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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.

Yes: use superdense coding, with a shared Bell state. Let the sender and receiver initially hold the first and second qubits of $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$. To encode bits $(a,b)$, the sender applies $Z^a{X}^b$ to her qubit. These Pauli operators produce four orthogonal states:She sends this qubit through the noiseless quantum channel. The receiver then possesses both qubits. He applies a controlled-NOT gate from the received qubit to his original qubit, followed by a Hadamard gate on the received qubit. This maps the four states in the table to $|ab⟩$. A measurement in the computational basis therefore recovers the two bits with certainty. No classical message is sent.

The resource assumption deserves care: one-qubit superdense coding needs a Bell pair already shared before encoding. If none is initially available, the sender can prepare a Bell pair locally and send one half through the same noiseless quantum channel in advance. The protocol then uses two transmitted qubits in total, including distribution of the entanglement, still with no classical communication. Alternatively, sending two computational basis qubits directly also communicates two bits. With a pre-shared Bell pair, one transmitted qubit suffices; without it, the entanglement-distribution cost must be counted. An unassisted single transmitted qubit cannot perfectly encode four classical messages: its Holevo bound is at most ${\log }_{2}2=1$ bit.

A secure key can be produced, at one ideal secret bit per copy. The shared Bell state is pure, so the preceding factorization shows that Eve's purifying system is independent of it. Measuring both qubits in the computational basis gives matching outcomes $00$ or $11$, each with probability $1/2$. Neither party announces these key outcomes. Eve's state is the same for both possibilities, so each copy supplies a shared uniform private bit.

No secure key can be produced. The off-diagonal terms in the two Bell state projectors cancel, giving the separable quantum stateA possible purification of a density operator is $(|00{⟩}_{AB}|0{⟩}_E+|11{⟩}_{AB}|1{⟩}_E)/\sqrt{2}$. Eve measures her register in the computational basis and learns whether Alice and Bob have $00$ or $11$. Their apparently shared random bit is therefore completely known to Eve. Conditioned on her record the Alice–Bob state is a known product state; public discussion and local private randomness cannot create a shared secret unknown to that record. The state has no distillable secret key in the stated purification model.

They need a pre-shared entanglement resource that permits deterministic production of a Bell pair. In particular, one shared Bell pair and two classical bits suffice for exact quantum teleportation. This condition is stronger than merely having some entanglement.

Let Alice's input be $|\psi {⟩}_C=\alpha |0⟩+\beta |1⟩$, and let her system $A$ and Bob's system $B$ share $|{\Phi }^{+}{⟩}_{AB}=(|00⟩+|11⟩)/\sqrt{2}$. Use the Bell states $|{\Phi }^{\pm }⟩=(|00⟩\pm |11⟩)/\sqrt{2}$ and $|{\Psi }^{\pm }⟩=(|01⟩\pm |10⟩)/\sqrt{2}$. Direct expansion givesAlice performs a projective measurement in this Bell basis and sends its two-bit outcome. Each outcome has probability $1/4$. Bob applies respectively $I,Z,X,ZX$, recovering $|\psi ⟩$ exactly, without either party knowing $\alpha ,\beta$. Before he receives the outcome, his averaged state is $I/2$, so the protocol does not communicate without the classical message. Alice's measurement consumes the input and the shared entanglement.

For necessity, exact transmission of every pure qubit means the induced quantum channel is the identity: pure-state projectors span the operator space, and a physical channel is linear. Alice can therefore prepare a Bell pair locally and transmit one half by the proposed protocol, leaving a Bell pair shared with Bob. The protocol would itself be a deterministic local operations and classical communication conversion of their resource into a Bell pair. This proves the exact teleportation resource criterion. With no shared entanglement, it is impossible: local operations and classical communication preserve separable quantum states, whereas the resulting Bell pair would be entangled.

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.

