Quantum Inspire is a quantum computing platform provided by QuTech. It consists of two real quantum processors - Starmon-5 and Spin-2. Whereas it is possible to use rotation gates $Rx$, $Ry$ and $Rz$ on Spin-2 processor, Starmon-5 gate set consist only of:

  • Pauli gates $\text{X}$, $\text{Y}$, $\text{Z}$ and $\text{I}$
  • Hadamard gate $\text{H}$
  • Phase gate ($\text{S}$) and $\pi/8$ ($\text{T}$) gate and their conjugate transpose gates
  • rotation around axes $x$ and $y$ for angle $\pm\frac{\pi}{2}$ gates
  • $\text{CNOT}$, $\text{CZ}$ and $\text{SWAP}$ gates

My question is how can I construct gates $Rx$, $Ry$ and $Rz$ with rotation angle $\theta$ from gate listed above.


Based on advice from JSdJ and the Nielsen and Chuang book, I found out that:

  • $HTH = Rx(\pi/4)$ which together with the fact that $T=Rz(\pi/4)$ allows to build gate $$R_{\hat{n}}=Rz(\pi/4)Rx(\pi/4)= \text{exp}(-i\frac{\pi}{8}Z)\text{exp}(-i\frac{\pi}{8}X) = \\ \cos^2\frac{\pi}{8}I-i(\cos\frac{\pi}{8}(X+Z)+\sin\frac{\pi}{8}Y)\sin\frac{\pi}{8}$$ (note that $Y=-iZX$)
  • this is a rotation around axis defined by vector $\hat{n}=(\cos(\pi/8);\sin(\pi/8);\cos(\pi/8))$. Rotation angle is given by equation $\cos(\theta/2) =\cos^2(\pi/8)$
  • angle $\theta$ is $2\arccos[\cos^2(\pi/8)] \approx 1.096$ which is irrational multiple of $2\pi$
  • since $\theta$ is irrational, repeated application of the gate $R_{\hat{n}}$ never leads to rotation by an angle which sum up to $2\pi$
  • as a result we are able to reach any rotation by angle $\alpha$ around $\hat{n}$ axis by repeated application of $R_{\hat{n}}$, i.e. we can construct $R_{\hat{n}}(\alpha)=R_{\hat{n}}^{n_1}$, where $n_1$ is an integer
  • also it is true that $HR_{\hat{n}}(\alpha)H = R_{\hat{m}}(\alpha)$ where $\hat{m}$ is axis defined by vector $(\cos(\pi/8);-\sin(\pi/8);\cos(\pi/8))$, so the same procedure allows to find rotation for $\alpha$ around axis $\hat{m}$
  • it can also be proven that any arbitrary single qubit unitary matrix $U$ can be written (up to global phase) as $U = R_{\hat{n}}(\beta_1)R_{\hat{m}}(\gamma_1)R_{\hat{n}}(\beta_2)R_{\hat{m}}(\gamma_2)\dots$
  • together this leads to conclusion that any $U$ (including $Rx$, $Ry$ and $Rz$ rotations) can be approximated by repeated application of $R_{\hat{n}}$ and Hadamards, i.e. $U \approx R_{\hat{n}}^{n_1}HR_{\hat{n}}^{n_2}HR_{\hat{n}}^{n_3}HR_{\hat{n}}^{n_4}H \dots$, where $n_i$ are integers

To sum up, only with $T$ and $H$ gates we can construct any rotation.

Only question is how to find integers $n_i$ for construction of arbitrary $Rx$, $Ry$ and $Rz$ rotation. Can anybody give me an example of such circuit, for exaple gate $Ry(\pi/8)$?


3 Answers 3


Ross and Selinger have solved this problem here: Optimal ancilla-free Clifford+T approximation of z-rotations

and provide a command-line tool for generating Clifford+T approximations to Rz gates, which you can download here: https://www.mathstat.dal.ca/~selinger/newsynth/

  • 1
    $\begingroup$ Thanks, that is really helpful. $\endgroup$ Commented May 23, 2020 at 7:09

Recently the Starmon-5 system was upgraded. Single qubit rotations Rx, Ry and Rz are now available, see https://www.quantum-inspire.com/kbase/starmon-5-operational-specifics/

  • $\begingroup$ Good to know, thanks for the info. $\endgroup$ Commented Jun 17, 2020 at 8:48

According to the answer of Simon Crane and an algorithm provided in the question, it is possible to implement any $Rz(\theta)$ gate with gates $X$, $H$, $S$ and $T$.

The provided algorithm is based on number theory (in particular on solving specific Diophanine equation). According to the paper Optimal ancilla-free Clifford+T approximation of z-rotations, the algorithm is optimal and faster than general Solovay-Kitaev algorithm for the gate set mentioned above. So, now we can prepare any $Rz(\theta)$.

Based on identities

  • $X=HZH$
  • $Y=-SXS^\dagger$
  • $\mathrm{e}^{-i\frac{\theta}{2}A}=\cos\frac{\theta}{2}I-i\sin\frac{\theta}{2}A$, where $A \in \{X;Y;Z\}$

it can be shown that

  • $Rx(\theta) = HRz(\theta)H$
  • $Ry(\theta) = S HRz(-\theta)HS^\dagger$ (note that $S^\dagger$ is not in the gate set above but it holds that $S^\dagger = S^3$)

Hence, we can implement any $x$ and $y$ rotation with $z$ rotation. Consequently we can create any rotation with gates $X$, $H$, $S$ and $T$ only.

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    $\begingroup$ I think that, in the last equation, the $S^\dagger$ and $S$ gates should be swapped. $\endgroup$
    – tigerjack
    Commented May 29, 2023 at 9:03
  • 1
    $\begingroup$ @tigerjack: Thanks, swapped. $\endgroup$ Commented May 29, 2023 at 20:58

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