Bell state ($|{\Phi }^{\pm }⟩,|{\Psi }^{\pm }⟩$)

The four Bell states form an orthonormal maximally entangled basis for two qubits.

For $|{\Psi }^{+}⟩=(|01⟩+|10⟩)/\sqrt{2}$, the Pauli operator correlations are $⟨XX⟩=1$, $⟨ZZ⟩=-1$, and $⟨XZ⟩=⟨ZX⟩=0$. Substituting the measurement of linear photon polarization into $⟨M(\alpha )\otimes M(\beta )⟩$ gives $\sin 2\alpha \sin 2\beta -\cos 2\alpha \cos 2\beta =-\cos 2(\alpha +\beta )$. In particular, opposite physical angles give perfect anticorrelation. The sum of angles here depends on this choice of Bell state and basis conventions.

Uniformly averaging $(|01⟩+e^{i\phi }|10⟩)/\sqrt{2}$ over $|\phi |\le {\phi }_0$ leaves the two populations unchanged and multiplies the off-diagonal coherences by $\mathrm{sinc}{\phi }_0=\sin {\phi }_0/{\phi }_0$. For real-plane analyser settings giving the pure-state Bell signal $\sqrt{2}(1+\cos \phi )$, linearity of expectation gives the displayed mixed-state signal. Its local-bound violation requires $\mathrm{sinc}{\phi }_0>\sqrt{2}-1$, or $0\le {\phi }_0<2.092751\dots$ radians. This threshold concerns fixed measurement settings rather than all possible tests of entanglement.

A Bell basis with phases chosen so that local special-unitary transformations of the two qubits are represented by real four-dimensional rotations. For one standard ordering it is ${\Phi }_{+},i{\Phi }_{-},i{\Psi }_{+},{\Psi }_{-}$. Permuting these vectors changes generator conventions but preserves the invariant concurrence $|{\sum }_j{b}_j^2|$.^[ref]

The anti-Hermitian generators of local qubit rotations become real antisymmetric matrices in the magic basis. Their exponentials preserve $b^{\mathsf{T}}b$, and hence its modulus and pure-state concurrence. A common phase changes that bilinear quantity by a phase but not its modulus.

Suppose Alice holds a classical orthogonal flag $F$ for the phase of a Bell state:Her local controlled phase correction $U=I_{AB}\otimes |0⟩⟨0{|}_F+(Z_A\otimes I_B)\otimes |1⟩⟨1{|}_F$ maps this to $|{\Phi }^{+}⟩⟨{\Phi }^{+}{|}_{AB}\otimes [p|0⟩⟨0|+(1-p)|1⟩⟨1|{]}_F$. Thus Alice and Bob recover an exact Bell pair without revealing its bit value. Even if Eve knows the flag, purity decouples a subsystem from its purification, so the corrected pair can supply a private bit. Discarding the flag first can instead yield a separable quantum state, as happens at $p=1/2$.

A Bell pair consists of two qubits prepared jointly in a Bell state. It is a maximally entangled two-qubit resource with Schmidt rank two. Two independent Bell pairs have resource Schmidt rank four across the same party partition.

For real $-1\le \alpha \le 1$, this two-qubit density operator has equal $00$ and $11$ populations and off-diagonal coherences $\alpha /2$. It is a pure state exactly when $\alpha =\pm 1$, and a separable quantum state exactly when $\alpha =0$. Its partial transpose has eigenvalues $1/2,1/2,\alpha /2,-\alpha /2$, directly witnessing entanglement for every nonzero admissible coherence.

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.

A device implementing a Bell-basis measurement on two qubits. A complete analyzer distinguishes all four Bell states. A basic passive linear optics analyzer for two polarization qubits uses a balanced beam splitter, polarizing beam splitters and single-photon detectors. Its ideal unambiguous success probability is $1/2$: the two opposite-polarization Bell states have distinct click patterns, while the two same-polarization Bell states remain mutually ambiguous. Additional physical resources can change this restriction. A destructive analyzer absorbs the measured photons; it need not preserve the detected Bell state.

The Bell states are joint eigenstates of commuting $Z_A{Z}_B$ and $X_A{X}_B$. Two shared Bell state meter pairs realize their two entanglement-assisted nondemolition parity measurements using spacelike local circuits. Their projector products give the four rank-one Bell state outcomes and preserve each input Bell state. The globally identified outcome requires later local operations and classical communication.

