Use the thin singular value decomposition $a=Ud{V}^{†}$, with $U^{†}U=I_N$, $V$ unitary and $d=\mathrm{diag}(d_k)$ nonnegative. Expanding its entries givesThe columns of $U$ and of $V^{\ast }$ are orthonormal, establishing the Schmidt decomposition. Bob's basis uses the complex-conjugated columns of $V$. Omitting that conjugation would not reproduce the coefficient matrix. The squared Schmidt coefficients are the eigenvalues of $a^{†}a$; equivalently $d^2=V^{†}{a}^{†}aV$ and $d=U^{†}aV$. The two reduced density matrices are $a{a}^{†}$ and $(a^{†}a{)}^T$, which have the same nonzero eigenvalues.

The printed coefficient $0.41231$ slightly rounds a normalized state: each squared norm is $0.9999995361$. Normalize before calculating an entropy. For the two states, the resulting Gram matrices are approximatelyFor a trace-one two-dimensional Hermitian matrix with diagonal entry $u$ and off-diagonal entry $c$, its eigenvalues areSort them decreasingly and take square roots:The Von Neumann entropy is $S=-{\lambda }_0{\log }_2{\lambda }_0-{\lambda }_1{\log }_2{\lambda }_1$. $\Psi$ is more entangled: its Schmidt weights are closer to equal and its reduced-state entropy is larger. Neither state is a product state or maximally entangled on its two-dimensional support.

By Nielsen's pure-state conversion theorem, deterministic LOCC conversion requires the input vector of squared Schmidt coefficients to be majorized by the output vector. Here $0.699941<0.894238$ and both sums are one, so $\lambda (\Psi )\prec \lambda (\Phi )$. Therefore $\Psi \to \Phi$ is possible deterministically; $\Phi \to \Psi$ is not. Entanglement can be discarded by local operations but cannot increase deterministically for these pure inputs. Probabilistic concentration is a separate question and is allowed below.^[ref]

There is a dimensional qualification in the proposed measurement. Write $P=|0_A^{\prime }⟩⟨0_A^{\prime }|+|1_A^{\prime }⟩⟨1_A^{\prime }|$. The two supplied measurement operators obeyThus their effects form a POVM on the occupied support, where $P$ is the identity. Alice's full space is three-dimensional, so these two operators alone do not sum to $I_3$. A full-space completion is $M_3=|2_A^{\prime }⟩⟨2_A^{\prime }|$. It has zero probability on either input, so it changes none of the requested outcomes. Strictly, the POVM consists of the effects $E_r=M_r^{†}{M}_r$; the specified $M_r$ additionally fix a quantum instrument and therefore the conditional states.

In the appropriate Schmidt basis for either input, write $|\psi ⟩=\sqrt{{\lambda }_0}|00⟩+\sqrt{{\lambda }_1}|11⟩$. The two outcomes have probabilitiesand normalized conditional statesThese formulas give both outcomes for each state by substituting its row of the table. For example,The two inputs have different Schmidt bases; the primed measurement basis is adapted separately to the state, as the question specifies, not one common unexamined laboratory basis.

For the $M_1$ outcome to be a maximally entangled state, its two amplitudes must be equal. HenceThe requested numerical results areThus $\Psi$ has the larger success probability, consistent with its greater initial entanglement. The $M_2$ branch cannot be balanced in the allowed angle range for either nonmaximal input: it would require $\tan \theta =\sqrt{{\lambda }_0/{\lambda }_1}>1$.

The success probabilities belong to this particular binary Schmidt filter. They are not the optimum over every possible quantum instrument. The diagonal success operator $\mathrm{diag}(\sqrt{{\lambda }_1/{\lambda }_0},1)$ instead gives the standard single-copy entanglement concentration probability $2{\lambda }_1$. Postselection can increase entanglement in the retained branch, while the outcome-averaged entanglement cannot increase under LOCC.

An eigenvalue of an observable in a known corresponding eigenstate is a simple example satisfying the Einstein–Podolsky–Rosen criterion of reality: its value is predictable with certainty. For example, a spin angular momentum component in a prepared spin-up eigenstate has a definite predicted value. For a distant system, Albert Einstein, Boris Podolsky and Nathan Rosen invoked locality: a measurement performed on a sufficiently separated, noninteracting partner should not disturb that system. In the relativistic version here, the measurement events have spacelike separation.

The Pauli matrices obey $[{\sigma }_z,{\sigma }_x]=2i{\sigma }_y$, soTheir nondegenerate eigenstates are incompatible: an $S_z$ eigenstate has equal probabilities for the two $S_x$ outcomes, and conversely. Thus no single quantum state has both components sharply determined. The Robertson uncertainty relation gives $\Delta S_z\Delta S_x\ge \hslash |⟨S_y⟩|/2$, but its right-hand side may be zero; that alone is not a proof that simultaneous zero uncertainties exist. The absence of a common eigenstate supplies the stronger conclusion for this spin-half pair.

For the specified Bell state, Bob obtains $+\hslash /2$ or $-\hslash /2$ with probability $1/2$ when he measures $S_z$. Conditional on his result, Alice obtains the same sign with certainty if she measures $S_z$. Under the EPR locality premise, this certainty supplies an element of reality for her $S_z$; it is a conditional prediction after Bob learns his own random result, not a premeasurement prediction of its sign.

The normalized $S_x$ eigenstates are $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. Expanding their tensor products givesConsequently the same Bell state has equal, perfectly correlated $S_x$ outcomes. Bob's two outcomes again each have probability $1/2$, and the same locality-and-certainty argument applies separately to Alice's $S_x$.

The disputed step is to combine predictions conditional on two incompatible choices into two simultaneous, measurement-independent pre-existing values. Quantum theory provides the two conditional ensembles, not a joint sharp assignment of $S_x$ and $S_z$. This is quantum steering. Bob cannot measure both incompatible observables sharply on the same member of the pair. Nor does conditional steering allow signalling: ignoring his result leaves Alice's reduced density matrix equal to $I/2$ for either measurement choice, by the no-communication theorem.

To test the stronger claim of a local hidden-variable theory, Alice needs several correlation settings, not just perfect agreement along the two common axes. For the CHSH inequality, choose observables with outcomes $\pm 1$:In the given Bell state, $⟨{\sigma }_z\otimes {\sigma }_z⟩=⟨{\sigma }_x\otimes {\sigma }_x⟩=1$ and the crossed correlations vanish. Henceexceeding the local bound $|\mathcal{S}|\le 2$. The bound assumes Bell local causality and measurement independence, with the appropriate treatment of sampling and settings.

Experimental Bell tests support violations rather than that local assignment. The Aspect switching experiment changed analyzer settings during separated photon flights, while leaving sampling and switching limitations. By the time of this examination, the Delft electron-spin Bell test had used electron spins at stations $1.3\mathrm{km}$ apart, efficient spin readout and rapid random basis choices. The combination addressed both detection and locality loopholes, providing evidence against the tested local model class.^[ref]

This does not prove that every conceivable notion of reality is false, or that noncommutation alone rules out every hidden-variable theory. A nonlocal theory can evade the local premise. What Alice can challenge is Bob's claim that locality, measurement-independent simultaneous values and all observed correlations can be retained together. An ordinary measurement disturbance at Bob's lab is not by itself a local explanation of Alice's conditional ensemble.