloning theorem.

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.

Use an Ekert protocol. Alice and Bob choose their analyzer settings independently and randomly for each received pair, record the outcomes, and confirm reception before announcing settings over an authenticated public classical channel. The authentication prevents an adversary from impersonating either party; it does not conceal the setting announcements. Announcing settings after reception also avoids the premature basis announcement in quantum key distribution attack.

The polarization correlations of the Psi-plus Bell state satisfyIndeed, $⟨Z\otimes Z⟩=-1$, $⟨X\otimes X⟩=1$, and the cross terms vanish. In the specified setting order,

For key sifting, retain the setting pairs $(i,j)=(1,1)$ and $(3,2)$. Their outcomes are perfectly anticorrelated. Alice assigns bit $0$ to outcome $+1$ and bit $1$ to outcome $-1$; Bob uses the opposite assignment. Their raw keys then agree. With uniform independent choices of the three settings, these pairs supply a raw-key fraction $2/9$, before sacrificing a test sample. They must keep the remaining raw-key outcomes secret. The four pairs $(1,2),(1,3),(2,2),(2,3)$ supply Bell-test data, and the other three setting pairs can be discarded.

Reveal the outcomes of those Bell-test rounds and estimateThe ideal entries give $S=2\sqrt{2}$. The CHSH inequality gives $|S|\le 2$ for local hidden-variable theories: for predetermined outcomes $a_1,a_2,b_2,b_3\in \{\pm 1\}$, the signed combination is $-a_1(b_2+b_3)-a_2(b_2-b_3)$. One bracket is zero and the other is $\pm 2$, so averaging cannot exceed two in magnitude. The Tsirelson bound limits quantum correlations to $2\sqrt{2}$.

For the specified, trusted Pauli operators, the maximal CHSH witness for a fixed Bell state makes the ideal security statement precise. Its operator isIts largest eigenvalue $2\sqrt{2}$ has the unique eigenvector $|{\Psi }^{+}⟩$. Exact attainment therefore fixes the shared density matrix to this pure state. Since purity decouples a subsystem from its purification, Eve cannot hold information correlated with that pair. This rules out information-gaining eavesdropping in the ideal model, rather than proving that no observer exists. More generally, CHSH self-testing identifies the entangled qubit pair up to local changes of representation under its own assumptions.

A quantum measurement that reads Eve's chosen polarization generally disturbs correlations in an incompatible basis. The no-cloning theorem also prevents copying arbitrary unknown photon polarizations and retaining a perfect copy while forwarding an unchanged original. The Bell test checks several incompatible settings, so observing perfect correlation in only one key basis is insufficient.

Alice and Bob also reveal a randomly chosen sample of their raw keys to estimate the quantum bit error rate, then discard that sample. Information reconciliation corrects discrepancies, and privacy amplification removes residual information, including leakage from the public transcript. With finite data they need statistical bounds and a security analysis appropriate to their devices and noise. A measured violation $S>2$ alone does not assert zero information for Eve; the exact ideal argument above uses the maximal value.^[ref]

Use a classical target bit $t$ and initialized auxiliary bits. The NOT gate maps $t\mapsto 1\oplus t$, the CNOT gate maps $(a,t)\mapsto (a,t\oplus a)$, and the Toffoli gate maps $(a,b,t)\mapsto (a,b,t\oplus ab)$. They are bijections and are their own inverses. Explicit circuits with their retained input wires are:

These maps also give the truth tables for all input combinations. AND and OR are not themselves reversible maps if the inputs are discarded; the retained inputs are what make their implementations reversible. Classical CLONE copies a known orthogonal computational basis alphabet; the no-cloning theorem forbids interpreting it as a copier of arbitrary quantum superpositions.

