What gate should one use to perform $R_y$ using a single $R_z$ + Clifford gates?

Question

I know how to perform Rz rotations with the least amount of T gates, eg by using Efficient Clifford+T approximation of single-qubit operators by Peter Selinger. Similarly, one could use H Rz H to perform an Rx rotation. This seems to me like a simple question, what gate should one use to perform Ry using a single Rz + Clifford gates?

Answer 1

The phase gate $$S=\begin{pmatrix}1&0\\0&i\end{pmatrix}=\sqrt{\sigma_z}$$ satisfies $$S\sigma_x S^\dagger=\sigma_y,$$ where $\sigma_i$ are the Pauli matrices. So, once you know that $R_x=H R_z H$, you can immediately find $$R_y=SH R_z HS^\dagger.$$

You can also do $$R_y=-S^\dagger H R_z HS.$$

Answer 2

There's a bunch of single Clifford gate types that you can use.

$R_y(\theta) = \sqrt{X} \sqrt{X} \sqrt{X} R_z(\theta) \sqrt{X}$ where $\sqrt{X} = HSH$ is a 90 degree rotation around X.

$R_y(\theta) = C_{XYZ} R_z(\theta) C_{XYZ} C_{XYZ}$ where $C_{XYZ} = (iI + X + Y + Z)/2$ is a 120 degree rotation around X+Y+Z.

$R_y(\theta) = H_{YZ} R_z(\theta) H_{YZ}$ where $H_{YZ} = (Y + Z)/\sqrt{2}$ is a 180 degree rotation around Y+Z.

There's no Clifford gate $U$ with the property that $U R_z(\theta)$ = $R_y(\theta)$. This is obvious because $Y$ basis interactions should commute with $R_y(\theta)$ but when crossing $Y$ over $U R_z(\theta)$ the $R_z(\theta)$ moves the $Y$'s rotation axis to something arbitrary (not a Clifford operation anymore) and a Clifford $U$ can't get it back to where it should be.

You can also do it with two qubit gates. E.g. if you have an ancilla in the $|0\rangle$ state:

$R_{Y_{target}}(\theta) = \text{Control}(Y_{target}, X_{ancilla}) R_{Z_{ancilla}}(\theta) \text{Control}(Y_{target}, X_{ancilla})$ where $\text{Control}(Y, X)$ applies an X to the target when the control is in the $|-i\rangle$ state.