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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.

For a Bell pair $|{\varphi }_{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$, Alice's basic measurement in the entrywise complex conjugate of any desired Bob orthonormal basis prepares each Bob pure state with probability $1/2$. Indeed, if $|u⟩={\sum }_{j=0}^1{u}_j|j⟩$, then $⟨\overline{u}|j⟩=u_j$, which proves the displayed contraction identity. The undisclosed-outcome density matrix stays $I/2$; quantum steering changes the conditional quantum state ensemble without violating the no-communication theorem.

Alice uses conjugate-basis steering of a Bell pair. Write the desired orthonormal basis for Bob as $|\psi ⟩=a|0⟩+b|1⟩$ and $|{\psi }^{\perp }⟩=c|0⟩+d|1⟩$. Entrywise complex conjugation in the computational basis gives Alice's orthonormal basisFor either $|u⟩$ in Bob's chosen orthonormal basis, the corresponding unnormalized conditional quantum state isIts squared norm is $1/2$. Thus Alice measures in the complex-conjugate basis, obtaining the required Bob pure states with probability $1/2$ each. Averaging over Alice's undisclosed outcome leaves Bob's density matrix equal to $I/2$, consistent with the no-communication 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.

In quantum dense coding, a shared Bell pair lets Alice send two classical bits by transmitting one qubit. Use photon polarization as the computational basis, $|0⟩=|H⟩$, $|1⟩=|V⟩$, and define the Bell statesAlice and Bob initially share $|{\varphi }^{+}{⟩}_{AB}$. Alice encodes a two-bit message $(m,n)$ by applying the Pauli operator $X^n{Z}^m$ to her qubit:The last minus sign is a global phase. Alice transmits her photon to Bob, who now holds both qubits and uses a Bell-basis measurement to identify the message. The four Bell states are orthogonal, so an ideal Bell state analyzer decodes them without error. The resource cost is one previously shared Bell pair and one transmitted qubit per two classical bits; the shared quantum entanglement is consumed. Before transmission Bob's reduced density matrix is $I/2$ for every message, consistent with the no-communication theorem.

The BBO crystal supplies the polarization-entangled photon pair. An ultraviolet laser pumps this nonlinear optical crystal; spontaneous parametric down-conversion produces signal and idler photons with ${\omega }_p={\omega }_s+{\omega }_i$ and phase matching ${\mathbf{k}}_p={\mathbf{k}}_s+{\mathbf{k}}_i$. In a noncollinear type-II phase matching arrangement, collect the intersections of the two emission cones. The alternatives $|H{⟩}_A|V{⟩}_B$ and $|V{⟩}_A|H{⟩}_B$ must be indistinguishable in arrival time, spectrum and spatial mode. Compensation of birefringence and adjustment of the relative quantum phase then give $(|HV⟩+|VH⟩)/\sqrt{2}$; a local Pauli X gate converts it to $|{\varphi }^{+}⟩$. Half-wave plates and other waveplates implement Alice's Pauli gates. The BBO is the pair source; the message is encoded subsequently by the local optical elements.^[ref]

A basic passive linear optics Bell state analyzer with no extra resources identifies only two of the four polarization Bell states unambiguously. Thus the ideal two-bit capacity presupposes complete Bell-basis measurement; that basic optical implementation alone does not attain it deterministically.

Keep the actual odd-$n$ GHZ state from the preceding calculation. Resolving its first qubit in the Hadamard basis givesThus the $+$ outcome leaves $|{\psi }^{-}{⟩}_{23}$ and the $-$ outcome leaves $|{\psi }^{+}{⟩}_{23}$, each with probability $1/2$. Conditional on a known result the remaining qubits share a maximally entangled state.

When the result is not revealed, the appropriate density matrix is the probability-weighted mixture from a nonselective quantum measurement:The cross terms cancel. The displayed convex combination of product states proves directly that this is a separable quantum state. It still has perfect computational-basis measurement anticorrelation, so it is not itself a product state. It also equals the partial trace ${\mathrm{Tr}}_1|G⟩⟨G|$, as required by the no-communication theorem: a local quantum measurement with an unobserved outcome cannot change the remote reduced density matrix.