A Boolean circuit can express any finite Boolean function using AND, OR and NOT. Replace its operations by the reversible versions, keeping inputs and intermediate bits. Copy the final output into a fresh answer register with CNOT gates, then run the entire computing network backward. Intermediate auxiliary bits return to zero, yieldingThis compute-copy-uncompute construction proves universality in the usual model with initialized auxiliary bits. If $f$ is a reversible permutation, compute $f(x)$, use a reversible implementation of $f^{-1}$ to clear $x$, and swap registers, leaving $f(x)$ with clean auxiliaries. The ancillary-resource qualification matters; a blanket claim about all permutations on an arbitrary fixed number of unassisted wires has parity restrictions.

Yes, access to the original qubits can supply information about the secret encoding basis once the bit is known. For bit zero the possibilities are $|0⟩$ and $|+⟩$, whose overlap is $1/\sqrt{2}$; for bit one they are $|1⟩$ and $|-⟩$ with the same overlap magnitude. A Helstrom measurement for two pure states can identify which candidate was supplied more reliably than a blind basis guess. Measuring and preparing a state based on that evidence can improve an additional counterfeit, although the discrimination is not perfect and the measurement can disturb the original.

The no-cloning theorem rules out a deterministic perfect extra copy of both nonorthogonal candidates; it does not imply that access gives no advantage. The task of producing an additional accepted note should also be distinguished from simply returning the unaltered original, which already passes Alice's test. Requiring both the original and the extra note to survive imposes a further information-disturbance tradeoff.

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.

There is a necessary distinction between a failure bound averaged over the unknown state, a worst-case bound, and a separate bound for each candidate state. The last interpretation is false: a forger can discard the inputs and always prepare $|{\psi }_0{⟩}^{\otimes N}$. If the actual state is $|{\psi }_0⟩$, every authenticity test passes with probability one. The meaningful uniform statement concerns worst-case failure, or average failure under any fixed prior giving both states positive probability. The PDF does not specify that prior; the argument below gives an explicit bound for either interpretation.

Put $c=|⟨{\psi }_0|{\psi }_1⟩|\in (0,1)$. Any unconditional strategy producing $N$ output qubits, including ancillas, adaptive measurement in quantum measurements and correlated outputs, is a quantum channel $\mathcal{E}$. Its inputs and ideal outputs are the density operatorsThe projective measurements on the different banknotes commute, and their all-pass effect is exactly $P_i$. Therefore the probability of at least one rejection, conditional on input $i$, isThis remains true for entangled outputs; no independence of their individual test outcomes is assumed.

Use trace distance $D(\rho ,\sigma )=\frac{1}{2}\Vert \rho -\sigma {\Vert }_1$. The pure-state trace distance and fidelity identity givesFor completeness, two pure-state projectors differ by a matrix with nonzero eigenvalues $\pm \sqrt{1-|⟨u|v⟩{|}^2}$, which proves this identity. Also the pure-target upper bound on trace distance gives $D({\rho }_i,P_i)\le \sqrt{e_i}$. To see the mixed-state step, write ${\rho }_i={\sum }_j{q}_j|u_j⟩⟨u_j|$ and use convexity of the trace norm followed by the Cauchy-Schwarz inequality:Finally, trace-distance contraction under quantum channels gives $D({\rho }_0,{\rho }_1)\le D({\sigma }_0,{\sigma }_1)$. This follows from $D(\rho ,\sigma )={\max }_{0\le E\le I}\mathrm{Tr}[E(\rho -\sigma )]$: the adjoint channel is positive and unital, so its image of any effect $E$ is still an effect.

The triangle inequality now yieldsThus the cloning failure bound from trace-distance contraction isand in particular $\max (e_0,e_1)\ge {\Delta }^2/4$. For positive priors ${\pi }_0+{\pi }_1=1$, another Cauchy-Schwarz inequality givesso the average rejection probability satisfiesFor equal priors, one may take $p_0={\Delta }^2/8$; the same choice is a strict lower bound on worst-case rejection. This proves a quantitative version of the no-cloning theorem for the bank's all-pass test. It concerns an unconditional attempt: heralded success cannot be postselected while omitting failed attempts from the accounting.

