$\partial_{x_{a}}\rho=\frac{1}{2}\left(\rho L_{x_{a}}+L_{x_{a}}\rho\right),$
(3)

one can easily obtain

$\langle\lambda_{i}|\partial_{x_{a}}\rho|\lambda_{j}\rangle=\frac{1}{2}\langle\lambda_{i}|L_{x_{a}}|\lambda_{j}\rangle\left(\lambda_{i}+\lambda_{j}\right).$
(4)

The derivative of $\rho$ reads
$\partial_{x_{a}}\rho=\sum^{N-1}_{i=0}\partial_{x_{a}}\lambda_{i}|\lambda_{i}\rangle\langle\lambda_{i}|+\lambda_{i}|\partial_{x_{a}}\lambda_{i}\rangle\langle\lambda_{i}|+\lambda_{i}|\lambda_{i}\rangle\langle\partial_{x_{a}}\lambda_{i}|$,
which gives
$\langle\lambda_{i}|\partial_{x_{a}}\rho|\lambda_{j}\rangle=\partial_{x_{a}}\lambda_{i}\delta_{ij}+\lambda_{j}\langle\lambda_{j}|\partial_{x_{a}}\lambda_{i}\rangle+\lambda_{i}\langle\partial_{x_{a}}\lambda_{j}|\lambda_{i}\rangle.$
$\delta_{ij}$ is the Kronecker delta function ($\delta_{ij}=1$ for $i=j$ and
zero otherwise). Substituting this expression into above equation,
$\langle\lambda_{i}|L_{x_{a}}|\lambda_{j}\rangle$ can be calculated as

$\langle\lambda_{i}|L_{x_{a}}|\lambda_{j}\rangle\\!=\\!\\!\Bigg{\\{}\begin{array}[]{cc}\\!\\!\\!\delta_{ij}\frac{\partial_{x_{a}}\lambda_{i}}{\lambda_{i}}\\!+\\!\frac{2(\lambda_{j}-\lambda_{i})}{\lambda_{i}+\lambda_{j}}\\!\langle\lambda_{i}|\partial_{x_{a}}\lambda_{j}\rangle,\\!\\!\\!&\\!\\!i\,\mathrm{or}\,j\\!\in\\![0,N\\!-\\!1];\\\
\mathrm{arbitrary}~{}\mathrm{value},&i,j\in[N,\dim\rho-1].\\\
\end{array}\\!\\!$ (5)

With the solution of SLD, we can now further calculate the QFIM. Utlizing
equation (1), the QFIM can be rewritten into

$\mathcal{F}_{ab}=\frac{1}{2}\sum^{N-1}_{i=0}\lambda_{i}\left(\langle\lambda_{i}|L_{x_{a}}L_{x_{b}}|\lambda_{i}\rangle+\langle\lambda_{i}|L_{x_{b}}L_{x_{a}}|\lambda_{i}\rangle\right).$
(6)

Inserting the equation $\leavevmode\hbox{\small 1\kern-4.3pt\normalsize
1}=\sum^{\dim\rho-1}_{i=0}|\lambda_{i}\rangle\langle\lambda_{i}|$ into above
equation, one can obtain

$\mathcal{F}_{ab}=\sum^{N-1}_{i=0}\sum^{\dim\rho-1}_{j=0}\lambda_{i}\mathrm{Re}(\langle\lambda_{i}|L_{x_{a}}|\lambda_{j}\rangle\langle\lambda_{j}|L_{x_{b}}|\lambda_{i}\rangle).$
(7)

In this equation, it can be seen that the arbitrary part of
$\langle\lambda_{i}|L_{x_{a}}|\lambda_{j}\rangle$ does not affect the value of
QFIM since it is not involved in above equation, but it provides a freedom for
the optimal measurements, which will be further discussed later. Substituting
equation (5) into above equation, we have

$\displaystyle\mathcal{F}_{ab}$ $\displaystyle=$
$\displaystyle\sum^{N-1}_{i,j=0}\delta_{ij}\frac{(\partial_{x_{a}}\lambda_{i})(\partial_{x_{b}}\lambda_{i})}{\lambda_{i}}+\frac{4\lambda_{i}(\lambda_{i}-\lambda_{j})^{2}}{(\lambda_{i}+\lambda_{j})^{2}}\mathrm{Re}\left(\langle\lambda_{i}|\partial_{x_{a}}\lambda_{j}\rangle\langle\partial_{x_{b}}\lambda_{j}|\lambda_{i}\rangle\right)$
(8)
$\displaystyle+\sum^{N-1}_{i=0}\sum^{\dim\rho-1}_{j=N}4\lambda_{i}\mathrm{Re}\left(\langle\partial_{x_{a}}\lambda_{i}|\lambda_{j}\rangle\langle\lambda_{j}|\partial_{x_{b}}\lambda_{i}\rangle\right),$

where the fact
$\langle\lambda_{i}|\partial_{x_{a}}\lambda_{j}\rangle=-\langle\partial_{x_{a}}\lambda_{i}|\lambda_{j}\rangle$
has been applied. The third term in above equation can be further calculated
as

$\displaystyle\sum^{N-1}_{i=0}\sum^{\dim\rho-1}_{j=N}4\lambda_{i}\mathrm{Re}\left(\langle\partial_{x_{a}}\lambda_{i}|\lambda_{j}\rangle\langle\lambda_{j}|\partial_{x_{b}}\lambda_{i}\rangle\right)$
(9) $\displaystyle=$
$\displaystyle\sum^{N-1}_{i=0}4\lambda_{i}\mathrm{Re}\left(\langle\partial_{x_{a}}\lambda_{i}|\left(\sum^{\dim\rho-1}_{j=N}|\lambda_{j}\rangle\langle\lambda_{j}|\right)|\partial_{x_{b}}\lambda_{i}\rangle\right)$
$\displaystyle=$
$\displaystyle\sum^{N-1}_{i=0}4\lambda_{i}\mathrm{Re}\left(\langle\partial_{x_{a}}\lambda_{i}|\left(\leavevmode\hbox{\small
1\kern-4.3pt\normalsize
1}-\sum^{N-1}_{j=0}|\lambda_{j}\rangle\langle\lambda_{j}|\right)|\partial_{x_{b}}\lambda_{i}\rangle\right)$
$\displaystyle=$
$\displaystyle\sum^{N-1}_{i=0}4\lambda_{i}\mathrm{Re}\left(\langle\partial_{x_{a}}\lambda_{i}|\partial_{x_{b}}\lambda_{i}\rangle-\sum^{N-1}_{j=0}\langle\partial_{x_{a}}\lambda_{i}|\lambda_{j}\rangle\langle\lambda_{j}|\partial_{x_{b}}\lambda_{i}\rangle\right).$

Therefore equation (8) can be rewritten into

$\displaystyle\mathcal{F}_{ab}$ $\displaystyle=$
$\displaystyle\sum^{N-1}_{i,j=0}\delta_{ij}\frac{(\partial_{x_{a}}\lambda_{i})(\partial_{x_{b}}\lambda_{i})}{\lambda_{i}}+\frac{4\lambda_{i}(\lambda_{i}-\lambda_{j})^{2}}{(\lambda_{i}+\lambda_{j})^{2}}\mathrm{Re}\left(\langle\lambda_{i}|\partial_{x_{a}}\lambda_{j}\rangle\langle\partial_{x_{b}}\lambda_{j}|\lambda_{i}\rangle\right)$
$\displaystyle+\sum^{N-1}_{i=0}4\lambda_{i}\mathrm{Re}\left(\langle\partial_{x_{a}}\lambda_{i}|\partial_{x_{b}}\lambda_{i}\rangle\right)-\sum^{N-1}_{i,j=0}4\lambda_{i}\mathrm{Re}\left(\langle\partial_{x_{a}}\lambda_{i}|\lambda_{j}\rangle\langle\lambda_{j}|\partial_{x_{b}}\lambda_{i}\rangle\right).$

Exchange the subscripts $i$ and $j$ in the second and fourth terms of above
equation, it reduces into a symmetric form as shown in equation (2).

##  Appendix C QFIM in Bloch representation

Bloch representation is a well-used representation of quantum states in
quantum mechanics and quantum information theory. For a $d$-dimensional
quantum state $\rho$, it can be expressed by

$\rho=\frac{1}{d}\left(\leavevmode\hbox{\small 1\kern-4.3pt\normalsize
1}+\sqrt{\frac{d(d-1)}{2}}\vec{r}\cdot\vec{\kappa}\right),$ (10)

where $\vec{r}=(r_{1},r_{2}...,r_{k},...)^{\mathrm{T}}$ is the Bloch vector
($|\vec{r}|^{2}\leq 1$) and $\vec{\kappa}$ is a $(d^{2}-1)$-dimensional vector
of $\mathfrak{su}(d)$ generator satisfying $\mathrm{Tr}(\kappa_{i})=0$ and

$\displaystyle\\{\kappa_{i},\kappa_{j}\\}$ $\displaystyle=$
$\displaystyle\frac{4}{d}\delta_{ij}\leavevmode\hbox{\small
1\kern-4.3pt\normalsize 1}+\sum^{d^{2}-1}_{m=1}\mu_{ijm}\kappa_{m},$ (11)
$\displaystyle\left[\kappa_{i},\kappa_{j}\right]$ $\displaystyle=$
$\displaystyle i\sum^{d^{2}-1}_{m=1}\epsilon_{ijm}\kappa_{m},$ (12)

where $\delta_{ij}$ is the Kronecker delta function. $\mu_{ijm}$ and
$\epsilon_{ijm}$ are the symmetric and antisymmetric structure constants,
which can be calculated via the following equations

$\displaystyle\mu_{ijl}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\mathrm{Tr}\left(\\{\kappa_{i},\kappa_{j}\\}\kappa_{l}\right),$
(13) $\displaystyle\epsilon_{ijl}$ $\displaystyle=$
$\displaystyle-\frac{i}{2}\mathrm{Tr}\left([\kappa_{i},\kappa_{j}]\kappa_{l}\right).$
(14)

It is easy to find $\mathrm{Tr}(\kappa_{i}\kappa_{j})=2\delta_{ij}$ and
$\mu_{ijl}$ keeps still for any order of $ijl$, which will be used in the
following.

Now we express the SLD operator with the $\mathfrak{su}(d)$ generators as [43,
44]

$L_{x_{a}}=z_{a}\leavevmode\hbox{\small 1\kern-4.3pt\normalsize
1}+\vec{y}_{a}\cdot\vec{\kappa},$ (15)

where $z_{a}$ is a number and $\vec{y}_{a}$ is a vector. In the following we
only use $z$ and $\vec{y}$ for convenience. From the equation
$\partial_{x_{a}}\rho=\frac{1}{2}(\rho L+L\rho)$, one can find

$c\partial_{x_{a}}\vec{r}\cdot\vec{\kappa}=\frac{z}{d}\leavevmode\hbox{\small
1\kern-4.3pt\normalsize
1}+\frac{1}{d}\vec{y}\cdot\vec{\kappa}+cz\vec{r}\cdot\vec{\kappa}+\frac{c}{2}\\{\vec{r}\cdot\vec{\kappa},\vec{y}\cdot\vec{\kappa}\\},$
(16)

where $c=\sqrt{(d-1)/(2d)}$. Furthermore,

$\frac{c}{2}\\{\vec{r}\cdot\vec{\kappa},\vec{y}\cdot\vec{\kappa}\\}=\sum_{ij}r_{i}y_{j}\frac{c}{2}\\{\kappa_{i},\kappa_{j}\\}=\frac{2c}{d}\vec{y}\cdot\vec{r}\leavevmode\hbox{\small
1\kern-4.3pt\normalsize
1}+\frac{c}{2}\sum_{ijm}r_{i}y_{j}\mu_{ijm}\kappa_{m}.$ (17)

The coefficients of 11 in equation (16) from both sides have to be the same,
which gives

$z=-2c\vec{y}\cdot\vec{r}=-2c\vec{y}^{\,\mathrm{T}}\vec{r},$ (18)

which can also be obtained from the equation $\mathrm{Tr}(\rho L_{x_{a}})=0$.
Similarly, the coefficients of $\kappa_{i}$ in equation (16) from both sides
have to be the same, which gives

$c\partial_{x_{a}}r_{i}=\frac{y_{i}}{d}+czr_{i}+\frac{c}{2}\sum_{jm}\mu_{jmi}y_{m}r_{j}.$
(19)

This can also be obtained by multipling $\kappa_{i}$ on both sides of equation
(16) and then taking the trace. Now we introduce the matrix $G$ with the entry

$G_{ij}=\frac{1}{2}\mathrm{Tr}(\rho\\{\kappa_{i},\kappa_{j}\\}).$ (20)

It is easy to see $G$ is real symmetric. Substituting equation (11) into the
equation above, it can be further written into

$G_{ij}=\frac{2}{d}\delta_{ij}+c\sum_{m}\mu_{ijm}r_{m}.$ (21)

Hence, the last term in equation (19) can be rewritten into

$\displaystyle\frac{c}{2}\sum_{jm}\mu_{jmi}y_{m}r_{j}$ $\displaystyle=$
$\displaystyle\frac{c}{2}\sum_{m}y_{m}\sum_{j}\mu_{jmi}r_{j}$ (22)
$\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{m}y_{m}\left(G_{im}-\frac{2}{d}\delta_{im}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{m}G_{im}y_{m}-\frac{y_{i}}{d}.$

Substituting this equation into equation (19), one has
$c\partial_{x_{a}}r_{i}=czr_{i}+\frac{1}{2}\sum_{m}G_{im}y_{m}$, which
immediately leads to