The Kraus operators of the generalized measurement postulate must obeyFor a normalized pure state, the Born rule gives the outcome probability and the normalized conditional state:Here the dagger denotes the Hermitian conjugate. Completeness ensures ${\sum }_i{p}_i=1$. An outcome with $p_i=0$ never occurs for that input, so no conditional state is required there. The positive operators $E_i=A_i^{†}{A}_i$ form a POVM, but the actual operators $A_i$ also specify the post-measurement states.

Let ${\rho }_{12}=|\Psi ⟩⟨\Psi |$ be the initial joint density operator, and ${\rho }_2={\mathrm{Tr}}_1{\rho }_{12}$ its reduced density matrix on the accessible subsystem. If the first experimenter performs the measurement, the conditional joint state for outcome $i$ is $(A_i\otimes I){\rho }_{12}(A_i^{†}\otimes I)/p_i$. The second experimenter does not know that outcome. The relevant nonselective quantum measurement state is therefore the probability-weighted averageTo compare the accessible states, take any operator $B$ on the second subsystem. Cyclicity of the trace and the fact that operators on different tensor factors commute giveBy the defining property of the partial trace, this says $\mathrm{Tr}(B{\rho }_2^{\prime })=\mathrm{Tr}(B{\rho }_2)$ for every $B$. Matrix units, or arbitrary Hermitian test operators, distinguish any two different density matrices, soThis proves remote state invariance under a local trace-preserving operation directly from completeness.

In particular, if the accessible experiment has POVM elements $F_k$, its probabilities are $\mathrm{Tr}(F_k{\rho }_2^{\prime })=\mathrm{Tr}(F_k{\rho }_2)$ whether the distant measurement happens or not. Attaching an independent local ancilla, evolving locally, or performing a sequence of local measurements cannot help: these operations act on the identical input density operator and therefore give identical outcome distributions. No experiment confined to the second subsystem can distinguish whether the unannounced measurement occurred. This is a derivation of the no-communication theorem, not an appeal to a signalling postulate.

The qualification is that the outcome is not communicated. A selective quantum measurement can give different conditional remote states, as happens when measuring one half of a Bell state; their probability-weighted average is still ${\rho }_2$. An observer supplied with the distant outcome has extra classical information and is no longer restricted to the unannounced local experiment considered here.

Take the labels $r,s\in \{0,\dots ,n-1\}$, with basis indices read modulo $n$. There is a missing hypothesis in the printed claim: the root of unity must be primitive. For example, when $n=2$ and $\omega =1$, the two labels $(0,0)$ and $(1,0)$ produce the identical state $(|00⟩+|11⟩)/\sqrt{2}$, so they are not orthogonal. More generally, if the order of $\omega$ is $d<n$, changing $r$ to $r+d$ repeats a state. The orthonormality claim is false for a nonprimitive root.

For the intended result, choose a primitive root of unity, for example $\omega =e^{2\pi i/n}$. Since $|\omega |=1$, expansion in the orthonormal basis givesIf $r=r^{\prime }$, the sum is $n$. If $r\ne r^{\prime }$ modulo $n$, primitiveness gives $q={\omega }^{r^{\prime }-r}\ne 1$ while $q^n=1$, and the finite geometric series is $(1-q^n)/(1-q)=0$. ThusThere are $n^2$ vectors in an $n^2$-dimensional Hilbert space, so they form the generalized Bell basis. Each has equal Schmidt coefficients $1/\sqrt{n}$ and is a maximally entangled state.

For qudit teleportation, let $C$ be Alice's unknown input, write $|\chi {⟩}_C={\sum }_j{\alpha }_j|j{⟩}_C$, and let Alice and Bob share $|\Phi {⟩}_{AB}=n^{-1/2}{\sum }_k|k{⟩}_A|k{⟩}_B$. Alice holds $C,A$ and Bob holds $B$. Define the qudit shift and phase operatorsBoth are unitary. Alice makes a joint projective measurement of $C,A$ in the generalized Bell basis, with projectors $|{\psi }_{rs}⟩⟨{\psi }_{rs}|$. Completeness follows from the orthonormality just proved.