Include all fixed resource states and all measurement records in an environment, so the entire physical protocol has an linear isometry $U$ from the unknown input to a larger pure output. If every recipient receives a perfect copy, its reduced density matrix is the pure projector $|\psi ⟩⟨\psi |$. A pure marginal forces factorization: write a joint vector in a basis beginning with $|\psi ⟩$; its squared components along every orthogonal basis vector vanish because their marginal probabilities are zero. Applying this successively to all recipients giveswhere $|R⟩$ is the fixed initial resource and $|E_{\psi }⟩$ may depend on the input. This includes Alice's discarded systems and every possible outcome record.

Choose two distinct nonorthogonal pure states, for example $|0⟩$ and $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, and write their overlap as $s$. Preservation of the inner product by the linear isometry requiresNormalized environment vectors have overlap of absolute value at most one. Thus $0<|s|<1$ would imply $|s|\le |s{|}^N$, impossible for $N>1$. Hence perfect transmission of a copy to every recipient is impossible for $N>1$. This proves the no-cloning theorem obstruction even though Alice keeps no copy herself.

Quantum teleportation transfers an arbitrary unknown qubit state to a separated receiver while using local operations and classical communication, rather than sending that input qubit. Alice and Bob first share one maximally entangled state of two qubits, such as a Bell pair; a shared spin singlet state is equivalent after a known local unitary operation.

Alice performs a joint Bell-basis measurement on the input qubit and her half of the pair. There are four possible outcomes. Bob's conditional qubit is the input state transformed by a known Pauli operator determined by that outcome. Alice sends the outcome as two classical bits, and Bob applies the inverse Pauli operator. Thus one shared Bell pair and two classical bits, together with local measurement and correction, transmit one unknown qubit exactly. The entangled resource is consumed.

The procedure does not produce a classical description of the unknown amplitudes and does not leave a second copy at Alice, so it respects the no-cloning theorem. Before receiving the classical outcome, Bob's reduced density matrix is $I/2$, independent of the input; this respects the no-communication theorem. The next part proves the full identity-channel action, including inputs entangled with another system.

Use the BB84 protocol to establish a secret random key, then encrypt the actual message with a one-time pad. Alice chooses independent random bits and independently chooses between the computational orthonormal basis $\{|0⟩,|1⟩\}$ and the diagonal basis $\{|+⟩,|-⟩\}$, where $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$. She sends one photon or other qubit in each corresponding state. Bob independently chooses one of the two bases for each measurement.

Over an authenticated public classical channel, they announce their bases but not their bit values. They retain only positions where their bases match; absent disturbance these sifted bits agree. They disclose and discard a randomly selected test sample to estimate the error rate. If it is too large they abort. Otherwise they perform information reconciliation and privacy amplification to produce a shared key about which an eavesdropper has negligible information. Authentication is essential: without it, an active attacker can impersonate the two parties. It may use a previously shared short authentication key, subsequently replenished from the new secret key.

The no-cloning theorem prevents an attacker from copying an unknown nonorthogonal state perfectly: preservation of inner products would require $⟨\psi |\varphi ⟩=⟨\psi |\varphi {⟩}^2$, impossible when its modulus is strictly between zero and one. In a simple intercept-and-resend attack, Eve chooses the wrong basis with probability $1/2$, and then introduces a wrong sifted bit with probability $1/2$. Thus the test error rate is $1/4$, and the probability that $m$ independent tested bits show no error is $(3/4{)}^m$. More general attacks are handled by bounding Eve's information from the observed statistics and shortening the reconciled key through privacy amplification.

Finally Alice sends $C=M\oplus K$, using a uniform independent key of the message's length exactly once. The one-time pad makes the ciphertext independent of the message. The security claim is secrecy with arbitrarily small failure probability after testing and privacy amplification, not that a finite quantum transmission can never be disturbed or intercepted. The scheme assumes trusted devices and an authenticated classical channel; it does not rely on a computational hardness assumption.