$c\partial_{x_{a}}\vec{r}=cz\vec{r}+\frac{1}{2}G\vec{y}.$ (23)

Futhermore, one can check that

$(\vec{r}\,\vec{r}^{\,\mathrm{T}})\vec{y}=\left(\vec{y}^{\,\mathrm{T}}\vec{r}\right)\vec{r}=-\frac{1}{2c}z\vec{r}.$
(24)

Utilizing this equation, equation (23) reduces to

$c\partial_{x_{a}}\vec{r}=-2c^{2}\left(\vec{r}\,\vec{r}^{\,\mathrm{T}}\right)\vec{y}+\frac{1}{2}G\vec{y}.$
(25)

Therefore,

$\vec{y}=\left(\frac{1}{2c}G-2c\,\vec{r}\,\vec{r}^{\,\mathrm{T}}\right)^{-1}\partial_{x_{a}}\vec{r}.$
(26)

Next, we calculate the entry of QFIM and we will use the full notation $z_{a}$
and $\vec{y}_{a}$ instead of $z$, $\vec{y}$. The entry of QFIM reads

$\displaystyle\mathcal{F}_{ab}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\mathrm{Tr}(\rho\\{L_{a},L_{b}\\})$ (27)
$\displaystyle=$ $\displaystyle
z_{a}z_{b}+2c(z_{a}\vec{y}_{b}\cdot\vec{r}+z_{b}\vec{y}_{a}\cdot\vec{r})+\frac{1}{2}\sum_{ij}y_{a,i}y_{b,j}\mathrm{Tr}\left(\rho\\{\kappa_{i},\kappa_{j}\\}\right)$
$\displaystyle=$ $\displaystyle-
z_{a}z_{b}+\vec{y}^{\,\mathrm{T}}_{a}G\vec{y}_{b}.$

Based on equations (18) and (23), one can obtain

$c\vec{y}^{\,\mathrm{T}}_{a}\partial_{x_{b}}\vec{r}=cz_{b}\vec{y}^{\,\mathrm{T}}_{a}\vec{r}+\frac{1}{2}\vec{y}^{\,\mathrm{T}}_{a}G\vec{y}_{b}=-\frac{1}{2}z_{a}z_{b}+\frac{1}{2}\vec{y}^{\,\mathrm{T}}_{a}G\vec{y}_{b}.$
(28)

Hence, the QFIM can be written as

$\mathcal{F}_{ab}=2c\vec{y}^{\,\mathrm{T}}_{a}\partial_{x_{b}}\vec{r}=2c(\partial_{x_{b}}\vec{r})^{\mathrm{T}}\vec{y}_{a}.$
(29)

Utilizing equation (26), the QFIM can be finally written into

$\displaystyle\mathcal{F}_{ab}$ $\displaystyle=$ $\displaystyle
2c(\partial_{x_{b}}\vec{r})^{\mathrm{T}}\left(\frac{1}{2c}G-2c\,\vec{r}\,\vec{r}^{\,\mathrm{T}}\right)^{-1}\partial_{x_{a}}\vec{r}$
(30) $\displaystyle=$
$\displaystyle(\partial_{x_{b}}\vec{r})^{\mathrm{T}}\left(\frac{1}{4c^{2}}G-\vec{r}\,\vec{r}^{\,\mathrm{T}}\right)^{-1}\partial_{x_{a}}\vec{r}$
$\displaystyle=$
$\displaystyle(\partial_{x_{b}}\vec{r})^{\mathrm{T}}\left(\frac{d}{2(d-1)}G-\vec{r}\,\vec{r}^{\,\mathrm{T}}\right)^{-1}\partial_{x_{a}}\vec{r}.$

The theorem has been proved.

The simplest case here is single-qubit systems. The corresponding density
matrix is $\rho=\frac{1}{2}(\leavevmode\hbox{\small 1\kern-4.3pt\normalsize
1}_{2}+\vec{r}\cdot\vec{\sigma})$ with
$\vec{\sigma}=(\sigma_{1},\sigma_{2},\sigma_{3})$ the vector of Pauli matrices
and $\leavevmode\hbox{\small 1\kern-4.3pt\normalsize 1}_{2}$ is the
2-dimensional identity matrix. In this case, $G=\leavevmode\hbox{\small
1\kern-4.3pt\normalsize 1}_{3}$ ($\leavevmode\hbox{\small
1\kern-4.3pt\normalsize 1}_{3}$ is the 3-dimensional identity matrix) due to
the fact $\\{\sigma_{i},\sigma_{j}\\}=2\delta_{ij}\leavevmode\hbox{\small
1\kern-4.3pt\normalsize 1}_{2}$. Equation (30) then reduces to

$\mathcal{F}_{ab}=(\partial_{x_{b}}\vec{r})^{\mathrm{T}}\left(\leavevmode\hbox{\small
1\kern-4.3pt\normalsize
1}_{3}-\vec{r}\,\vec{r}^{\,\mathrm{T}}\right)^{-1}\partial_{x_{a}}\vec{r}.$
(31)

It can be checked that

$\left(\leavevmode\hbox{\small 1\kern-4.3pt\normalsize
1}_{3}-\vec{r}\,\vec{r}^{\,\mathrm{T}}\right)^{-1}=\leavevmode\hbox{\small
1\kern-4.3pt\normalsize
1}_{3}+\frac{1}{1-|\vec{r}|^{2}}\vec{r}\,\vec{r}^{\,\mathrm{T}}.$ (32)

The QFIM then reads

$\mathcal{F}_{ab}=(\partial_{x_{b}}\vec{r})^{\mathrm{T}}(\partial_{x_{a}}\vec{r})+\frac{\left[(\partial_{x_{a}}\vec{r})^{\mathrm{T}}\vec{r}\right]\left[(\partial_{x_{b}}\vec{r})^{\mathrm{T}}\vec{r}\right]}{1-|\vec{r}|^{2}},$
(33)

or equivalently,

$\mathcal{F}_{ab}=(\partial_{x_{a}}\vec{r})\cdot(\partial_{x_{b}}\vec{r})+\frac{(\vec{r}\cdot\partial_{x_{a}}\vec{r})(\vec{r}\cdot\partial_{x_{b}}\vec{r})}{1-|\vec{r}|^{2}}.$
(34)

##  Appendix D One-qubit basis-independent expression of QFIM

This appendix shows the proof of Theorem 2.6. We first prove that the QFIM for
one-qubit mixed state can be written as [46]

$\mathcal{F}_{ab}=\mathrm{Tr}\left[(\partial_{x_{a}}\rho)(\partial_{x_{b}}\rho)\right]+\frac{1}{\det\rho}\mathrm{Tr}\left[(\partial_{x_{a}}\rho-\rho\partial_{x_{a}}\rho)(\partial_{x_{b}}\rho-\rho\partial_{x_{b}}\rho)\right].$
(1)

Utilizing the spectral decomposition
$\rho=\lambda_{0}|\lambda_{0}\rangle\langle\lambda_{0}|+\lambda_{1}|\lambda_{1}\rangle\langle\lambda_{1}|$,
the first term reads

$\displaystyle\mathrm{Tr}[(\partial_{x_{a}}\rho)(\partial_{x_{b}}\rho)]$
$\displaystyle=$
$\displaystyle\langle\lambda_{0}|\partial_{x_{a}}\rho|\lambda_{0}\rangle\langle\lambda_{0}|\partial_{x_{b}}\rho|\lambda_{0}\rangle+\langle\lambda_{1}|\partial_{x_{a}}\rho|\lambda_{1}\rangle\langle\lambda_{1}|\partial_{x_{b}}\rho|\lambda_{1}\rangle$
(2)
$\displaystyle+2\mathrm{Re}(\langle\lambda_{0}|\partial_{x_{a}}\rho|\lambda_{1}\rangle\langle\lambda_{1}|\partial_{x_{b}}\rho|\lambda_{0}\rangle).$

The second term

$\displaystyle\frac{1}{\det\rho}\mathrm{Tr}\left[(\partial_{x_{a}}\rho-\rho\partial_{x_{a}}\rho)(\partial_{x_{b}}\rho-\rho\partial_{x_{b}}\rho)\right]$
(3) $\displaystyle=$
$\displaystyle\frac{1}{\det\rho}\mathrm{Tr}\left[(\partial_{x_{a}}\rho)(\partial_{x_{b}}\rho)+\rho(\partial_{x_{a}}\rho)\rho(\partial_{x_{b}}\rho)-\rho\\{\partial_{x_{a}}\rho,\partial_{x_{b}}\rho\\}\right].$

Since

$\displaystyle\frac{1}{\det\rho}\mathrm{Tr}\left[\rho(\partial_{x_{a}}\rho)\rho(\partial_{x_{b}}\rho)\right]$
(4) $\displaystyle=$
$\displaystyle\frac{\lambda_{0}}{\lambda_{1}}\langle\lambda_{0}|\partial_{x_{a}}\rho|\lambda_{0}\rangle\langle\lambda_{0}|\partial_{x_{b}}\rho|\lambda_{0}\rangle+\frac{\lambda_{1}}{\lambda_{0}}\langle\lambda_{1}|\partial_{x_{a}}\rho|\lambda_{1}\rangle\langle\lambda_{1}|\partial_{x_{b}}\rho|\lambda_{1}\rangle$
$\displaystyle+2\mathrm{Re}(\langle\lambda_{0}|\partial_{x_{a}}\rho|\lambda_{1}\rangle\langle\lambda_{1}|\partial_{x_{b}}\rho|\lambda_{0}\rangle),$

and

$\displaystyle-\frac{1}{\det\rho}\mathrm{Tr}\left[\rho\\{\partial_{x_{a}}\rho,\partial_{x_{b}}\rho\\}\right]$
(5) $\displaystyle=$
$\displaystyle-\frac{2}{\lambda_{1}}\langle\lambda_{0}|\partial_{x_{a}}\rho|\lambda_{0}\rangle\langle\lambda_{0}|\partial_{x_{b}}\rho|\lambda_{0}\rangle-\frac{2}{\lambda_{0}}\langle\lambda_{1}|\partial_{x_{a}}\rho|\lambda_{1}\rangle\langle\lambda_{1}|\partial_{x_{b}}\rho|\lambda_{1}\rangle$
$\displaystyle-\frac{2}{\lambda_{0}\lambda_{1}}\mathrm{Re}(\langle\lambda_{0}|\partial_{x_{a}}\rho|\lambda_{1}\rangle\langle\lambda_{1}|\partial_{x_{b}}\rho|\lambda_{0}\rangle),$

one can finally obtain

$\displaystyle\mathrm{Tr}\left[(\partial_{x_{a}}\rho)(\partial_{x_{b}}\rho)\right]+\frac{1}{\det\rho}\mathrm{Tr}\left[(\partial_{x_{a}}\rho-\rho\partial_{x_{a}}\rho)(\partial_{x_{b}}\rho-\rho\partial_{x_{b}}\rho)\right]$
(6) $\displaystyle=$
$\displaystyle\frac{1}{\lambda_{0}}\langle\lambda_{0}|\partial_{x_{a}}\rho|\lambda_{0}\rangle\langle\lambda_{0}|\partial_{x_{b}}\rho|\lambda_{0}\rangle+\frac{1}{\lambda_{1}}\langle\lambda_{1}|\partial_{x_{a}}\rho|\lambda_{1}\rangle\langle\lambda_{1}|\partial_{x_{b}}\rho|\lambda_{1}\rangle$
$\displaystyle+4\mathrm{Re}(\langle\lambda_{0}|\partial_{x_{a}}\rho|\lambda_{1}\rangle\langle\lambda_{1}|\partial_{x_{b}}\rho|\lambda_{0}\rangle),$

which coincides with the traditional formula of QFIM in Theorem 2.1. Equation
(1) is then proved. Furthermore, one may notice that
$\langle\lambda_{0}|\partial_{x_{a}}\rho|\lambda_{0}\rangle=\partial_{x_{a}}\lambda_{0}$,
$\langle\lambda_{1}|\partial_{x_{a}}\rho|\lambda_{1}\rangle=\partial_{x_{a}}\lambda_{1}$,
and
$\langle\lambda_{0}|\partial_{x_{a}}\rho|\lambda_{1}\rangle=\langle\partial_{x_{a}}\lambda_{0}|\lambda_{1}\rangle+\langle\lambda_{0}|\partial_{x_{a}}\lambda_{1}\rangle=0$,
which gives

$\displaystyle\mathrm{Tr}\left[(\partial_{x_{a}}\rho)(\partial_{x_{b}}\rho)-\rho\\{\partial_{x_{a}}\rho,\partial_{x_{b}}\rho)\\}\right]$
(7) $\displaystyle=$
$\displaystyle(\partial_{x_{a}}\lambda_{0})(\partial_{x_{b}}\lambda_{0})-\lambda_{0}(\partial_{x_{a}}\lambda_{0})(\partial_{x_{b}}\lambda_{0})-\lambda_{1}(\partial_{x_{a}}\lambda_{1})(\partial_{x_{b}}\lambda_{1})$
$\displaystyle=$ $\displaystyle 0,$

namely, the equality
$\mathrm{Tr}\left[(\partial_{x_{a}}\rho)(\partial_{x_{b}}\rho)\right]=\mathrm{Tr}\left(\rho\\{\partial_{x_{a}}\rho,\partial_{x_{b}}\rho)\\}\right)$
holds for single-qubit mixed states. Thus, equation (1) can further reduce to

$\mathcal{F}_{ab}=\mathrm{Tr}\left[(\partial_{x_{a}}\rho)(\partial_{x_{b}}\rho)\right]+\frac{1}{\det\rho}\mathrm{Tr}\left[(\rho(\partial_{x_{a}}\rho)\rho(\partial_{x_{b}}\rho)\right].$
(8)

The theorem has been proved.

