# CNOT gate CNOT gate has two qubit input: control and target. $\text{CNOT}|00\rangle=|00\rangle$ $\text{CNOT}|01\rangle=|01\rangle$ $\text{CNOT}|10\rangle=|11\rangle$ $\text{CNOT}|11\rangle=|10\rangle$ note: another way to represent it: $\text{CNOT}|a,b\rangle=|a, a\oplus b\rangle$ for $a,b\in\{0,1\}$ ## CNOT on general 2-qubits assume $|a\rangle=\alpha|0\rangle+\beta|1\rangle$ , $|b\rangle=\gamma|0\rangle+\delta|1\rangle$ . 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 \\ &= |0\rangle\otimes(\alpha\gamma|0\rangle + \alpha\delta|1\rangle) + |1\rangle\otimes(\beta\gamma|0\rangle+\beta\delta|1\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