Quantum teleportation transfers the quantum state of an unknown qubit from a sender's carrier to a receiver's carrier. It also transfers correlations with any external reference. It does not determine classical amplitudes or send the original particle.

The standard deterministic protocol consumes one shared maximally entangled Bell pair and two classical bits from Alice to Bob. Alice performs a local Bell-basis measurement on the input and her half of the pair; Bob applies the corresponding local Pauli gate. The classical message is essential: before receiving it, Bob has the outcome-averaged density operator $I/2$ and cannot recover or detect the unknown state. The original input is consumed by the measurement, so quantum teleportation is compatible with the no-cloning theorem.

Quantum cryptography includes quantum key distribution, which establishes a shared secret key rather than transmitting an arbitrary plaintext directly. In BB84, Alice randomly prepares one of the two computational-basis or two complementary-basis qubit states; Bob randomly measures in either basis. They announce bases over an authenticated classical channel and retain the positions where their choices agree, about half in the ideal unbiased case. They test a subset for errors, reconcile the remaining strings, and apply privacy amplification to remove a bounded eavesdropper's information. A protocol can abort when the observed data do not support a secret key.^[ref]

Nonorthogonal states cannot be perfectly copied by a state-independent device, by the no-cloning theorem. An intercept-resend attacker who measures each signal in a random BB84 basis chooses the wrong basis half the time. On those sifted positions Bob has a one-half error probability, giving the ideal BB84 intercept-resend error rate $1/4$. This simple calculation demonstrates detectable disturbance for that attack; it is not a complete security proof for arbitrary attacks.

Authentication is essential: otherwise an adversary can impersonate both participants. Error correction reveals information that must be included in the privacy budget, and losses, finite sample sizes, multiphoton emission and detector behavior belong in the security model. Entanglement-based variants can use Bell correlations, but observing some entanglement or a Bell violation alone does not replace a quantitatively specified key-distillation protocol. The final key can subsequently be used in a suitable classical encryption scheme.

For quantum teleportation, Alice holds an unknown input $|\psi {⟩}_1$ and one member of a shared $|{\Phi }^{+}{⟩}_{23}$ pair; Bob holds qubit 3. Alice applies a CNOT gate from 1 to 2 and a Hadamard gate on 1, then measures both in the computational basis. Expansion gives the Bell-basis teleportation identityEvery outcome pair $(m,n)$ has probability $1/4$, independently of the input. Alice sends these two classical bits, and Bob appliesThe order of the two corrections matters only by a global sign for the doubly nonzero outcome. The transformation also transfers entanglement with an untouched reference, because the identity is linear in the input amplitudes.

Before receiving the bits, Bob must average over all four Pauli-related states, obtaining $I/2$ for any input. Thus teleportation requires classical communication and does not send information faster than light. The original input is consumed by Alice's measurement, so no second copy remains and the no-cloning theorem is respected. Ideal qubit teleportation consumes one shared Bell pair and two classical bits, as well as local gates and measurements. Imperfect entanglement, gate error, detection error or lost classical records reduce practical fidelity; teleportation transfers the quantum state, not the underlying material particle.

Quantum cryptography includes quantum key distribution, which uses quantum transmissions and authenticated public discussion to establish a secret random key. In BB84, the sender independently chooses a random bit and either the computational or Hadamard basis, sending one of $|0⟩,|1⟩,|+⟩,|-⟩$. A receiver independently chooses one of those two bases. They announce bases after receipt, retain matching-basis positions and reveal a random sample of those bits to estimate noise or eavesdropping. Ideal uniform independent basis choices retain about half of the detected signals before testing and further processing.^[ref]

The no-cloning theorem prevents an adversary from copying arbitrary unknown nonorthogonal signal states perfectly. In a simple intercept–resend attack, the adversary chooses a random protocol basis. On half of the sifted positions it is the wrong basis, and the resent state then yields an erroneous receiver bit with probability one half. Thus the BB84 intercept-resend error rate isif every signal is intercepted, or $Q=q/4$ when an independently selected fraction $q$ is attacked this way. This is an illustrative attack calculation, not a security proof against all possible quantum attacks.

