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

I recently came to know that there is a connection between Bures Fidelity $$(F_B)$$ and Quantum Fisher Information $$(F_Q)$$ given by

$$[F_{B}(\rho, \rho_\theta)]^2 = 1 - \frac{\theta^2}{4} F_Q[\rho, A] + O(\theta^3),$$ where $$\rho_\theta = \exp(-iA\theta)\rho \exp(iA\theta)$$ and $$F_B(\rho_1, \rho_2) = \mathrm{Tr}\Big[ \sqrt{\sqrt{\rho_1}~ \rho_2~ \sqrt{\rho_1}}\Big]$$.

How can one derive this result?

Here is the link to my source

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 $$\frac{d^2 \rho_\theta}{d \theta^2} = -\big[[\rho_\theta, A], A\big]$$.

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:

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

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

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

• Are there some references for your deduction(or is there some deduction from some papers)? Feb 17 at 11:57