Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 57 / 2 / c / Solution

Take ${\alpha }^2+{\beta }^2=1$ and label the first vector of each measurement basis by outcome $0$. Assign the Pauli measurement value $+1$ to outcome $0$ and $-1$ to outcome $1$. The four observables areHere $X$ and $Z$ are the Pauli X gate and Pauli Z gate matrices. These follow by subtracting the two rank-one basis projectors; a real basis rotated through $\gamma$ has observable $\sin (2\gamma )X+\cos (2\gamma )Z$.

The Schmidt-basis Pauli correlation tensor of the state givesConsequently $E_{11}=1/2$, $E_{21}=E_{22}=\sqrt{3}\alpha \beta$ and $E_{12}=-1/2$, where $E_{jk}=⟨A_j\otimes B_k⟩$. For binary outcomes, $\mathbb{E}([A-B{]}_2)=P(A\ne B)=(1-E_{AB})/2$, whereas the final offset term has $\mathbb{E}([B-A-1{]}_2)=P(A=B)=(1+E_{AB})/2$. The chained modular Bell inequality left side is thereforeThe local bound is $I\ge 1$, so the exact violation condition isEquality saturates the bound. The maximal violation for these fixed measurements occurs at $\alpha =\beta =1/\sqrt{2}$ or their common negative, giving $I=(3-\sqrt{3})/2$. Opposite signs do not violate this particular inequality with these fixed bases, although other measurement choices can reveal the state's entanglement. If unnormalized real amplitudes are used, replace $\alpha \beta$ throughout by $\alpha \beta /({\alpha }^2+{\beta }^2)$.