Charlie measures his qubit in the $X$ basis, $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$. Rewriting the GHZ state in that basis givesEach outcome occurs with probability $1/2$. Charlie records $c=0$ for $+$ and $c=1$ for $-$ and sends that bit to Bob. Alice and Bob's conditional Bell state is $(I\otimes Z^c)|{\Phi }^{+}⟩$. Bob applies $X{Z}^c$, with $Z^c$ applied first. ThenThus every measurement branch gives the requested state after a known local correction: success is deterministic. All quantum operations act on only one party, and the only nonlocal resource used during the protocol is classical communication. This is localizing a Bell pair from a GHZ state by LOCC.

Ignoring Charlie's outcome instead gives Alice and Bob the average state $\frac{1}{2}(|{\Phi }^{+}⟩⟨{\Phi }^{+}|+|{\Phi }^{-}⟩⟨{\Phi }^{-}|)=\frac{1}{2}(|00⟩⟨00|+|11⟩⟨11|)$, which is separable. The classical record supplies the correction needed to use the conditional entanglement; no communication-free preparation of a known entangled state is being claimed.

Use the first copy to determine computational-basis parity and the second to determine phase parity. On the first pair, Alice and Bob each measure the Pauli operator $Z$ and communicate their signs $z_A,z_B$. On the second, each measures $X$ in the $|+⟩,|-⟩$ basis and communicates the signs $x_A,x_B$.

Every Bell state is an eigenstate of both $Z\otimes Z$ and $X\otimes X$. Direct application to its two basis terms gives the tableFor example, $Z\otimes Z$ acts positively on $|00⟩,|11⟩$ and negatively on $|01⟩,|10⟩$; $X\otimes X$ exchanges the two terms in each pair, detecting their relative sign. Thus individual outcomes are random, but the products in the table are certain. The Bell parity and phase observables label the state uniquely.

The two communicated parities identify all four states with probability one. Each measurement is local and all remaining operations are classical, so the protocol is LOCC discrimination of four Bell states using two copies. Alice can simply send her two signs to Bob, who combines them with his own. Measuring $Z$ on the first copy destroys its phase information, but the second identical copy still supplies it; this is why the two-copy assumption enables this protocol.

Put $|0⟩=|{\uparrow }_z⟩$ and $|1⟩=|{\downarrow }_z⟩$. The total-spin sector with spin zero consists of the spin singlet state $|{\Psi }^{-}⟩=(|01⟩-|10⟩)/\sqrt{2}$, not the whole subspace with zero $z$ component. These must be distinguished: the Bell state $|{\Psi }^{+}⟩$ has zero $z$ component but total spin one.

There are two meanings of verification to separate here. A one-pair entanglement-assisted statistical singlet verification is possible: choose a shared random axis from $x,y,z$, measure that axis's two-spin parity by the meter circuit below, and accept anticorrelation. An $x$ or $y$ test uses local basis changes before and after the computational-axis circuit. The singlet always passes and is unchanged. Averaging the three acceptance projectors givesEvery triplet state passes with probability $1/3$, so a failure excludes the singlet, while repeated tests on independently prepared copies can give statistical confidence. A single pass does not certify the singlet with certainty.

Exact single-shot verification cannot be implemented with just one shared Bell pair. Here exact verification means a yes outcome with probability one on the spin singlet state, zero on its orthogonal complement, and preservation of the singlet on yes. If that stronger meaning is intended, the question needs an extra resource. The Bell-pair cost of exact nondemolition singlet verification gives a short proof. For a fully resolved tuple $\mu ,\nu$ of local measurement records, contraction of the shared $|{\Phi }^{+}⟩$ gives a system Kraus operatorLocal ancillas and locally adaptive operations are included in these operators. Refining any unobserved local environment gives the same form, so the operator Schmidt rank is at most two. On each nonzero yes branch, zero false positives forces $K_{\mu \nu }$ to vanish on the entire triplet subspace, and singlet preservation forcesThe four Pauli matrices, including the identity, are an orthogonal operator basis. Thus $P_s$ has operator Schmidt rank four, contradicting the rank-two bound. At least one yes branch is nonzero because the singlet must pass with certainty. Shared classical randomness cannot evade this branchwise argument.

