Stabilizer state QFI lower limit query

Question

On page 1 of this paper it states that the QFI (Quantum Fisher Information) for pure states $\psi$ is $$\mathcal{Q}(\psi) = \sum_{i,j=1}^n\text{Tr}(X_iX_j\psi)-\text{Tr}(X_i \psi)\text{Tr}(X_j \psi)~~~~~~~~~~(3)$$ Further down it states:

It is clear from Eq. (3) that if the generators are chosen from the Pauli group such that there are no stabilizers of the form $\pm X_i$ or $-X_iX_j$, then the QFI of the stabilizer state is equal to the number of stabilizers of the form $X_iX_j$.

How does this conclusion follows from equation (3)? What I get as a start is that if $X_iX_j$ are stabilizers then $$\sum_{i,j=1}^n[\text{Tr}(X_iX_j\psi)-\text{Tr}(X_i \psi)\text{Tr}(X_j \psi)] = \sum_{i,j=1}^n[1-\text{Tr}(X_i \psi)\text{Tr}(X_j \psi)]$$

Answer 1

The state $\psi$ (this is denoting the density matrix, even though it's a pure state) can be described as a sum of all the products of the stabilizers. We are promised that $X_i$ is not in the stabilizer, so every term in the sum of $\psi$, when multiplied by $X_i$, returns a tensor product of terms that is not just identity. Hence, it has zero trace. Thus, $\text{Tr}(X_i\psi)=0$, and you are just left with $\sum 1$, where the sum is taken over all stabilizers of the form $X_iX_j$.