##  Appendix E Derivation of SLD operator for Gaussian states

###  E.1 SLD operator for multimode Gaussian states

The derivation in this appendix is majorly based on the calculation in
references [62, 67]. Let us first recall the notations before the derivation.
The vector of quadrature operators are defined as
$\vec{R}=(\hat{q}_{1},\hat{p}_{1},...,\hat{q}_{m},\hat{p}_{m})^{\mathrm{T}}$.
$\Omega$ is a symplectic matrix $\Omega=i\sigma^{\oplus m}_{y}$.
$\chi(\vec{s})$ is the characteristic function. Furthermore, denote
$d=\langle\vec{R}\rangle$. To keep the calculation neat, we use $\chi$,
$\rho$, $L$ instead of $\chi(\vec{s})$, $\rho$, $L_{x_{a}}$ in this appendix.
Some other notations are $\langle\cdot\rangle=\mathrm{Tr}(\rho\cdot)$ and
$\dot{A}=\partial_{x_{a}}A$.

The characteristic function $\chi=\langle D\rangle$, where
$D=e^{i\vec{R}^{\mathrm{T}}\Omega\vec{s}}$ with $\vec{s}\in\mathbb{R}^{2m}$.
Substituting $D$ into the equation $\partial_{x_{a}}\rho=\frac{1}{2}(\rho
L+L\rho)$ and taking the trace, we obtain

$\partial_{x_{a}}\chi=\frac{1}{2}\langle\\{L,D\\}\rangle.$ (1)

Next, assume the SLD operator is in the following form

$L=L^{(0)}\leavevmode\hbox{\small 1\kern-4.3pt\normalsize
1}+\vec{L}^{(1),\mathrm{T}}\vec{R}+\vec{R}^{\,\mathrm{T}}G\vec{R},$ (2)

where $L^{(0)}$ is a real number, $\vec{L}^{(1)}$ is a $2m$-dimensional real
vector and $G$ is a $2m$-dimensional real symmetric matrix. These conditions
promise the Hermiticity of SLD. With this ansatz, equation (1) can then be
rewritten into

$\partial_{x_{a}}\chi=L^{(0)}\chi+\frac{1}{2}\sum_{i}L^{(1)}_{i}\langle\\{R_{i},D\\}\rangle+\frac{1}{4}\sum_{ij}G_{ij}\langle\\{\\{R_{i},R_{j}\\},D\\}\rangle.$
(3)

Now we calculate $\langle\\{R_{i},D\\}\rangle$ in equation (3). Denote
$[A,\cdot]=A^{\times}(\cdot)$, one can have

$\displaystyle\partial_{s_{k}}D$ $\displaystyle=$
$\displaystyle\sum_{i}i\Omega_{ik}\int^{1}_{0}e^{iy\vec{R}^{\mathrm{T}}\Omega\vec{s}}R_{i}e^{-iy\vec{R}^{\mathrm{T}}\Omega\vec{s}}\mathrm{d}yD$
(4) $\displaystyle=$
$\displaystyle\sum_{i}i\Omega_{ik}\int^{1}_{0}\sum_{n}\frac{(iy\vec{R}^{\mathrm{T}}\Omega\vec{s})^{\times,n}}{n!}R_{i}\mathrm{d}yD$
$\displaystyle=$
$\displaystyle\sum_{i}i\Omega_{ik}\int^{1}_{0}R_{i}+y\left[i\vec{R}^{\mathrm{T}}\Omega\vec{s},R_{i}\right]\mathrm{d}yD$
$\displaystyle=$
$\displaystyle\sum_{i}i\Omega_{ik}\left(R_{i}+\frac{1}{2}\sum_{jl}\Omega_{ij}\Omega_{jl}s_{l}\right)D$
$\displaystyle=$ $\displaystyle
i\sum_{i}\Omega_{ki}\left(\frac{1}{2}s_{i}-R_{i}\right)D.$

where we have used the fact $\Omega_{ij}=-\Omega_{ji}$ and
$\sum_{j}\Omega_{ij}\Omega_{jk}=-\delta_{ik}$ which comes from the equation
$\Omega^{2}=-\leavevmode\hbox{\small 1\kern-4.3pt\normalsize 1}_{2m}$.
Substituting the equation

$\int^{1}_{0}e^{iy\vec{R}^{\mathrm{T}}\Omega\vec{s}}R_{i}e^{-iy\vec{R}^{\mathrm{T}}\Omega\vec{s}}\mathrm{d}yD=D\int^{1}_{0}e^{-iy\vec{R}^{\mathrm{T}}\Omega\vec{s}}R_{i}e^{iy\vec{R}^{\mathrm{T}}\Omega\vec{s}}\mathrm{d}y,$
(5)

into the first line of equation (4) and repeat the calculation, one can obtain
$\partial_{s_{k}}D=-i\sum_{i}\Omega_{ki}D\left(\frac{1}{2}s_{i}+R_{i}\right)$.
Combining this equation with equation (4), $\partial_{s_{k}}D$ can be finally
written as

$\partial_{s_{k}}D=-\frac{i}{2}\sum_{i}\Omega_{ki}\\{R_{i},D\\},$ (6)

which further gives
$\langle\partial_{s_{k}}D\rangle=-\frac{i}{2}\sum_{i}\Omega_{ki}\langle\\{R_{i},D\\}\rangle$.
Based on this equation, it can be found
$\sum_{k}\Omega_{jk}\langle\partial_{s_{k}}D\rangle=-\frac{i}{2}\sum_{ik}\Omega_{jk}\Omega_{ki}\langle\\{R_{i},D\\}\rangle$.
Again since $\sum_{k}\Omega_{jk}\Omega_{ki}=-\delta_{ij}$, it reduces to

$\langle\\{R_{j},D\\}\rangle=-i2\sum_{k}\Omega_{jk}\langle\partial_{s_{k}}D\rangle=-i2\sum_{k}\Omega_{jk}\partial_{s_{k}}\chi.$
(7)

Next, continue to take the derivative on $\partial_{s_{k}}D$, we have

$\partial_{s_{k^{\prime}}}\partial_{s_{k}}D=-\frac{i}{2}\sum_{i}\Omega_{ki}\\{R_{i},\partial_{s_{k^{\prime}}}D\\}-\frac{1}{4}\sum_{ij}\Omega_{k^{\prime}j}\Omega_{ki}\\{R_{i},\\{R_{j},D\\}\\}.$
(8)

Due to the Baker-Campbell-Hausdorff formula,
$D^{\dagger}R_{i}D=R_{i}+[-i\vec{R}^{\mathrm{T}}\Omega\vec{s},R_{i}]=R_{i}+s_{i}$,
the commutation between $R_{i}$ and $D$ can be obtained as $[R_{i},D]=s_{i}D$,
which further gives
$\\{R_{i},\\{R_{j},D\\}\\}=\\{\\{R_{i},R_{j}\\},D\\}-s_{i}s_{j}D$. Then
equation (8) can be rewritten into

$\partial_{s_{k^{\prime}}}\partial_{s_{k}}D=-\frac{1}{4}\sum_{ij}\Omega_{k^{\prime}j}\Omega_{ki}\\{\\{R_{i},R_{j}\\},D\\}+\frac{1}{4}\sum_{ij}\Omega_{k^{\prime}j}\Omega_{ki}s_{i}s_{j}D.$
(9)

And one can finally obtain

$\displaystyle\langle\\{R_{i},\\{R_{j},D\\}\\}\rangle$ $\displaystyle=$
$\displaystyle-4\sum_{kk^{\prime}}\Omega_{ik}\Omega_{jk^{\prime}}\langle\partial_{s_{k^{\prime}}}\partial_{s_{k}}D\rangle+s_{i}s_{j}\chi$
(10) $\displaystyle=$
$\displaystyle-4\sum_{kk^{\prime}}\Omega_{ik}\Omega_{jk^{\prime}}\left(\partial_{s_{k^{\prime}}}\partial_{s_{k}}\chi\right)+s_{i}s_{j}\chi.$

With equations (7) and (10), equation (3) can be expressed by

$\displaystyle\partial_{x_{a}}\chi$ $\displaystyle=$
$\displaystyle\left(L^{(0)}+\frac{1}{4}\sum_{ij}G_{ij}s_{i}s_{j}\right)\chi-i\sum_{ik}L^{(1)}_{i}\Omega_{ik}(\partial_{s_{k}}\chi)$
(11)
$\displaystyle-\sum_{ijkk^{\prime}}G_{ij}\Omega_{ik}\Omega_{jk^{\prime}}(\partial_{s_{k^{\prime}}}\partial_{s_{k}}\chi).$

On the other hand, from the expression of characteristic function

$\chi=e^{-\frac{1}{2}\vec{s}^{\mathrm{T}}\Omega
C\Omega^{\mathrm{T}}\vec{s}-i(\Omega d)^{\mathrm{T}}\vec{s}}.$ (12)

it can be found that

$\partial_{x_{a}}\chi=\left[-\frac{1}{2}\vec{s}^{\,\mathrm{T}}\Omega\dot{C}\Omega^{\mathrm{T}}\vec{s}-i(\Omega\dot{d})^{\mathrm{T}}\vec{s}\right]\chi,$
(13)

and

$\displaystyle\partial_{s_{k}}\chi$ $\displaystyle=$
$\displaystyle-\sum_{ijl}\Omega_{kl}\Omega_{ij}C_{jl}s_{i}\chi-i\sum_{i}\Omega_{ki}d_{i}\chi,$
$\displaystyle\partial_{s_{k^{\prime}}}\partial_{s_{k}}\chi$ $\displaystyle=$
$\displaystyle-\sum_{i_{1}j_{1}}\Omega_{ki_{1}}\Omega_{k^{\prime}j_{1}}C_{i_{1}j_{1}}\chi-\sum_{i_{1}j_{1}}\Omega_{ki_{1}}\Omega_{k^{\prime}j_{1}}d_{i_{1}}d_{j_{1}}\chi$
$\displaystyle+\\!\\!\sum_{i_{1}j_{1}l_{1}i_{2}j_{2}l_{2}}\\!\\!\\!\\!\Omega_{kl_{1}}\Omega_{k^{\prime}l_{2}}\Omega_{i_{1}j_{1}}\Omega_{i_{2}j_{2}}C_{j_{1}l_{1}}C_{j_{2}l_{2}}s_{i_{1}}s_{i_{2}}\chi$
$\displaystyle+i\\!\\!\sum_{i_{1}j_{1}l_{1}i_{2}}\\!\\!\\!\\!\left(\Omega_{ki_{2}}\Omega_{k^{\prime}l_{1}}+\Omega_{k^{\prime}i_{2}}\Omega_{kl_{1}}\right)\Omega_{i_{1}j_{1}}C_{j_{1}l_{1}}s_{i_{1}}d_{i_{2}}\chi.$

Then we have
$-i\sum_{ik}L^{(1)}_{i}\Omega_{ik}(\partial_{s_{k}}\chi)=\sum_{i}L^{(1)}_{i}\left(-i\sum_{jk}\Omega_{kj}C_{ij}s_{k}+d_{i}\right)\chi$,
and

$\displaystyle-\sum_{ijkk^{\prime}}G_{ij}\Omega_{ik}\Omega_{jk^{\prime}}(\partial_{s_{k^{\prime}}}\partial_{s_{k}}\chi)$
$\displaystyle=$
$\displaystyle\sum_{ij}G_{ij}\left(C_{ij}+d_{i}d_{j}\right)\chi-\\!\\!\sum_{iji_{1}j_{1}i_{2}j_{2}}\\!\\!G_{ij}\Omega_{i_{1}j_{1}}\Omega_{i_{2}j_{2}}C_{ij_{1}}C_{jj_{2}}s_{i_{1}}s_{i_{2}}\chi$
$\displaystyle-i\sum_{iji_{1}j_{1}}G_{ij}\Omega_{i_{1}j_{1}}s_{i_{1}}\left(C_{jj_{1}}d_{i}+C_{ij_{1}}d_{j}\right)\chi.$

Substituting these equations into equation (11), $\partial_{x_{a}}\chi$ can be
expressed by

$\displaystyle\partial_{x_{a}}\chi$ $\displaystyle=$
$\displaystyle\left(L^{(0)}+\frac{1}{4}\sum_{ij}G_{ij}s_{i}s_{j}\right)\chi+\sum_{i}L^{(1)}_{i}\left(-i\sum_{jk}\Omega_{kj}C_{ij}s_{k}+d_{i}\right)\chi$
(14)
$\displaystyle+\sum_{ij}G_{ij}\left(C_{ij}+d_{i}d_{j}\right)\chi-\\!\\!\sum_{iji_{1}j_{1}i_{2}j_{2}}\\!\\!G_{ij}\Omega_{i_{1}j_{1}}\Omega_{i_{2}j_{2}}C_{ij_{1}}C_{jj_{2}}s_{i_{1}}s_{i_{2}}\chi$
$\displaystyle-i\sum_{iji_{1}j_{1}}G_{ij}\Omega_{i_{1}j_{1}}s_{i_{1}}\left(C_{jj_{1}}d_{i}+C_{ij_{1}}d_{j}\right)\chi.$

This equation can be written into a more compact way as below

$\displaystyle\partial_{x_{a}}\chi$ $\displaystyle=$ $\displaystyle
L^{(0)}\chi+\vec{L}^{(1),\mathrm{T}}d\chi+\mathrm{Tr}(GC)\chi+d^{\mathrm{T}}Gd\chi$
(15)
$\displaystyle+i\vec{L}^{(1),\mathrm{T}}C\Omega\vec{s}\chi+i2d^{\mathrm{T}}GC\Omega\vec{s}\chi+\vec{s}^{\mathrm{T}}\Omega
CGC\Omega\vec{s}\chi+\frac{1}{4}\vec{s}^{\mathrm{T}}G\vec{s}\chi.$