For outcome $(r,s)$, contracting the corresponding bra with the joint input and resource gives Bob's unnormalized state:The Born rule therefore gives $p_{rs}=1/n^2$, since $X^s{Z}^{-r}$ preserves the input norm. Conditional on this outcome, Bob holds $X^s{Z}^{-r}|\chi ⟩$. Alice sends him the classical label $(r,s)$, and he applies its inverse in the displayed order:The corrected receiver state is exactly the unknown input state for every outcome. Alice needs no knowledge of the amplitudes. The label has $n^2$ possibilities and can be encoded in $\lceil {\log }_2(n^2)\rceil$ classical bits; when $n$ is a power of two this is exactly $2{\log }_{2}n$. The protocol consumes the shared maximally entangled state and the original carrier is measured, so it does not retain an additional copy of the unknown state.

Equivalently, the complete joint quantum state has the teleportation expansionIt follows from the contraction calculation and completeness of the generalized Bell basis. After correction the branch operators are all $I/n$, so their sum defines the identity quantum channel: ${\sum }_{r,s}(I/n)\rho (I/n)=\rho$. Acting on the input part of a joint state with a reference gives the same identity, which proves teleportation as an identity channel on a reference and extends the protocol to mixed inputs.

Before receiving the label, Bob instead has the outcome-averaged stateTo check the equality, averaging over $r$ kills the off-diagonal matrix entries by the same root-of-unity sum; averaging the cyclic shifts $s$ makes every diagonal entry the total trace divided by $n$. This is the Heisenberg-Weyl twirling channel. Thus the classical message is necessary to recover the input, consistently with the no-communication theorem derived in Question 1.

Let $R=|\psi ⟩⟨\psi |$ and choose an orthonormal basis $\{|e_{\mu }⟩\}$ of ${\mathcal{H}}_2$. The reduced density matrix on $S_1$ is the partial traceHere $I_1$ is the identity on ${\mathcal{H}}_1$, and the bras and kets in the sum contract only the second tensor factor. Equivalently, if $|\psi ⟩={\sum }_{a,\mu }{c}_{a\mu }|f_a⟩\otimes |e_{\mu }⟩$, then $({\rho }_1{)}_{ab}={\sum }_{\mu }{c}_{a\mu }{c}_{b\mu }^{\ast }$. The definition is independent of the chosen basis: it is characterized by $\mathrm{Tr}(M{\rho }_1)=\mathrm{Tr}[(M\otimes I_2)R]$ for every operator $M$ on ${\mathcal{H}}_1$.

The generalized measurement postulate takes $A_i$ to be operators on ${\mathcal{H}}_2$ satisfying ${\sum }_i{A}_i^{†}{A}_i=I_2$. The Born rule and the conditional state update giveIf the outcome is not supplied to $S_1$, the appropriate post-measurement density matrix is $R^{\prime }={\sum }_i(I_1\otimes A_i)R(I_1\otimes A_i^{†})$. For any local operator $M$, cyclicity of the full trace givesSince this holds for every $M$, $\boxed{{\mathrm{Tr}}_2{R}^{\prime }={\rho }_1}$. This is the no-communication theorem.

The invariance concerns the nonselective measurement in quantum mechanics. A conditional reduced density matrix ${\rho }_{1|i}={\mathrm{Tr}}_2|{\psi }_i⟩⟨{\psi }_i|$ can change, although ${\sum }_i{p}_i{\rho }_{1|i}={\rho }_1$. For example, measuring one half of a Bell pair in the computational basis changes the other half conditionally to $|0⟩$ or $|1⟩$; before learning the outcome, its density matrix remains $I/2$. Thus distant measurement in quantum measurements cannot transmit a controllable signal without classical communication. This quantum no-signalling is compatible with special relativity, even though some entangled states have correlations incompatible with a local hidden-variable theory.

