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Classical communication transmits a distinguishable classical record, such as measurement bits, between parties. In quantum teleportation it tells the receiver which local correction to apply. Quantum no-signalling prevents shared entanglement alone from transmitting the record. A classical message cannot propagate across a spacelike separation in a relativistic protocol.

A rank-one projective measurement whose Bob basis is computational when Alice is zero and rotated by $\theta$ when Alice is one can signal. Bob starts in zero. Alice zero leaves Bob zero; Alice one dephases Bob in the rotated basis, giving the displayed nonzero flip probability for $0<\theta \le \pi /4$. This violates quantum no-signalling. The conclusion applies to a measurement distinguishing the four product-basis states, rather than a degenerate observable that does not resolve them.

Use the nonnegative representative of reduction modulo four. The four product eigenstates and eigenvalues arebecause $-2\equiv 2(\mathrm{mod}4)$. Consequently the observable is $I+Z_A{Z}_B$. It has the even sector $\mathrm{span}\{|00⟩,|11⟩\}$ with value two and the odd sector $\mathrm{span}\{|01⟩,|10⟩\}$ with value zero.

The entanglement-assisted nondemolition parity measurement needs just the single shared $|{\Phi }^{+}⟩$ provided in the question. Apply ${\mathrm{CNOT}}_{A\to d_1}$ and ${\mathrm{CNOT}}_{B\to d_2}$. For a computational-basis input $|ab⟩$, the meter becomesMeasuring the meters gives $u\oplus v=a\oplus b$. For an arbitrary coherent system input, its conditional Kraus operator isIt preserves every superposition within the measured sector, so this realizes the Lüders rule for the two degenerate eigenvalues. The answer is two for equal meter bits, zero for unequal meter bits. Each individual meter bit is uniform; only later comparison reveals the parity, respecting quantum no-signalling. This construction is independent of the defective one-pair singlet-verification request in part (a).

Interpret the requested measurement as distinguishing all four stated orthogonal eigenstates, hence as their rank-one projective measurement. If they have indistinguishable degenerate eigenvalues, this assumption need not hold; the identity observable, for example, cannot reveal the basis and gives no contradiction.

Let Bob initially prepare $|0{⟩}_B$. Alice encodes a bit by preparing $|0{⟩}_A$ or $|1{⟩}_A$, without changing Bob's initial reduced density matrix. If Alice prepares zero, the global input is the first eigenstate, so nondemolition leaves Bob in $|0⟩$ with certainty. If Alice prepares one, the rank-one Lüders rule, after discarding the outcome, dephases Bob in the rotated basisHis output is ${\rho }_B={\cos }^2\theta |u⟩⟨u|+{\sin }^2\theta |v⟩⟨v|$. A computational-basis measurement then givesIt is positive for every $0<\theta \le \pi /4$. Repeating the experiment would transmit Alice's choice across a spacelike interval, violating quantum no-signalling. The controlled-basis measurement causality obstruction therefore excludes the entire nonzero interval, including $\pi /4$.

At $\theta =0$, the basis is the computational product basis, up to an irrelevant sign on the fourth vector. Alice and Bob each measure $Z$ locally, then compare their results later. Each product eigenstate is preserved. Thus being a product basis is insufficient for an instantaneous quantum nondemolition measurement: a remote party's choice of which local basis is measured can still cause signalling.

The EPR criterion of reality says that a physical quantity has an element of reality if its value can be predicted with certainty without physically disturbing the system. The proposed criterion distinguishes what the system possesses from what one happens to measure.

A local hidden-variable theory supplements the preparation by a variable $\lambda$, drawn from a distribution $\mu (\lambda )$ independent of the later measurement settings. Its locality assumption is conditional factorization:Alice's local response does not depend on Bob's setting $y$, and Bob's does not depend on Alice's $x$. Correlations can arise from the shared past variable $\lambda$. In a deterministic local hidden-variable model, the response probabilities are point masses at functions $a=A_x(\lambda )$ and $b=B_y(\lambda )$. In particular, all possible local settings have definite values for a fixed $\lambda$, even when only one setting is used in a trial.

The setting-independence assumption is needed for comparing these predetermined values across different experiments. Conditional factorization is stronger than quantum no-signalling: no-signalling constrains observed marginal probabilities, whereas local hidden-variable theory constrains their decomposition at fixed $\lambda$. This distinction is what Bell theorem tests.

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).

Assume the proposed device distinguishes the four displayed eigenstates, as an ideal rank-one projective measurement. This assumption matters: if all four had the same eigenvalue, the identity observable would admit that eigenbasis but its Lüders rule measurement would do nothing and could not signal. The printed eigenvectors alone do not exclude this degeneracy.

