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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.

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.

Let $B={\sum }_k{g}_k{Z}_k$. In the Zurek spin-bath model, $H=-Z_D\otimes B$, so unitary time evolution with $\hslash =1$ is $U(t)=e^{it{Z}_D\otimes B}$. The bath Hamiltonian terms commute, and device states $|0⟩,|1⟩$ have $Z_D$ eigenvalues $+1,-1$. Hencewhere the two normalized conditional bath states areTake each bath factor normalized, $|{\alpha }_k{|}^2+|{\beta }_k{|}^2=1$, and $|a{|}^2+|b{|}^2=1$. This entails no restriction: if only the product is initially normalized, divide each nonzero factor by its norm; the product of these norms is one.

Taking the partial trace over the bath gives the reduced density matrixThe orientation of this conditional environment overlap fixes the sign of the phase in the upper-right entry. Factorizing the overlap yields the decoherence factorThe populations are conserved because $[H,Z_D]=0$. Only phase coherence can be reduced. In particular,Quantum decoherence here results from distinguishable conditional bath states, although the complete system remains in a pure state under unitary time evolution.

For these three couplings, the decoherence factor becomesIt is periodic with period $2$. To locate its extrema, put $c=\cos \pi t$; then $z=c(1+c)/2$. Its maximum is $1$ at $t=0,2,4$, and its minimum is $-1/8$ where $c=-1/2$, namely $t=2/3,4/3,8/3,10/3$. The zeros on the displayed interval are $t=1/2,1,3/2,5/2,3,7/2$. The half-integer zeros are crossings; the zeros at $1$ and $3$ are double zeros, where the curve touches zero from below.

The finite spin-bath coherence recurrence returns the device to full coherence every two time units. Negative $z$ indicates a relative phase change and is not a negative probability. When $ab\ne 0$, the device reduced density matrix becomes diagonal at each zero, but quantum decoherence is not irreversible in this finite bath. The sketch explicitly displays both loss of coherence and its revival.

The Schmidt decomposition theorem states that a normalized vector in a finite-dimensional tensor product ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$ has the form ${\sum }_{j=1}^r{s}_j|u_j⟩|v_j⟩$, where the two families are orthonormal, $s_j>0$, ${\sum }_j{s}_j^2=1$, and $r\le \min (\dim {\mathcal{H}}_A,\dim {\mathcal{H}}_B)$. Extending each family to an orthonormal basis for two qubits givesThe nonzero Schmidt coefficients can be made positive by absorbing phases into the basis vectors. Both are positive exactly when the pure state is entangled. A product state has Schmidt rank one, so one coefficient is zero; the source's assertion of two positive coefficients for every pure state needs this exception.

Choose each local orthonormal basis independently as the computational basis. In that basis ${\sigma }_z=|0⟩⟨0|-|1⟩⟨1|$. This is a local change of coordinates, implemented by separate unitary matrices, rather than a physical restriction on the original quantum state. The associated Pauli operators supply the other two local axes. Up to an irrelevant common phase, each unitary change of qubit basis corresponds to a rotation of its Bloch sphere.

Put $s=2{\lambda }_0{\lambda }_1$. The Schmidt-basis Pauli correlation tensor is diagonal. The Pauli operators ${\sigma }_x\otimes {\sigma }_x$ exchange $|00⟩$ and $|11⟩$, while ${\sigma }_y\otimes {\sigma }_y$ do so with minus signs, and ${\sigma }_z\otimes {\sigma }_z$ leaves both fixed. HenceMixed components vanish: those containing one $z$ and one transverse Pauli operator map the occupied basis vectors outside their span, while the $xy$ and $yx$ matrix elements are purely imaginary and cancel for real Schmidt coefficients. By bilinearity, for arbitrary real vectors,For a CHSH inequality test take unit quantum measurement axesThese CHSH axes for an entangled pure two-qubit state giveThe local bound is two: for each hidden state, $A(B+B^{\prime })+A^{\prime }(B-B^{\prime })$ is $\pm 2$ when all four outcomes are $\pm 1$, and averaging cannot increase its absolute value. Thus every entangled pure two-qubit state violates a CHSH inequality, the content of Gisin's theorem. A product state has $s=0$ and does not violate it, so the unqualified final claim in the question is false for that case. The maximum $2\sqrt{2}$ occurs for equal Schmidt coefficients. This excludes local hidden-variable theories satisfying measurement independence, but does not permit faster-than-light signalling: the local reduced density matrix and its quantum measurement probabilities are unchanged by the remote choice of axis.

Use the Schmidt decomposition of the bipartite pure state:The two reduced density matrices, obtained by partial trace, areTheir nonzero eigenvalues are identical, even if the two ambient dimensions differ. Zero eigenvalues contribute no Von Neumann entropy, soThis common Von Neumann entropy is the entanglement entropy of the bipartite pure state.

The coefficient matrix in the computational basis isFor a bipartite pure state, the partial traces give ${\psi }_A=M{M}^{†}$ and ${\psi }_B=M^T\overline{M}$. Therefore the requested reduced density matrices areThe columns of $M$ are orthogonal. Their norms are $\sqrt{2/3}$ and $\sqrt{1/3}$, so normalizing them gives the orthonormal setWith $|v_0⟩=|0⟩$ and $|v_1⟩=|1⟩$, a Schmidt decomposition isExpanding this expression reproduces the minus sign in the $|01⟩$ coefficient. The two reduced density matrices both have eigenvalues $2/3$ and $1/3$, consistent with these Schmidt coefficients.

For any normalized pure state on finite-dimensional Hilbert spaces ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$, the Schmidt decomposition givesThe $u_j$ and $v_j$ form orthonormal sets in their respective spaces. The positive numbers $\sqrt{p_j}$ are the Schmidt coefficients, and $r$ is the Schmidt rank. Taking a partial trace gives the reduced density matricesThus the nonzero eigenvalues of the two reduced density matrices agree.

One construction is a singular value decomposition of the coefficient matrix $M$ in $|\psi ⟩={\sum }_{a,b}{M}_{ab}|a⟩|b⟩$. If $M=U\Sigma V^{†}$, the columns of $U$ give $u_j$, the complex conjugates of the columns of $V$ give $v_j$, and the singular values give $\sqrt{p_j}$. This also explains why the Schmidt coefficients are real and nonnegative.