With two shared Bell pairs, an explicit corrected protocol is available. On the first meter pair, apply local system-to-meter CNOT gates and measure both meters in the computational basis. If their records are $u,v$, the Kraus operator is $P_z/\sqrt{2}$, where $z=(-1{)}^{u\oplus v}$ and $P_z=(I+z{Z}_A{Z}_B)/2$. On the second pair, measure $X_A{X}_B$ by the same circuit conjugated with local Hadamard gates on the system. The two parities commute, and their joint projectors are the four Bell-state nondemolition measurement projectors. The singlet is exactly the outcome $(z,x)=(-1,-1)$ and remains unchanged. Local interactions and readouts fit within the stated interval; the combined verdict becomes available only after the records are compared by classical communication.

This corrected exact protocol resolves the triplet into three Bell states. It preserves the singlet and every Bell state, but generally destroys triplet superpositions. A binary Lüders rule measurement preserving all triplet coherence is a different operation: it violates the singlet-triplet measurement causality obstruction. The one-pair protocol used below measures a single parity and supplies the statistical test above; it does not give exact single-shot singlet verification.

Use one shared $|{\Phi }^{+}⟩$ pair to measure $Z_A{Z}_B$, and the other to measure $X_A{X}_B$. The latter entanglement-assisted nondemolition parity measurement is obtained by applying Hadamard gates to both system qubits before and after the $Z$-parity circuit. Since $ZX=-XZ$ at each site, the two minus signs cancel, giving $[Z_A{Z}_B,X_A{X}_B]=0$.

The Bell states have joint eigenvaluesand are therefore distinguished uniquely by the two meter parities. For records $(u,v)$ and $(r,s)$, the combined Kraus operator isEach $P_{z,x}$ is the rank-one projector onto the corresponding Bell state. This is a Bell-state nondemolition measurement: an input Bell state remains exactly that state for every possible local record. The local circuits can be scheduled without communication; the identity of the Bell state is known only when the classical records are brought together. Two parity bits distinguish all four Bell states without disturbing them.

Write the arbitrary joint pure state of the input $A$ and an external reference $R$ asThe reference vectors need not be normalized or orthogonal. Share $|{\Phi }^{+}{⟩}_{aB}$. Label Alice's Bell states byProjecting onto this Bell basis gives the unnormalized stateEach outcome has probability $1/4$. Alice sends $(r,s)$ and Bob applies $Z_B^r{X}_B^s$, reversing the two Pauli gates. The final state is exactly $|\Omega {⟩}_{BR}$ for every outcome. In particular all input entanglement and other correlations with $R$ now belong to $B$, while the reference marginal is unchanged. This is teleportation as an identity channel on a reference.

Conditioned on the record, Alice's two measured carriers are in $|{\beta }_{rs}⟩$ and factor from $BR$. Thus $A$ no longer retains its original correlations with $R$. There is no second copy, even when the teleported qubit was initially entangled with a larger system.

Write the system's $z$ basis as $|0⟩,|1⟩$, and use a separate pair of meter qubits in the Bell state $|{\Phi }^{+}⟩$. Alice applies a CNOT gate from system $A$ to her meter qubit; Bob simultaneously applies a CNOT gate from $B$ to his meter qubit. Flipping neither or both meter qubits preserves $|{\Phi }^{+}⟩$, while flipping exactly one gives $|{\Psi }^{+}⟩$. Hence the entanglement-assisted nondemolition parity measurement interaction produceswhere $P_{\mathrm{e}}=|00⟩⟨00|+|11⟩⟨11|$ and $P_{\mathrm{o}}=|01⟩⟨01|+|10⟩⟨10|$. Each party now measures only their meter qubit in the $z$ basis. If their binary records are $u,v$, the system Kraus operator isUnequal records verify zero total $z$ spin, since $S_z^{\mathrm{tot}}=\hslash (Z_A+Z_B)/2$ vanishes precisely on the odd sector. The probability of success is $⟨\psi |P_{\mathrm{o}}|\psi ⟩$, and the successful conditional state is $P_{\mathrm{o}}|\psi ⟩/\sqrt{⟨P_{\mathrm{o}}⟩}$. Every zero-total-$z$-spin state is left unchanged, including any coherent superposition of $|01⟩$ and $|10⟩$. Similarly the even-sector coherence is preserved. This is a quantum nondemolition measurement of the parity, rather than separate measurements of both system spins.