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.

Yes: deterministic entanglement swapping gives the required singlet. Denote Bob's two local qubits by $B_1,B_2$. Known local unitary operations first convert the two given spin singlet states into $|{\Phi }^{+}{⟩}_{A{B}_1}$ and $|{\Phi }^{+}{⟩}_{B_{2}C}$. For example applying $ZX$ to the second qubit of a singlet gives $|{\Phi }^{+}⟩$, where $X,Z$ are Pauli operators.

Use the binary-labelled Bell statesThey are ${\Phi }^{+},{\Phi }^{-},{\Psi }^{+},{\Psi }^{-}$ for labels $(0,0),(1,0),(0,1),(1,1)$. Reordering registers to group Bob's qubits, direct expansion givesAll four basis vectors are real in this convention; otherwise the paired expansion must account for complex conjugation. Bob performs the Bell-basis measurement on $B_1{B}_2$. This joint operation is local to him, so it is allowed by LOCC. Each outcome occurs with probability $1/4$ and leaves $AC$ in the corresponding Bell state.

Bob sends $(p,q)$ to Charlie. Charlie applies $Z^p{X}^{-q}$ to obtain $|{\Phi }^{+}{⟩}_{AC}$, and then $XZ$ to obtainThe correction works for every outcome, so no postselection or successful-outcome assumption is needed. This is entanglement swapping, equivalently teleportation of $B_1$ to $C$ while preserving its entanglement with $A$. If Bob's measurement result is ignored, the average $AC$ state is $I_4/4$; the protocol therefore does not supply entanglement usable without the classical outcome or violate the no-communication theorem.

No. Before any operation, Bob's reduced density matrix is $I/2$, independent of $|\alpha ⟩$. Any trace-preserving operation performed only by Alice leaves Bob's reduced state unchanged when her uncommunicated outcomes are averaged. This is the no-communication theorem, or quantum no-signalling.

If Bob received the unknown state, his reduced state would instead be the pure state $|\alpha ⟩⟨\alpha |$, which depends on Alice's input. That would let Alice signal by choosing different inputs, contradicting no-signalling. The two classical outcome bits are therefore essential.

The pure state $|\Psi {⟩}_{AB}$ is entangled when it cannot be written as a product state $|a{⟩}_A|b{⟩}_B$, equivalently when its Schmidt rank exceeds one. Its subsystem states are the reduced density matriceswhere the partial trace is characterized by $\mathrm{Tr}(M_A{\rho }_A)=⟨\Psi |M_A\otimes I_B|\Psi ⟩$ for every observable $M_A$.

Let Bob's measurement have Kraus operators $M_b$ satisfying ${\sum }_b{M}_b^{†}{M}_b=I_B$. If Alice does not learn Bob's random outcome, her state after the measurement isThis no-communication theorem means that Bob can change Alice's conditional state after she learns his outcome, but cannot change any local outcome distribution available to her alone. Standard nonrelativistic quantum mechanics is therefore operationally compatible with the prohibition of superluminal signalling in special relativity, despite its nonlocal conditional-state updates.

An exact nondisturbing state readout would destroy this protection if it reported the globally collapsed state on an absolute-time slice. For example, Alice and Bob may share Bell states. At a prearranged time Bob encodes a bit by measuring his qubit in either the computational or Hadamard basis. The usual projection postulate assigns Alice respectively one of $|0⟩,|1⟩$ or one of $|+⟩,|-⟩$. A device returning the exact pure state lets Alice identify which basis Bob chose without waiting for his outcome, producing a superluminal signal. Equivalently, Bob may choose whether to measure, and the device distinguishes the resulting proper pure state from the original improper maximally mixed local state.

