 Solution

By the Schmidt decomposition theorem, suitable local unitary operations put the normalized pure state into the formIt is entangled precisely when both Schmidt coefficients are nonzero. Put $s=2{\lambda }_0{\lambda }_1$, so $0<s\le 1$ for the case of interest. The Pauli operatorssatisfy $⟨Z\otimes Z⟩=1$ and $⟨X\otimes X⟩=s$ in this pure state: $Z\otimes Z$ fixes both basis terms, and $X\otimes X$ exchanges them.

Choose Alice's two observables as $A_0=Z$, $A_1=X$, and Bob's asThese are valid dichotomic projective measurements. Indeed, $XZ+ZX=0$ and $X^2=Z^2=I$, so $B_0^2=B_1^2=I$; each observable has outcomes $\pm 1$. For the CHSH inequality convention $|⟨A_0\otimes (B_0+B_1)+A_1\otimes (B_0-B_1)⟩|\le 2$, the quantum value isFor the original pure state, conjugate these four observables by the corresponding local unitary operators. Their outcomes and Pauli correlators are unchanged, and each party still measures only their own qubit. Thus every entangled pure two-qubit state violates CHSH for suitable local measurements, proving Gisin's theorem directly. The CHSH axes for an entangled pure two-qubit state approach a nonviolating configuration continuously as one Schmidt coefficient tends to zero; the strict violation therefore need not be large for weak entanglement.