Let $G = (V, E)$ be a $\sqrt{n} \times \sqrt{n}$ grid graph. Consider preparing a 2D cluster state as follows.

$$|\psi_{\mathsf{C}}\rangle = \underset{(i, j) \in E}{\Pi} (\mathsf{CZ})_{i, j} | + \rangle^{\otimes n},$$

where $\mathsf{CZ}$ is the controlled-$\mathsf{Z}$ operator acting on qubits $i$ and $j$. We know that this state is a universal resource for measurement based quantum computing (using postselection, we can implement any $\mathsf{BQP}$ circuit with this state.)

Let's say we generalize this state.

Let's say for a particular angle $\theta$, I replace $\mathsf{CZ}$ with the following:

$$|\psi_{\mathsf{C, \theta}}\rangle = \underset{(i, j) \in E}{\Pi} e^{i \theta~ \mathsf{Z}_i \mathsf{Z}_j} | + \rangle^{\otimes n},$$

where $\mathsf{Z}_i$ and $\mathsf{Z}_j$ are $n$ qubit Pauli-$\mathsf{Z}$ operators each, acting non-trivially on qubits $i$ and $j$ respectively.

Does $|\psi_{\mathsf{C, \theta}}\rangle$ also remain a universal resource for measurement based quantum computing, for any value of $\theta >0$?

  • $\begingroup$ There is quite some work on "computational phases of matter" (by Raussendorf, Tzu-Chieh Wei, David Stephen, and others). Among others, this asks how much you can deform the cluster state to still have a resource for MBQC. One key insight is that this is related to certain symmetries (subsystem symmetries). I don't think your deformation respects them, so I suspect the answer is no, but I didn't check carefully. -- But that literature might be a good place to check. $\endgroup$ Sep 4, 2022 at 11:32

1 Answer 1


I think that a generalized 2D weighted graph state is a resource with $\textbf{specific}$ values in angles $\theta$ of the entangling gates. Not any pattern of weighted graph state is a universal resource. This paper may give insights to your question. https://arxiv.org/abs/1704.06504


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