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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.

For $n\ge 2$, the Pauli X gate and Pauli Z gate act on different qubits in $X_0{Z}_1$, so they commute. Their product is a Hermitian matrix and squares to the identity operator. Its eigenvalues have modulus one; equivalently it is a unitary operator. Therefore its spectral norm isThe $n\ge 2$ qualification is necessary because the qubit labelled one does not exist for $n=1$.

The product of the data Pauli Z gates has computational-basis eigenvalueUse the parity circuit $W$ from part (ii), apply the target unitary gate $R=e^{itZ}=\mathrm{diag}(e^{it},e^{-it})$, and then uncompute the parity with $W^{†}$. For every basis input,This equals $e^{it{Z}^{\otimes n}}$ on the data, with the ancilla qubit returned to zero. The compute-phase-uncompute construction therefore gives an exact linear-size circuit:This is a Pauli-string phase by parity computation, an instance of simulation of a computable diagonal Hamiltonian. Part (i) also explains it by $W^{†}(I\otimes Z)W=Z^{\otimes n}\otimes Z$, whose exponential restricts correctly to the target-$|0⟩$ subspace. If no extra line is desired, accumulate parity into the last data qubit, apply $e^{itZ}$ there, and undo the $n-1$ CNOT gates, for $2n-1$ gates. Both constructions implement the global phase as well as the relative phases exactly.

Prepare the two data qubits in $|s⟩=\frac{1}{2}{\sum }_{x\in B_2}|x⟩$ and a phase ancilla qubit in $|-⟩$. A single Boolean quantum oracle query flips only the marked amplitude. Its success fraction is $p=1/4$, so $\theta =\pi /6$ in amplitude amplification. The fixed diffusion reflection $D=2|s⟩⟨s|-I$ gives $\sin (3\theta )=1$ after one iteration.

Directly, after the query the marked amplitude is $-1/2$ and the other three are $1/2$. Their mean is $1/4$. The diffusion reflection replaces each amplitude $c_x$ by $2(1/4)-c_x$, yielding one at the marked input and zero elsewhere. ThusHere $U_f$ is understood with its target ancilla qubit, and $D$ acts only on the data. It is independent of $f$: $D=H^{\otimes 2}(2|00⟩⟨00|-I)H^{\otimes 2}$, with the central diagonal gate implementable using two Pauli Z gates and one Controlled-Z gate. Measuring the data in the computational basis therefore finds the unique marked string with certainty after one oracle query.

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.

The computational-basis projectors on the first qubit areThe Pauli Z gate acts as $+1$ on the first subspace and $-1$ on the second, so $Z_1={\Pi }_0-{\Pi }_1$. By the Born rule, $p_b=⟨\psi |{\Pi }_b|\psi ⟩$. Therefore the expectation determines the output probabilities:The normalization supplies $p_0+p_1=1$; no product-state assumption is needed for this identity.

Label the measured vertex of the four-cycle by $0$, its two neighbours by $1,3$, and the opposite vertex by $2$. The edges are $01,12,23,30$. By the definition of a graph state,All Controlled-Z gates commute. Conditioning on the computational basis value $r$ of vertex $0$, the factors on edges $01$ and $03$ act on their other endpoints as $Z_1^r$ and $Z_3^r$, since $E|r,a⟩=(-1{)}^{ra}|r,a⟩$. The remaining edges form the three-vertex path $1-2-3$:The Born rule gives probability one half for either result. Normalizing provesThis is computational-basis measurement of a graph-state vertex: remove the measured vertex and apply a known Pauli Z gate byproduct to each neighbour. The chosen labels specify precisely which factors occur; any vertex of the square can serve as vertex $0$ by relabelling.

The basic one-bit teleportation primitive uses the input as qubit 1 and a fresh $|+⟩$ quantum ancilla as qubit 2. Apply the Controlled-Z gate $E_{12}$ and measure qubit 1 in the equatorial qubit measurement basis with angle $\alpha$. If $s\in \{0,1\}$ is the result, the normalized post-measurement state on qubit 2 isBoth results have probability one half. For example, writing $|\psi ⟩=a|0⟩+b|1⟩$, the unnormalized branch is $(a|+⟩+(-1{)}^s{e}^{i\alpha }b|-⟩)/\sqrt{2}=X^{s}J(\alpha )|\psi ⟩/\sqrt{2}$. Thus $J(\alpha )$ is applied deterministically as a logical gate with known Pauli frame $X^s$; a permitted conditional Pauli X gate would remove this byproduct physically.

The literal resource restriction does not need an unstated direct Pauli X gate. A Pauli Z gate can be enacted by applying $E$ to the data and a fixed $|1⟩$ quantum ancilla. Two consecutive angle-zero one-bit teleportations, with results $p,q$, map an arbitrary current state $|\chi ⟩$ to $X^{q}H{X}^{p}H|\chi ⟩=X^q{Z}^p|\chi ⟩$. Apply the available $Z^p$ to remove the latter factor, up to an irrelevant global phase. This implements a heralded $X^q$: if $q=0$, the input is unchanged; if $q=1$, the required Pauli X gate has been applied.

