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I'm reading about cluster states using this online source. It is explained that a CNOT can be performed on cluster states using as little as $4$ qubits. However, the standard implementation is with $15$ qubits. Why is that the case?

Is it also possible with any number of qubits beyond $4$? So can I implement a CNOT in a cluster state with $5$ qubits for instance, and if yes, how?

glS
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nippon
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1 Answers1

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In answer to your second question, remember that you can always add any extra vertices (with arbitrary edges) to your graph. You just have to remove those vertices by using a $Z$ measurement on those qubits. It's kind of a cheat, which you perhaps didn't intend....

I think the reason for the "standard" implementation is that this is something that you can build out of the standard cluster state formed on the 2D square lattice, where the minimal one is not. Note that to have your separate qubit lines running from left to right, they have to be separated by another row that you (mostly) apply $Z$-measurements to. So the vertical line has two edges. I assume most of the extra qubits on the horizontal lines are there to compensate for the effect of those two vertical edges instead of one.

DaftWullie
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