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I was reading the paper An introduction to measurement based quantum computation (Josza, 2005) and on page 13 they say the following:

Theorem: Any gate array using gates from the set $\{CX,R_x(\theta) \text{ all } θ\}$ or from the set $\{CX,R_z(\theta) \text{ all } θ\}$ can be implemented with just two measurement layers.

Remark: Neither of these sets is believed to be universal although it is known that $CX$ with all $y$-rotations is universal

Has anyone since proved/disproved these two gate sets are not universal?

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  • $\begingroup$ I haven't read those papers, specifically citation $[13]$, but when they say $y$-rotations, do they mean $R_x(\theta_1)$ followed by $R_z(\theta_2)$ operation (or the other way around)? Because otherwise, I don't see why rotation around the $Y$-axis is special. I am very curious about that remark! $\endgroup$
    – FDGod
    Nov 14, 2023 at 8:47
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    $\begingroup$ @FDGod It's just a normal $R_Y(\theta)$ operation. It's special because controlled-not picks out two special bases: $X$ and $Z$. To see that it's universal, you can prove that $Y$ rotations + controlled-not let you build controlled-$\sqrt{Y}$. In arxiv.org/pdf/quant-ph/0512058.pdf they argue that gate alone is universal. $\endgroup$
    – DaftWullie
    Nov 14, 2023 at 9:10
  • $\begingroup$ @DaftWullie I see. Thank you very much for your comment. Naively, I would have never guessed {Controlled-$\sqrt{Y}$} would be universal. Very interesting! $\endgroup$
    – FDGod
    Nov 14, 2023 at 15:50
  • $\begingroup$ I think it is controlled- $\sqrt{R_{Y}}$ and not controlled-$\sqrt{𝑌}$ as pointed out by @DaftWullie. $\endgroup$
    – R.G.J
    Nov 14, 2023 at 19:44

2 Answers 2

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I think the authors haven't tried to prove it, hence the formulation.

In fact, it is simple to see that $CX$ and $R_z(\theta)$ is not universal as both gates map computational basis states to computational basis states, up to a phase. An analogous argument applies to $CX$ and $R_x(\theta)$ by noting that $CX$ also permutes the $X$ eigenbasis.

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There is another way to see that CNOT+ $R_Y(\theta)$ is universal while CNOT+ $R_X(\theta)$ or CNOT+ $R_Z(\theta)$ is not. The general mathematical result says that one has to satisfy Theorem 3.1 of https://arxiv.org/abs/quant-ph/0205115 for universal gate set with CNOT. Theorem reads - 'CNOT + any single qubit real gate that is basis changing after squaring is universal'.

One can show that only CNOT+ $R_Y(\theta)$ satisfies this condition.

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