If the original result is $s=1$ and a physically corrected output is required, repeat this heralded Pauli X correction using controlled-Z and measurements until $q=1$. Each attempt succeeds with probability one half independent of the input, so it terminates with probability one, using two attempts on average. Only $E$, fixed ancillary quantum states and single-qubit quantum measurements are used. This exact physical correction has no finite worst-case measurement bound; the standard finite deterministic realization is the logical Pauli frame version, which suffices for the later output-simulation parts.

Let the ideal two-qubit output before final quantum measurements beThe PDF places $J(\alpha )$ only on the upper wire, followed by the Controlled-Z gate and a quantum measurement in the computational basis on each output wire. Measure vertex $0$ of the square in the computational basis, obtaining $r$, then vertex $1$ of the surviving path in the fixed equatorial qubit measurement basis at angle $\alpha$, obtaining $s$. Put $t=r\oplus s$. The previous part gives $E_{23}{X}_2^{t}J(\alpha {)}_2{Z}_3^r|++⟩$.

Commute its Pauli frame through the Controlled-Z gate. Since $E_{23}{X}_2^t=X_2^t{Z}_3^t{E}_{23}$ and $E_{23}$ commutes with $Z_3^r$, the actual output isFinally measure vertices $2,3$ in the computational basis, with raw results $d_2,d_3$. A Pauli X gate flips a computational basis result, whereas a Pauli Z gate changes only its phase. Thus the purely classical correction isThis four-cycle graph-state simulation of an entangle-and-measure circuit reproduces the full joint output distribution, not just each marginal. All measurement bases are fixed beforehand, and no physical byproduct correction is necessary. As a check, the ideal Controlled-Z gate is diagonal in the output basis, so $P(k,l)=\frac{1}{2}{\cos }^2(\alpha /2)$ for $k=0$ and $\frac{1}{2}{\sin }^2(\alpha /2)$ for $k=1$, independently of $l$. The measurement procedure gives exactly these probabilities after its classical relabelling.

Let the known phase gate on the answer register bewhere the comparison uses ordinary integers, not modular arithmetic. A reversible circuit computes the predicate, applies a Pauli Z gate to its flag, then performs uncomputation.

The compute-phase-uncompute construction now givesUse modular-oracle inversion by negation to realize the last operation with one further $U_f$ query. Exactly two oracle queries implement $I_g$, returning the answer register and comparison workspace to their initial states.

Use a clean quantum ancilla and apply the quantum circuit $C$, then the Pauli Z gate on the ancilla, then the inverse quantum circuit $C^{†}$:This compute-phase-uncompute construction implements $A\otimes I$ on every input with the ancilla initially $|0⟩$, and by linearity on any superposition of those inputs. The action on ancilla-$|1⟩$ inputs need not coincide with $A\otimes I$; the question only requires the clean-ancilla subspace.

The Hadamard gate conjugates the Pauli X gate into the Pauli Z gate:When the control qubit is $|0⟩$, all three circuits act as the identity on the target. When it is $|1⟩$, the right-hand circuit acts on the target as $HXH=Z$. It therefore agrees with the Controlled-Z gate on every computational basis state, provingConjugating both sides by $I\otimes H$ and using $H^2=I$ immediately gives the reverse identity

Let the unknown qubit be $|\alpha ⟩=a|0⟩+b|1⟩$. Alice holds this qubit and her half of the shared Bell state $|{\Psi }^{+}⟩$, while Bob holds the other half.

Alice performs a measurement of her two qubits in the Bell basis and sends the two-bit outcome to Bob through the classical channel. For the shared state $|{\Psi }^{+}⟩$, the teleportation with the psi-plus Bell state correction table isup to physically irrelevant global phases. Here $X$ and $Z$ are the Pauli X gate and Pauli Z gate. After the indicated correction, Bob's qubit is $|\alpha ⟩$. This is quantum teleportation; no physical copy of the unknown qubit is sent.

In the ordered orthonormal basisof $\mathcal{S}$, the Pauli Z gate acts asIt fixes the line $|g{⟩}^{\perp }\cap \mathcal{S}$ and negates $|g⟩$, so it is the reflection in a hyperplane orthogonal to the good state.

A Pauli gate applies a Pauli matrix as a unitary operator to a qubit. The Pauli X gate, Pauli Y gate and Pauli Z gate are common correction operations in quantum teleportation.

The Pauli Z gates give eigenvalue $(-1{)}^{{\sum }_j{x}_j}$ on $|x⟩$. Use parity computation by CNOT gates to put the parity bit in a zero ancilla, apply $e^{itZ}$ to it, and uncompute. The data acquire exactly $e^{it(-1{)}^{{\sum }_j{x}_j}}$, and the ancillary line returns to zero. This compute-phase-uncompute construction needs $2n+1$ one- and two-qubit gates. Accumulating parity into the last data line instead uses $2n-1$ gates without an extra ancilla. Neither implementation drops the global phase.

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.