For the complete resolving measurement, let $c=\cos \theta$, $s=\sin \theta$, and ${\Pi }_j$ be the four rank-one projectors. If the nonlocal outcome is ignored, the nonselective projective measurement channel is ${\mathcal{D}}_{\theta }(\rho )={\sum }_j{\Pi }_j\rho {\Pi }_j$. Each listed state's local reduced density matrix is diagonal in $Z_A$, with $Z_A$ expectation $\cos 2\theta$ for either plus state and $-\cos 2\theta$ for either minus state. ConsequentlyCompute $C_{\theta }$ in its even and odd two-dimensional blocks: both have diagonal entries $\cos 2\theta ,-\cos 2\theta$ and off-diagonal entries $\sin 2\theta$. ThusandTo signal, prepare Alice in $|+⟩$ and Bob initially in $|+⟩$. Bob encodes a bit by either doing nothing or applying a local Pauli Z gate, which changes his state to $|-⟩$. Alice's input reduced density matrix is identical in both cases, but after the hypothetical instantaneous measurement her local $Z_A$ expectation isThe two probabilities for Alice's outcome $0$ are $(1\pm \sin 2\theta \cos 2\theta )/2$, so their difference is $\sin 2\theta \cos 2\theta$. It is strictly positive for $0<\theta <\pi /4$. Repeated trials let Alice infer Bob's bit while their operations are still spacelike, violating quantum no-signalling and relativistic causality. The relativistic causality constraint on an ideal nonlocal measurement therefore permits onlywithin the specified interval. The argument requires no rapid communication of the hypothetical nonlocal outcome: Alice reads her own changed local statistics. It rules out the full ideal instrument, not merely the later classical comparison of locally obtained records.

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.

Quantum no-signalling says that, without communicating a measurement outcome, any local trace-preserving operation performed on subsystem $A$ leaves all outcome probabilities for measurements on subsystem $B$ unchanged; equivalently, it leaves the reduced density matrix ${\rho }_B$ unchanged.

The two Bell states are orthogonal. Applying the inverse Bell-state preparation circuit, namely a Controlled-NOT gate from $A$ to $B$ followed by a Hadamard gate on $A$, givesA computational-basis measurement therefore distinguishes them perfectly, soIf only $B$ is accessible, however,Every measurement on $B$ consequently has the same outcome distribution under both hypotheses. With equal prior probabilities no information is available, andthe error probability of a random guess.

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.

Without Alice's operation, Bob obtains outcome $b$ with probabilityIf Alice applies $U$ and then measures, but her outcome is not communicated, Bob's probability isbecause ${\sum }_a{\Pi }_a=I_A$. Equivalently, a trace-preserving local operation leaves Bob's reduced density matrix unchanged. This is the no-communication theorem: local operations cannot signal without communication of their outcomes.

The no-signalling theorem states that any trace-preserving quantum operation performed locally by Bob leaves Alice's reduced density matrix unchanged when Bob's outcome is not communicated:Thus Bob's choice cannot change the statistics of any measurement made by Alice.

Let a projective measurement have mutually orthogonal projections $P_i$ satisfying ${\sum }_i{P}_i=I$. On a density operator $\rho$, the Born rule and the Lüders rule giveconditional on outcome $i$. On the first part of a bipartite quantum system, replace $P_i$ by $P_i\otimes I_2$.

The subsystems are isolated when there is no interaction term coupling them. Their Hamiltonian operator has the formso their subsequent unitary time evolution factorizes as $U_1\otimes U_2$.

The reduced density matrix of subsystem 1 iswhere the partial trace is characterized byfor every local observable $M_1$. Thus ${\rho }_1$ contains exactly the statistics accessible by measurements on subsystem 1.

Suppose a projective measurement $\{Q_j\}$ is performed on subsystem 2 and its outcome is not communicated. The resulting nonselective state isFor every $M_1$, the cyclic property of the trace and ${\sum }_j{Q}_j^2=I_2$ giveHence ${\rho }_1^{\prime }={\rho }_1$. No local projective measurement on subsystem 1 can reveal whether the remote unreported measurement occurred. This is quantum no-signalling, which prevents a choice made at a spacelike separation from transmitting information faster than light and makes the measurement formalism compatible with relativistic causality.

The proposed nondisturbing device would violate no information without disturbance. In an ordinary quantum instrument, let $M_{i\alpha }$ be the Kraus operators associated with classical output $i$. If every pure state $|\psi ⟩$ remains unchanged even conditional on the displayed outcome, every nonzero $M_{i\alpha }|\psi ⟩$ must be parallel to $|\psi ⟩$. A linear operator for which every vector is an eigenvector is a scalar multiple of the identity, so $M_{i\alpha }=c_{i\alpha }I$. Its output probabilityis independent of the state. It cannot equal $⟨\psi |P_i|\psi ⟩$ for arbitrary projectors. Equivalently, repeated nondisturbing samples would permit quantum state tomography of one specimen and then quantum cloning, contradicting ordinary quantum theory.

Such devices would not necessarily enable superluminal signalling. One consistent operational extension could make every sequence of outputs depend only on the local reduced density matrix and local settings. Since an unreported remote measurement leaves that matrix unchanged, all local device statistics would remain unchanged too. Other extensions could add nonlocal outcome-dependent rules and permit signalling, but that behavior is additional to the device specification rather than forced by it.

The quantum no-signalling principle says Alice's uncommunicated local choice cannot alter Bob's reduced state. Expanding the two tensor products cancels the cross terms and givesAlice obtains either result with probability $1/2$; conditionally Bob has probability ${\cos }^2\theta$ or ${\sin }^2\theta$ of $0$, respectively. Averaging,independent of $\theta$, exactly as no-signalling requires.

The nonselective channel of a hypothetical spacelike nonlocal quantum measurement must not allow a distant input change to alter local output statistics. For a rank-one measurement in the rotated even and odd two-qubit bases with angle $\theta$, Alice's output satisfiesBob can change the last correlation using only his local Pauli Z gate while Alice keeps the input $|+⟩$. The resulting signal excludes $0<\theta <\pi /4$. Degenerate observables that do not distinguish this basis are not covered by the rank-one argument. Locally recording a jointly encoded result, which is decoded later by local operations and classical communication, is compatible with quantum no-signalling.