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.

We can create the CNOT-gate from the two-qubit identity matrix by interchanging the order of the last two elements, like this:

$$CNOT=\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{bmatrix}$$

The CNOT-gate takes two inputs and gives two outputs. The first input is called the control qubit. The second input is called the target qubit.

The result of the CNOT-gate is straightforward if the control qubit is in a basic state $|0\rangle$ or $|1\rangle$. If the control qubit is $|0\rangle$, then nothing happens. The output equals the input. If the control qubit is $|1\rangle$, the CNOT-gate applies the $X$-gate (NOT-gate) on the target qubit. It flips the state of the target qubit.

The following figure depicts the truth table of the CNOT-gate.

It gets interesting when the control qubit is in superposition—for instance, when we apply the Hadamard gate to the first qubit before we apply the CNOT-gate.

The following equation denotes the state of our two-qubit system.

$CNOT\cdot(H\otimes I)|00\rangle$

$\qquad\qquad=\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{bmatrix}\cdot \frac{1}{\sqrt{2}}\begin{bmatrix}1&0&1&0\\0&1&0&1\\1&0&-1&0\\0&1&0&-1\end{bmatrix}\begin{bmatrix} 1\\ 0\\0 \\ 0\end{bmatrix}$

$\qquad\qquad=\begin{bmatrix} \tfrac{1}{\sqrt{2}} & 0 & \tfrac{1}{\sqrt{2}} & 0 \\ 0 & \tfrac{1}{\sqrt{2}} & 0 & \tfrac{1}{\sqrt{2}} \\ 0 & \tfrac{1}{\sqrt{2}} & 0 & -\tfrac{1}{\sqrt{2}} \\ \tfrac{1}{\sqrt{2}} & 0 & -\tfrac{1}{\sqrt{2}} & 0 \\ \end{bmatrix}\cdot\begin{bmatrix} 1\\ 0\\0 \\ 0\end{bmatrix}=\begin{bmatrix} \tfrac{1}{\sqrt{2}}\\ 0\\0 \\ \tfrac{1}{\sqrt{2}}\end{bmatrix}$ ...