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

  • 2
    $\begingroup$ 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 May 27 at 6:37
  • 1
    $\begingroup$ 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 May 27 at 10:43
  • $\begingroup$ Major typos, sorry about all of that - edits should suffice $\endgroup$ – Quantum Mechanic May 27 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.