Superdense coding uses a shared Bell state to communicate two classical bits by transmitting one qubit. The sender selects one of four orthogonal Bell states with a local Pauli operator, and the receiver performs a Bell-basis measurement. For an ideal shared Bell pair, the transmitted qubit alone has a maximally mixed state for every message and reveals no information about the chosen encoding.

With phase/parity labels $(u,v)$ encoded by $X^v{Z}^u$, a resource $a|{\Phi }^{+}⟩+b|{\Phi }^{-}⟩$ gives intended-label probability $|a{|}^2$ and phase-label-flip probability $|b{|}^2$ under a Bell-basis measurement. This follows because the local Pauli operator maps the two orthogonal components into the two Bell states whose phase label differs. The resource is a coherent pure state, even though its measured label statistics form a classical noisy channel.

If a wrong Bell-label partner occurs with probability $ϵ$ and differs from the intended two-bit label in $w$ positions, its contribution to the error probability of a uniformly selected bit is $ϵw/2$. Relabeling the Bell states can change $w$ while preserving the physical wrong-state probability. Thus symbol error probability and bit error probability are different quantities.

For four uniformly distributed Bell states, measure one qubit along Bloch vector $\mathbf{n}$ and the other along $z$. The sign within each ${\Phi }^{\pm }$ or ${\Psi }^{\pm }$ pair is invisible and the displayed mutual information is at most one bit. A local Hadamard gate before a computational detector changes $n_z$ from one to zero and destroys the remaining information. An entangling Bell decoder falls outside this product-measurement restriction and recovers two bits.

The two-spin singlet isSpin measurements along axes $\mathbf{a}$ and $\mathbf{b}$ have correlation $E(\mathbf{a},\mathbf{b})=-\mathbf{a}\cdot \mathbf{b}$, and equal axes give perfect anticorrelation.

This is the spin-one-half singlet state, an example of an angular momentum singlet state. The general singlet condition is zero total angular momentum, not the choice of a particular two-qubit Bell basis.

For $|s⟩=(|01⟩-|10⟩)/\sqrt{2}$ and any two-by-two unitary matrix, expand the two columns of $U$. The equal-index coefficients cancel, and the coefficient of $|01⟩-|10⟩$ is the determinant. Its modulus is one, so a common change of qubit basis preserves the spin singlet state up to a global phase. The density projector is exactly invariant.

For a bipartite density operator ${\rho }_{AB}$, the reduced state of subsystem $A$ is the partial trace ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$. Every measurement performed only on $A$ has outcome probabilities determined entirely by ${\rho }_A$.

Suppose two density operators on $R\otimes E$ have the same reduced density matrix ${\sigma }_R$. Every measurement confined to $R$ then has the same outcome probabilities for the two preparations: a POVM effect $F_R$ has probability $\mathrm{Tr}(F_R{\sigma }_R)$ in either case. This remains true when $R$ contains several laboratories brought together for a joint measurement.

For independent copies, the accessible states are both ${\sigma }_R^{\otimes n}$. No collective measurement on those copies can distinguish the original preparations. Adaptive local processing also cannot help, since applying the same quantum operation to equal inputs gives equal outputs. Access to an additional environment measurement record is a change of resources, and can reveal differences hidden by the partial trace.

A trace-preserving operation on subsystem $B$ cannot change the reduced density matrix of a disjoint subsystem $A$ when its unobserved outcomes are averaged. Local measurements on $A$ therefore cannot reveal which operation was chosen on $B$.

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.

A bipartite density operator is a product state when it factors as ${\rho }_A\otimes {\rho }_B$. A pure product state factors as $|{\psi }_A⟩\otimes |{\psi }_B⟩$; its reduced density matrices have rank one and zero entanglement entropy. Mixed product states can have nonzero marginal Von Neumann entropy.

A nonzero bipartite pure state is a product state if and only if all its partial inner products on the first subsystem lie in a common one-dimensional vector subspace of the second. For a product $|a⟩\otimes |b⟩$, every contraction equals $⟨u|a⟩|b⟩$. Conversely, contract against each vector of an orthonormal basis to write $|\Psi ⟩={\sum }_i|i⟩\otimes |{\chi }_i⟩$. If $|{\chi }_i⟩=c_i|b⟩$ for a common nonzero $|b⟩$, then $|\Psi ⟩=({\sum }_i{c}_i|i⟩)\otimes |b⟩$. Zero conditional vectors are allowed. A nontrivial linear dependence relation requires multipliers not both zero; allowing both multipliers to vanish would give no test at all.