Compare this equation with equation (13), it can be found that

$\displaystyle
L^{(0)}+L^{(1),\mathrm{T}}d+\mathrm{Tr}(GC)+d^{\mathrm{T}}Gd=0,$ (16)
$\displaystyle\vec{L}^{(1),\mathrm{T}}C\Omega+2d^{\mathrm{T}}GC\Omega=-\dot{d}^{\mathrm{T}}\Omega^{\mathrm{T}},$
(17) $\displaystyle\frac{1}{4}G+\Omega
CGC\Omega=-\frac{1}{2}\Omega\dot{C}\Omega^{\mathrm{T}}.$ (18)

From the second equation above, one can obtain

$\vec{L}^{(1)}=C^{-1}\dot{d}-2Gd.$ (19)

Using this equation and the equation (16), it can be found that

$L^{(0)}=d^{\mathrm{T}}Gd-\dot{d}^{\mathrm{T}}C^{-1}d-\mathrm{Tr}(GC).$ (20)

Once we obtain the expression of $G$, $L^{(0)}$ and $\vec{L}^{(1)}$ can be
obtained correspondingly, which means we need to solve equation (18), which
can be rewritten into following form

$\Omega G\Omega+4CGC=2\dot{C}.$ (21)

Since $C=SC_{\mathrm{d}}S^{\mathrm{T}}$ and $S\Omega S^{\mathrm{T}}=\Omega$,
above equation can be rewritten into

$\Omega
G_{\mathrm{s}}\Omega+4C_{\mathrm{d}}G_{\mathrm{s}}C_{\mathrm{d}}=2S^{-1}\dot{C}\left(S^{\mathrm{T}}\right)^{-1},$
(22)

where $G_{\mathrm{s}}=S^{\mathrm{T}}GS$. Denote the map
$\mathcal{E}(G_{\mathrm{s}}):=\Omega
G_{\mathrm{s}}\Omega+4C_{\mathrm{d}}G_{\mathrm{s}}C_{\mathrm{d}}$, then
$G_{\mathrm{s}}=\mathcal{E}^{-1}(\mathcal{E}(G_{\mathrm{s}}))$. The map
$\mathcal{E}(G_{\mathrm{s}})$ can be decomposed via the generators
$\\{A^{(jk)}_{l}\\}$ ($j,k=1,...,m$), where

$A^{(jk)}_{l}=\frac{1}{\sqrt{2}}i\sigma^{(jk)}_{y},~{}\frac{1}{\sqrt{2}}\sigma^{(jk)}_{z},~{}\frac{1}{\sqrt{2}}\leavevmode\hbox{\small
1\kern-4.3pt\normalsize 1}^{(jk)}_{2},~{}\frac{1}{\sqrt{2}}\sigma^{(jk)}_{x}$
(23)

for $l=0,1,2,3.$ $\sigma^{(jk)}_{i}$ is a $2m$-dimensional matrix with all the
entries zero expect the $2\times 2$ block shown as below

$\sigma^{(jk)}_{i}=\left(\begin{array}[]{ccccc}&1\mathrm{st}&\cdots&k\mathrm{th}&\cdots\\\
1\mathrm{st}&0_{2\times 2}&0_{2\times 2}&0_{2\times 2}&0_{2\times 2}\\\
\vdots&0_{2\times 2}&\vdots&\vdots&\vdots\\\ j\mathrm{th}&0_{2\times
2}&\cdots&\sigma_{i}&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\vdots\end{array}\right),$ (24)

where $0_{2\times 2}$ represents a 2 by 2 block with zero entries.
$\leavevmode\hbox{\small 1\kern-4.3pt\normalsize 1}^{(jk)}_{2}$ is similar to
$\sigma^{(jk)}_{i}$ but replace the block $\sigma_{i}$ with
$\leavevmode\hbox{\small 1\kern-4.3pt\normalsize 1}_{2}$. $A^{(jk)}_{l}$
satisfies the orthogonal relation
$\mathrm{Tr}(A^{(jk)}_{l}A^{(j^{\prime}k^{\prime})}_{l^{\prime}})=\delta_{jj^{\prime}}\delta_{kk^{\prime}}\delta_{ll^{\prime}}$.
Recall that $C_{\mathrm{d}}=\bigoplus^{m}_{k=1}c_{k}\leavevmode\hbox{\small
1\kern-4.3pt\normalsize 1}_{2}$, it is easy to check that

$\displaystyle\Omega A^{(jk)}_{l}\Omega$ $\displaystyle=$
$\displaystyle(-1)^{l+1}A^{(jk)}_{l},$ (25) $\displaystyle
C_{\mathrm{d}}A^{(jk)}_{l}C_{\mathrm{d}}$ $\displaystyle=$ $\displaystyle
c_{j}c_{k}A^{(jk)}_{l}.$ (26)

Next decompose $S^{-1}\dot{C}(S^{\mathrm{T}})^{-1}$ with $\\{A^{(jk)}_{l}\\}$
as

$S^{-1}\dot{C}(S^{\mathrm{T}})^{-1}=\sum_{jkl}g^{(jk)}_{l}A^{(jk)}_{l},$ (27)

where
$g^{(jk)}_{l}=\mathrm{Tr}[S^{-1}\dot{C}(S^{\mathrm{T}})^{-1}A^{(jk)}_{l}]$.
Decomposing $G_{\mathrm{s}}$ as
$G_{\mathrm{s}}=\sum_{jkl}\tilde{g}^{(jk)}_{l}A^{(jk)}_{l}$, and substituting
it into equation (22), we have

$\sum_{jkl}g^{(jk)}_{l}\left(\Omega
A^{(jk)}_{l}\Omega+4C_{\mathrm{d}}A^{(jk)}_{l}C_{\mathrm{d}}\right)=\sum_{jkl}g^{(jk)}_{l}A^{(jk)}_{l}.$
(28)

Utilizing equations (25) and (26), above equation reduces to

$\sum_{jkl}\tilde{g}^{(jk)}_{l}\left[4c_{j}c_{k}+(-1)^{l+1}\right]A^{(jk)}_{l}=\sum_{jkl}g^{(jk)}_{l}A^{(jk)}_{l},$
(29)

which indicates

$\tilde{g}^{(jk)}_{l}=\frac{g^{(jk)}_{l}}{4c_{j}c_{k}+(-1)^{l+1}}.$ (30)

Thus, $G=(S^{\mathrm{T}})^{-1}G_{\mathrm{s}}S^{-1}$ can be solved as below

$G=\sum_{jkl}\frac{g^{(jk)}_{l}}{4c_{j}c_{k}+(-1)^{l+1}}\left(S^{\mathrm{T}}\right)^{-1}A^{(jk)}_{l}S^{-1}.$
(31)

In summary, the SLD can be expressed by

$L=L^{(0)}\leavevmode\hbox{\small 1\kern-4.3pt\normalsize
1}+\vec{L}^{(1),\mathrm{T}}\vec{R}+\vec{R}^{\mathrm{T}}G\vec{R},$ (32)

where

$G=\sum^{m}_{j,k=1}\sum^{3}_{l=0}\frac{g^{(jk)}_{l}}{4c_{j}c_{k}+(-1)^{l+1}}\left(S^{\mathrm{T}}\right)^{-1}A^{(jk)}_{l}S^{-1}$
(33)

with

$A^{(jk)}_{l}=\frac{1}{\sqrt{2}}i\sigma^{(jk)}_{y},~{}\frac{1}{\sqrt{2}}\sigma^{(jk)}_{z},~{}\frac{1}{\sqrt{2}}\leavevmode\hbox{\small
1\kern-4.3pt\normalsize 1}^{(jk)}_{2},~{}\frac{1}{\sqrt{2}}\sigma^{(jk)}_{x}$
(34)

for $l=0,1,2,3$ and
$g^{(jk)}_{l}=\mathrm{Tr}[S^{-1}\dot{C}(S^{\mathrm{T}})^{-1}A^{(jk)}_{l}]$.
And

$\displaystyle\vec{L}^{(1)}$ $\displaystyle=$ $\displaystyle
C^{-1}\dot{d}-2Gd,$ (35) $\displaystyle L^{(0)}$ $\displaystyle=$
$\displaystyle d^{\mathrm{T}}Gd-\dot{d}^{\mathrm{T}}C^{-1}d-\mathrm{Tr}(GC).$
(36)

###  E.2 SLD operator for single-mode Gaussian state

For a single-mode case, a 2-dimensional symplectic $S$ matrix satisfies $\det
S=1$. Based on the equation $C=SC_{\mathrm{d}}S^{\mathrm{T}}=cSS^{\mathrm{T}}$
($c$ is the symplectic value), the following equations can be obtained

$\displaystyle C_{00}$ $\displaystyle=$ $\displaystyle
c(S^{2}_{00}+S^{2}_{01}),$ $\displaystyle C_{11}$ $\displaystyle=$
$\displaystyle c(S^{2}_{10}+S^{2}_{11}),$ (37) $\displaystyle C_{01}$
$\displaystyle=$ $\displaystyle c(S_{00}S_{10}+S_{01}S_{11}).$ (38)

Together with the equation $\det S=S_{00}S_{11}-S_{01}S_{10}=1$, the
symplectic value $c$ can be solved as $c=\sqrt{\det C}$. Assume the form

$\displaystyle S_{00}$ $\displaystyle=$
$\displaystyle\sqrt{\frac{C_{00}}{c}}\cos\theta,\quad
S_{01}=\sqrt{\frac{C_{00}}{c}}\sin\theta,$ (39) $\displaystyle S_{10}$
$\displaystyle=$ $\displaystyle\sqrt{\frac{C_{11}}{c}}\cos\phi,\quad
S_{11}=\sqrt{\frac{C_{11}}{c}}\sin\phi.$ (40)

Substituting above expressions into equation (37), it can be found
$\theta,\phi$ need to satisfy

$\cos(\phi-\theta)=\frac{C_{01}}{\sqrt{C_{00}C_{11}}},~{}~{}\sin(\phi-\theta)=\frac{c}{\sqrt{C_{00}C_{11}}}.$
(41)

Since there is only four constrains for these five variables, one of them is
free. We take $\theta=\pi/2-\phi$, and the symplectic matrix reduces to

$S=\frac{1}{\sqrt{c}}\left(\begin{array}[]{cc}\sqrt{C_{00}}\sin\phi&\sqrt{C_{00}}\cos\phi\\\
\sqrt{C_{11}}\cos\phi&\sqrt{C_{11}}\cos\phi\end{array}\right),$ (42)

where $\phi$ satisfies $\sin(2\phi)=\frac{C_{01}}{\sqrt{C_{00}C_{11}}}$ and
$\cos(2\phi)=-\frac{c}{\sqrt{C_{00}C_{11}}}$. Based on Theorem 2.9, we obtain

$\displaystyle g_{0}$ $\displaystyle=$ $\displaystyle 0,$ (43) $\displaystyle
g_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2C_{00}C_{11}}}C_{11}\dot{C}_{00}-C_{00}\dot{C}_{11},$
(44) $\displaystyle g_{2}$ $\displaystyle=$ $\displaystyle\sqrt{2}\dot{c},$
(45) $\displaystyle g_{3}$ $\displaystyle=$
$\displaystyle\frac{\sqrt{2}(c\dot{C}_{01}-\dot{c}C_{01})}{\sqrt{C_{00}C_{11}}},$
(46)

where $\dot{C}$, $\dot{c}$ are short for $\partial_{x_{a}}C$ and
$\partial_{x_{a}}c$. Meanwhile, we have

$\displaystyle(S^{\mathrm{T}})^{-1}A_{1}S^{-1}$ $\displaystyle=$
$\displaystyle\frac{\sqrt{2C_{00}C_{11}}}{2}\left(\begin{array}[]{cc}\frac{1}{C_{00}}&0\\\
0&-\frac{1}{C_{11}}\end{array}\right),$ (49)
$\displaystyle(S^{\mathrm{T}})^{-1}A_{2}S^{-1}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}c}\left(\begin{array}[]{cc}C_{11}&-C_{01}\\\
-C_{01}&C_{00}\end{array}\right),$ (52)
$\displaystyle(S^{\mathrm{T}})^{-1}A_{3}S^{-1}$ $\displaystyle=$
$\displaystyle\frac{\sqrt{C_{00}C_{11}}}{\sqrt{2}c}\left(\begin{array}[]{cc}-\frac{C_{01}}{C_{00}}&1\\\
1&-\frac{C_{01}}{C_{11}}\end{array}\right).$ (55)

These expressions immediately give $G_{x_{a}}$ as below

$\displaystyle[G_{x_{a}}]_{00}$ $\displaystyle=$
$\displaystyle\frac{1}{4c^{2}+1}\left[\frac{C_{11}\dot{C}_{00}-C_{00}\dot{C}_{11}}{2C_{00}}-\frac{C_{01}}{C_{00}}(\dot{C}_{01}-\frac{\dot{c}}{c}C_{01})\right]+\frac{1}{4c^{2}-1}\frac{\dot{c}}{c}C_{11}$
(56) $\displaystyle=$
$\displaystyle\frac{1}{4c^{2}+1}\left(\frac{\dot{c}}{c}C_{11}-\dot{C}_{11}\right)+\frac{1}{4c^{2}-1}\frac{\dot{c}}{c}C_{11}$
$\displaystyle=$
$\displaystyle-\frac{1}{16c^{4}-1}\left[(4c^{2}-1)C_{11}-8c\dot{c}C_{11}\right].$

