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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.

The Schmidt decomposition and local unitary operations put any entangled two-qubit pure state intoBoth Schmidt coefficients are nonzero. Alice performs a two-outcome generalized measurement with Kraus operatorsTheir effects sum to the identity, so this is a legitimate local measurement in quantum mechanics. On success,The normalized output is the Bell state ${\Phi }_{+}$. Bob's unitary matrix $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ maps it to the spin singlet state. Alice communicates her success flag; no shared quantum operation is needed. At $\lambda =1/2$ the filter is the identity and success is certain. This is the singlet target case of optimal stochastic conversion of a two-qubit pure state.

For dichotomic local outcomes $A_x,B_y\in \{-1,1\}$, the CHSH inequality isIt assumes a local hidden-variable theory with freely chosen settings independent of the hidden-variable distribution. For the spin singlet state, direct evaluation of its Pauli matrices correlations givesChoose $a_0=e_z$, $a_1=e_x$, $b_0=(e_z+e_x)/\sqrt{2}$, and $b_1=(e_z-e_x)/\sqrt{2}$. The four correlations are $-1/\sqrt{2},-1/\sqrt{2},-1/\sqrt{2},+1/\sqrt{2}$, respectively, giving

The success flag must be recorded and communicated before the subsequent independent choice of Bell-test settings. This conditioning is legitimate because setting-independent heralding preserves Bell locality: a putative local hidden-variable theory for the entire sequential experiment would, after successful filtering, still have local response functions with a setting-independent conditional hidden-variable distribution. More explicitly, if the hidden variable is $\xi$ and $h$ is the already determined success flag, conditioning changes its distribution to $\rho (\xi \mid h=1)$ but leavesClassical messages used in preparation can be included in $\xi$; none are exchanged during the spacelike Bell measurements. Such a conditioned model still obeys the CHSH inequality, contradicting the displayed singlet correlations. Therefore no local hidden-variable theory reproduces all sequential measurement correlations of any entangled two-qubit pure state. This argument does not discard outcomes on the basis of the later measurement settings.

By the Schmidt decomposition theorem, suitable local unitary operations put the normalized pure state into the formIt is entangled precisely when both Schmidt coefficients are nonzero. Put $s=2{\lambda }_0{\lambda }_1$, so $0<s\le 1$ for the case of interest. The Pauli operatorssatisfy $⟨Z\otimes Z⟩=1$ and $⟨X\otimes X⟩=s$ in this pure state: $Z\otimes Z$ fixes both basis terms, and $X\otimes X$ exchanges them.

Choose Alice's two observables as $A_0=Z$, $A_1=X$, and Bob's asThese are valid dichotomic projective measurements. Indeed, $XZ+ZX=0$ and $X^2=Z^2=I$, so $B_0^2=B_1^2=I$; each observable has outcomes $\pm 1$. For the CHSH inequality convention $|⟨A_0\otimes (B_0+B_1)+A_1\otimes (B_0-B_1)⟩|\le 2$, the quantum value isFor the original pure state, conjugate these four observables by the corresponding local unitary operators. Their outcomes and Pauli correlators are unchanged, and each party still measures only their own qubit. Thus every entangled pure two-qubit state violates CHSH for suitable local measurements, proving Gisin's theorem directly. The CHSH axes for an entangled pure two-qubit state approach a nonviolating configuration continuously as one Schmidt coefficient tends to zero; the strict violation therefore need not be large for weak entanglement.

Yes: a deterministic LOCC protocol exists for every allowed value of the parameter. Here is an explicit implementation, so existence does not depend on merely quoting a pure-state conversion theorem.

First convert the input Bell state to $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$ by local unitary operations:Write $c=\cos \varphi$ and $s=\sin \varphi$. Alice performs the two-outcome measurement in quantum mechanics with Kraus operatorsThey define a physical measurement in quantum mechanics becauseActing on $|{\Phi }^{+}⟩$, the unnormalized branch vectors areBoth branches have probability $1/2$. Alice communicates the outcome to Bob. If it is one, both apply the Pauli X gate; this turns the normalized second branch into $c|00⟩+s|11⟩$. If it is zero, neither applies this correction. Finally Bob applies a Pauli X gate in either branch, obtainingEvery outcome produces the same target pure state, so this is deterministic local operations and classical communication, not a successful branch selected by postselection. This two-outcome LOCC dilution of a Bell pair remains valid at $\varphi =0$, where the output is a product state, and at $\varphi =\pi /4$, where it is another Bell state.

The target Schmidt coefficients are $c,s$, while those of the input are both $1/\sqrt{2}$. For an interior value $0<\varphi <\pi /4$, local unitary operations alone could not change these Schmidt coefficients; the local measurement in quantum mechanics is the step that reduces the entanglement.

