# How can one check whether a given quantum state is a graph state?

We can build a quantum state from a graph, which is a mathematical concept. But, vice versa, how can one check whether or not a given quantum state is a graph state?

• You might want to have a look at stabilizer testing which can probably be adapted for graph states ... See e.g. Sec. 1.3 in [Gross et al. CMP 2021], arxiv.org/abs/1712.08628 Apr 12, 2021 at 8:03

Remember that a graph state is simply the $$|+\rangle$$ state on every qubit together with a bunch of controlled phases enacted between them. So, assuming you have a list of the probability amplitudes of your state, you first check that, if there are $$n$$ qubits, every amplitude is $$\pm1/\sqrt{2^n}$$.

Once you have done this, you need to determine the pattern of controlled phases. This is easy. Find the amplitude of a term $$x$$ which is all 0s except for two 1s. The sign of that amplitude tells you whether $$(-1)$$ or not $$(+1)$$ a controlled phase gate was applied between that particular pair of qubits. So, go through every possible pair of qubits, determine the controlled-phase gates. Then you just have to verify whether all the other $$\pm$$ signs on the amplitudes are compatible with that assignment.

• But how to explain that GHZ state is a kind of Graph state? "Here we report the creation of two special instances of graph states, the six-photon Greenberger-Horne-Zeilinger states". references Apr 10, 2021 at 12:39
• @ZhaoyiZhou The GHZ state is not a graph state, strictly speaking, because you cannot obtain it by the process described in the answer. However, the GHZ state (for any number of qubits) is locally equivalent to a particular graph state - there exist a set of single-qubit operations/rotations that when you perform them on the qubits of a GHZ state, you get a graph state. Moreover, these operations are Clifford operations, making the GHZ state LC-equivalent (for Local Clifford).
– JSdJ
Apr 10, 2021 at 13:25
• As @JSdJ says, the GHZ state is only locally equivalent to a graph state. It is not a graph state itself. Identifying if a state is locally equivalent to a graph state is, I suspect, a much harder question. Apr 11, 2021 at 12:16
• @DaftWullie Well, any stabiliser state is locally Clifford equivalent to a graph state and vice versa. If you ask about local unitary equivalence, then this is basically answered by entanglement theory. Apr 12, 2021 at 7:56
• @MarkusHeinrich I'm curious to know if there's an algorithm to go from a stabilizer state to a locally Clifford equivalent graph state. I just posted a question on this. Nov 4, 2023 at 19:32

If you want to find the answer experimentally, a possible (not necessarily optimal) way to do so is the following. Note that a graph state is stabilized by the generators $$g_v=X_v \prod_{w\in N(v)} Z_w,$$ where $$v$$ is a vertex of the graph and $$N(v)$$ is its neighborhood. You can find the neighborhood of a vertex by measuring $$X$$ on qubit $$v$$ and $$Z$$ on all others. The outcomes of the neighborhood will be perfectly correlated with the $$X$$-outcome, because $$\langle \psi | g_v |\psi \rangle = 1$$. This can be repeated for each qubit. Probably this could be done more efficiently, but at least this should be better than full tomography.