= Solution
The <Schmidt decomposition> and <local unitary operations> put any entangled two-<qubit> <pure state> into
$$
|\psi\rangle=\sqrt\lambda\,|00\rangle+\sqrt{1-\lambda}\,|11\rangle,
\qquad \frac12\leq\lambda<1.
$$
Both <Schmidt coefficients> are nonzero. Alice performs a two-outcome <generalized measurement> with <Kraus operators>
$$
K_s=\begin{pmatrix}\sqrt{(1-\lambda)/\lambda}&0\\0&1\end{pmatrix},
\qquad
K_f=\begin{pmatrix}\sqrt{(2\lambda-1)/\lambda}&0\\0&0\end{pmatrix}.
$$
Their effects sum to the identity, so this is a legitimate local <quantum measurement>. On success,
$$
(K_s\otimes I)|\psi\rangle=\sqrt{1-\lambda}(|00\rangle+|11\rangle),
\qquad
\boxed{p_s=2(1-\lambda)>0.}
$$
The normalized output is the <Bell state> $\Phi_+$. Bob's <unitary matrix> $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ maps it to the <spin singlet state>. Alice communicates her success flag; no shared quantum operation is needed. At $\lambda=1/2$ the filter is the identity and success is certain. This is the singlet target case of <optimal stochastic conversion of a two-qubit pure state>.
For dichotomic local outcomes $A_x,B_y\in\{-1,1\}$, the <CHSH inequality> is
$$
\boxed{|E_{00}+E_{01}+E_{10}-E_{11}|\leq2,\qquad
E_{xy}=\mathbb E[A_xB_y].}
$$
It assumes a <local hidden-variable theory> with freely chosen settings independent of the hidden-variable distribution. For the <spin singlet state>, direct evaluation of its <Pauli matrices> correlations gives
$$
\langle\sigma_i\otimes\sigma_j\rangle=-\delta_{ij},\qquad
E(a,b)=-a\cdot b.
$$
Choose $a_0=e_z$, $a_1=e_x$, $b_0=(e_z+e_x)/\sqrt2$, and $b_1=(e_z-e_x)/\sqrt2$. The four correlations are $-1/\sqrt2,-1/\sqrt2,-1/\sqrt2,+1/\sqrt2$, respectively, giving
$$
\boxed{|E_{00}+E_{01}+E_{10}-E_{11}|=2\sqrt2>2.}
$$
The success flag must be recorded and communicated before the subsequent independent choice of Bell-test settings. This conditioning is legitimate because <setting-independent heralding preserves Bell locality>: a putative <local hidden-variable theory> for the entire sequential experiment would, after successful filtering, still have local response functions with a setting-independent conditional hidden-variable distribution. More explicitly, if the hidden variable is $\xi$ and $h$ is the already determined success flag, conditioning changes its distribution to $\rho(\xi\mid h=1)$ but leaves
$$
P(A,B\mid x,y,h=1)=\int \rho(\xi\mid h=1)
P_A(A\mid x,\xi,h=1)P_B(B\mid y,\xi,h=1)\,d\xi.
$$
Classical messages used in preparation can be included in $\xi$; none are exchanged during the spacelike Bell measurements. Such a conditioned model still obeys the <CHSH inequality>, contradicting the displayed singlet correlations. Therefore \b[no <local hidden-variable theory> reproduces all sequential measurement correlations of any entangled two-<qubit> <pure state>]. This argument does not discard outcomes on the basis of the later measurement settings.
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