Quantum Computing 3.3 -- Correlated measurements

We apply the operations $G \otimes I$ , CNOT and $G \otimes I$ one after the other to the state $| 00 \rangle$ : $\begin{aligned} |\psi\rangle &= (G \otimes I)\cdot \text{CNOT} \cdot (G \otimes I) |00\rangle \\ &= (G \otimes I)\cdot \text{CNOT} (G|0\rangle) \otimes |0\rangle\\ &= (G \otimes I)\cdot \text{CNOT} \left(\frac{\sqrt{3}}{2} |00\rangle + \frac{1}{2} |10\rangle\right) \\ &= (G \otimes I) \left(\frac{\sqrt{3}}{2} |00 \rangle + \frac{1}{2} |11 \rangle \right) \\ &= \left(\frac{\sqrt{3}}{2} (G |0\rangle) \otimes 0 \rangle + \frac{1}{2}(G |1\rangle) \otimes |1 \rangle \right) \\ &= \left(\frac{\sqrt{3}}{2}\left(\frac{\sqrt{3}}{2} |00\rangle + \frac{1}{2} |10\rangle\right) + \frac{1}{2} \left(\frac{1}{2} |01\rangle - \frac{\sqrt{3}}{2} |11\rangle\right) \right) \\ &= \frac{3}{4} |00\rangle + \frac{1}{4} |01\rangle + \frac{\sqrt{3}}{4} |10\rangle - \frac{\sqrt{3}}{4} |11\rangle \end{aligned}$ The probability $p$ of obtaining the result $| 00 \rangle$ or $| 11 \rangle$ during the measurement results accordingly $\begin{aligned} p &= |\langle 00 |\psi\rangle |^2 + |\langle 11 |\psi\rangle |^2 \\ &= \left( \frac{3}{4} \right)^2 + \left(- \frac{\sqrt{3}}{4} \right)^2 \\ &= \frac{9 + 3}{16} = \frac{3}{4} = 75\,\% \end{aligned}$ Thus, the states of the two qubits are correlated. In contrast to the Bell states, however, there is only a partial entanglement.