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?


2 Answers 2


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

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    $\begingroup$ 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) $\endgroup$
    – DaftWullie
    Commented May 27, 2021 at 6:37
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    $\begingroup$ 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. $\endgroup$
    – Pablo
    Commented May 27, 2021 at 10:43
  • $\begingroup$ Major typos, sorry about all of that - edits should suffice $\endgroup$ Commented 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.


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