# CNOT gate controlled-NOT gate has two qubit input: control and target. ## CNOT on computational basis | a | b | $CNOT\ket a\ket b$ | $XOR(a,b)$ | | --- | --- | ------------------ | ---------- | | 0 | 0 | $\ket {00}$ | 0 | | 0 | 1 | $\ket {01}$ | 1 | | 1 | 0 | $\ket {11}$ | 1 | | 1 | 1 | $\ket {10}$ | 0 | CNOT is the same as logic XOR : $\text{CNOT}|a,b\rangle=|a, a\oplus b\rangle$ for $a,b\in\{0,1\}$ note: CNOT can be applied on qubits in superposition and not only on computational basis. ## CNOT quantum circuit ``` a: ─────●───── │ out: ───⊕───── ``` ## Exercise assume $|a\rangle=\alpha|0\rangle+\beta|1\rangle$ , $|b\rangle=\gamma|0\rangle+\delta|1\rangle$ . compute $cnot\ket{a,b}$. Solution: we have, $ \begin{align} |a,b\rangle &= (\alpha|0\rangle+\beta|1\rangle)\otimes(\gamma|0\rangle+\delta|1\rangle) \\ &=\alpha\gamma|00\rangle+\alpha\delta|01\rangle+\beta\gamma|10\rangle+\beta\delta|11\rangle \\ \end{align} $ And so, $ \begin{align} \text{CNOT}|a,b\rangle&=\text{CNOT}(\alpha\gamma|00\rangle+\alpha\delta|01\rangle+\beta\gamma|10\rangle+\beta\delta|11\rangle) \\ &=\text{CNOT}\big(|0\rangle(\alpha\gamma|0\rangle+\alpha\delta|1\rangle)+|1\rangle(\beta\gamma|0\rangle+\beta\delta|1\rangle)\big)\\ &= \alpha\gamma|0\rangle+\alpha\delta|1\rangle + X(\beta\gamma|0\rangle+\beta\delta|1\rangle) \\ &=\alpha\gamma|0\rangle+\alpha\delta|1\rangle + \beta\gamma|1\rangle+\beta\delta|0\rangle \\ &=(\alpha\gamma+\beta\delta)|0\rangle +(\alpha\delta+\beta\gamma)|1\rangle \end{align} $ Related: [[Bell state]] Tag: #tech ## Created 2026-01-14 15:25