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?

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    $\begingroup$ 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 $\endgroup$ Apr 12, 2021 at 8:03

2 Answers 2


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.

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    $\begingroup$ 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 $\endgroup$
    – narip
    Apr 10, 2021 at 12:39
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    $\begingroup$ @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). $\endgroup$
    – JSdJ
    Apr 10, 2021 at 13:25
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    $\begingroup$ 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. $\endgroup$
    – DaftWullie
    Apr 11, 2021 at 12:16
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    $\begingroup$ @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. $\endgroup$ Apr 12, 2021 at 7:56
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    $\begingroup$ @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. $\endgroup$
    – unknown
    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.


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