Define a matrix

$J:=\frac{1}{4c^{2}-1}C,$ (57)

$[G_{x_{a}}]_{00}$ can be rewritten into

$[G_{x_{a}}]_{00}=-\frac{4c^{2}-1}{4c^{2}+1}\partial_{x_{a}}J_{11},\\\ $ (58)

Similarly, one can obtain

$\displaystyle\left[G_{x_{a}}\right]_{11}$ $\displaystyle=$
$\displaystyle-\frac{4c^{2}-1}{4c^{2}+1}\partial_{x_{a}}J_{00},$ (59)

and

$\left[G_{x_{a}}\right]_{01}=\left[G_{x_{a}}\right]_{10}=\frac{4c^{2}-1}{4c^{2}+1}\partial_{x_{a}}J_{01}.$
(60)

These expressions indicate

$G_{x_{a}}=\frac{4c^{2}-1}{4c^{2}+1}\Omega(\partial_{x_{a}}J)\Omega,$ (61)

with

$\Omega=\left(\begin{array}[]{cc}0&1\\\ -1&0\end{array}\right).$ (62)

##  Appendix F QFIM and Bures metric

The Bures distance between two quantum states $\rho_{1}$ and $\rho_{2}$ is
defined as

$D^{2}_{\mathrm{B}}(\rho_{1},\rho_{2})=2-2f(\rho_{1},\rho_{2}),$ (1)

where
$f(\rho_{1},\rho_{2})=\mathrm{Tr}\sqrt{\sqrt{\rho_{1}}\rho_{2}\sqrt{\rho_{1}}}$
is the quantum fidelity. Now we calcuate the fidelity for two close quantum
states $\rho(\vec{x})$ and $\rho(\vec{x}+\mathrm{d}\vec{x})$. The Taylor
series of $\rho(\vec{x}+\mathrm{d}\vec{x})$ (up to the second order) reads

$\rho(\vec{x}+\mathrm{d}\vec{x})=\rho(\vec{x})+\sum_{a}\partial_{x_{a}}\rho(\vec{x})\mathrm{d}x_{a}+\frac{1}{2}\sum_{ab}\frac{\partial^{2}\rho(\vec{x})}{\partial
x_{a}\partial x_{b}}\mathrm{d}x_{a}\mathrm{d}x_{b},$ (2)

In the following $\rho$ will be used as the abbreviation of $\rho(\vec{x})$.
Utilizing the equation above, one can obtain

$\displaystyle\sqrt{\rho}\rho(\vec{x}+\mathrm{d}\vec{x})\sqrt{\rho}$ (3)
$\displaystyle=$
$\displaystyle\rho^{2}+\sum_{a}\sqrt{\rho}\partial_{x_{a}}\rho\sqrt{\rho}\mathrm{d}x_{a}+\frac{1}{2}\sum_{ab}\left(\sqrt{\rho}\frac{\partial^{2}\rho}{\partial
x_{a}\partial x_{b}}\sqrt{\rho}\right)\mathrm{d}x_{a}\mathrm{d}x_{b}.$

Now we assume

$\sqrt{\sqrt{\rho}\rho(\vec{x}+\mathrm{d}\vec{x})\sqrt{\rho}}=\rho+\sum_{a}W_{a}\mathrm{d}x_{a}+\sum_{ab}Y_{ab}\mathrm{d}x_{a}\mathrm{d}x_{b}.$
(4)

Taking the square of above equation and compare it to equation (3), one can
obtain

$\displaystyle\sqrt{\rho}\partial_{x_{a}}\rho\sqrt{\rho}$ $\displaystyle=$
$\displaystyle\rho W_{a}+W_{a}\rho,$ (5)
$\displaystyle\frac{1}{2}\sqrt{\rho}\partial^{2}\rho\sqrt{\rho}$
$\displaystyle=$ $\displaystyle\rho
Y_{ab}+Y_{ab}\rho+\frac{1}{2}\\{W_{a},W_{b}\\},$ (6)

where $\partial^{2}\rho$ is short for $\frac{\partial^{2}\rho}{\partial
x_{a}\partial x_{b}}$. $\frac{1}{2}\\{W_{a},W_{b}\\}$ (not $W_{a}W_{b}$) is
used to make sure the equation is unchanged when the subscripts $a$ and $b$
exchange. Meawhile, from equation (4), the Bures metric
$D_{\mathrm{B}}(\rho(\vec{x}),\rho(\vec{x}+\mathrm{d}\vec{x}))$ (the
abbreviation $D_{\mathrm{B}}$ will bu used below) can be calculated as

$D^{2}_{\mathrm{B}}=-2\sum_{a}(\mathrm{Tr}W_{a})\mathrm{d}x_{a}-2\sum_{ab}(\mathrm{Tr}Y_{ab})\mathrm{d}x_{a}\mathrm{d}x_{b}.$
(7)

As long as we obtain the specific expressions of $W_{a}$ and $Y_{ab}$ from
equations (5) and (6), $D_{\mathrm{B}}$ can be obtained immediately. To do
that, we utilize the spectral decomposition
$\rho=\sum_{i}\lambda_{i}|\lambda_{i}\rangle\langle\lambda_{i}|$. In the basis
$\\{|\lambda_{i}\rangle\\}$, the matrix entries
($[\cdot]_{ij}=\langle\lambda_{i}|\cdot|\lambda_{j}\rangle$) read

$\displaystyle\left[\sqrt{\rho}\partial_{x_{a}}\rho\sqrt{\rho}\right]_{ij}$
$\displaystyle=$
$\displaystyle\sqrt{\lambda_{i}\lambda_{j}}[\partial_{x_{a}}\rho]_{ij},$ (8)
$\displaystyle\left[\sqrt{\rho}\partial^{2}\rho\sqrt{\rho}\right]_{ij}$
$\displaystyle=$
$\displaystyle\sqrt{\lambda_{i}\lambda_{j}}\left[\partial^{2}\rho\right]_{ij}.$
(9)

Here
$[\partial_{x_{a}}\rho]_{ij}=\delta_{ij}\partial_{x_{a}}\lambda_{i}-(\lambda_{i}-\lambda_{j})\langle\lambda_{i}|\partial_{x_{a}}\lambda_{j}\rangle$,
and

$\displaystyle\left[\partial^{2}\rho\right]_{ij}$ $\displaystyle=$
$\displaystyle\partial^{2}\lambda_{i}\delta_{ij}+\lambda_{i}\langle\partial^{2}\lambda_{i}|\lambda_{j}\rangle+\lambda_{j}\langle\lambda_{i}|\partial^{2}\lambda_{j}\rangle$
(10)
$\displaystyle+(\partial_{x_{a}}\lambda_{j}-\partial_{x_{a}}\lambda_{i})\langle\lambda_{i}|\partial_{x_{b}}\lambda_{j}\rangle+(\partial_{x_{b}}\lambda_{j}-\partial_{x_{b}}\lambda_{i})\langle\lambda_{i}|\partial_{x_{a}}\lambda_{j}\rangle$
$\displaystyle+\sum_{k}\lambda_{k}(\langle\lambda_{i}|\partial_{x_{a}}\lambda_{k}\rangle\langle\partial_{x_{b}}\lambda_{k}|\lambda_{j}\rangle+\langle\lambda_{i}|\partial_{x_{b}}\lambda_{k}\rangle\langle\partial_{x_{a}}\lambda_{k}|\lambda_{j}\rangle).$

where $\partial^{2}\lambda_{i}$ and $|\partial^{2}\lambda_{i}\rangle$ are
short for $\frac{\partial^{2}\lambda_{i}}{\partial x_{a}\partial x_{b}}$ and
$\frac{\partial^{2}}{\partial x_{a}\partial x_{b}}|\lambda_{i}\rangle$. Now we
denote $\rho$’s dimension as $N$ and $\lambda_{i}\in\mathcal{S}$ for
$i=0,1,2...,M-1$. Under the assumption that the support $\mathcal{S}$ is not
affected by the values of $\vec{x}$, i.e., the rank of $\rho(\vec{x})$ equals
to that of $\rho(\vec{x}+\mathrm{d}\vec{x})$,
$\sqrt{\rho}\partial_{x_{a}}\rho\sqrt{\rho}$ and
$\sqrt{\rho}\partial^{2}\rho\sqrt{\rho}$ are both block diagonal. Based on
equation (5), the $ij$th matrix entry of $W_{a}$ can be calculated as

$\displaystyle[W_{a}]_{ij}=\frac{\left[\sqrt{\rho}\partial_{x_{a}}\rho\sqrt{\rho}\right]_{ij}}{\lambda_{i}+\lambda_{j}}=\frac{1}{2}\partial_{x_{a}}\lambda_{i}\delta_{ij}-\frac{\sqrt{\lambda_{i}\lambda_{j}}(\lambda_{i}-\lambda_{j})}{\lambda_{i}+\lambda_{j}}\langle\lambda_{i}|\partial_{x_{a}}\lambda_{j}\rangle,$
(11)

for $i,j\in[0,M-1]$ and $[W_{a}]_{ij}=0$ for others. One can observe that
$W_{a}$ is a Hermitian matrix, and

$\mathrm{Tr}W_{a}=\sum_{ii}[W_{a}]_{ii}=\frac{1}{2}\sum_{i}\partial_{x_{a}}\lambda_{i}=0,$
(12)

which means there is no first order term in Bures metric. With respect to the
second order term, we need to know the value of $[Y_{ab}]_{ii}$. From equation
(6), one can obtain

$\mathrm{Tr}Y_{ab}=\frac{1}{4}\sum_{i}[\partial^{2}\rho]_{ii}-\sum_{ik}\frac{1}{2\lambda_{i}}\mathrm{Re}\left([W_{a}]_{ik}[W_{b}]_{ki}\right).$
(13)

Due to the fact
$\langle\partial^{2}\lambda_{i}|\lambda_{i}\rangle+\langle\lambda_{i}|\partial^{2}\lambda_{i}\rangle=-2\mathrm{Re}(\langle\partial_{x_{a}}\lambda_{i}|\partial_{x_{b}}\lambda_{i}\rangle)$,
one can have

$[\partial^{2}\rho]_{ii}=\partial^{2}\lambda_{i}-2\lambda_{i}\mathrm{Re}(\langle\partial_{x_{a}}\lambda_{i}|\partial_{x_{b}}\lambda_{i}\rangle)+\sum_{k}2\lambda_{k}\mathrm{Re}(\langle\lambda_{i}|\partial_{x_{a}}\lambda_{k}\rangle\langle\partial_{x_{b}}\lambda_{k}|\lambda_{i}\rangle),$
(14)

which further gives

$\displaystyle\frac{1}{4}\sum_{i}[\partial^{2}\rho]_{ii}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\sum_{i}\lambda_{i}\mathrm{Re}(\langle\partial_{x_{a}}\lambda_{i}|\partial_{x_{b}}\lambda_{i}\rangle)$
(15)
$\displaystyle+\sum_{ik}\frac{1}{4}(\lambda_{i}+\lambda_{k})\mathrm{Re}(\langle\lambda_{i}|\partial_{x_{a}}\lambda_{k}\rangle\langle\partial_{x_{b}}\lambda_{k}|\lambda_{i}\rangle),$

where the fact $\sum_{i}\partial^{2}\lambda_{i}=0$ has been applied. Next,
from equation (11) one can obtain

$[W_{a}]_{ik}[W_{b}]_{ki}=\frac{1}{4}(\partial_{x_{a}}\lambda_{i})(\partial_{x_{b}}\lambda_{i})\\!+\\!\sum_{k}\frac{\lambda_{i}\lambda_{k}(\lambda_{i}-\lambda_{k})^{2}}{(\lambda_{i}+\lambda_{k})^{2}}\langle\lambda_{i}|\partial_{x_{a}}\lambda_{k}\rangle\langle\partial_{x_{b}}\lambda_{k}|\lambda_{i}\rangle,$

which means

$\displaystyle\sum_{ik}\frac{1}{2\lambda_{i}}\mathrm{Re}([W_{a}]_{ik}[W_{b}]_{ki})$
$\displaystyle=$
$\displaystyle\sum_{i}\frac{1}{8\lambda_{i}}(\partial_{x_{a}}\lambda_{i})(\partial_{x_{b}}\lambda_{i})+\sum_{ik}\frac{\lambda_{k}(\lambda_{i}-\lambda_{k})^{2}}{2(\lambda_{i}+\lambda_{k})^{2}}\mathrm{Re}(\langle\lambda_{i}|\partial_{x_{a}}\lambda_{k}\rangle\langle\partial_{x_{b}}\lambda_{k}|\lambda_{i}\rangle)$
$\displaystyle=$
$\displaystyle\sum_{i}\frac{1}{8\lambda_{i}}(\partial_{x_{a}}\lambda_{i})(\partial_{x_{b}}\lambda_{i})+\sum_{ik}\frac{(\lambda_{i}-\lambda_{k})^{2}}{4(\lambda_{i}+\lambda_{k})}\mathrm{Re}(\langle\lambda_{i}|\partial_{x_{a}}\lambda_{k}\rangle\langle\partial_{x_{b}}\lambda_{k}|\lambda_{i}\rangle)$

Thus, $\mathrm{Tr}Y_{ab}$ can then be expressed by

$\displaystyle\mathrm{Tr}Y_{ab}$ $\displaystyle=$
$\displaystyle-\frac{1}{8}\Bigg{[}\sum_{i}\frac{(\partial_{x_{a}}\lambda_{i})(\partial_{x_{b}}\lambda_{i})}{\lambda_{i}}+\sum_{i}4\lambda_{i}\mathrm{Re}(\langle\partial_{x_{a}}\lambda_{i}|\partial_{x_{b}}\lambda_{i}\rangle)$
(16)
$\displaystyle-\sum_{ik}\frac{8\lambda_{i}\lambda_{k}}{\lambda_{i}+\lambda_{k}}\mathrm{Re}\langle\lambda_{i}|\partial_{x_{a}}\lambda_{k}\rangle\langle\partial_{x_{b}}\lambda_{k}|\lambda_{i}\rangle\Bigg{]}$
$\displaystyle=$ $\displaystyle-\frac{1}{8}\mathcal{F}_{ab}.$

With this equation, we finally obtain

$D_{\mathrm{B}}(\rho(\vec{x}),\rho(\vec{x}+\mathrm{d}\vec{x}))=\sum_{ab}\frac{1}{4}\mathcal{F}_{ab}\mathrm{d}x_{a}x_{b}.$
(17)

One should notice that this proof shows that this relation is established for
density matrices with any rank as long as the rank of $\rho(\vec{x})$ is
unchanged with the varying of $\vec{x}$. In the case the rank can change, a
thorough discussion can be found in reference [87].