After parameter estimation, classical error reconciliation aligns the remaining keys; privacy amplification compresses them to remove the adversary's bounded information, including information leaked during public reconciliation. The classical channel requires message authentication, otherwise an attacker could impersonate each party to the other. Real sources, losses, multiphoton pulses and detector behavior must fit the security model. A low observed error rate alone is insufficient to certify arbitrary hardware. Finally the protocol distributes a key; encrypting a later message is a separate step.

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.

Quantum teleportation transfers an unknown qubit state using a previously shared Bell pair, local operations and two classical bits. It requires no description of the unknown complex amplitudes and produces no extra copy.

Let the input be $|\chi {⟩}_S=\alpha |0⟩+\beta |1⟩$, and let Alice and Bob share $|{\Phi }^{+}{⟩}_{AB}$. Alice applies a controlled-NOT gate from $S$ to $A$, then a Hadamard gate to $S$. Expanding the circuit gives the Bell-basis teleportation identity in computational-output form:Her projective measurement yields each bit pair $(a,b)$ with probability $1/4$. She sends the two classical bits, and Bob applies $Z^a{X}^b$ to recover $|\chi ⟩$. The order is the inverse of the displayed conditional operator; interchanging these Pauli corrections changes only an irrelevant global phase when both bits are one.

Before receiving the bits, Bob hasThus teleportation respects the no-communication theorem and cannot transmit the usable unknown state faster than the classical message. Alice's input has been measured, and the shared entanglement resource has been consumed, in agreement with the no-cloning theorem. The protocol transfers the state, not the material particle initially carrying it.

A physical implementation must distribute and preserve an entangled pair, implement a Bell-basis measurement and transmit its result to a controllable output system. Imperfect entanglement, lost outcomes and noisy corrections reduce quantum fidelity; a heralded probabilistic experiment must report its success probability separately from conditional fidelity. Ideal deterministic teleportation assumes every Bell outcome can be identified and corrected.

Quantum error correction encodes a logical qubit into a larger Hilbert space, diagnoses a specified error family with an error syndrome, and reverses it without learning the protected logical amplitudes. It must preserve arbitrary superpositions, not just distinguish two classical codewords.

The bit-flip repetition code encodesMeasure the commuting stabilizer generators $Z_1{Z}_2$ and $Z_2{Z}_3$ using ancillary qubits. Their outcomes reveal the parity relations, with the three-qubit bit-flip syndrome tableBoth codewords have the same parity checks, so the syndrome does not reveal $\alpha$ or $\beta$. For example, the $X_2$ error produces $\alpha |010⟩+\beta |101⟩$; applying $X_2$ restores the encoded superposition. A coherent combination of correctable errors is also recoverable after the syndrome is recorded in an ancilla, by linearity.

This three-qubit code corrects one bit flip, not every one-qubit error. A physical $Z_1$ changes the relative logical phase without a nontrivial bit-flip syndrome. General single-qubit noise is a linear combination of $I,X,Y,Z$; protecting arbitrary single-qubit errors therefore needs a code such as the Shor code, the Steane code or the five-qubit code, with the corresponding enlarged syndrome structure.

The general exact-recovery criterion is the Knill--Laflamme condition. For code projector $P$ and error operators $E_r$, there must be constants $C_{rs}$ such thatIt expresses that the syndrome information can depend on the error but not on the logical state. Encoding entangles several physical qubits; it does not create independent copies of an unknown input, so it respects the no-cloning theorem. Fault-tolerant implementations must also prevent noisy syndrome extraction and recovery gates from spreading an initially correctable error into an uncorrectable one.

A cloning operation for a family of pure states would mapfor every state in that family. A single physical operation cannot clone every unknown quantum state.

The no-cloning theorem forbids a universal unknown-state cloning operation, while orthogonal families can be cloned.