I have a state $$\dfrac{1}{2}(|00000\rangle+|00111\rangle+|11101\rangle+|11010\rangle).$$ How does one create this state? In general, how does one create for instance an $n$-bit cluster state, is there any particular rule? And by looking at a particular cluster state, how can one construct the quantum circuit that created the state?

  • $\begingroup$ Do you know the underlying graph? if you do one path would be to go from graph to stabilizer code (simple construction) and from that construct an encoder for that stabilizer code (well known process but requires some work) $\endgroup$
    – unknown
    Jun 28, 2020 at 18:23

1 Answer 1


You can think of a cluster state as a graph state, where the graph's vertices are on some $d$-dimensional lattice (normally just $2$-dimensional). Each vertex represents a qubit in the $|+\rangle$ state, and two vertices that are next to each other on the lattice may be connected, which means that a $CZ$ or controlled-$Z$ gate has been applied to them.

Within this picture, it's relatively straightforward to prepare a cluster state. For a cluster state that is equivalent to some graph with vertices $V = (1,2,3...n)$ and edges $\{a,b\} \in E$, where $E$ is the set of all connected edges and obvsiouly $E$ is a subset of $V^{2}$. Starting from $n$ qubits in the $|0\rangle$ state:

  • Apply a Hadamard operation $H$ to all $n$ qubits

  • Apply a controlled-$Z$ operation to all pairs $\{a,b\}$ of qubits if $\{a,b\} \in E$

This results in that particular cluster state. Note that there can be multiple different cluster states for a particular number of qubits.

Concerning the particular state that you wrote down; I am not sure if this is a cluster state. The operators $ZZIII$ and $IIZIZ$ are in the stabilizer of this state, meaning that the state is a $+1$ eigenstate of both these operators. However, the stabilizer of a graph state can easily be calculated from the graph, and all the operators in the stabilizer need to have at least one $X$ or $Y$ Pauli in them, which does not hold for the above two operators.

If you're up for it, this paper contains a very detailed and thorough introduction to graph states, of which cluster states are a strict subset. It is a tough read though, so be prepared:)

  • 2
    $\begingroup$ One might also point out that for a graph/cluster state of $n$ qubits, the amplitude for every basis vector is $\pm1/\sqrt{2^n}$, so the given state is obviously not a graph/cluster state by the standard definition of these things. $\endgroup$
    – DaftWullie
    Jun 30, 2020 at 7:39
  • $\begingroup$ That's a good point, especially when you're accustomed to writing down states instead of stabilizers. Thanks for pointing that out! $\endgroup$
    – JSdJ
    Jun 30, 2020 at 9:29

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