##  Appendix G Relation between QFIM and cross-correlation functions

This appendix gives the thorough calculation of the relation between QFIM and
dynamic susceptibility in reference [109]. Consider the unitary
parameterization $U=\exp(i\sum_{a}x_{a}O_{a})$ with a thermal state
$\rho=\frac{1}{Z}e^{-\beta H}$. Here $Z=\mathrm{Tr}(e^{-\beta H})$ is the
partition function. $O_{a}$ is a Hermitian generator for $x_{a}$. In the
following we set $k_{B}=1$ and assume all $O_{a}$ are commutative, i.e.,
$[O_{a},O_{b}]=0$ for any $a$ and $b$. Denote $O_{a}(t)=e^{iHt}O_{a}e^{-iHt}$,
and $\langle\cdot\rangle=\mathrm{Tr}(\rho\cdot)$, the symmetric cross-
correlation spectrum in this case reads

$S_{ab}(\omega)=\frac{1}{2}\int^{\infty}_{-\infty}\langle\\{Q_{a}(t),O_{b}\\}\rangle
e^{i\omega t}\mathrm{d}t.$ (1)

Utilizing the energy basis $\\{|E_{i}\rangle\\}$ (with $E_{i}$ the $i$th
energy), it can be rewritten into

$S_{ab}(\omega)=\frac{1}{2Z}\sum_{ij}(e^{-\beta E_{i}}+e^{-\beta
E_{j}})\langle E_{i}|O_{a}|E_{j}\rangle\langle
E_{j}|O_{b}|E_{i}\rangle\\!\\!\int^{\infty}_{-\infty}\\!\\!e^{i(\omega+E_{i}-E_{j})}\mathrm{d}t.$

Further use the equation
$\int^{\infty}_{-\infty}e^{i(\omega+E_{i}-E_{j})}\mathrm{d}t=2\pi\delta(\omega+E_{i}-E_{j})$,
$S_{ab}(\omega)$ can reduce to

$S_{ab}(\omega)=\frac{\pi}{Z}\sum_{ij}(e^{-\beta E_{i}}+e^{-\beta
E_{j}})\delta(\omega+E_{i}-E_{j})\langle E_{i}|O_{a}|E_{j}\rangle\langle
E_{j}|O_{b}|E_{i}\rangle.$ (2)

With this expression, one can find

$\displaystyle\int^{\infty}_{-\infty}\tanh^{2}\left(\frac{\omega}{2T}\right)\mathrm{Re}(S_{ab}(\omega))\mathrm{d}\omega$
$\displaystyle=$
$\displaystyle\sum_{ij}\int^{\infty}_{-\infty}\tanh^{2}\left(\frac{\omega}{2T}\right)\delta(\omega+E_{i}-E_{j})\mathrm{d}\omega$
$\displaystyle\times\frac{\pi}{Z}(e^{-\beta E_{i}}+e^{-\beta
E_{j}})\mathrm{Re}(\langle E_{i}|O_{a}|E_{j}\rangle\langle
E_{j}|O_{b}|E_{i}\rangle)$ $\displaystyle=$
$\displaystyle\sum_{ij}\tanh^{2}\left(\frac{E_{j}-E_{i}}{2T}\right)\frac{\pi}{Z}(e^{-\beta
E_{i}}+e^{-\beta E_{j}})\mathrm{Re}(\langle E_{i}|O_{a}|E_{j}\rangle\langle
E_{j}|O_{b}|E_{i}\rangle).$

Since

$\displaystyle\tanh\left(\frac{E_{i}-E_{j}}{2T}\right)$ $\displaystyle=$
$\displaystyle\frac{[e^{\frac{1}{2}\beta(E_{i}-E_{j})}-e^{-\frac{1}{2}\beta(E_{i}-E_{j})}]e^{-\frac{1}{2}\beta(E_{i}+E_{j})}}{[e^{\frac{1}{2}\beta(E_{i}-E_{j})}+e^{-\frac{1}{2}\beta(E_{i}-E_{j})}]e^{-\frac{1}{2}\beta(E_{i}+E_{j})}}$
(3) $\displaystyle=$ $\displaystyle\frac{e^{-\beta E_{j}}-e^{-\beta
E_{i}}}{e^{-\beta E_{i}}+e^{-\beta E_{j}}},$

one can obtain

$\displaystyle\int^{\infty}_{-\infty}\tanh^{2}\left(\frac{\omega}{2T}\right)\mathrm{Re}(S_{ab}(\omega))\mathrm{d}\omega$
(4) $\displaystyle=$ $\displaystyle\pi\sum_{ij}\frac{\left(\frac{1}{Z}e^{\beta
E_{i}}-\frac{1}{Z}e^{-\beta E_{j}}\right)^{2}}{\frac{1}{Z}e^{-\beta
E_{i}}+\frac{1}{Z}e^{-\beta E_{j}}}\mathrm{Re}(\langle
E_{i}|O_{a}|E_{j}\rangle\langle E_{j}|O_{b}|E_{i}\rangle).$ (5)

From the expression of QFIM, it can be found that

$\mathcal{F}_{ab}=\frac{4}{\pi}\int^{\infty}_{-\infty}\tanh^{2}\left(\frac{\omega}{2T}\right)\mathrm{Re}(S_{ab}(\omega))\mathrm{d}\omega.$
(6)

It can also be checked that

$\mathrm{Re}(S_{ab}(\omega))=\frac{1}{2}\int^{\infty}_{-\infty}\langle
Q_{a}(t)O_{b}+O_{b}(t)O_{a}\rangle e^{i\omega t}\mathrm{d}t.$ (7)

In the mean time, the asymmetric cross-correlation spectrum is in the form

$\displaystyle\chi_{ab}(\omega)$ $\displaystyle=$
$\displaystyle\frac{i}{2}\int^{\infty}_{-\infty}e^{i\omega
t}\langle[O_{a}(t),O_{b}]\rangle\mathrm{d}t$ $\displaystyle=$ $\displaystyle
i\frac{\pi}{Z}\sum_{ij}\delta(\omega+E_{i}-E_{j})(e^{-\beta E_{i}}-e^{-\beta
E_{j}})\langle E_{i}|O_{a}|E_{j}\rangle\langle E_{j}|O_{b}|E_{i}\rangle,$

which directly gives the relation between $\chi_{ab}$ and $S_{ab}$ as below

$\displaystyle\mathrm{Im}(\chi_{ab}(\omega))$ $\displaystyle=$
$\displaystyle\frac{\pi}{Z}\sum_{ij}\tanh\left(\frac{E_{j}-E_{i}}{2T}\right)(e^{-\beta
E_{i}}+e^{-\beta E_{j}})\delta(\omega+E_{i}-E_{j})$ (8)
$\displaystyle\times\mathrm{Re}(\langle E_{i}|O_{a}|E_{j}\rangle\langle
E_{j}|O_{b}|E_{i}\rangle)$ $\displaystyle=$
$\displaystyle\frac{\pi}{Z}\tanh\left(\frac{\omega}{2T}\right)\sum_{ij}(e^{-\beta
E_{i}}+e^{-\beta E_{j}})\delta(\omega+E_{i}-E_{j})$
$\displaystyle\times\mathrm{Re}(\langle E_{i}|O_{a}|E_{j}\rangle\langle
E_{j}|O_{b}|E_{i}\rangle)$ $\displaystyle=$
$\displaystyle\tanh\left(\frac{\omega}{2T}\right)\mathrm{Re}(S_{ab}(\omega)),$

namely,

$\mathrm{Im}(\chi_{ab}(\omega))=\tanh\left(\frac{\omega}{2T}\right)\mathrm{Re}(S_{ab}(\omega)),$
(9)

which is just the fluctuation-dissipation theorem. Using this relation, one
can further obtain the result in reference [109] as below

$\mathcal{F}_{ab}=\frac{4}{\pi}\int^{\infty}_{-\infty}\tanh\left(\frac{\omega}{2T}\right)\mathrm{Im}(\chi_{ab}(\omega))\mathrm{d}\omega.$
(10)

##  Appendix H Derivation of quantum multiparameter Cramér-Rao bound

The derivation of quantum multiparameter Cramér-Rao bound is based on the
Cauchy-Schwarz inequality below

$\mathrm{Tr}(X^{\dagger}X)\mathrm{Tr}(Y^{\dagger}Y)\geq\frac{1}{4}|\mathrm{Tr}(X^{\dagger}Y+XY^{\dagger})|^{2},$
(1)

which comes from the complete form

$\mathrm{Tr}(X^{\dagger}X)\mathrm{Tr}(Y^{\dagger}Y)\geq\frac{1}{4}|\mathrm{Tr}(X^{\dagger}Y+XY^{\dagger})|^{2}+\frac{1}{4}|\mathrm{Tr}(X^{\dagger}Y-XY^{\dagger})|^{2}.$
(2)

Define $X$ and $Y$ as

$\displaystyle X$ $\displaystyle:=$
$\displaystyle\sum_{m}f_{m}L_{m}\sqrt{\rho},$ (3) $\displaystyle Y$
$\displaystyle:=$ $\displaystyle\sum_{m}g_{m}(O_{m}-\langle
O_{m}\rangle)\sqrt{\rho},$ (4)

where $f_{m},g_{m}$ are real numbers for any $m$, and
$\langle\cdot\rangle=\mathrm{Tr}(\cdot\rho)$. Here $O_{m}$ is an observable
defined as

$O_{m}:=\sum_{k}\hat{x}_{m}(k)\Pi_{k}.$ (5)

$\hat{x}_{m}(k)$ is the estimator of $x_{m}$ and is the function of $k$th
result. Based on equation (3), it can be calculated that
$\mathrm{Tr}(X^{\dagger}X)=\sum_{ml}f_{m}f_{l}\mathrm{Tr}(L_{m}L_{l}\rho)$,
which can be symmetrized into
$\mathrm{Tr}(X^{\dagger}X)=\frac{1}{2}\sum_{ml}f_{m}f_{l}\mathrm{Tr}(\\{L_{m},L_{l}\\}\rho)=\sum_{ml}f_{m}f_{l}\mathcal{F}_{ab}$.
Define $\vec{f}=(f_{0},f_{1},...,f_{m},...)^{\mathrm{T}}$,
$\mathrm{Tr}(X^{\dagger}X)$ can be further rewritten into

$\mathrm{Tr}(X^{\dagger}X)=\vec{f}^{\,\mathrm{T}}\mathcal{F}\vec{f}.$ (6)

Similarly, define $\vec{g}=(g_{0},g_{1},...,g_{m},...)^{\mathrm{T}}$ and
through some algebra, $\mathrm{Tr}(Y^{\dagger}Y)$ can be calculated as

$\mathrm{Tr}(Y^{\dagger}Y)=\vec{g}^{\,\mathrm{T}}C\vec{g},$ (7)

where $C$ is the covariance matrix for $\\{O_{m}\\}$, and is defined as
$C_{ml}=\frac{1}{2}\langle\\{O_{m},O_{l}\\}\rangle-\langle O_{m}\rangle\langle
O_{l}\rangle$. Furthermore, one can also obtain

$\frac{1}{2}\mathrm{Tr}(X^{\dagger}Y+XY^{\dagger})=\vec{f}^{\,\mathrm{T}}B\vec{g},$
(8)

where the entry of $B$ is defined as
$B_{ml}=\frac{1}{2}\mathrm{Tr}(\rho\\{L_{m},\Pi_{l}\\})=\frac{1}{2}\mathrm{Tr}(\\{\rho,L_{m}\\}O_{l})$.
Utilizing $\partial_{x_{a}}\rho=\frac{1}{2}\\{\rho,L_{m}\\}$, $B_{ml}$ reduces
to $B_{ml}=\mathrm{Tr}(\Pi_{l}\partial_{x_{m}}\rho)$. Since we consider
unbiased estimators, i.e., $\langle
O_{m}\rangle=\sum_{m}\hat{x}_{m}\mathrm{Tr}(\rho\Pi_{k})=x_{m}$, $B_{ml}$
further reduces to $B_{ml}=\partial_{x_{m}}\langle O_{l}\rangle=\delta_{ml}$,
with $\delta_{ml}$ the Kronecker delta function. Hence, for unbiased
estimators $B$ is actually the identity matrix 11.

