Entanglement

What if we constructed the two-qubit system differently? When we discard the factors, the four basis states ($a_0,b_0, ...$) and replace them with general variables. We can state the following equation for an arbitrary two-qubit system.

$$|\psi\rangle=\alpha|0\rangle|0\rangle+\beta|0\rangle|1\rangle+\gamma|1\rangle|0\rangle+\delta|1\rangle|1\rangle=\begin{bmatrix}\alpha \\ \beta \\ \gamma \\ \delta \end{bmatrix}$$

We’re holding on to the normalization of the sum of all probabilities must be 1, but we do not insist that $\alpha\delta$=$\beta\gamma$.

In the lesson Implementation of CNOT gate, we learned about the CNOT-gate. It applies the X-gate to the target qubit only if we measure the control qubit as a 1.