= Solution
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 $\sum_{j=1}^r s_j|u_j\rangle|v_j\rangle$, where the two families are orthonormal, $s_j>0$, $\sum_js_j^2=1$, and $r\leq\min(\dim\mathcal H_A,\dim\mathcal H_B)$. Extending each family to an <orthonormal basis> for two <qubits> gives
$$
|\psi\rangle=\lambda_0|00\rangle+\lambda_1|11\rangle,\qquad
\lambda_i\geq0,\quad\lambda_0^2+\lambda_1^2=1.
$$
The 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 $\sigma_z=|0\rangle\langle0|-|1\rangle\langle1|$. 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 $|00\rangle$ and $|11\rangle$, while $\sigma_y\otimes\sigma_y$ do so with minus signs, and $\sigma_z\otimes\sigma_z$ leaves both fixed. Hence
$$
\langle\sigma_x\otimes\sigma_x\rangle=s,\qquad
\langle\sigma_y\otimes\sigma_y\rangle=-s,\qquad
\langle\sigma_z\otimes\sigma_z\rangle=1.
$$
Mixed 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,
$$
\boxed{P(\mathbf a,\mathbf b)=s(a_xb_x-a_yb_y)+a_zb_z.}
$$
For a <CHSH inequality> test take unit <quantum measurement> axes
$$
\mathbf a=\mathbf e_z,\quad\mathbf a'=\mathbf e_x,\qquad
\mathbf b=\frac{\mathbf e_z+s\mathbf e_x}{\sqrt{1+s^2}},\quad
\mathbf b'=\frac{\mathbf e_z-s\mathbf e_x}{\sqrt{1+s^2}}.
$$
These <CHSH axes for an entangled pure two-qubit state> give
$$
\boxed{S=P(\mathbf a,\mathbf b)+P(\mathbf a,\mathbf b')+P(\mathbf a',\mathbf b)-P(\mathbf a',\mathbf b')
=2\sqrt{1+s^2}>2\quad(s>0).}
$$
The local bound is two: for each hidden state, $A(B+B')+A'(B-B')$ is $\pm2$ when all four outcomes are $\pm1$, and averaging cannot increase its absolute value. Thus \b[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 $2\sqrt2$ 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.
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