Now substituting equations (6), (7) and (8) into the Cauchy-Schwarz inequality
(1), one can obtain

$\vec{f}^{\,\mathrm{T}}\mathcal{F}\vec{f}\vec{g}^{\,\mathrm{T}}C\vec{g}\geq\left(\vec{f}^{~{}\mathrm{T}}\vec{g}\right)^{2}.$
(9)

Assuming $\mathcal{F}$ is invertable, i.e., it is positive-definite, and
taking $\vec{f}=\mathcal{F}^{-1}\vec{g}$, above inequality reduces to
$\vec{g}^{\,\mathrm{T}}\mathcal{F}^{-1}\vec{g}\vec{g}^{\,\mathrm{T}}C\vec{g}\geq\left(\vec{g}^{\,\mathrm{T}}\mathcal{F}^{-1}\vec{g}\right)^{2}$.
Since $\mathcal{F}$ is positive-definite, $\mathcal{F}^{-1}$ is also positive-
definite, which means $\vec{g}^{\,\mathrm{T}}\mathcal{F}^{-1}\vec{g}$ is a
positive number, thus, the above equation can further reduce to
$\vec{g}^{\,\mathrm{T}}C\vec{g}\geq\vec{g}^{\,\mathrm{T}}\mathcal{F}^{-1}\vec{g}$,
namely,

$C\geq\mathcal{F}^{-1}.$ (10)

Next we discuss the relation between $C$ and
$\mathrm{cov}(\hat{\vec{x}},\\{\Pi_{k}\\})$. Utilizing the defintion of
$O_{m}$, $C_{ml}$ can be written as

$\displaystyle C_{ml}$ $\displaystyle=$
$\displaystyle\sum_{kk^{\prime}}\hat{x}_{m}(k)\hat{x}_{l}(k^{\prime})\frac{1}{2}\mathrm{Tr}(\rho\\{\Pi_{k},\Pi_{k^{\prime}}\\})$
(11)
$\displaystyle-\left[\sum_{k}\hat{x}_{m}\mathrm{Tr}(\rho\Pi_{k})\right]\left[\sum_{k}\hat{x}_{l}\mathrm{Tr}(\rho\Pi_{k})\right],$

and $\mathrm{cov}(\hat{\vec{x}},\\{\Pi_{k}\\})$ for unbiased estimators reads

$\mathrm{cov}(\hat{\vec{x}},\\{\Pi_{k}\\})=\sum_{k}\hat{x}_{m}\hat{x}_{l}\mathrm{Tr}(\rho\Pi_{k})-x_{m}x_{l}.$
(12)

If $\\{\Pi_{k}\\}$ is a set of projection operators, it satisfies
$\Pi_{k}\Pi_{k^{\prime}}=\Pi_{k}\delta_{kk^{\prime}}$. For unbiased
estimators, $\sum_{k}\hat{x}_{m(l)}\mathrm{Tr}(\rho\Pi_{k})=x_{m(l)}$, which
gives
$C_{ml}=\sum_{k}\hat{x}_{m}\hat{x}_{l}\mathrm{Tr}(\rho\Pi_{k})-x_{m}x_{l}$,
i.e.,

$C=\mathrm{cov}(\hat{\vec{x}},\\{\Pi_{k}\\}).$ (13)

If $\\{\Pi_{k}\\}$ is a set of POVM, one can see

$\displaystyle\vec{g}^{\,\mathrm{T}}\mathrm{cov}(\hat{\vec{x}},\\{\Pi_{k}\\})\vec{g}$
$\displaystyle=$
$\displaystyle\sum_{k}\sum_{ml}g_{m}g_{l}\hat{x}_{m}\hat{x}_{l}\mathrm{Tr}(\rho\Pi_{k})-\sum_{ml}g_{m}g_{l}x_{m}x_{l}$
(14) $\displaystyle=$
$\displaystyle\sum_{k}\left(\sum_{m}g_{m}\hat{x}_{m}\right)^{\\!\\!2}\mathrm{Tr}(\rho\Pi_{k})-\sum_{ml}g_{m}g_{l}x_{m}x_{l}.$

In the mean time,

$\vec{g}^{\,\mathrm{T}}C\vec{g}=\frac{1}{2}\sum_{ml}g_{m}g_{l}\mathrm{Tr}\left(\rho
O_{m}O_{l}+\rho O_{l}O_{m}\right)-\sum_{ml}g_{m}g_{l}x_{m}x_{l}.$ (15)

Now define
$\mathcal{B}_{k}:=\sqrt{\Pi_{k}}\left(\sum_{m}g_{m}\hat{x}_{m}-\sum_{m}g_{m}O_{m}\right)$.
Based on the Cauchy-Schwarz inequality $\mathrm{Tr}(A^{\dagger}A)\geq 0$,
which is valid for any operator $A$, one can immediately obtain
$\mathrm{Tr}(\rho\mathcal{B}^{\dagger}_{k}\mathcal{B}_{k})=\mathrm{Tr}(\sqrt{\rho}\mathcal{B}^{\dagger}_{k}\mathcal{B}_{k}\sqrt{\rho})\geq
0$, which further gives
$\sum_{k}\mathrm{Tr}(\rho\mathcal{B}^{\dagger}_{k}\mathcal{B}_{k})\geq 0$.
Through some calculations, the expression of
$\sum_{k}\mathrm{Tr}(\rho\mathcal{B}^{\dagger}_{k}\mathcal{B}_{k})$ is in the
form

$\sum_{k}\mathrm{Tr}(\rho\mathcal{B}^{\dagger}_{k}\mathcal{B}_{k})=\sum_{k}\\!\left(\\!\sum_{m}g_{m}\hat{x}_{m}\\!\right)^{\\!2}\\!\\!\mathrm{Tr}(\rho\Pi_{k})-\sum_{ml}g_{m}g_{l}\mathrm{Tr}(\rho
O_{m}O_{l}).$ (16)

Now taking the difference of equations (14) and (15), it can be found

$\sum_{k}\mathrm{Tr}(\rho\mathcal{B}^{\dagger}_{k}\mathcal{B}_{k})=\vec{g}^{\,\mathrm{T}}\left(\mathrm{cov}(\hat{\vec{x}},\\{\Pi_{k}\\})-C\right)\vec{g}.$
(17)

Finally, we obtain the following inequality
$\vec{g}^{\,\mathrm{T}}(\mathrm{cov}(\hat{\vec{x}},\\{\Pi_{k}\\})-C)\vec{g}\geq
0$, namely, for any POVM measurement,

$\mathrm{cov}(\hat{\vec{x}},\\{\Pi_{k}\\})\geq C.$ (18)

Based on inequality (10) and the property of quadratic form, we finally obtain
$\mathrm{cov}(\hat{\vec{x}},\\{\Pi_{k}\\})\geq\mathcal{F}^{-1}$. Consider the
repetition of experiments (denoted as $n$), above bound needs to add a factor
of $1/n$. Hence the quantum multiparameter Cramér-Rao bound can be finally
expressed by

$\mathrm{cov}(\hat{\vec{x}},\\{\Pi_{k}\\})\geq\frac{1}{n}\mathcal{F}^{-1}.$
(19)

##  Appendix I Construction of optimal measurement for pure states

Assume the true values of the vector of unknown parameters $\vec{x}$ is
$\vec{x}_{\mathrm{true}}$, we now provide the proof that for a pure
parameterized state $|\psi\rangle$, a set of projectors containing the state
$|\psi_{\vec{x}_{\mathrm{true}}}\rangle:=|\psi(\vec{x}=\vec{x}_{\mathrm{true}})\rangle$,
i.e., $\\{|m_{k}\rangle\langle
m_{k},|m_{0}\rangle=|\psi_{\vec{x}_{\mathrm{true}}}\rangle\\}$ is possible to
be an optimal measurement to attain the quantum Cramér-Rao bound, as shown in
reference [151, 136]. The calculation in this appendix basically coincides
with the appendix in reference [136].

Since $\\{|m_{k}\rangle\langle m_{k}|\\}$ contains the information of the true
value, in practice one need to use the estimated value of $\vec{x}$ (denoted
as $\hat{\vec{x}}$) to construct $|m_{0}\rangle=|\psi(\hat{\vec{x}})\rangle$
to perform this measurement, then improve the accuracy of $\hat{\vec{x}}$
adaptively. Thus, it is reasonable to assume
$|\psi_{\vec{x}_{\mathrm{true}}}\rangle=|m_{0}\rangle+\sum_{x_{a}}\delta
x_{a}|\partial_{x_{a}}\psi\rangle|_{\vec{x}=\hat{\vec{x}}}$. The probability
for $|m_{k}\rangle\langle m_{k}|$ is $p_{k}=|\langle\psi|m_{k}\rangle|^{2}$,
which gives the CFIM as below

$\mathcal{I}_{ab}(\hat{\vec{x}})=\sum_{k}\frac{4\mathrm{Re}(\langle
m_{k}|\partial_{x_{a}}\psi\rangle\langle\psi|m_{k}\rangle)\mathrm{Re}(\langle
m_{k}|\partial_{x_{b}}\psi\rangle\langle\psi|m_{k}\rangle)}{|\langle\psi|m_{k}\rangle|^{2}}.$
(20)

At the limit $\hat{\vec{x}}\rightarrow\vec{x}_{\mathrm{true}}$, it is

$\mathcal{I}_{ab}(\vec{x}_{\mathrm{true}})=\lim_{\hat{\vec{x}}\rightarrow\vec{x}_{\mathrm{true}}}\\!\sum_{k}\frac{4\mathrm{Re}(\langle
m_{k}|\partial_{x_{a}}\psi\rangle\langle\psi|m_{k}\rangle)\mathrm{Re}(\langle
m_{k}|\partial_{x_{b}}\psi\rangle\langle\psi|m_{k}\rangle)}{|\langle\psi|m_{k}\rangle|^{2}}.$
(21)

For the $k=0$ term,

$\displaystyle\lim_{\hat{\vec{x}}\rightarrow\vec{x}_{\mathrm{true}}}\mathrm{Re}(\langle
m_{0}|\partial_{x_{a}}\psi\rangle\langle\psi|m_{0}\rangle)$ $\displaystyle=$
$\displaystyle\mathrm{Re}(\langle\psi_{\vec{x}_{\mathrm{true}}}|\partial_{x_{a}}\psi\rangle)$
(22) $\displaystyle=$
$\displaystyle\mathrm{Re}(\langle\psi|\partial_{x_{a}}\psi\rangle)|_{\vec{x}=\vec{x}_{\mathrm{true}}}=0,$

and
$\lim_{\vec{x}\rightarrow\vec{x}_{\mathrm{true}}}\langle\psi|m_{0}\rangle=1$.
Thus, the CFIM is

$\mathcal{I}_{ab}(\vec{x}_{\mathrm{true}})=\lim_{\hat{\vec{x}}\rightarrow\vec{x}_{\mathrm{true}}}\\!\sum_{k\neq
0}\frac{4\mathrm{Re}(\langle
m_{k}|\partial_{x_{a}}\psi\rangle\langle\psi|m_{k}\rangle)\mathrm{Re}(\langle
m_{k}|\partial_{x_{b}}\psi\rangle\langle\psi|m_{k}\rangle)}{|\langle\psi|m_{k}\rangle|^{2}}\\!.$
(23)

Due to the fact

$\displaystyle 4\mathrm{Re}(\langle
m_{k}|\partial_{x_{a}}\psi\rangle\langle\psi|m_{k}\rangle)\mathrm{Re}(\langle
m_{k}|\partial_{x_{b}}\psi\rangle\langle\psi|m_{k}\rangle)$ (24)
$\displaystyle=$ $\displaystyle
2\mathrm{Re}(\langle\partial_{x_{a}}\psi|m_{k}\rangle\langle
m_{k}|\partial_{x_{b}}\psi\rangle)|\langle\psi|m_{k}\rangle|^{2}$
$\displaystyle+2\mathrm{Re}(\langle m_{k}|\partial_{x_{a}}\psi\rangle\langle
m_{k}|\partial_{x_{b}}\psi\rangle\langle\psi|m_{k}\rangle^{2}),$

one can have

$\displaystyle\mathcal{I}_{ab}(\vec{x}_{\mathrm{true}})=\mathcal{F}_{ab}(\vec{x}_{\mathrm{true}})-Q_{ab},$
(25)

where

$\displaystyle Q_{ab}$ $\displaystyle:=$
$\displaystyle\lim_{\hat{\vec{x}}\rightarrow\vec{x}_{\mathrm{true}}}\sum_{k\neq
0}\frac{2\mathrm{Re}(\langle
m_{k}|\partial_{x_{a}}\psi\rangle\langle\partial_{x_{b}}\psi|m_{k}\rangle)|\langle\psi|m_{k}\rangle|^{2}}{|\langle\psi|m_{k}\rangle|^{2}}$
(26) $\displaystyle-\frac{2\mathrm{Re}(\langle
m_{k}|\partial_{x_{a}}\psi\rangle\langle
m_{k}|\partial_{x_{b}}\psi\rangle\langle\psi|m_{k}\rangle^{2})}{|\langle\psi|m_{k}\rangle|^{2}}$
$\displaystyle=$
$\displaystyle\lim_{\hat{\vec{x}}\rightarrow\vec{x}_{\mathrm{true}}}\sum_{k\neq
0}\frac{4\mathrm{Im}(\langle\partial_{x_{a}}\psi|m_{k}\rangle\langle
m_{k}|\psi\rangle)\mathrm{Im}(\langle\partial_{x_{b}}\psi|m_{k}\rangle\langle
m_{k}|\psi\rangle)}{|\langle\psi|m_{k}\rangle|^{2}}.$

To let
$\mathcal{I}(\vec{x}_{\mathrm{true}})=\mathcal{F}(\vec{x}_{\mathrm{true}})$,
$Q$ has to be a zero matrix. Since the diagonal entry

