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In Nielsen and Chuang, there's the following paragraph:

The Hadamard operation is just a rotation of the sphere about the ˆy axis by 90◦, followed by a rotation about the ˆx axis by 180◦.

I am wondering how we were able to know that this is the transformation that the H gate creates? Is it implied from the matrix representation of the H gate? If so, how?

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We start with definition of $Rx(\theta_x)$ and $Ry(\theta_y)$ rotations: $$ Rx(\theta_x)= \begin{pmatrix} \cos(\theta_x/2) & -i\sin(\theta_x/2)\\ -i\sin(\theta_x/2) & \cos(\theta_x/2) \end{pmatrix} $$

$$ Ry(\theta_y)= \begin{pmatrix} \cos(\theta_y/2) & -\sin(\theta_y/2)\\ \sin(\theta_y/2) & \cos(\theta_y/2) \end{pmatrix} $$

Setting $\theta_x = \pi$ we have $$ Rx(\pi)= \begin{pmatrix} 0 & -i\\ -i & 0 \end{pmatrix} =-i \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} =-iX, $$ which is NOT gate ($X$), up to global phase represented by $-i$. However, the global phase can be neglected and we are left with $X$ gate.

If we put $\theta_y = \pi/2$ we get $$ Ry(\pi/2)= \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1\\ 1 & 1 \end{pmatrix} $$ Multiplying $XRy(\pi/2)$ we get $$ XRy(\pi/2)= \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1\\ 1 & 1 \end{pmatrix} =\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix} =H $$

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  • $\begingroup$ I think this should answers the OP question (+1). Also you probably meant to put $\theta_x$ instead of $\theta$ in your $R_X$ gate, similar with $R_Y$. Not a big deal but just thought I should mention it. $\endgroup$
    – KAJ226
    Aug 3 at 14:37
  • $\begingroup$ @KAJ226: Thanks, fixed. $\endgroup$ Aug 3 at 18:00

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