Write a two-qubit pure state as ${\sum }_{i,j=0}^1{c}_{ij}|ij⟩$ and form its coefficient matrix $C=(c_{ij})$. The state is a product state exactly when $C$ has rank one, equivalently when

A bipartite density operator is entangled when it is not a separable quantum state. For a pure state, this is equivalent to not being a product state, or to either reduced density matrix having rank greater than one.

For two systems of equal dimension $d\ge 2$, no joint pure state guarantees equal outcome labels whenever both systems undergo the same arbitrary basic measurement. Equal outcomes in the computational basis would force $|\Psi ⟩={\sum }_j{c}_j|jj⟩$. On any two coordinates j,k, changing both bases to the real sum/difference pair makes the different-outcome amplitude $(c_j-c_k)/2$, so all coefficients must agree. Changing both to the imaginary-phase pair makes a different-outcome amplitude $(c_j+c_k)/2$, so those coefficients must also be negatives. They must all vanish, contradicting normalization. The one-dimensional case is an exception.

The requirement of all complex bases matters. A maximally entangled vector $d^{-1/2}{\sum }_j|jj⟩$ does produce matching labels if one party uses a basis and the other its coordinatewise complex conjugate basis. Conjugate-state steering from a maximally entangled pair gives this conjugation explicitly. Identical real bases are a special case, not evidence for identical arbitrary complex bases.

Quantum entanglement between electromagnetic polarization degrees of freedom. For example, $(|HV⟩+|VH⟩)/\sqrt{2}$ is a Bell state of two photons; neither photon separately has a definite horizontal polarization or vertical polarization in this description.

A pair of photons whose joint electromagnetic polarization has polarization entanglement.

Entanglement distillation uses local operations and classical communication on many noisy entangled states to produce fewer high-fidelity Bell pairs. It is also called entanglement purification. The protocol may fail or abort; naming a purification step does not guarantee successful output for arbitrary inputs. If the output is an exact pure state of Bell pairs, every purification of a density operator factors between those pairs and the environment, so measuring each pair in the computational basis gives a shared uniform bit independent of an eavesdropper. Approximate privacy follows when the entire retained output is sufficiently close to that pure target. A separable quantum state cannot be distilled into a Bell pair by LOCC, because each local branch preserves separability.

Entanglement dilution uses local operations and classical communication to prepare partially entangled states from maximally entangled states, such as Bell pairs. For many identical bipartite pure targets, the asymptotic cost in Bell pairs per target, allowing a vanishing approximation error, is their entanglement entropy in bits. The quantum typical subspace of the target's Schmidt coefficients has almost all its weight and dimension at most $2^{m(S+\delta )}$. A maximally entangled state of this dimension can be converted to the normalized truncated target by Nielsen's pure-state conversion theorem. The resulting state approaches the full target as the retained weight tends to one. Average monotonicity of pure-state entanglement entropy and entropy continuity give the converse lower bound. This is the reverse asymptotic task to entanglement concentration.

For $0<p<1$, every bipartition of ${\Psi }_p^{\otimes n}$ has Schmidt rank $2^n$, while $m$ copies of a GHZ state have Schmidt rank $2^m$. Schmidt-rank contraction under product operators therefore forbids any exact successful LOCC branch when $m<n$. Approximate entanglement dilution avoids this obstruction by retaining only a high-weight quantum typical subspace, whose dimension has logarithm $n{h}_2(p)+o(n)$. Exact finite-block conversion and asymptotically faithful conversion are different tasks.

For nonnegative $u,v$ with $u^2+v^2=1$, Alice applies Kraus operators $M_0=\mathrm{diag}(u,v)$ and $M_1=\mathrm{diag}(v,u)$ to one half of a Bell state. Their squared effects sum to the identity, and both outcomes have probability one half. The first branch is the target; the second becomes the target after both parties apply Pauli X gates. Sending the outcome by classical communication makes this exact deterministic entanglement dilution, rather than a postselected conversion.

An entanglement monotone quantifies entanglement without increasing on average under local operations and classical communication. The average is over recorded measurement in quantum mechanics outcomes: individual successful branches can have more entanglement, provided their probabilities compensate for that increase. This distinction makes an entanglement monotone useful for bounding probabilistic state conversion. A pure-state entanglement monotone need only satisfy this inequality on pure input states and pure conditional branches, with local measurement outcomes refined to individual Kraus operators.