$Q_{aa}=\lim_{\hat{\vec{x}}\rightarrow\vec{x}_{\mathrm{true}}}\sum_{k\neq
0}\frac{4\mathrm{Im}^{2}(\langle\partial_{x_{a}}\psi|m_{k}\rangle\langle
m_{k}|\psi\rangle)}{|\langle\psi|m_{k}\rangle|^{2}},$ (27)

in which all the terms within the summation are non-negative. Therefore, its
value is zero if and only if

$\lim_{\hat{\vec{x}}\rightarrow\vec{x}_{\mathrm{true}}}\frac{\mathrm{Im}(\langle\partial_{x_{a}}\psi|m_{k}\rangle\langle
m_{k}|\psi\rangle)}{|\langle\psi|m_{k}\rangle|}=0,~{}\forall x_{a},k\neq 0.$
(28)

Furthermore, this condition simultaneously makes the non-diagonal entries of
$Q$ vanish. Thus, it is the necessary and sufficient condition for $Q=0$,
which means it is also the necessary and sufficient condition for
$\mathcal{I}=\mathcal{F}$ at the point of true value. One may notice that the
limitation in equation (28) is a $0/0$ type. Thus, we use the formula
$|\psi_{\vec{x}_{\mathrm{true}}}\rangle=|m_{0}\rangle+\sum_{x_{j}}\delta
x_{j}|\partial_{x_{j}}\psi\rangle|_{\vec{x}=\hat{\vec{x}}}$ to further
calculate above equation. With this formula, one has

$\frac{\mathrm{Im}(\langle\partial_{x_{a}}\psi|m_{k}\rangle\langle
m_{k}|\psi\rangle)}{|\langle\psi|m_{k}\rangle|}=\frac{\sum_{j}\delta
x_{j}\mathrm{Im}(\langle\partial_{x_{a}}\psi|m_{k}\rangle\langle
m_{k}|\partial_{x_{j}}\psi\rangle)}{|\sum_{j}\delta
x_{j}\langle\partial_{x_{j}}\psi|m_{k}\rangle|}.$ (29)

$\langle\partial_{x_{j}}\psi|m_{k}\rangle$ cannot generally be zero for all
$x_{j}$ since all $\partial_{x_{j}}\psi\rangle$ are not orthogonal in general.
Thus, equation (28) is equivalent to

$\mathrm{Im}(\langle\partial_{x_{a}}\psi|m_{k}\rangle\langle
m_{k}|\partial_{x_{b}}\psi\rangle)=0,~{}\forall x_{a},x_{b},k\neq 0.$ (30)

##  Appendix J Gradient in GRAPE for Hamiltonian estimation

###  J.1 Gradient of CFIM

The core of GRAPE algorithm is to obtain the expression of gradient. The
dynamics of the system is described by

$\partial_{t}\rho=\mathcal{E}_{\vec{x}}\rho.$ (1)

For the Hamiltonian estimation under control, the dynamics is

$\partial_{t}\rho=-i[H_{0}(\vec{x})+H_{\mathrm{c}},\rho]+\mathcal{L}\rho,$ (2)

where $H_{\mathrm{c}}=\sum^{p}_{k=1}V_{k}(t)H_{k}$ is the control Hamiltonian
with $H_{k}$ the $k$th control and $V_{k}(t)$ the corresponding time-dependent
control amplitude. To perform the algorithm, the entire evolution time $T$ is
cut into $m$ parts with time interval $\Delta t$, i.e., $m\Delta t=T$.
$V_{k}(t)$ within the $j$th time interval is denoted as $V_{k}(j)$ and is
assumed to be a constant.

For a set of probability distribution $p(y|\vec{x})=\mathrm{Tr}(\rho\Pi_{y})$
with $\\{\Pi_{y}\\}$ a set of POVM. The gradient of $\mathcal{I}_{ab}$ at
target time $T$ reads [172]

$\displaystyle\frac{\delta\mathcal{I}_{ab}(T)}{\delta V_{k}(j)}$
$\displaystyle=$ $\displaystyle\Delta
t\mathrm{Tr}\left(\tilde{L}_{2,ab}\mathcal{M}^{(1)}_{j}\right)-\Delta^{2}t\mathrm{Tr}\left[\tilde{L}_{1,b}\left(\mathcal{M}^{(2)}_{j,a}+\mathcal{M}^{(3)}_{j,a}\right)\right]$
(3)
$\displaystyle-\Delta^{2}t\mathrm{Tr}\left[\tilde{L}_{1,a}\left(\mathcal{M}^{(2)}_{j,b}+\mathcal{M}^{(3)}_{j,b}\right)\right],$

where

$\displaystyle\tilde{L}_{1,a(b)}$ $\displaystyle=$
$\displaystyle\sum_{y}\left[\partial_{x_{a(b)}}\ln
p(y|\vec{x})\right]\Pi_{y},$ (4) $\displaystyle\tilde{L}_{2,ab}$
$\displaystyle=$ $\displaystyle\sum_{y}\left[\partial_{x_{a}}\ln
p(y|\vec{x})\right]\left[\partial_{x_{b}}\ln p(y|\vec{x})\right]\Pi_{y},$ (5)

and $\mathcal{M}^{(1)}_{j}$, $\mathcal{M}^{(2)}_{j,a(b)}$ and
$\mathcal{M}^{(3)}_{j,a(b)}$ are Hermitian operators and can be expressed by

$\displaystyle\mathcal{M}^{(1)}_{j}$ $\displaystyle=$ $\displaystyle
i\mathcal{D}^{m}_{j+1}H^{\times}_{k}(\rho_{j}),$ (6)
$\displaystyle\mathcal{M}^{(2)}_{j,a(b)}$ $\displaystyle=$
$\displaystyle\sum^{j}_{i=1}\mathcal{D}^{m}_{j+1}H^{\times}_{k}\mathcal{D}^{j}_{i+1}(\partial_{x_{a(b)}}H_{0})^{\times}(\rho_{i}),$
(7) $\displaystyle\mathcal{M}^{(3)}_{j,a(b)}$ $\displaystyle=$
$\displaystyle(1-\delta_{jm})\sum^{m}_{i=j+1}\mathcal{D}^{m}_{i+1}(\partial_{x_{a(b)}}H_{0})^{\times}\mathcal{D}^{i}_{j+1}H^{\times}_{k}(\rho_{j}).$
(8)

The notation $A^{\times}(\cdot):=[A,\cdot]$ is a superoperator. $\delta_{jm}$
is the Kronecker delta function. $D^{j^{\prime}}_{j}$ is the propagating
superoperator from the $j$th time point to the $j^{\prime}$th time with the
definition $D^{j^{\prime}}_{j}:=\prod^{j^{\prime}}_{i=j}\exp(\Delta
t\mathcal{E}_{i})$ for $j\leq j^{\prime}$. We define
$D^{j^{\prime}}_{j}=\leavevmode\hbox{\small 1\kern-4.3pt\normalsize 1}$ for
$j>j^{\prime}$. $\rho_{j}=D^{j}_{1}\rho(0)$ is the quantum state at $j$th
time.

For a two-parameter case $\vec{x}=(x_{0},x_{1})$, the objective function can
be chosen as $F_{\mathrm{eff}}(T)$ according to corollary 3.1.3, and the
corresponding gradient is

$\frac{\delta\mathcal{F}_{\mathrm{eff}}(T)}{\delta
V_{k}(j)}=\frac{\mathcal{I}^{2}_{11}+\mathcal{I}^{2}_{01}}{(\mathcal{I}_{00}+\mathcal{I}_{11})^{2}}\frac{\delta\mathcal{I}_{00}}{\delta
V_{k}(j)}+\frac{\mathcal{I}^{2}_{00}+\mathcal{I}^{2}_{01}}{(\mathcal{I}_{00}+\mathcal{I}_{11})^{2}}\frac{\delta\mathcal{I}_{11}}{\delta
V_{k}(j)}-\frac{2\mathcal{I}_{01}}{\mathcal{I}_{00}+\mathcal{I}_{11}}\frac{\delta\mathcal{I}_{01}}{\delta
V_{k}(j)}.$ (9)

For the objective function

$f_{0}(T)=\left(\sum_{a}\frac{1}{\mathcal{I}_{aa}(T)}\right)^{-1},$ (10)

the gradient reads

$\frac{\delta f_{0}(T)}{\delta
V_{k}(j)}=\sum_{a}\left(\frac{f_{0}}{\mathcal{I}_{aa}}\right)^{2}\frac{\delta\mathcal{I}_{aa}}{\delta
V_{k}(j)}.$ (11)

###  J.2 Gradient of QFIM

Now we calculate the gradient of the QFIM. Based on the equation

$\partial_{x_{a}}\rho(T)=\frac{1}{2}(\rho(T)L_{x_{a}}(T)+L_{x_{a}}(T)\rho(T)),$
(12)

we can obtain

$\displaystyle\mathrm{Tr}\left(\partial_{x_{a}}\frac{\delta\rho(T)}{\delta
V_{k}(j)}L_{x_{b}}\right)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\mathrm{Tr}\left(\frac{\delta\rho(T)}{\delta
V_{k}(j)}\left\\{L_{x_{a}}(T),L_{x_{b}}(T)\right\\}\right)$ (13)
$\displaystyle+\frac{1}{2}\mathrm{Tr}\left(\frac{\delta L_{x_{a}}(T)}{\delta
V_{k}(j)}\left\\{\rho(T),L_{x_{b}}(T)\right\\}\right).$

Similarly, we have

$\displaystyle\mathrm{Tr}\left(\partial_{x_{b}}\frac{\delta\rho(T)}{\delta
V_{k}(j)}L_{x_{a}}\right)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\mathrm{Tr}\left(\frac{\delta\rho(T)}{\delta
V_{k}(j)}\left\\{L_{x_{a}}(T),L_{x_{b}}(T)\right\\}\right)$ (14)
$\displaystyle+\frac{1}{2}\mathrm{Tr}\left(\frac{\delta L_{x_{b}}(T)}{\delta
V_{k}(j)}\left\\{\rho(T),L_{x_{a}}(T)\right\\}\right).$

Next, from the definition of QFIM, the gradient for $\mathcal{F}_{ab}$ at
target time $T$ reads [172]

$\displaystyle\frac{\delta\mathcal{F}_{ab}(T)}{\delta V_{k}(j)}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\mathrm{Tr}\\!\left(\\!\frac{\delta\rho(T)}{\delta
V_{k}(j)}\\!\left\\{L_{x_{a}}(T),L_{x_{b}}(T)\right\\}\\!\\!\right)\\!+\\!\frac{1}{2}\mathrm{Tr}\\!\left(\\!\frac{\delta
L_{x_{a}}(T)}{\delta
V_{k}(j)}\\!\left\\{\rho(T),L_{x_{b}}(T)\right\\}\\!\\!\right)$ (15)
$\displaystyle+\frac{1}{2}\mathrm{Tr}\\!\left(\frac{\delta
L_{x_{b}}(T)}{\delta V_{k}(j)}\left\\{\rho(T),L_{x_{a}}(T)\right\\}\right).$

Combing equations (13), (14) and (15), one can obtain

$\displaystyle\frac{\delta\mathcal{F}_{ab}(T)}{\delta V_{k}(j)}$
$\displaystyle=$
$\displaystyle\mathrm{Tr}\left(\partial_{x_{a}}\frac{\delta\rho(T)}{\delta
V_{k}(j)}L_{x_{b}}(T)\right)+\mathrm{Tr}\left(\partial_{x_{b}}\frac{\delta\rho(T)}{\delta
V_{k}(j)}L_{x_{a}}(T)\right)$ (16)
$\displaystyle-\frac{1}{2}\mathrm{Tr}\left(\frac{\delta\rho(T)}{\delta
V_{k}(j)}\left\\{L_{x_{a}}(T),L_{x_{b}}(T)\right\\}\right).$

Substituing the specific expressions of $\frac{\delta\rho(T)}{\delta
V_{k}(j)}$ given in reference [172], one can obtain the gradient of
$\mathcal{F}_{ab}$ as below

$\displaystyle\frac{\delta\mathcal{F}_{ab}(T)}{\delta V_{k}(j)}$
$\displaystyle=$ $\displaystyle\frac{1}{2}\Delta
t\mathrm{Tr}\left(\left\\{L_{x_{a}}(T),L_{x_{b}}(T)\right\\}\mathcal{M}^{(1)}_{j}\right)$
(17)
$\displaystyle-\Delta^{2}t\mathrm{Tr}\left[L_{x_{b}}(T)\left(\mathcal{M}^{(2)}_{j,a}+\mathcal{M}^{(3)}_{j,a}\right)\right]$
$\displaystyle-\Delta^{2}t\mathrm{Tr}\left[L_{x_{a}}(T)\left(\mathcal{M}^{(2)}_{j,b}+\mathcal{M}^{(3)}_{j,b}\right)\right].$

The gradient for the diagonal entry $\mathcal{F}_{aa}$ reduces to the form in
reference [172], i.e.,

$\displaystyle\frac{\delta\mathcal{F}_{aa}(T)}{\delta V_{k}(j)}$
$\displaystyle=$ $\displaystyle\Delta
t\mathrm{Tr}\\!\left(L^{2}_{x_{a}}(T)\mathcal{M}^{(1)}_{j}\right)-2\Delta^{2}t\mathrm{Tr}\\!\left[\\!L_{x_{a}}(T)\\!\left(\mathcal{M}^{(2)}_{j,a}+\mathcal{M}^{(3)}_{j,a}\right)\\!\right]\\!.$
(18)

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