All quantum operations and local meter measurements can finish within the spacelike time window. Nevertheless each local meter record is individually uniform: $K_{u0}^{†}{K}_{u0}+K_{u1}^{†}{K}_{u1}=I/2$. The verification result is obtained only by comparing the records using local operations and classical communication. Thus “instantaneous” refers to the local completion of the joint measurement instrument, not instant access to its nonlocal outcome; quantum no-signalling remains intact.

At $\theta =0$ the four eigenstates are the computational product basis, with irrelevant signs on two vectors. Alice and Bob measure their own system qubits in $Z$. These local projective measurements preserve each product eigenstate; their pair of records identifies the global outcome after local operations and classical communication.

At $\theta =\pi /4$ the four states are the Bell states. They are simultaneous eigenstates of the commuting Pauli operators $Z_A{Z}_B$ and $X_A{X}_B$: ${\Phi }^{+},{\Phi }^{-},{\Psi }^{+},{\Psi }^{-}$ have respective pairs $(+,+),(+,-),(-,+),(-,-)$. Their nonlocal parity measurements can be performed without directly distinguishing the local system spins.

Use the first shared Bell state pair to perform the entanglement-assisted nondemolition parity measurement of $Z_A{Z}_B$. Use the second shared pair for $X_A{X}_B$: both parties apply a local Hadamard gate to their system qubit, execute the same local system-to-meter CNOT gates and $z$-meter measurements, then undo the Hadamard gates. This measures $X_A{X}_B$ because $HZH=X$. The two system parity projectors commute, since anticommutation at both sites cancels:Each $P_{z,x}$ is the corresponding rank-one Bell state projector. Every complete tuple of four local meter records has system Kraus operator $P_{z,x}/2$ for its two parities. Summing the four record tuples compatible with $(z,x)$ gives the ideal outcome map $\rho \mapsto P_{z,x}\rho P_{z,x}$. The protocol is a Bell-state nondemolition measurement: an input Bell state is preserved, while an arbitrary input is projected onto the reported Bell state with the Born rule probability.

The local circuits need no adaptive communication between the laboratories, so both parties can finish inside the specified time window. Global identification of $(z,x)$ still requires later local operations and classical communication. This endpoint protocol respects quantum no-signalling, unlike the hypothetical intermediate-angle instrument in part (i).

For a bipartite density operator ${\rho }_{AB}$, let $S(A{)}_{\rho }=S({\rho }_A)$, with ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$, and similarly for other systems. The quantum conditional entropy and coherent information areUnlike classical conditional entropy, quantum conditional entropy can be negative. For a Bell state, $S(AB)=0$ and $S(B)=1$, giving $H(A|B)=-1$ and coherent information equal to one bit.

Use one-bit remote preparation of real qubit states. Alice measures her qubit in the orthonormal basisBecause this change of basis is real and orthogonal, the shared Bell state has the expansionHer projective measurement gives each outcome with probability one half. She sends $m=0$ for the first outcome and $m=1$ for the second. Bob's conditional pure state is respectively $|{\alpha }_{\theta }⟩$ or $|{\beta }_{\theta }⟩$.

