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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 $$\sum_{i,j=1}^n[\text{Tr}(X_iX_j\psi)-\text{Tr}(X_i \psi)\text{Tr}(X_j \psi)] = \sum_{i,j=1}^n[1-\text{Tr}(X_i \psi)\text{Tr}(X_j \psi)]$$

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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 $X_iX_j$.

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  • $\begingroup$ Thanks for your response. Could you maybe elaborate on the reasoning behind your conclusion "Hence, it has zero trace". Are you stating that $X_i\psi \neq I \implies \text{Tr}(X_i \psi) = 0$? $\endgroup$
    – John Doe
    Oct 18, 2019 at 13:17
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    $\begingroup$ @JohnDoe Not exactly - I'm saying that since $X_i\psi$ does not have an $I$ term in its sum, that implies that trace is 0. $\endgroup$
    – DaftWullie
    Oct 18, 2019 at 14:35
  • $\begingroup$ Okay thanks, I'm just trying to find out what mathematical result you are using to come to the final conclusion. Are you stating the following: We can write $X_i\psi$ as $$X_i\psi = X_i\bigg[2^{-n}\sum_{s \in \mathcal{S}}s\bigg] = 2^{-n}\sum_{s \in \mathcal{S}}X_is,$$where $\mathcal{S}$ is the stabilizer group, then since $X_is \neq I~~\forall s \in \mathcal{S} \implies \text{Tr}[X_i\psi] = 0?$ $\endgroup$
    – John Doe
    Oct 18, 2019 at 16:44
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    $\begingroup$ @JohnDoe yes, coupled with the fact that all members of the Pauli group, except I, are traceless $\endgroup$
    – DaftWullie
    Oct 19, 2019 at 5:34
  • $\begingroup$ Is it clear to you why it is assumed that there are no stabilizers of the form $-X_iX_j$, it doesn't seem to be required for the statement that the QFI of the stabilizer state is equal to the number of stabilizers of the form $X_iX_j$. It seems that we only need the assumption that there are no stabilizers of the form $\pm X_i$. $\endgroup$
    – John Doe
    Jan 18, 2020 at 10:55

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