How to represent Hadamard gate as product of Rx and Ry gates? As simple as question asks - how to represent Hadamard gate as product of Rx and Ry gates assuming that values for rotation of Rx and Ry gate can be different?
 A: This is not possible, and one can easily see this via the determinant. All rotation operator gates
$$
R_x(\theta)=\begin{pmatrix} \cos\frac\theta2&-i\sin\frac\theta2\\-i\sin\frac\theta2&\cos\frac\theta2 \end{pmatrix}\quad R_y(\theta)=\begin{pmatrix} \cos\frac\theta2&-\sin\frac\theta2\\\sin\frac\theta2&\cos\frac\theta2 \end{pmatrix}\quad R_z(\theta)=\begin{pmatrix} e^{-i\theta/2}&0\\0&e^{i\theta/2} \end{pmatrix}
$$
have determinant $1$, regardless of the angle $\theta\in\mathbb R$. The Hadamard gate
$$
H=\frac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}
$$
however, has determinant $-1$. Now if there existed $k\in\mathbb N$, angles $\theta_1,\ldots,\theta_k\in\mathbb R$ and a gate combination $i_1,\ldots,i_k\in\{x,y,z\}$ such that $H=R_{i_1}(\theta_1)\cdot\ldots\cdot R_{i_k}(\theta_k)$ then
$$
-1=\operatorname{det}(H)=\operatorname{det}(R_{i_1}(\theta_1)\cdot\ldots\cdot R_{i_k}(\theta_k))=\operatorname{det}(R_{i_1}(\theta_1))\cdot\ldots\cdot \operatorname{det}(R_{i_k}(\theta_k))=1
$$
because the determinant is multiplicative. But this is a contradiction, hence no decomposition of the Hadamard gate into rotation gates exists.
A: We have
$$R_x(\pi)R_y(\pi/2)=\left(
\begin{array}{cc}
 0 & -i \\
 -i & 0 \\
\end{array}
\right)\left(
\begin{array}{cc}
 \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\
 \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\end{array}
\right)$$
$$=\left(
\begin{array}{cc}
 -\frac{i}{\sqrt{2}} & -\frac{i}{\sqrt{2}} \\
 -\frac{i}{\sqrt{2}} & \frac{i}{\sqrt{2}} \\
\end{array}
\right)=-i\left(
\begin{array}{cc}
 \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
 \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\
\end{array}
\right)=-iH$$
This suffices for the purposes of quantum computing since the overall phase of $\frac{3\pi}{2}$ (represented by $-i$) doesn't effect the final measurement output.
