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I'm looking for a representation of Hadamard gate that uses only $R_x(x)$ and $R_y(y)$ gates. The values $x$ and $y$ may be the same, but they don't necessarily need to be.

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If you're not concerned with global phase then the following works using only two rotation gates: \begin{align} R_y\left(-\frac{\pi}{2}\right) R_x\left(\pi\right) &= \exp \left(i\frac{\pi}{4}Y\right) \exp \left(-i\frac{\pi}{2}X\right) \\&= \left(\cos \frac{\pi}{4} I + i\sin \frac{\pi}{4} Y\right) \left(-iX\right) \\&=\begin{pmatrix} \cos \frac{\pi}{4} & \sin\frac{\pi}{4} \\ -\sin \frac{\pi}{4} & \cos \frac{\pi}{4} \end{pmatrix} \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} \\&= -\frac{i}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \\&= -iH \end{align} where $R_x(\theta) = \exp(-i\theta X/2)$ and $R_y(\theta) = \exp(-i\theta Y/2)$.

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    $\begingroup$ Nice answer! Just a quick addition: if you do care about the global phase then it is actually not possible $\endgroup$ Jul 8 at 9:43

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