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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?

Adam Zalcman
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Upstart
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1 Answers1

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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:)

JSdJ
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