Under the intended assumption that one use of the quantum channel transmits one qubit without noise, Alice can communicate two classical bits with certainty if she and Bob already share a Bell pair. She uses superdense coding. The prior distribution of the Bell pair must be a resource available before the permitted channel use; distributing it through that same channel would consume another use.

Take the shared Bell state to be $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$, with Alice holding the first qubit. For a two-bit message $(a,b)\in \{0,1{\}}^2$, Alice applies the local unitary operation $U_{ab}=X^a{Z}^b$ to her qubit. The four encoded pure states are, up to irrelevant global phases,They are orthogonal: for any matrix $V$ on the first qubit, $⟨{\Phi }^{+}|(V\otimes I)|{\Phi }^{+}⟩=\mathrm{Tr}V/2$, and the four Pauli operators are orthogonal in the trace inner product. Explicitly,Alice sends her qubit through the quantum channel once. Bob now holds both qubits and performs a Bell-basis measurement. Its four orthogonal projectors identify $(a,b)$ with certainty, givingBefore the transmission, Bob's reduced density matrix is $I/2$ for every message, so shared entanglement alone does not convey the message.

Without shared entanglement, a noiseless single-qubit transmission cannot perfectly distinguish four messages: perfect discrimination requires orthogonal state supports, and its two-dimensional Hilbert space has room for at most two nonzero mutually orthogonal supports. The Holevo bound also limits its accessible classical information to one bit. For the usual protocol starting with a shared two-qubit pure state and using local unitary encodings, exact two-bit transmission requires a maximally mixed Bob marginal. Indeed all encoded joint states are pure and have the same ${\rho }_B$, while Subadditivity of Von Neumann entropy gives their average entropy at most $1+S({\rho }_B)$. Four equally likely perfectly distinguishable messages have average Von Neumann entropy two, forcing $S({\rho }_B)=1$ and hence a maximally entangled state.

The specification “quantum channel” alone does not determine its dimension or noise, so the one-qubit, noiseless interpretation is needed for this particular conclusion. A noiseless four-dimensional channel could send four orthogonal states without any shared entanglement; a completely depolarizing qubit channel cannot transmit the message even with a shared Bell pair, since its output is independent of Alice's encoding. Shared entanglement is therefore the extra resource for the intended single-qubit channel, not a guarantee for every unspecified quantum channel.

Here is a tripartite Schmidt projection entanglement concentration protocol. For $0<p<1$, set $q=1-p$. Regrouping the $n$ local qubits of each party giveswhere $|x|$ counts ones. Alice performs a projective measurement of that count, without learning the individual string. The orthogonal projector for outcome $k$ is ${\Pi }_k={\sum }_{|x|=k}|x⟩⟨x|$. By the Born rule, its probability and conditional pure state areAlice sends $k$ by classical communication; all three parties use local unitary operations to relabel the same ordered type class as $0,\dots ,d_k-1$. There are now $d_k$ equal common-label amplitudes. Measuring the individual qubits instead would have destroyed precisely this coherence.

To obtain an integer number $f$ of target GHZ states, choose $m=2^f\le d_k$. The following exact LOCC step handles type classes whose sizes are not powers of two. For every $m$-element subset $S\subseteq \{0,\dots ,d_k-1\}$, Alice uses the Kraus operatorEvery label occurs in $\left(\genfrac{}{}{0ex}{}{d_k-1}{m-1}\right)$ such subsets, proving ${\sum }_S{K}_S^{†}{K}_S=I$ on the occupied subspace. Thus this is a complete measurement in quantum mechanics. Each outcome has probability $m/[d_k(\genfrac{}{}{0ex}{}{d_k-1}{m-1})]=1/\left(\genfrac{}{}{0ex}{}{d_k}{m}\right)$, and leaves the normalized pure state $m^{-1/2}{\sum }_{r\in S}|r,r,r⟩$. After Alice communicates $S$, all parties relabel its elements in the same order. Binary expansion of the labels then factors the state into exactly $f$ independent copies of the target GHZ state. Local extensions of these operations handle unused basis vectors and do not affect the protocol.

For example, taking $f(k)=\lfloor {\log }_2{d}_k\rfloor$ gives an outcome-dependent yield without any failure branch. Since $K$ has a binomial distribution with parameters $(n,q)$, the weak law of large numbers gives $K/n\to q$ in probability. Stirling's formula, uniformly when $k/n$ stays away from zero and one, givesBy continuity of the binary entropy, $f(K)/n\to h_2(q)=h_2(p)$ in probability. The integer rounding and logarithmic correction have no first-order effect.

A fixed yield with success probability tending to one is also explicit. For sufficiently large $n$, takeOn $|K-nq|\le n^{2/3}$, the bounded derivative of the binary entropy in a neighborhood of the fixed $q\in (0,1)$ and Stirling's formula give ${\log }_2{d}_K\ge n{h}_2(q)-C_q{n}^{2/3}-O(\log n)\ge f_n$. Chebyshev's inequality bounds the complementary probability by $q(1-q)n^{-1/3}$. Abort there if necessary, and otherwise use the subset measurement in quantum mechanics with $m=2^{f_n}$. ThusThis matches the first-order rate in part (i), with exact outputs on every successful forward branch. If $p=1/2$, simply retain the original $n$ target copies; if $p$ is an endpoint, output zero target copies.