The relative entropy of entanglement is the smallest quantum relative entropy from a density operator to a separable quantum state, with the separable set specified for the relevant division into parties. Nonnegativity of quantum relative entropy implies $E_R\ge 0$. For a separable quantum state, choosing $\sigma =\rho$ gives $E_R=0$. For finite-dimensional bipartite states, compactness of the separable set and lower semicontinuity of the relative entropy give the converse. The support convention is $D(\rho \Vert \sigma )=+\infty$ unless $\mathrm{supp}\rho \subseteq \mathrm{supp}\sigma$.

For a bipartite pure state, the squared Schmidt coefficients are the nonzero eigenvalues of either reduced density matrix. Their sum except the largest defines $E_2=1-{\lambda }_{\max }({\rho }_A)$. The largest eigenvalue is a convex function, since ${\lambda }_{\max }(\rho )={\max }_{\Vert v\Vert =1}⟨v|\rho |v⟩$; consequently $E_2$ is concave. If Alice measures locally and records outcome $r$, Bob's conditional reductions satisfy ${\sum }_r{p}_r{\rho }_{B,r}={\rho }_B$. Concavity gives ${\sum }_r{p}_r{E}_2({\chi }_r)\le E_2(\chi )$. For Bob's measurements use Alice's unchanged average reduction instead. Iteration proves average monotonicity under LOCC. For Schmidt rank at most two, $E_2$ is the smaller squared Schmidt coefficient. A conversion succeeding with probability $p$ therefore obeys $p{E}_2(\varphi )\le E_2(\psi )$, since failure branches contribute nonnegative amounts.

For a normalized pure state of two qubits with coefficient matrix $C$, the concurrence is $2|\det C|$. It ranges from zero for a product state to one for a maximally entangled state.

For a normalized pure two-qubit state written in a magic basis, concurrence is $\mathcal{C}=|{\sum }_j{b}_j^2|$, while $\det {\rho }_A={\mathcal{C}}^2/4$. This bilinear sum uses no complex conjugation inside its square; replacing it by ${\sum }_j|b_j{|}^2$ would incorrectly give maximal concurrence for every state.

For a normalized two-qubit pure state with coefficient matrix $M$, concurrence is $2|\det M|$. The displayed convex roof extension defines its mixed-state version. Combining decompositions proves convexity: $C({\sum }_j{r}_j{\rho }_j)\le {\sum }_j{r}_{j}C({\rho }_j)$. Every separable quantum state has a decomposition into product vectors, each with zero determinant, so its concurrence is zero.

For ${\rho }^{\prime }=(A\otimes B)\rho (A^{†}\otimes B^{†})/p$, with invertible $A,B$, a pure coefficient matrix transforms as $M\mapsto AM{B}^{\mathsf{T}}$. Its unnormalized determinant is multiplied by $\det A\det B$. The invertible filter bijects pure decompositions before and after conditioning, with branch weights rescaled by their success probabilities. Taking both infima in the convex roof extension gives the displayed exact scaling. If either filter has rank one, the retained state is separable across that party.

The state $a|00⟩+b|11⟩$ is entangled exactly when $ab\ne 0$. Its reduced density matrix has nonzero eigenvalues $|a{|}^2$ and $|b{|}^2$.

Entanglement concentration probabilistically converts partially entangled pure states into fewer maximally entangled states by local operations and classical communication.

For one pure rank-two entangled state with descending squared Schmidt coefficients $({\lambda }_0,{\lambda }_1)$, the success operator $F=\mathrm{diag}(\sqrt{{\lambda }_1/{\lambda }_0},1)$ gives a Bell pair with probability $2{\lambda }_1$. The complementary operator $\sqrt{I-F^{†}F}$ makes a valid quantum instrument. Optimality follows from the Schmidt-tail entanglement monotone: a successful Bell branch has smallest weight $1/2$, so its probability cannot exceed $2{\lambda }_1$. Failure branches have nonnegative monotone and cannot loosen this bound.

For a pure state with two descending squared Schmidt coefficients ${\lambda }_0\ge {\lambda }_1$, this quantum instrument acts in one party's Schmidt basis. The first outcome balances the two Schmidt amplitudes when $\tan \theta =\sqrt{{\lambda }_1/{\lambda }_0}$, with probability $2{\lambda }_0{\lambda }_1$. This is not the optimal unrestricted success probability $2{\lambda }_1$ in single-copy entanglement concentration. An embedding in a larger subsystem requires a POVM on the occupied support or a full-space completion.

