2
$\begingroup$

In Nielsen and Chuang, there's the following paragraph: > The H gate is sometimes described as being like a ‘square-root of NOT’ gate, in that it turns a |0> into (|0> + |1>)/ √ 2 (first column of H), ‘halfway’ between |0? and |1?, and turns |1? into (|0? − |1?)/ √ 2 (second column of H), which is also ‘halfway’ between |0? and |1?.

I understand that \begin{align*} \sqrt{NOT} = \frac{1}{2}\left( {\begin{array}{*{20}{c}} \sqrt 2 e^{i\pi / 4}&\sqrt 2 e^{-i\pi / 4}\\ \sqrt 2 e^{-i\pi / 4}&\sqrt 2 e^{i\pi / 4} \end{array}} \right) \\ \end{align*}

But, I'm still confused how the H gate is related to the ‘square-root of NOT’ gate.

$\endgroup$
1
  • 1
    $\begingroup$ It is already explained in the paragraph you copied to your question. Hadamard is "like" square root in sense that turns a qubit to half-way between states $|0\rangle$ and $|1\rangle$. There is also stated that $H^2$ is not $X$. Hence there is no relation to square root of $X$. $\endgroup$ Aug 3 at 9:45
2
$\begingroup$

I don't think the Hadamard gate is ever actually called "square root of NOT". I'd guess the emphasis in this sentence is on the word "like", in that it possesses the same property of converting the basis states into states halfway between the basis states, though the Hadamard gate does it without adding complex phases, and the $\sqrt{NOT}$ throws in complex numbers.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.