A causal alternative is a quantum state readout device located at a spacetime point $x$ that reports the local quantum state under objective collapse: take reduced states on spacelike hypersurfaces through $x$ and let those hypersurfaces approach the past light cone of $x$. The result includes localized collapses in $J^{-}(x)$ and excludes spacelike-separated collapses. This assumes a fixed Minkowski spacetime, localized collapse events, ordinary local unitary dynamics between them, and outputs that may control only operations in their causal future. The postulate is logically consistent because it adds a classical record of this local state without changing the state or any standard measurement probability. It also cannot support an indirect signalling algorithm. Inductively through the algorithm's events, every readout at $x$ depends only on operations, collapses, and earlier readouts in $J^{-}(x)$; any operation selected from that output lies in the readout's future light cone. Composing readouts, unitary evolutions, and measurements therefore never carries a controllable dependence outside a future light cone, so relativistic causality is preserved.

A bipartite pure state is entangled when it is not a product state. Its Schmidt decomposition is $|\psi ⟩={\sum }_i\sqrt{{\lambda }_i}|i_A{i}_B⟩$, where ${\lambda }_i\ge 0$ and ${\sum }_i{\lambda }_i=1$. More than one nonzero coefficient means entanglement. The two reduced density matrices have the same nonzero spectrum, and their Von Neumann entropy $-{\sum }_i{\lambda }_i{\log }_2{\lambda }_i$ is the pure-state entanglement entropy. A maximally entangled pair of qubits has one bit of this entropy.

For mixed states, classical correlations are not sufficient: a separable quantum state is a convex mixture ${\sum }_j{p}_j{\rho }_A^{(j)}\otimes {\rho }_B^{(j)}$, and entanglement means that no such decomposition exists. A reduced entropy alone no longer measures mixed-state entanglement. An entanglement witness or a suitable separability criterion can detect it.

Local unitary transformations preserve the Schmidt spectrum. More general local operations and classical communication can discard entanglement but cannot freely create it from a separable input. Nielsen's pure-state conversion theorem gives the exact deterministic criterion: $|\psi ⟩\to |\varphi ⟩$ is possible precisely when ${\lambda }^{\psi }\prec {\lambda }^{\varphi }$. In decreasing order, every partial sum for the input is no larger than the corresponding output sum, with equal totals. The direction matters: a uniform spectrum can be converted to a concentrated one; a product state cannot deterministically become a Bell pair.

A probabilistic local filter can concentrate a partially entangled pair. For $|\psi ⟩=\sqrt{\lambda }|00⟩+\sqrt{1-\lambda }|11⟩$ with $1/2\le \lambda <1$, Alice's successful operator $F=\mathrm{diag}(\sqrt{(1-\lambda )/\lambda },1)$ produces equal amplitudes. The outcome is a Bell pair with probability $2(1-\lambda )$; a complementary operator supplies a complete measurement. Failure is heralded, so there is no deterministic creation of a more entangled pair. This illustrates entanglement concentration.

Many-copy entanglement distillation and entanglement dilution connect noisy or nonuniform resources with standard Bell pairs. Entanglement enables quantum teleportation and other protocols, but classical communication is still required; the no-communication theorem excludes message transmission merely by choosing a distant local measurement.

An eigenvalue of an observable in a known corresponding eigenstate is a simple example satisfying the Einstein–Podolsky–Rosen criterion of reality: its value is predictable with certainty. For example, a spin angular momentum component in a prepared spin-up eigenstate has a definite predicted value. For a distant system, Albert Einstein, Boris Podolsky and Nathan Rosen invoked locality: a measurement performed on a sufficiently separated, noninteracting partner should not disturb that system. In the relativistic version here, the measurement events have spacelike separation.

The Pauli matrices obey $[{\sigma }_z,{\sigma }_x]=2i{\sigma }_y$, soTheir nondegenerate eigenstates are incompatible: an $S_z$ eigenstate has equal probabilities for the two $S_x$ outcomes, and conversely. Thus no single quantum state has both components sharply determined. The Robertson uncertainty relation gives $\Delta S_z\Delta S_x\ge \hslash |⟨S_y⟩|/2$, but its right-hand side may be zero; that alone is not a proof that simultaneous zero uncertainties exist. The absence of a common eigenstate supplies the stronger conclusion for this spin-half pair.