For $n$ copies of a normalized pure state $\alpha |00⟩+\beta |11⟩$, regroup Alice's and Bob's tensor factors. Alice performs a projective measurement of the total number $k$ of zeroes, preserving all coherence among strings with that count. The outcome has binomial distributionEach retained string has the same complex amplitude ${\alpha }^k{\beta }^{n-k}$, so, up to a common phase, the conditional quantum state isIts equal Schmidt coefficients make it a maximally entangled state of Schmidt rank $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. Alice sends the count to Bob. They can relabel the corresponding local computational basis strings with local unitary operations, and convert the resulting resource to Bell pairs. Measuring the individual bits instead would reveal the entire string and leave a product state.

For $\sqrt{p}|000⟩+\sqrt{1-p}|111⟩$, a projective measurement of the number of ones in one party's $n$ local qubits gives a common type class of size $d_k=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. The conditional pure state is $d_k^{-1/2}{\sum }_{|x|=k}|x,x,x⟩$. It can be converted exactly and deterministically to a uniform common-label state of any dimension $m\le d_k$: one party measures an $m$-element subset $S$ with Kraus operator ${\Pi }_S/\sqrt{\left(\genfrac{}{}{0ex}{}{d_k-1}{m-1}\right)}$, and all parties relabel their common retained labels after receiving $S$. Choosing $m=2^f$ gives $f$ independent GHZ states. Stirling's formula and the weak law of large numbers give $f/n\to h_2(p)$ in probability. Measuring every local bit separately would instead destroy the coherence and leave a product state.

The Bell states $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$ and $|{\Phi }^{-}⟩=(|00⟩-|11⟩)/\sqrt{2}$ both have reduced state $I/2$ on either qubit. No measurement on one qubit alone can distinguish their relative sign.

Local operations and classical communication allow separated parties to measure or transform their own subsystems and exchange ordinary classical messages.

An auxiliary entangled pure state can enable an otherwise forbidden deterministic bipartite LOCC transformation while being returned exactly and uncorrelated with the output. Nielsen's pure-state conversion theorem reduces this to the displayed majorization condition on sorted squared Schmidt coefficients. It does not follow from a scalar comparison of entanglement entropies: tensoring changes which products occupy the largest cumulative-weight sums. An exact returned catalyst can be reused; merely restoring its marginal while allowing correlations is a different resource task.^[ref]

Let $c,s\ge 0$ and $c^2+s^2=1$. Starting from the Bell state $(|00⟩+|11⟩)/\sqrt{2}$, Alice applies the measurement in quantum mechanics with Kraus operators $M_0,M_1$ displayed above. Completeness follows from $M_0^{†}{M}_0+M_1^{†}{M}_1=I$. Each outcome has probability $1/2$. The normalized outputs are $c|00⟩+s|11⟩$ and $s|00⟩+c|11⟩$. Communicating the outcome allows both parties to apply a Pauli X gate in the second branch, making both outputs equal to the first. Thus this LOCC protocol deterministically converts one Bell pair into any two-qubit pure state with these Schmidt coefficients, after appropriate local unitary operations. It includes the product state endpoint without postselection.

Let $|{\Phi }_d⟩=d^{-1/2}{\sum }_{j=0}^{d-1}|j,j⟩$ and $1\le m\le d$. It can be converted deterministically by LOCC to $|{\Phi }_m⟩$. For each $m$-element subset $S$ of the local basis labels, let ${\Pi }_S$ be its projector and let Alice's Kraus operator be $K_S={\Pi }_S/\sqrt{\left(\genfrac{}{}{0ex}{}{d-1}{m-1}\right)}$. Every label occurs in $\left(\genfrac{}{}{0ex}{}{d-1}{m-1}\right)$ subsets, so ${\sum }_S{K}_S^{†}{K}_S=I$. Each outcome has probability $m/[d(\genfrac{}{}{0ex}{}{d-1}{m-1})]=1/\left(\genfrac{}{}{0ex}{}{d}{m}\right)$ and leaves the maximally entangled state $m^{-1/2}{\sum }_{j\in S}|j,j⟩$. Alice reports $S$, and both parties relabel these $m$ basis states to $0,\dots ,m-1$ by local unitary operations.

