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Quantum Fisher Information is proportional to Fidelity susceptibility.

Mathematically the equation is:

$QFI=-\frac{\partial^2 d_B(\epsilon) }{\partial \epsilon^2}$

where above equation shows QFI is equal to second derivative of ($d_B$) Bures Distance wrt to the parameter $\epsilon$. For simplicity let us consider pure states.

$d_B=2(1-\sqrt{F})$

where $F$ is Fidelity. The Bures Distance is just replaced with fidelity to connect QFI to some distance measure and nothing is lost.

Now my question is Bures distance is not a monotonically decreasing function of the parameter (\epsilon). Then why is QFI always positive ? It is infact oscillatory for unitary evolutions. Then the QFI can turn out to be positive as well as negative.

Why do we say that Quantum Fisher Information is always positive then ?

Links 1, 2, 3

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In your link for the quantum fisher information matrix(QFIM), it states that $$ D_{\mathrm{B}}^2(\rho(\vec{x}), \rho(\vec{x}+\mathrm{d} \vec{x}))=\frac{1}{4} \sum_{\mu \nu} \mathcal{F}_{\mu \nu} \mathrm{d} x_\mu \mathrm{d} x_\nu \tag{1} $$ where $\mathcal{F} $ stands for QFIM, $D_{\mathrm{B}}^2\left(\rho_1, \rho_2\right)=2-2 f\left(\rho_1, \rho_2\right)$ is the bures distance and $f\left(\rho_1, \rho_2\right):=\operatorname{Tr} \sqrt{\sqrt{\rho_1} \rho_2 \sqrt{\rho_1}}$ is the fidelity.

For quantum fisher information(QFI), eq(1) becomes $$D_{\mathrm{B}}^{2}(\rho (\vec{x}),\rho (\vec{x}+\mathrm{d}\vec{x}))=\frac{1}{4}\mathcal{F} \mathrm{d}{x_{\mu}}^2 $$ and $\mathcal{F} $ is QFI now. Easy to see $D_{\mathrm{B}}^{2}\ge0$ hence we have $\mathcal{F} \ge0$.

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  • $\begingroup$ but you see the last equation is d_B^2=Fdx^2/4. Hence, F=4*d_B^2/dx^2 no this quantity need not be always positive. For example let d_B^2=Cos(x)^2 this is always positive, but second derivative of it will not always be positive. $\endgroup$ Commented Feb 27, 2023 at 18:19
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    $\begingroup$ @ChetanWaghela in this comment you are confusing derivatives with distances. $D_B$ is a quantity, not a derivative, and $dx_\mu$ is a differential (note narip was even extra careful to make the $d$ non-italicized). Fisher information is given here as the ratio of two squares, not as a second derivative. $\endgroup$ Commented Mar 2, 2023 at 13:53
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    $\begingroup$ @ChetanWaghela other definitions for QFI make its positivity more clear. One easy way is to show that Fisher information is always positive (expectation value of a positive quantity) and then show QFI to be an upper bound for Fisher information. FI is $\langle (\partial \log p/\partial \theta)^2\rangle$, clearly positive, which under certain regularity conditions can be rewritten as $-\langle \partial^2 \log p/\partial \theta^2\rangle$; the regularity conditions guarantee that the latter is always positive, even though it is not obvious from the formula. $\endgroup$ Commented Mar 2, 2023 at 13:57

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