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

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?

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 $$R_y=-S^\dagger H R_z HS.$$

• Have you mixed up your subscripts on the last couple of lines? ($R_x=HR_zH$ and $R_y=SR_xS^\dagger$? (Note, also, that I don't think you need the -ve sign) May 27, 2021 at 6:37
• Thanks a lot for your answer! I am confused by something though, if $S = \sqrt{\sigma_z}$ then doing $S R_z S$ should make $R_z$ commute with $S$ and therefore the result is diagonal, which need not to be $R_x$. I think what you meant is $$S^\dagger R_y S = R_x$$ (I've checked it computationally). Had not seen the previous comment, sorry. May 27, 2021 at 10:43
• Major typos, sorry about all of that - edits should suffice May 27, 2021 at 13:58

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.