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##  Appendix A Unitary Equivalence of Encoding Dynamics

In this appendix we go over the proofs of the interplay between the dynamics
and the initial state. Starting on the first theorem:

See 2.1

###### Proof.

Since $G$ and $G^{\prime}$ are generators for single qubit unitary operators,
then they can be decomposed into a linear combination of Pauli operators. Let
them be $G=\vec{a}\cdot\vec{\sigma}$ and
$G^{\prime}=\vec{b}\cdot\vec{\sigma}$, such that $|\vec{a}|=|\vec{b}|=1$.
Using the expression for the adjoint action on the Pauli vector

$R_{\vec{n}}(-\alpha)\ \vec{\sigma}\
R_{\vec{n}}(\alpha)=\vec{\sigma}\cos(\alpha)+\vec{n}\times\vec{\sigma}\sin(\alpha)+\vec{n}\
\vec{n}\cdot\vec{\sigma}(1-\cos(\alpha))$ (57)

Taking the inner product with $\vec{a}$ we get that:

$\displaystyle R_{\vec{n}}(-\alpha)\ \vec{a}\cdot\vec{\sigma}\
R_{\vec{n}}(\alpha)$
$\displaystyle=\vec{a}\cdot\vec{\sigma}\cos(\alpha)+\vec{a}\cdot(\vec{n}\times\vec{\sigma})\sin(\alpha)+\vec{a}\cdot\vec{n}\
\vec{n}\cdot\vec{\sigma}(1-\cos(\alpha))$
$\displaystyle=\left[\vec{a}\cos(\alpha)+\vec{a}\times\vec{n}\sin(\alpha)+\vec{n}\
\vec{a}\cdot\vec{n}(1-\cos(\alpha))\right]\cdot\vec{\sigma}$
$\displaystyle=\left[\mathbb{1}\cos(\alpha)+\vec{n}\vec{n}^{T}(1-\cos(\alpha))+[\vec{n}]_{\times}\sin(\alpha)\right]\vec{a}\cdot\vec{\sigma}$
$\displaystyle=(R_{\vec{n}}^{xyz}(\alpha)\vec{a})\cdot\vec{\sigma}$
$\displaystyle=\vec{b}\cdot\vec{\sigma},$

where $R_{\vec{n}}^{xyz}(\alpha)$ is the rotation in $\mathbb{R}^{3}$ around
the axis $\vec{n}$, by an angle of $\alpha$. This in turn means that we can
always find values for $\vec{n}$ and $\alpha$ such that one can transform the
vector $\vec{a}$ into vector $\vec{b}$. Setting $W=R_{\vec{n}}(\alpha)$ we can
always find a unitary transformation that transforms one generator into
another. ∎

Which consequently leads to:

See 2.1

###### Proof.

Follows trivially from previous theorem:

$\displaystyle W^{\dagger}U_{\theta}W$ $\displaystyle=W^{\dagger}e^{-i\theta
G}W$
$\displaystyle=W^{\dagger}\left[\mathbb{1}+\frac{-i\theta}{1!}G+\frac{(-i\theta)^{2}}{2!}G^{2}+...\right]W$
$\displaystyle=W^{\dagger}W+\frac{-i\theta}{1!}W^{\dagger}GW+\frac{(-i\theta)^{2}}{2!}W^{\dagger}G^{2}W+...$
$\displaystyle=\mathbb{1}+\frac{-i\theta}{1!}W^{\dagger}GW+\frac{(-i\theta)^{2}}{2!}W^{\dagger}GWW^{\dagger}GW+...$
$\displaystyle=\mathbb{1}+\frac{-i\theta}{1!}G^{\prime}+\frac{(-i\theta)^{2}}{2!}G^{\prime
2}+...$ $\displaystyle=e^{-i\theta G^{\prime}}.$

∎

##  Appendix B Privacy Measure

In this appendix we look at the measure of privacy introduced in Def. 2.3, and
prove the properties it should follow, introduced in the main text. Looking at
the second and third properties, they should come by construction, as we
verify below. For the last two, it is not obvious, but we still get them:

  1. 2.