For $m=0$ Bob does nothing. For $m=1$ he applies the fixed unitary operatorwhere $Y$ is the Pauli Y gate. Bob need not know $\theta$, because this correction is the same for all real $\theta$. Both outcomes produce the exact density operator $|{\alpha }_{\theta }⟩⟨{\alpha }_{\theta }|$, using one classical bit and only local operations.

This remote state preparation consumes the shared entangled state. Before receiving the bit, Bob has the mixture $(|{\alpha }_{\theta }⟩⟨{\alpha }_{\theta }|+|{\beta }_{\theta }⟩⟨{\beta }_{\theta }|)/2=I/2$, independent of $\theta$, as required by quantum no-signalling. The real-state restriction is what permits a fixed unitary correction of the orthogonal outcome.

Use superdense coding. For the two-bit message $(a,b)\in \{0,1{\}}^2$, Alice applies $Z^a{X}^b$ to her half of the shared Bell state, where $X,Z$ are the Pauli X gate and Pauli Z gate. The encoded states areThese four Bell states are orthonormal. Alice then transmits her one qubit to Bob, who now holds both qubits. A Bell-basis measurement identifies the message with probability one.

More explicitly, Bob can apply a CNOT gate with the received qubit as control and his original qubit as target, followed by a Hadamard gate on the control. The CNOT gate sends the encoded state toand the Hadamard gate sends this to $|a⟩\otimes |b⟩$. Computational-basis projective measurements therefore read out $a,b$ exactly. One transmitted qubit, assisted by one previously shared Bell pair, communicates two classical bits.

A density operator is a pure state exactly when its matrix rank is one. Within the admissible interval from part (a), one of $(1+\alpha )/2$ and $(1-\alpha )/2$ must vanish. Therefore the pure-state parameters and vectors arewhere the Bell states are $|{\Phi }^{\pm }⟩=(|00⟩\pm |11⟩)/\sqrt{2}$. Equivalently, the purity of a density operator is $\mathrm{Tr}{\rho }_{AB}^2=(1+{\alpha }^2)/2$, reaching one exactly at these two endpoints.

This is superdense coding. To encode $00,01,10,11$, Alice applies respectively $I,X,Z,XZ$ to her half of $|{\Phi }^{+}⟩$. These local Pauli operators transform the shared state into the four mutually orthogonal Bell states $|{\Phi }^{+}⟩,|{\Psi }^{+}⟩,|{\Phi }^{-}⟩,|{\Psi }^{-}⟩$, up to irrelevant global phases. Alice sends her one qubit to Bob. Bob now holds both qubits and performs a Bell-basis measurement, whose outcome identifies Alice's two-bit message with certainty. One transmitted qubit communicates two classical bits because one shared Bell pair was available beforehand.

Let $|\alpha ⟩=a|0⟩+b|1⟩$ be Alice's unknown qubit, and label her half of the shared Bell state by $A$ and Bob's by $B$. Alice applies a Controlled-NOT gate from the unknown qubit to $A$, then a Hadamard gate to the unknown qubit. The three-qubit state becomesAlice measures her two qubits in the computational basis and sends the two classical outcomes $(j,k)$ to Bob. Bob applies $Z^j{X}^k$, recovering $|\alpha ⟩$ with certainty. This is quantum teleportation; Alice never needs to know the state.

For the shared GHZ state, Charlie first applies a Hadamard gate to his qubit. ThenCharlie measures in the computational basis. Outcome $0$ leaves Alice and Bob with $|{\varphi }^{+}⟩$; outcome $1$ leaves $|{\varphi }^{-}⟩$, which either party converts to $|{\varphi }^{+}⟩$ by applying $Z$ after Charlie reports the outcome. Thus

The partial transpose of the spin singlet state projector is the Hermitian operatorFor every separable quantum state $\tau$, the positive partial transpose criterion and the trace-adjoint identity for partial transpose giveFor the specified Bell state, however,Thus $W$ is an entanglement witness detecting $|{\Phi }^{+}⟩$. In fact, it detects any state whose overlap with that Bell state exceeds $1/2$.