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

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

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

Write $a=|\alpha {|}^2$, $b=|\beta {|}^2=1-a$, so $1/2<a<1$ and $0<b<1/2$. Regrouping the four copies by party giveswhere $z(x)$ counts zeroes. Alice must measure only this count, preserving coherence inside each degenerate eigenspace, rather than measuring her four bits separately. Her projective measurement has projectors ${\Pi }_k={\sum }_{z(x)=k}|x⟩⟨x|$, and the Born rule givesAlice sends $k$ to Bob. Up to the common phase of ${\alpha }^k{\beta }^{4-k}$, their conditional pure state isThis is Schmidt projection entanglement concentration: its equal Schmidt coefficients give a maximally entangled state of Schmidt rank $d_k$.

For $k=0$ or $4$, $d_k=1$, and the branch is a product state. No Bell pair can be extracted using local operations and classical communication. The respective probabilities are $b^4$ and $a^4$.

For $k=1$ or $3$, $d_k=4$. List the four surviving strings in lexicographic order as $x_0,x_1,x_2,x_3$. For $k=1$ this list is $0111,1011,1101,1110$; for $k=3$ it is $0001,0010,0100,1000$. Each party applies a local unitary operation extending the basis relabellingSuch a unitary exists because these are four distinct orthonormal basis vectors mapped to four distinct basis vectors; extend the mapping to a permutation of all sixteen. Discarding the two fixed local ancillary bits leavesThus these branches give exactly two Bell pairs, with probabilities $4a{b}^3$ and $4a^{3}b$, and no further measurement in quantum mechanics is needed.

For $k=2$, the six strings in lexicographic order are $x_0=0011$, $x_1=0101$, $x_2=0110$, $x_3=1001$, $x_4=1010$, $x_5=1100$. A suitable local POVM gives two Bell pairs with certainty, rather than losing some of these branches to a simple subspace projection. Partition these labels into the three pairs $T_0=\{0,1\}$, $T_1=\{2,3\}$, $T_2=\{4,5\}$. Alice uses the Kraus operatorsOn the six-dimensional support, each label is retained by two operators, so ${\sum }_{r=0}^2{K}_r^{†}{K}_r={\Pi }_2$. Add $K_{\perp }=I-{\Pi }_2$ to define a complete measurement in quantum mechanics on Alice's four qubits; that extra outcome has probability zero on this branch. Each nonzero outcome has conditional probability $(1/2)(4/6)=1/3$, and gives the normalized stateAlice sends $r$ to Bob. Both parties relabel the four surviving strings in increasing order to $|00⟩,|01⟩,|10⟩,|11⟩$ on two output qubits, with two fixed ancillary bits, just as above. Every $r$ therefore gives two Bell pairs. The joint probability of $(k,r)=(2,r)$ is $p_2/3=2a^2{b}^2$ for each $r$.

For an explicit local implementation of this POVM, Alice can append an ancilla in a fixed state and apply an isometry on the six-dimensional support,It preserves inner products, extends to a unitary on system and ancilla, and measuring the ancilla realizes precisely the operators $K_r$. No nonlocal operation has been used. This is a deterministic reduction of maximally entangled Schmidt rank from six to four.

The complete output distribution for this strategy is consequentlyThe mean yield is $2(1-a^4-b^4)$ pairs per four-copy block. These are the maximum possible pair counts after the stipulated count measurement: any nonzero branch has Schmidt rank at most six, whereas three independent Bell pairs require Schmidt rank eight. In any refined LOCC branch the coefficient matrix is multiplied by local matrices on the left and right, which cannot increase its rank.

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.

Introduce the orthonormal local vectors $|+⟩=(|1⟩+|2⟩)/\sqrt{2}$ and $|-⟩=(|1⟩-|2⟩)/\sqrt{2}$. Reexpressing the second state's factors in this basis gives its Schmidt decompositionThe squared Schmidt coefficients are therefore $1/3,2/3$, whereas those of the first state are $1/2,1/2$.

To see directly why these distinguish the states, a local unitary operation $U_A\otimes U_B$ changes the reduced density matrix toThis is local-unitary invariance of reduced-state spectra: the unitary on the traced subsystem cancels, and conjugation on $A$ leaves its eigenvalues unchanged. Here the reductions areTheir spectra differ. Hence no pair of local unitary operations carries the first state to the second. More generally, normalized bipartite pure states are locally unitarily equivalent exactly when their squared Schmidt coefficients agree, including multiplicities; sufficiency follows by mapping one pair of Schmidt bases to the other.

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.