From the fact that $\mathcal{Q}$ is a positive semi-definite matrix, we get
that $\mathcal{P}(\mathcal{Q},\vec{a})\leq 1$. The minimum value follows from:

$\displaystyle\frac{\vec{a}^{T}\mathcal{Q}\vec{a}}{\operatorname{Tr}\mathcal{Q}}$
$\displaystyle=\frac{\vec{a}^{T}\sum_{i}\beta_{i}\vec{b}_{i}\vec{b}_{i}^{T}\vec{a}}{\operatorname{Tr}\mathcal{Q}}$
(58)
$\displaystyle=\frac{\sum_{i}\beta_{i}(\vec{a}\cdot\vec{b}_{i})^{2}}{\operatorname{Tr}\left[\sum_{i}\beta_{i}\vec{b}_{i}\vec{b}_{i}^{T}\right]}$
$\displaystyle=\frac{\sum_{i}\beta_{i}(\vec{a}\cdot\vec{b}_{i})^{2}}{\sum_{i}\beta_{i}}$
$\displaystyle\leq 1,$

as $\vec{a}\cdot\vec{b}_{i}\leq 1$ always, choosing normalized vectors
$\vec{b}_{i}$.

  2. 3.

The fact that only a QFI of the type $\mathcal{Q}=\alpha\vec{a}\vec{a}^{T}$
saturates the bounds comes trivially from before, as
$\vec{a}\cdot\vec{b}_{i}=1$ iff $\vec{b}_{i}=\vec{a}$. Moreover, even if the
QFI does not come in a diagonal form, we can still show that:

$\displaystyle\frac{\vec{a}^{T}\mathcal{Q}\vec{a}}{\operatorname{Tr}\mathcal{Q}}$
$\displaystyle=\frac{\vec{a}^{T}\sum_{ij}\beta_{ij}\vec{b}_{i}\vec{b}_{j}^{T}\vec{a}}{\operatorname{Tr}\mathcal{Q}}$
(59)
$\displaystyle=\frac{\vec{a}^{T}\sum_{i}\tilde{\beta}_{i}\vec{\tilde{b}}_{i}\vec{\tilde{b}}_{i}^{T}\vec{a}}{\operatorname{Tr}\mathcal{Q}}$
$\displaystyle=\frac{\sum_{i}\tilde{\beta}_{i}(\vec{a}\cdot\vec{\tilde{b}}_{i})^{2}}{\sum_{i}\tilde{\beta}_{i}}$
$\displaystyle\leq 1,$

as the matrix of the QFI $\mathcal{Q}$ is always real and symmetric, so there
is always a diagonalization for it. Moreover, since it is positive matrix all
$\tilde{\beta}_{i}$ are positive. The only way the bound is saturated is if
$\exists i:\vec{\tilde{b}}_{i}=\vec{a}$ and $\tilde{\beta}_{j}=0\forall j\neq
i$.

  3. 4.

$\displaystyle\mathcal{P}(B\mathcal{Q}B^{T},B\vec{a})$
$\displaystyle=\frac{\operatorname{Tr}B\mathcal{Q}B^{T}-\vec{a}^{T}B^{T}B\mathcal{Q}B^{T}B\vec{a}}{\operatorname{Tr}B\mathcal{Q}B^{T}}$
(60)
$\displaystyle=\frac{\operatorname{Tr}\mathcal{Q}B^{T}B-\vec{a}^{T}\mathcal{Q}\vec{a}}{\operatorname{Tr}\mathcal{Q}B^{T}B}$
$\displaystyle=\frac{\operatorname{Tr}\mathcal{Q}-\vec{a}^{T}\mathcal{Q}\vec{a}}{\operatorname{Tr}\mathcal{Q}}$
$\displaystyle=\mathcal{P}(\mathcal{Q},\vec{a}),$

where we made the assumption that $B$ is a orthonormal change of basis matrix,
$i.e.$ $BB^{T}=\mathbb{1}$. In case it is not, the two measures are not
equivalent, but they still present a valid skewed measure.

  4. 5.