For $d=6$, $m=4$, only three outcomes are needed: partition the six labels into disjoint pairs $T_0,T_1,T_2$ and use $K_r=(I-{\Pi }_{T_r})/\sqrt{2}$. Each label is retained twice, proving completeness, and each outcome has probability $1/3$. Every branch leaves four equal Schmidt coefficients, hence two independent Bell pairs after binary relabelling.

For bipartite pure source and target states with positive target entanglement entropy, the optimal rate of LOCC conversion with vanishing output error is the ratio of their entanglement entropies. Here vanishing error means that the trace distance between the output and $|\varphi ⟩⟨\varphi {|}^{\otimes m}$ tends to zero; a recorded failure probability must also tend to zero. Entanglement concentration first provides $nS({\rho }_A^{\psi })-o(n)$ Bell pairs, and entanglement dilution needs $mS({\rho }_A^{\varphi })+o(m)$ such pairs. This achieves the ratio. Conversely, refine the output into pure branches. Average monotonicity of pure-state entanglement entropy bounds their average entropy by the input entropy; continuity bound for quantum conditional entropy, with trivial conditioning, makes the output entropy loss per target tend to zero. This forbids a larger rate. The convention permits approximations for finite blocks: exact deterministic conversion of every finite block can obey stricter majorization constraints. The reversible concentration construction is given in the original concentration paper.

For two-qubit pure states with smaller squared Schmidt coefficients $s<t$, optimal stochastic conversion of a two-qubit pure state succeeds independently with probability $s/t$. The asymptotic pure-state entanglement conversion rate of collective LOCC is instead $h_2(s)/h_2(t)$, where $h_2$ is binary entropy. The difference is strict: differentiating gives $\frac{d}{dx}[h_2(x)/x]={\log }_2(1-x)/x^2<0$. Hence $h_2(s)/s>h_2(t)/t$. Independent filtering loses entanglement in its product-state failure branches, whereas collective concentration and dilution preserve entanglement asymptotically.

Write a source and target in Schmidt decomposition with squared coefficients $(1-s,s)$ and $(1-t,t)$, where $0<s,t\le 1/2$. If $s\ge t$, Nielsen's pure-state conversion theorem allows deterministic conversion. If $s<t$, the Schmidt-tail entanglement monotone gives $p\le s/t$. Alice attains this bound with a local two-outcome measurement in quantum mechanics whose success Kraus operator is $K_s=\mathrm{diag}(\sqrt{s(1-t)/(t(1-s))},1)$ and whose failure operator is $K_f=\mathrm{diag}(\sqrt{1-s(1-t)/(t(1-s))},0)$. They obey $K_s^{†}{K}_s+K_f^{†}{K}_f=I$. The successful unnormalized state is $\sqrt{s/t}$ times the target, and failure leaves a product state. A classical message tells Bob which branch occurred. Known local changes of Schmidt bases allow the same protocol for arbitrary two-qubit pure states with these coefficients.

An operation that applies a unitary operator to one party's subsystem and the identity to the other is a local unitary operation. Several separated parties may each apply a known unitary, resulting in a product of their local unitaries. Such operations preserve the squared Schmidt coefficients across the relevant bipartition.

Two bipartite pure states on fixed subsystem spaces are local-unitary equivalent when related by the displayed product of unitary operators. Identity operators give reflexivity, inverse operators give symmetry, and products of local operators give transitivity. Thus this is an equivalence relation, whose classes are the orbits under local unitaries.

Every pair of normalized product states is local-unitary equivalent: extend each factor to an orthonormal basis and use a unitary operator carrying one basis to the other. Local operators preserve a product factorization, and their inverses do too, so quantum entanglement cannot be created or removed this way. More generally, the tensor-product coefficient matrix changes as $C\mapsto UC{V}^T$, which preserves its singular values and therefore its Schmidt coefficients. The transpose on the second factor is ordinary, not a conjugate transpose, because coefficients of a tensor product depend bilinearly on the two input vectors.

A local unitary sends a reduced density matrix to a unitary conjugate on the acted-on subsystem and leaves the other subsystem unchanged. Its eigenvalues and von Neumann entropy therefore do not change. This distinguishes a deterministic change of local basis from a selective measurement or nonunitary local filter.

Under a product unitary operator $U_A\otimes U_B$, the reduced density matrix of $A$ becomes $U_A{\rho }_A{U}_A^{†}$. The unitary on $B$ cancels under the partial trace. Unitary conjugation preserves eigenvalues, so the squared Schmidt coefficients of a bipartite pure state are unchanged. Conversely, two normalized bipartite pure states with the same squared Schmidt coefficients, including multiplicities and zero padding, have Schmidt decompositions carried into each other by local unitaries mapping their Schmidt bases. Thus reduced-state spectra completely classify bipartite pure states up to local unitary operations.