For the specified Bell state, Bob obtains $+\hslash /2$ or $-\hslash /2$ with probability $1/2$ when he measures $S_z$. Conditional on his result, Alice obtains the same sign with certainty if she measures $S_z$. Under the EPR locality premise, this certainty supplies an element of reality for her $S_z$; it is a conditional prediction after Bob learns his own random result, not a premeasurement prediction of its sign.

The normalized $S_x$ eigenstates are $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. Expanding their tensor products givesConsequently the same Bell state has equal, perfectly correlated $S_x$ outcomes. Bob's two outcomes again each have probability $1/2$, and the same locality-and-certainty argument applies separately to Alice's $S_x$.

The disputed step is to combine predictions conditional on two incompatible choices into two simultaneous, measurement-independent pre-existing values. Quantum theory provides the two conditional ensembles, not a joint sharp assignment of $S_x$ and $S_z$. This is quantum steering. Bob cannot measure both incompatible observables sharply on the same member of the pair. Nor does conditional steering allow signalling: ignoring his result leaves Alice's reduced density matrix equal to $I/2$ for either measurement choice, by the no-communication theorem.

To test the stronger claim of a local hidden-variable theory, Alice needs several correlation settings, not just perfect agreement along the two common axes. For the CHSH inequality, choose observables with outcomes $\pm 1$:In the given Bell state, $⟨{\sigma }_z\otimes {\sigma }_z⟩=⟨{\sigma }_x\otimes {\sigma }_x⟩=1$ and the crossed correlations vanish. Henceexceeding the local bound $|\mathcal{S}|\le 2$. The bound assumes Bell local causality and measurement independence, with the appropriate treatment of sampling and settings.

Experimental Bell tests support violations rather than that local assignment. The Aspect switching experiment changed analyzer settings during separated photon flights, while leaving sampling and switching limitations. By the time of this examination, the Delft electron-spin Bell test had used electron spins at stations $1.3\mathrm{km}$ apart, efficient spin readout and rapid random basis choices. The combination addressed both detection and locality loopholes, providing evidence against the tested local model class.^[ref]

This does not prove that every conceivable notion of reality is false, or that noncommutation alone rules out every hidden-variable theory. A nonlocal theory can evade the local premise. What Alice can challenge is Bob's claim that locality, measurement-independent simultaneous values and all observed correlations can be retained together. An ordinary measurement disturbance at Bob's lab is not by itself a local explanation of Alice's conditional ensemble.

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.

Alice's outcome is random. Conditional on knowing it, Bob may assign a different quantum state, but without that classical record he sees the same density matrix. For the displayed pure two-qubit state, his reduced density matrix before Alice's measurement isAfter Alice's computational-basis measurement with the outcome unannounced, the weighted mixture $p_0{\rho }_{B|0}+p_1{\rho }_{B|1}$ is exactly this same matrix. More generally, for any local trace-preserving operation with measurement operators $K_a$,The equality follows from the cyclic property of the partial trace for operators acting on Alice's subsystem. Bob's every local outcome probability therefore remains unchanged. This is the no-communication theorem: $\boxed{\text{entanglement alone cannot send information faster than light}}$. Conditional state changes become usable correlations only when the classical outcomes are compared.

For local Kraus operators $A_i$ with ${\sum }_i{A}_i^{†}{A}_i=I$, let ${\rho }_{AB}^{\prime }={\sum }_i(A_i\otimes I){\rho }_{AB}(A_i^{†}\otimes I)$. For every remote operator $B$, cyclicity of the trace gives $\mathrm{Tr}[(I\otimes B){\rho }_{AB}^{\prime }]={\sum }_i\mathrm{Tr}[(A_i^{†}{A}_i\otimes B){\rho }_{AB}]=\mathrm{Tr}[(I\otimes B){\rho }_{AB}]$. Hence the remote reduced density matrix is unchanged. In particular every remote POVM has the same outcome probabilities, proving the no-communication theorem from the measurement rule rather than assuming it as a physical postulate.