$\displaystyle\mathcal{P}(\mathcal{Q}+\epsilon
A,\vec{a})-\mathcal{P}(\mathcal{Q},\vec{a})$
$\displaystyle=\frac{\operatorname{Tr}\left[\mathcal{Q}+\epsilon
A\right]-\vec{a}^{T}(\mathcal{Q}+\epsilon
A)\vec{a}}{\operatorname{Tr}\left[\mathcal{Q}+\epsilon
A\right]}-\frac{\operatorname{Tr}\mathcal{Q}-\vec{a}^{T}\mathcal{Q}\vec{a}}{\operatorname{Tr}\mathcal{Q}}$
(61) $\displaystyle=1-\frac{\vec{a}(\mathcal{Q}+\epsilon
A)\vec{a}^{T}}{\operatorname{Tr}\mathcal{Q}+\epsilon\operatorname{Tr}A}-\left(1-\frac{\vec{a}\mathcal{Q}\vec{a}^{T}}{\operatorname{Tr}\mathcal{Q}}\right)$
$\displaystyle=\frac{\vec{a}\mathcal{Q}\vec{a}^{T}}{\operatorname{Tr}\mathcal{Q}}-\frac{\vec{a}\mathcal{Q}\vec{a}^{T}}{\operatorname{Tr}\mathcal{Q}+\epsilon\operatorname{Tr}A}-\epsilon\frac{\vec{a}A\vec{a}^{T}}{\operatorname{Tr}\mathcal{Q}+\epsilon\operatorname{Tr}A}$
$\displaystyle|\mathcal{P}(\mathcal{Q}+\epsilon
A,\vec{a})-\mathcal{P}(\mathcal{Q},\vec{a})|$
$\displaystyle\leq\left|\frac{\vec{a}\mathcal{Q}\vec{a}^{T}}{\operatorname{Tr}\mathcal{Q}+\epsilon\operatorname{Tr}A}+\epsilon\frac{\vec{a}A\vec{a}^{T}}{\operatorname{Tr}\mathcal{Q}+\epsilon\operatorname{Tr}A}-\frac{\vec{a}\mathcal{Q}\vec{a}^{T}}{\operatorname{Tr}\mathcal{Q}}\right|$
$\displaystyle\leq\left|\frac{\vec{a}\mathcal{Q}\vec{a}^{T}}{\operatorname{Tr}\mathcal{Q}}+\epsilon\frac{\vec{a}A\vec{a}^{T}}{\operatorname{Tr}\mathcal{Q}}-\frac{\vec{a}\mathcal{Q}\vec{a}^{T}}{\operatorname{Tr}\mathcal{Q}}\right|$
$\displaystyle=\left|\epsilon\frac{\vec{a}A\vec{a}^{T}}{\operatorname{Tr}\mathcal{Q}}\right|\equiv\delta,$

providing the continuity of the privacy upon change of the initial state.

Using this measure of private one can try and understand what the different
values of $\mathcal{P}$ mean. The complete private case $\mathcal{P}=1$ can be
verified to naturally satisfy all conditions for the privacy. If the QFI
matrix is of the form $\mathcal{Q}=\alpha\vec{a}\vec{a}^{T}$ then is only
possible to find an estimator for the function
$f({\boldsymbol{\theta}})=\vec{a}\cdot\vec{{\boldsymbol{\theta}}}$. This also
means that whatever estimators for other functions would be ill-defined and in
not accessible (infinite variance). Consequently all parties can not estimate
but the target function, and the dishonest parties having access to all the
values in the dishonest subset, could at most gather information about the
honest parties function $f({\boldsymbol{\theta}}_{H})$ but not their local
values.

For the case of $\mathcal{P}=\epsilon$ let us provide two examples to
understand how $\epsilon$ would affect the information that might be leaked in
the case of dishonest parties:

######  Example B.1.

Suppose the QFI matrix is given by:

$\mathcal{Q}=\epsilon\vec{a}\vec{a}^{T}+(1-\epsilon)\vec{b}\vec{b}^{T},$

such that $\vec{a}_{H}\cdot\vec{b}_{H}=0$, $\vec{a}\cdot\vec{b}=0$, denoting
by $\vec{x}_{H}$ the vector restricted to the honest parties. First verify
that $\mathcal{P}=\epsilon$ for the estimation of the function
$f({\boldsymbol{\theta}})=\vec{a}\cdot\vec{{\boldsymbol{\theta}}}$. Then,
since the dishonest parties can share their own values and therefore have
access to all $\theta_{\mu}\in D$, they also have access to
$\vec{b}_{H}\cdot\vec{{\boldsymbol{\theta}}}$ which is a different from
$f({\boldsymbol{\theta}}_{H})$, which is the private function. This is not
private, since it violates the third condition of privacy. If $\epsilon=0$,
then their information about this function is the maximum possible for this
estimation $(\operatorname{Tr}\mathcal{Q})$.

