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The stabilizers of a given graph all commute, thus it must be possible to diagonalize them simultaneously. If I start with one graph state and write down all its stabilizers is there an easy way to derive a complete eigenbasis of the stabilizers in terms of graph states?

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If you start with one graph state, which is an eigenstate of stabilizer (each of which comprises an X tensored with a bunch of Zs), then the other eigenstates of those stabilizers are the original state acted on by Z rotations (take all possible combinations). To see this note the commutation and anticommutation relations between a tensor product of Zs and each stabilizer.

Consider a stabilizer of the form $K_n=X_nZ_iZ_j\ldots Z_k$, i.e. a graph-state stabilizer with a Pauli $X$ on qubit $n$, and a bunch of $Z$s on some other qubits (related to the graph structure). By definition, the graph state satisfies $$ K_n|\psi\rangle=|\psi\rangle $$ for all $n$. Now, let $$ |\Psi_x\rangle=\left(\bigotimes_{i=1}^nZ^{x_i}\right)|\psi\rangle, $$ for $x\in\{0,1\}^n$, a binary string. If we apply the stabilizers to such a state, we get $$ K_n|\Psi_x\rangle=K_n\left(\bigotimes_{i=1}^nZ^{x_i}\right)|\psi\rangle=\left\{\begin{array}{cc} \left(\bigotimes_{i=1}^nZ^{x_i}\right)K_n|\psi\rangle & x_n=0 \\ -\left(\bigotimes_{i=1}^nZ^{x_i}\right)K_n|\psi\rangle & x_n=1 \end{array}\right. $$ Thus, every $|\Psi_x\rangle$ is an eigenstate of every stabilizer, and with a different pattern of $\pm 1$ eigenvalues (and must therefore be mutually orthogonal).

While it must be the case that the different states $|\Psi_x\rangle$ are orthogonal, I think it's useful to think about this in terms of the way that the graph states are created. Remember that $|\psi\rangle$ is created from qubits in the $|+\rangle$ state and then have controlled-phase gates applied between them. Now, if I evaluate $\langle\Psi_y|\Psi_x\rangle$, think of this as a calculation $$ \langle+|^{\otimes n}CP\cdot Z_{x\oplus y}\cdot CP|+\rangle^{\otimes n} $$ where $CP$ represents the collection of controlled phase gates. Pauli $Z$ commutes with controlled phase, which means we can bring the two CPs together and annihilate them, leaving behind $$ \langle+|^{\otimes n} Z_{x\oplus y}|+\rangle^{\otimes n}. $$ Given that $x\neq y$, there is at least one $Z$, and we know $\langle +|Z|+\rangle=\langle -|+\rangle=0$.

This way of thinking gives another way of describing the basis of graph states: prepare each qubit in either $|+\rangle$ or $|-\rangle$, and combine with controlled-phases. Each different choice of $\pm$ gives a different graph state in the basis.

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  • $\begingroup$ Ok, this is cool as independently I got more or less the same solution, although it took me much longer than you :-) $\endgroup$ – sycramore Apr 3 at 16:55
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I'm not sure I understand the question, since this seems quite straightforward.

Graph states are Clifford states, so for a state on $n$ qubits, the set of stabilizers has $n$ generators looking like

$$i^k P_1\otimes\ldots\otimes P_n$$

where $k\in\{0,1,2,3\}$ and the $P$'s are Pauli $X,Y,$ or $Z$ operators. (For graph states, these stabilizers look like $\{X_v Z_{\text{Nbhd}(v)} : v \in V\}$ )

Each Pauli has two eigenvectors. If we choose one of them and call it $|\psi_i\rangle$, and call its corresponding eigenvalue $\lambda_i$, for each $P_i$, then $|\psi_1\rangle\otimes\ldots\otimes|\psi_n\rangle$ will be an eigenvector of the stabilizer with eigenvalue $i^k\prod_i \lambda_i$.

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