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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.

To use $S$ as the apparatus, prepare it in the ready state $|+{⟩}_S$ and later read it in the Hadamard basis. For an arbitrary input $\alpha |+{⟩}_D+\beta |-{⟩}_D$, part (b) givesThe two apparatus states are orthogonal, so its readout measures the corresponding projectors on $D$ and preserves each input eigenstate. The observable iswith outcomes $+1$ and $-1$. This is the measurement direction of a CNOT interaction in the complementary basis.

The ready state and readout basis matter. Keeping $S$ in the original $|0{⟩}_S$ would make the old-basis controlled gate act as the identity on $D$, producing no record. Thus the same interaction can serve either measurement direction, with different apparatus preparation and pointer readout; the Hamiltonian operator alone does not assign an intrinsic system/apparatus role.

Put $M=2^m$ and prepare $m$ control qubits in $|0{⟩}^{\otimes m}$ and the target in the supplied eigenstate $|\psi ⟩$. Apply a Hadamard gate to every control qubit. Let $k={\sum }_{j=0}^{m-1}{2}^j{k}_j$ denote the integer encoded in that quantum register.

For every $j$, apply a controlled unitary gate $U^{2^j}$ with control $k_j$ and the same target. The quantum phase kickback yieldsApplying ${\mathrm{QFT}}_M^{-1}$ to the controls therefore produces $|x⟩$ exactly. A computational-basis measurement returns $x$ with certainty, and the algorithm outputsThis is exact quantum phase estimation; exactness uses the finite dyadic phase promise, not a rounding argument.

Only a controlled-U gate is supplied as a primitive. Its $2^j$ repetitions implement controlled-$U^{2^j}$, so the total number of black-box uses is ${\sum }_{j=0}^{m-1}{2}^j=2^m-1$. Access to powers as unit-cost primitives would be a different quantum query complexity model. This distinction is the cost of exact phase estimation on a dyadic spectrum.

For the separated stationary state, restore its time factor $e^{-iEt/\hslash }$. Away from radial nodes, the quantum phase is $S=k\varphi -Et/\hslash$, up to a constant $0$ or $\pi$ where the real radial function has fixed sign. The guidance equation in plane polar coordinates therefore gives the Bohmian circulation of an angular-momentum eigenstateThe direction is $\hat{\mathbf{\varphi }}$ for $k>0$ and $-\hat{\mathbf{\varphi }}$ for $k<0$; the velocity is zero for $k=0$. Each admissible trajectory is a circle:The orbital angular momentum along the trajectory is $m{r}_0{v}_{\varphi }=\hslash k$. The origin or any zero-amplitude circle is excluded from this local formula. For the real degenerate superposition constructed above, the spatial quantum phase is constant on each nodal sector, so its Bohmian mechanics velocity is instead zero. Thus the circular motion is a conclusion about the separated angular-momentum eigenstate, not an arbitrary energy eigenstate.

The rotational invariance of a central-potential Hamiltonian allows a simultaneous eigenstate of energy and the two-dimensional orbital angular momentum $L_z=-i\hslash {\partial }_{\varphi }$. For a separated wavefunction $F(r)G(\varphi )$, the Laplacian in polar coordinates givesThe $\varphi$ term must be a constant; write $G^{\prime \prime }/G=-k^2$. The angular eigenfunctions can be chosen as $e^{ik\varphi }$. Single-valuedness under $\varphi \mapsto \varphi +2\pi$ imposes $e^{2\pi ik}=1$, hence $k\in \mathbb{Z}$. The radial equation in the separation of a two-dimensional central-potential eigenstate isFor a real central potential and the usual real self-adjoint radial boundary conditions, the radial equation admits a basis of real solutions: real and imaginary parts of a complex solution obey the same equation and boundary conditions. Choose a real normalized radial eigenfunction. Since the plane area element in plane polar coordinates is $rdrd\varphi$, the normalization isThis establishes the intended separated simultaneous eigenstate form. It is not the form of every stationary state. The radial equation depends on $k^2$, so the $k$ and $-k$ sectors have the same energy. For $k\ge 1$, their normalized superpositionis a single-valued stationary state with that energy but is not a single angular exponential. Quantum degeneracy is precisely why separation of variables selects a convenient eigenstate basis rather than all vectors in an energy eigenspace. The printed assertion needs this qualification. The printed polar-coordinate aid also labels a gradient component tuple as a divergence; the gradient used below is the two-dimensional vector $\hat{\mathbf{r}}{\partial }_r+\hat{\mathbf{\varphi }}{r}^{-1}{\partial }_{\varphi }$.

Apply the modulo operation to the eigenvalues of $Z_A+Z_B$. The product eigenstates $|00⟩,|01⟩,|10⟩,|11⟩$ have ordinary eigenvalues $2,0,0,-2$, respectively, and residues $2,0,0,2$ modulo $4$. The resulting observable isUse the entanglement-assisted nondemolition parity measurement from part (a). Equal local meter records give $O=2$; unequal records give $O=0$. Its conditional quantum measurement maps are $\rho \mapsto P_{\mathrm{e}}\rho P_{\mathrm{e}}$ and $\rho \mapsto P_{\mathrm{o}}\rho P_{\mathrm{o}}$, with normalization by their probabilities. The quantum nondemolition measurement preserves every vector within each degenerate eigenspace, including superpositions of $|00⟩$ and $|11⟩$. Measuring the two system spins separately would destroy that even-sector coherence and would therefore not realize the same Lüders rule instrument. The nonlocal eigenvalue again becomes known only after local operations and classical communication compares the local records.

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.

With every bath spin up, the two conditional bath states in the Zurek spin-bath model differ only by global phases. The decoherence factor isThere is no quantum decoherence, even for a very large bath. The device evolves as the pure state $a{e}^{it\sum g_k}|0⟩+b{e}^{-it\sum g_k}|1⟩$, while the bath stays in its original product eigenstate up to phase. Since no device information is imprinted in distinguishable bath states, the conditional environment overlap has unit modulus. The phase rotation must not be mistaken for decay of off-diagonal magnitude.

Use spin one-half states $|\uparrow ⟩,|\downarrow ⟩$ with $S^z=\pm 1/2$, so ${\sigma }^i=2S^i$ and $H=-J{\sum }_n({\sigma }_n^x{\sigma }_{n+1}^x+g{\sigma }_n^z)$. At $g=0$, every ferromagnetic bond has its lowest energy $-J$ when adjacent $x$ spins agree. On the connected periodic chain the normalized ground states arewhere $|\pm x⟩=(|\uparrow ⟩\pm |\downarrow ⟩)/\sqrt{2}$. Their span is the two-dimensional ground-state subspace; any normalized superposition is another ground state. They are exchanged by the global spin-flip discrete symmetry generated by ${\prod }_n{\sigma }_n^z$.

As $g\to +\infty$, the field dominates and selectsThe limiting polarized state is unique. At large finite $g$, exchange admixes virtual spin flips, so the product state is the limiting wavefunction rather than an exact finite-field eigenstate. The bulk expansion below gives $E_0/N=-J[g+1/(4g)+O(g^{-3})]$. The exact twofold degeneracy stated at $g=0$ should be distinguished from thermodynamic spontaneous symmetry breaking in the ordered phase: a finite chain at nonzero field can have a split pair of symmetry eigenstates.