An exact deterministic LOCC conversion of a bipartite pure state to another is possible exactly when the input vector of squared Schmidt coefficients is majorized by the output vector. Sort the two probability vectors decreasingly and pad with zeros to a common length before applying majorization. A maximally entangled state may therefore be converted to a product state, while the reverse conversion is forbidden. The criterion does not assert conversion by postselection or with a catalyst.

Two parties can distinguish $|{\Phi }^{+}⟩$ from $|{\Phi }^{-}⟩$ by both measuring in the Hadamard basis and comparing outcomes: equal outcomes identify $|{\Phi }^{+}⟩$, while unequal outcomes identify $|{\Phi }^{-}⟩$.

Quantum teleportation transfers an unknown qubit using one shared Bell pair, a Bell-basis measurement by the sender, two classical bits, and a Pauli correction by the receiver.

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.

Share a GHZ state on $A,B,C$ and let the unknown input $Q$ be $|\chi ⟩$. Alice applies CNOT from $Q$ to $A$, then a Hadamard gate to $Q$, and measures $Q,A$ computationally, giving $s,t$. Charlie measures $C$ in the Hadamard basis, giving $c$. Bob's conditional state isup to a global phase. Alice's two bits and Charlie's one bit therefore specify Bob's inverse correction $Z^{s\oplus c}{X}^t$.

To prove the phase rule, write $|\chi ⟩=\alpha |0⟩+\beta |1⟩$. After Alice's result, $BC$ is $\alpha |tt⟩+(-1{)}^s\beta |(1\oplus t)(1\oplus t)⟩$. Taking Charlie's Hadamard basis partial inner product multiplies its second term relative to its first by $(-1{)}^c$. The remaining common phase is $(-1{)}^{ct}$, which is physically irrelevant. All eight result triples have probability $1/8$. Charlie's record localizes a known-phase Bell pair, allowing the standard Bell-basis teleportation identity to complete the transfer.

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.

Suppose a perfect $|{\varphi }^{+}⟩$ Bell pair is used for quantum teleportation, but the sender cannot distinguish the ${\varphi }^{+}$ and ${\varphi }^{-}$ Bell-basis measurement outcomes and guesses their labels uniformly. Each of the four actual outcomes has probability $1/4$. Either ambiguous outcome gives an incorrect Pauli Z gate correction with conditional probability $1/2$, so the total phase-flip probability is $1/4$. The unconditional output is therefore the displayed phase-flip channel. For $|\psi ⟩=\alpha |0⟩+\beta |1⟩$, its teleportation fidelity is $⟨\psi |\mathcal{E}(|\psi ⟩⟨\psi |)|\psi ⟩=1-|\alpha {|}^2|\beta {|}^2$. It ranges from $3/4$ on the Bloch sphere equator to $1$ at its poles. This averages the output density operators, including the ambiguous rounds.

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]

For a pure input target $|\psi ⟩$ and normalized output density operator ${\rho }_{\mathrm{out}}$, the displayed squared quantum fidelity measures how well a teleportation channel reproduces that target. A mean value additionally specifies the input ensemble and whether it is conditioned on success. Detection efficiency, conditional fidelity, heralding and overall success probability are distinct experimental quantities.

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.

Two initial maximally entangled pairs $A{B}_1$ and $B_{2}C$ can be turned into an entangled pair $AC$ by a generalized Bell basis measurement on $B_1{B}_2$, followed by a local correction at an endpoint using its classical label. This is qudit teleportation of $B_1$, including its correlations with $A$, to $C$. The initial two pairs are consumed. Without the classical label, averaging the conditional endpoint states need not leave entanglement.

Take two identical normalized entangled qubit pairs ${\sum }_{r=0}^1{b}_r|{\xi }_r{\eta }_r⟩$. Project their two $\eta$ systems onto $|\Phi ⟩=2^{-1/2}{\sum }_k|{\eta }_k{\eta }_k⟩$. The overlap with $|{\eta }_r{\eta }_s⟩$ is ${\delta }_{rs}/\sqrt{2}$, so the unnormalized remote state is ${\sum }_r{b}_r^2|{\xi }_r{\xi }_r⟩/\sqrt{2}$. Its squared norm gives success probability $\frac{1}{2}{\sum }_r|b_r{|}^4$, and its normalized Schmidt coefficients are $b_r^2/({\sum }_s|b_s{|}^4{)}^{1/2}$. At least two survive whenever the original pair was entangled. This is a conditional entanglement swapping result; disregarding the local measurement outcome leaves the remote reduced state unchanged. The coefficients are squared amplitudes, not their absolute squares, because the two input kets multiply before the Bell-state contraction.

