Controlled-Hadamard Gate

Last updated Jun 23, 2022

• Type atom
• Status needs revision

In this gate, Hadamard Gate is applied to the right qubit if the left qubit is $1$.

It acts on $Z$-basis states as: $$\begin{aligned}
CH\ket{00} &= \ket{00}, \
CH\ket{01} &= \ket{01}, \
CH\ket{10} &= \ket{1+} = \frac{1}{\sqrt 2}(\ket{10} + \ket{11}), \
CH\ket{11} &= \ket{1-} = \frac{1}{\sqrt 2}(\ket{10} - \ket{11}).
\end{aligned}$$
It can simulate both the Toffoli Gate and the Hadamard Gate.