######  Example B.2.

Suppose we have $k$ parties and $d\leq k/2$ dishonest parties among them. Le
the QFI matrix be given by:

$\mathcal{Q}=\epsilon\vec{a}\vec{a}^{T}+\frac{1-\epsilon}{d}\sum_{j=1}^{d}\vec{b}_{j}\vec{b}_{j}^{T},$

Such that $\vec{a}\cdot\vec{b}_{j}=0$ for all vectors $\vec{b}_{j}$. In
particular, one can choose these vectors $\vec{b}_{j}$ such that
$\vec{b}_{j_{H}}$, which is the correspondent to slicing the vectors
$\vec{b}_{j}$ for the indices correspondent to honest parties, form a
$d$-dimensional orthogonal subspace. Again, we first verify that
$\mathcal{P}=\epsilon$ as every $\vec{b}_{j}$ is orthogonal to $\vec{a}$.
Second, we verify that the set of information available by considering every
dishonest party knows the dishonest subspace, is given by an orthogonal basis
of $d$ vectors in the honest subspace. Namely they have information about all
$\vec{b}_{j_{H}}\cdot\vec{{\boldsymbol{\theta}}}$ after using the dishonest
information. In particular, if $d=k/2$, then they have $d$ linearly
independent vectors to estimate the honest subspace with $k-d=d$ functions.
This means they also have access to each individual parameter of the honest
parties. Again, if $\epsilon\rightarrow 0$ then they are more efficient, in
the sense the variance at which they can do this the same amount of resources
is smaller.

Note however that in both of these cases, for the parties to implement a non-
private strategy and acquire information about the honest parties, they would
have to distribute a non-private state. If one utilizes a state certification
protocol in conjunction with this [42], then one can guarantee privacy and
security.

##  Appendix C Integer Vectors Properties

See 3.3

###### Proof.

If $\alpha\in\mathbb{Z}$, then
$\vec{a}\in\mathbb{Z}^{k}\implies\vec{b}\in\mathbb{Z}^{k}$. To prove it is the
only option choose a $b_{i}$ such that $b_{i}=\alpha a_{i}$, where
$\alpha\notin\mathbb{Z}=c/d$. For $b_{j}$ to be integer for every other $j$,
then $gcd(\vec{a},d)>1$, as in, every $a_{i}$. However, $gcd$ is associative,
meaning $gcd(\vec{a},d)=gcd(gcd(\vec{a}),d)=1$. This means $\alpha$ has to be
integer. ∎

See 3.4

###### Proof.

Since $\vec{a},\vec{b}\prec\vec{m},\vec{0}\prec\vec{a}+\vec{b}\prec
2\vec{m},-\vec{m}\prec\vec{a}-\vec{b}\prec\vec{m}$. $gcd(\vec{m})=1\implies$
if $\alpha\vec{m}\propto\vec{m}$ then $\alpha\in\mathbb{Z}$.

$\displaystyle\Leftrightarrow\begin{cases}2\left[\vec{m}-\left(\vec{a}+\vec{b}\right)\right]\propto\vec{m}\\\
2\left(\vec{a}-\vec{b}\right)\propto\vec{m}\end{cases}\Leftrightarrow\begin{cases}\left[\vec{m}-\left(\vec{a}+\vec{b}\right)\right]\propto\vec{m}\\\
\left(\vec{a}-\vec{b}\right)\propto\vec{m}\end{cases}\text{ but
}\begin{cases}-\vec{m}\prec\vec{m}-\left(\vec{a}+\vec{b}\right)\prec\vec{m}\\\
-\vec{m}\prec\vec{a}-\vec{b}\prec\vec{m}\end{cases}.$ (62)

∎

See 3.5

###### Proof.

The proof follows similarly. Since
$\vec{a},\vec{b}\prec\vec{n},\vec{0}\prec\vec{a}+\vec{b}\prec
2\vec{n},-\vec{n}\prec\vec{a}-\vec{b}\prec\vec{n}$. $gcd(\vec{m})=1\implies$
if $\alpha\vec{m}\propto\vec{m}$ then $\alpha\in\mathbb{Z}$.