Start from three normalized Bell states ${\Phi }_{12}^{+}{\Phi }_{34}^{+}{\Phi }_{56}^{+}$. A polarizing beam splitter parity filter ${\Pi }_Z^{25}$ followed by $X$-basis outcome $++$, and a second filter ${\Pi }_X^{46}$ followed by $Z$-basis outcome $00$, have effective bras $⟨{\Phi }^{+}{|}_{25}/\sqrt{2}$ and $⟨{\Phi }^{+}{|}_{46}/\sqrt{2}$. Their contraction leaves ${\Phi }_{13}^{+}/8$. The herald probability is therefore $1/64$ per ideal three-pair input, or $1/16$ conditional on both parity-routing filters having succeeded. This distinction prevents renormalizing away the optical postselection cost.

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.

A sender and receiver share $n^{-1/2}{\sum }_j|jj⟩$. The sender measures the input and their resource half in the generalized Bell basis. Outcome $(r,s)$ gives the receiver $X^s{Z}^{-r}|\chi ⟩/n$, with probability $1/n^2$. After classical communication of the outcome, the correction $Z^r{X}^{-s}$ restores the unknown quantum state. The same calculation preserves correlations with a reference, as in teleportation as an identity channel on a reference.

Ideal corrected quantum teleportation transfers an input system to the receiver while preserving its entire joint density operator with an arbitrary reference. Expand a joint pure state as ${\sum }_j|j⟩|r_j⟩$. Every teleportation branch acts with the same known correction on each system basis vector and does nothing to the reference vectors; undoing that correction restores the complete sum. Linearity extends the result to mixed states. The sender's measured carrier no longer retains the original reference correlations.

Apply a Controlled-Z gate to input $a|0⟩+b|1⟩$ and a fresh $|+⟩$ quantum ancilla. Measuring the input in the equatorial qubit measurement basis at angle $\alpha$, with outcome $s$, leaves the unnormalized outputEach branch has probability $1/2$, independently of the input. The logical output is $J(\alpha )|\psi ⟩$ with known Pauli frame $X^s$; the input qubit has been measured, so this does not clone it.

With equatorial qubit measurement vectors $(|0⟩+(-1{)}^s{e}^{i\eta }|1⟩)/\sqrt{2}$, the gate in one-bit teleportation is $U(\eta )=H\mathrm{diag}(1,e^{-i\eta })=J(-\eta )$. Direct multiplication gives $U(\eta )Z=XU(\eta )$ and $U(\eta )X=e^{-i\eta }ZU(-\eta )$. Moving a Pauli frame through this gate therefore exchanges its $X$ and $Z$ exponents and reverses the angle if the old $X$ exponent is one, as displayed up to global phase. A measurement outcome $s$ adds a further $X^s$ factor. This calculation supplies adaptive angle choices and the final classical output-bit correction of a linear cluster state computation.

A two-vertex graph state starts with two $|+⟩$ states joined by a Controlled-Z gate. Measuring the first vertex in the $X$ basis leaves the second in $X^{s}H|+⟩=X^s|0⟩$, where $s$ is the measurement outcome. Thus a known zero input can be supplied to a measurement wire using only a graph-state resource and a known Pauli frame. This preparation link precedes the links implementing the desired logical J gate; later measurement angles and output-bit corrections absorb its byproduct.

Two angle-zero one-bit teleportations with outcomes $p,q$ implement $X^{q}H{X}^{p}H=X^q{Z}^p$ on an arbitrary input, up to global phase. A Controlled-Z gate with a fixed $|1⟩$ quantum ancilla implements $Z$, so $Z^p$ removes the known $Z$ factor. The result is $X^q$ up to phase. With probability one half the desired Pauli X gate is applied; otherwise the input is unchanged. Repeating until $q=1$ gives an exact heralded correction with two attempts on average and almost-sure termination. It has no finite worst-case measurement count. A Pauli frame avoids this repeat-until-success procedure when only logical action or classical output statistics are required.