0
$\begingroup$

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

$\endgroup$
0

1 Answer 1

0
$\begingroup$

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 $\mathcal{F} \ge0$.

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$ Feb 27 at 18:19
  • 1
    $\begingroup$ @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. $\endgroup$ Mar 2 at 13:53
  • 1
    $\begingroup$ @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. $\endgroup$ Mar 2 at 13:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.