What is the stabilizer group of a $|W\rangle$ state?

Question

This question is related and complementary to this one: How to get the stabilizer group for a given state?

What I want is to find the stabilizer group generators for the following state:

$$|W\rangle = \dfrac{1}{\sqrt{3}}\Big(|011\rangle + |101\rangle + |110\rangle \Big)$$

In theory, I should find $n-k = 3-0=3$ independent non trivial generators. But the only one I can find is $M_1 = Z\otimes Z\otimes Z$ because any other combination, like $-Z\otimes Z\otimes -Z$ or $iZ\otimes iZ\otimes -Z$ actually is equivalent to the first one, and there cannot be a combination with $X$ because it would alter the difference between 0s and 1s that is conserved in each sum.

Where are the other two generators?

Answer 1

To cite from my answer from over at physics.SE: The W state is not a stabilizer state - for a stabilizer state, the 1-site reduced density matrices must be maximally mixed or pure, which they aren't.

Or, to phrase it without reduced density matrices: For a stabilizer state, if you measure $X$, $Y$, or $Z$ for any single qubit, the probability of getting either outcome is either $0$, $1/2$, or $1$. This is clearly not the case for a $Z$ measurement on the W state above, where the probability of getting $0$ is $1/3$.

Answer 2

As stated in the Schuch's answer: the W state is not a stabilizer state.

We can also find some operators which can "stabilizer" $W_{3}$. You can refer to Entanglement detection in the stabilizer formalism Page 9 Eq. 59 for $W_{3}$, and Efficient estimation of multipartite quantum coherence Eq.A.8 - A.11 in Appendix A for $W_{3}$ and $W_{4}$. You can find “-ZZZ” is inside after calculating.