# Constructing an eigenbasis of graph states for a set of stabilizers

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?

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.

• Ok, this is cool as independently I got more or less the same solution, although it took me much longer than you :-) – sycramore Apr 3 at 16:55

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 $$i^k\prod_i \lambda_i$$.