How is the connection between Bures fidelity and quantum Fisher information derived?

Let $ rho = sum_n rho_n |psi_n rangle langle psi_n | $ be the eigendecomposition of $rho$. We will calculate everything in terms of $ |psi_n rangle$ basis.

Note that $ frac{d rho_theta}{d theta} = i [rho_theta, A] $ and $ frac{d^2 rho_theta}{d theta^2} = -big[[rho_theta, A], Abig] $.

Now we write $$ sqrt{ sqrt{rho} rho_theta sqrt{rho} } = rho + theta cdot X + theta^2 cdot Y + O(theta^3) $$ for some matrices $ X, Y $ to be determined.

Squaring the above equation we get $$ sqrt{rho} rho_theta sqrt{rho} = rho^2 + theta cdot (X rho + rho X) + theta^2 cdot (X^2 + Y rho + rho Y) + O(theta^3) $$ This means that:

• $ X rho + rho X = sqrt{rho} frac{d rho_theta}{d theta}Big|_{theta=0} sqrt{rho} = i sqrt{rho} [rho, A] sqrt{rho} implies X_{nm} = i frac{sqrt{rho_n rho_m}}{rho_n + rho_m} langle psi_n | [rho, A] |psi_m rangle = i frac{sqrt{rho_n rho_m}}{rho_n + rho_m} (rho_n - rho_m) langle psi_n | A |psi_m rangle $

and $ text{Tr}[X] = sum_n X_{nn} = 0 $.

Similarly:

• $ Y rho + rho Y + X^2 = sqrt{rho} frac{d^2 rho_theta}{d theta^2}Big|_{theta=0} sqrt{rho} = -sqrt{rho} Big[[rho, A], ABig] sqrt{rho} implies Y_{nm} = - frac{(X^2)_{nm}}{rho_n + rho_m} - frac{sqrt{rho_n rho_m}}{rho_n + rho_m} langle psi_n | Big[[rho, A], ABig] |psi_m rangle $

and $ text{Tr}[Y] = sum_n Y_{nn} = - sum_n frac{(X^2)_{nn}}{2rho_n} - frac{1}{2} text{Tr}Big[big[[rho, A], Abig]Big] = - sum_n frac{(X^2)_{nn}}{2rho_n} = - sum_{n,k} frac{X_{nk}X_{kn}}{2rho_n} = - frac{1}{2} sum_{n,k} frac{rho_k}{(rho_n + rho_k)^2} (rho_n - rho_k)^2 |langle psi_n | A |psi_k rangle|^2 = - frac{1}{4} sum_{n,k} frac{rho_k}{(rho_n + rho_k)^2} (rho_n - rho_k)^2 |langle psi_n | A |psi_k rangle|^2 - frac{1}{4} sum_{n,k} frac{rho_n}{(rho_n + rho_k)^2} (rho_n - rho_k)^2 |langle psi_n | A |psi_k rangle|^2 = - frac{1}{4} sum_{n,k} frac{(rho_n - rho_k)^2}{rho_n + rho_k} |langle psi_n | A |psi_k rangle|^2 = -frac{1}{8} F_Q[rho, A] $.

Finally: $ F_B(rho, rho_theta) = Big(text{Tr}big[sqrt{ sqrt{rho} rho_theta sqrt{rho} } big]Big)^2 = Big(text{Tr}rho + theta^2 cdot text{Tr}Y + O(theta^3) Big)^2 = Big(1 - theta^2 frac{F_Q[rho, A]}{8} + O(theta^3) Big)^2 = 1 - theta^2 frac{F_Q[rho, A]}{4} $