The authors of the paper Graph States for Quantum Secret Sharing https://journals.aps.org/pra/abstract/10.1103/PhysRevA.78.042309 define a 'labeled state' as $$|G_{\vec{l}}\rangle=\bigotimes_{i}X_i^{l_{i1}}Z_i^{l_{i2}}|G\rangle$$where G is a graph state. Then they define an 'encoded graph state' as $$ |G_{\vec{l}_{*2}}\rangle=\bigotimes_{i}Z_i^{l_{i2}}|G\rangle $$ and say that the encoded graph state is the labeled graph state with $l_{1i}=0, \forall i$.
Here $\vec{l}_{i*}=(l_{i1},l_{i2})$ for the i-th vertex, $\vec{l}_{*j}=(l_{1j},l_{2j},....l_{nj})$ for the $j$ th bit over all the $n$ vertices, and $\vec{l}=(\vec{l}_{1*}, \vec{l}_{2*},........\vec{l}_{n*}$, each $l_{ij}\in \{0,1\}$.
With this framework they try to define a secret sharing protocol. However before the actual protocol, they give an example. I am understanding bits of the theory but not all of it, Here it is
- For a 4-party scheme, they prepare a graph state $$Z^{l_{12}}\otimes Z^{l_{22}}\otimes Z^{l_{32}}\otimes Z^{l_{42}} \left(|0+++\rangle+|1---\rangle\right)$$, corresponding to this they have 4 stabilizers (generators) of the stabilizer group $K_1=XZZZ,K_2=ZXII,K_3=ZIXI,K_4=ZIIX$, with eigenvalues $(l_{12},l_{22},l_{32},l_{42})$.
My first understanding about the eigenvalues is that $l_{i2}$ are eigenvalues because those are the indices (powers) where we have possible commuting issues between the $X$ and $Z$ gate, so when they are zero the eigenvalues are $1$ and when they are $1$ they eigenvalues are $-1$, and correspondingly the the graph states are orthogonal (since they correspond to different eigenvalues). Is my undertanding okay or atleast near okay? Shouldn't the eigenvalues be $(-1)^{l_{i2}} \forall~i$?
