Why can't Quantum Fisher Information be negative?

Question

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

Answer 1

An easy way to see it is that the QFI can be written as $F(\theta)=\operatorname{tr}(\rho_\theta L_\theta^2)$, given some parameterization $\theta\mapsto\rho_\theta$, with $L_\theta$ the symmetric logarithmic derivative (SLD), defined by the relation $\partial_\theta \rho_\theta=\frac12\{\rho_\theta,L_\theta\}$.

Both the SLD and $\rho_\theta$ are positive semidefinite, and thus $F(\theta)\ge0$.

Another possible approach is to remember that the QFI is the maximum of the classical Fisher information over choices of POVMs, and the classical Fisher information is easily seen to be positive from its definition: $$I_\mu(\theta) = \mathbb{E}\left[(\partial_\theta\log p_\mu(X|\theta)^2\right] \equiv\sum_x \frac{(\partial_\theta p_\mu(x|\theta))^2}{p_\mu(x|\theta)},$$ where here $x$ denote the possible outcomes and $p_\mu(x|\theta)\equiv \langle\mu_x,\rho_\theta\rangle\equiv\operatorname{tr}(\mu_x \rho_\theta)$, for some POVM $\mu:x\mapsto \mu_x$.

The multiparameter case can be handled similarly.

Answer 2

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 $\mathcal{F} \ge0$.