$\displaystyle\Leftrightarrow\begin{cases}2\left[\vec{n}-\left(\vec{a}+\vec{b}\right)\right]\propto\vec{m}\\\
2\left(\vec{a}-\vec{b}\right)\propto\vec{m}\end{cases}\Leftrightarrow\begin{cases}\left[\vec{n}-\left(\vec{a}+\vec{b}\right)\right]\propto\vec{m}\\\
\left(\vec{a}-\vec{b}\right)\propto\vec{m}\end{cases}\text{ but
}\begin{cases}-\vec{n}\prec\vec{n}-\left(\vec{a}+\vec{b}\right)\prec\vec{n}\\\
-\vec{n}\prec\vec{a}-\vec{b}\prec\vec{n}\end{cases}.$ (63)

From here we verify that as $\vec{n}\not\preceq\vec{m},\exists j:n_{j}<m_{j}$
and $gcd(\vec{m})=1$ and therefore no proportionality can be found. ∎

##  Appendix D QFI Matrix for Arbitrary States

Starting from the definition of an arbitrary pure state of Eq. 25 can write
the elements of the QFI matrix (see Eq. 6) as:

$\displaystyle\mathcal{Q}_{\mu\nu}(\rho_{{\boldsymbol{\theta}}})$
$\displaystyle=4\text{Re}\left\\{\sum_{l,m=1}^{2^{n}}\alpha_{l}^{*}\alpha_{m}\braket{\mathcal{G}_{l}}{\boldsymbol{G}_{\mu}\boldsymbol{G}_{\nu}}{\mathcal{G}_{m}}-\sum_{l,m=1}^{2^{n}}\alpha_{l}^{*}\alpha_{m}\braket{\mathcal{G}_{l}}{\boldsymbol{G}_{\mu}}{\mathcal{G}_{m}}\sum_{p,q=1}^{2^{n}}\alpha_{p}^{*}\alpha_{q}\braket{\mathcal{G}_{p}}{\boldsymbol{G}_{\nu}}{\mathcal{G}_{q}}\right\\},$
(64)

with $\boldsymbol{G}_{\mu}=\sum_{j\in\mu}G_{j}$. To calculate the different
elements above consider the following labeling for the basis states: let
$\tilde{s}_{l}^{\pm}=(G_{1}^{\perp})^{l_{1}}\otimes(G_{2}^{\perp})^{l_{2}}\otimes\cdots\otimes(G_{n-1}^{\perp})^{l_{n-1}}\otimes(G_{n})^{\pm}$,
where $G_{n}^{+}=\mathbb{1}$ and $G_{n}^{-}=G_{n}$. Let
$l\in\\{0,1,\cdots,2^{n-1}-1\\}$ and $l_{1}l_{2}\cdots l_{n-1}$ be the binary
string associated to $l$. From our definition of $G,G^{\perp}$, we also have
that:

$\displaystyle G_{j}\circ G_{j}^{\perp}\circ G_{j}$
$\displaystyle=-G_{j}^{\perp},\qquad G_{j}\circ G_{j}\circ G_{j}=G_{j},$ (65)
$\displaystyle G_{j}^{\perp}\circ G_{j}\circ G_{j}^{\perp}$
$\displaystyle=-G_{j},\qquad G_{j}^{\perp}\circ G_{j}^{\perp}\circ
G_{j}^{\perp}=G_{j}^{\perp}.$

Using this when applying $\tilde{s}_{l}^{\pm}$ to the stabilizer table in Eq.
22, we get a minus sign for the lines $j$ such that $l_{j}=1$. This yields:

$\displaystyle\braket{\mathcal{G}_{l}^{\pm}}{\sum_{j\in\mu}\sum_{k\in\nu}G_{j}G_{k}}{\mathcal{G}_{m}^{\pm}}$
$\displaystyle=h^{*}_{\mathcal{N}_{\mu}}(l)h^{*}_{\mathcal{N}_{\nu}}(m)\braket{\mathcal{G}_{l}^{\pm}}{\mathcal{G}_{m}^{\pm}}$
(66)
$\displaystyle\braket{\mathcal{G}_{l}^{\pm}}{\sum_{j\in\mu}G_{j}}{\mathcal{G}_{m}^{\pm}}$
$\displaystyle=h^{*}_{\mathcal{N}_{\mu}}(l)\braket{\mathcal{G}_{l}^{\pm}}{\mathcal{G}_{m}^{\mp}},$

where $h^{*}_{\mathcal{N}_{\mu}}$ is the $\mu$ element of the vector-Hamming
weight in Def. 3.3. Note that states differing on $\pm$ have the same Hamming-
weight, as expected since $G_{n}$ commutes with all $G_{j}G_{k}$ terms. We can
rewrite then Eq. 64 the QFI in the following way:

$\displaystyle\mathcal{Q}_{\mu\nu}(\rho_{{\boldsymbol{\theta}}})$
$\displaystyle=\sum_{l=0}^{2^{n-1}-1}\left[|\alpha_{l}^{+}|^{2}+|\alpha_{l}^{-}|^{2}\right]h^{*}_{\mathcal{N}_{\mu}}(l)h^{*}_{\mathcal{N}_{\nu}}(l)-$
(67)
$\displaystyle-\left[\sum_{l=0}^{2^{n-1}-1}2\text{Re}\left\\{\alpha_{l}^{+^{*}}\alpha_{l}^{-}\right\\}h^{*}_{\mathcal{N}_{\mu}}(l)\right]\left[\sum_{q=0}^{2^{n-1}-1}2\text{Re}\left\\{\alpha_{q}^{+^{*}}\alpha_{q}^{-}\right\\}h^{*}_{\mathcal{N}_{\nu}}(q)\right]$
$\displaystyle=\sum_{l}\lambda_{l}h^{*}_{\mathcal{N}_{\mu}}(l)h^{*}_{\mathcal{N}_{\nu}}(l)-\left[\sum_{l}v_{l}h^{*}_{\mathcal{N}_{\mu}}(l)\right]\left[\sum_{q}v_{q}h^{*}_{\mathcal{N}_{\nu}}(q)\right].$

Let $\boldsymbol{\mathcal{N}}=\\{\mathcal{N}_{\mu}\\}_{\mu\in V}$ be the
partition of the set of qubits into each of the nodes. Assembling the last
equation for all entries $\mu\nu$ we get exactly:

$\displaystyle\boldsymbol{\mathcal{Q}}(\rho_{{\boldsymbol{\theta}}})$
$\displaystyle=\sum_{m}\lambda_{m}\vec{h}^{*}_{\boldsymbol{\mathcal{N}}}(m)\vec{h}^{*^{T}}_{\boldsymbol{\mathcal{N}}}(m)-\sum_{m}v_{m}\vec{h}^{*}_{\boldsymbol{\mathcal{N}}}(m)\sum_{q}v_{q}\vec{h}^{*^{T}}_{\boldsymbol{\mathcal{N}}}(q)$
(68)
$\displaystyle=\sum_{m,q}\mathfrak{Q}_{mq}\vec{h}^{*}_{\boldsymbol{\mathcal{N}}}(m)\vec{h}^{*}_{\boldsymbol{\mathcal{N}}}(q)^{T}$
$\displaystyle=C\boldsymbol{\mathfrak{Q}}C^{T},$

where $C$ is a matrix where each line is given by a vector
$\vec{n}-2\vec{c}_{i}$, where $\vec{0}\preceq\vec{c}_{i}\prec\vec{n}$. This is
equivalent, under the choice of $G=Z,G^{\perp}=X$ to the GHZ state on $n$
qubits, and the basis formed by
$\ket{\mathcal{G}_{m}^{\pm}}\propto\ket{m}\pm\ket{\overline{m}}$ where $m$
ranges between 0 and $2^{n-1}-1\equiv N$, is represented by the binary string
associated to the number $m$ and $\overline{m}$ is the binary negation of $m$
($e.g.$ $m=00101,\overline{m}=11010$). Furthermore, we also get two additional
important properties: from Eq. 68 we can see that the QFI matrix can be
decomposed in $\mathfrak{Q}=\Lambda-\vec{v}\vec{v}^{T}$, where
$\Lambda=diag(\lambda_{0},...,\lambda_{m},...,\lambda_{N})$ and
$\vec{v}=(v_{0},...,v_{m},...,v_{N})$. On top of this, we can verify that:

$\displaystyle|\alpha_{j}^{+}\pm\alpha_{j}^{-}|^{2}\geq
0\implies|\alpha_{j}^{+}|^{2}+|\alpha_{j}^{-}|^{2}\geq\pm
2\text{Re}\left\\{\alpha_{j}^{+}\alpha_{j}^{-^{*}}\right\\}\implies\lambda_{j}\geq|v_{j}|.$
(69)

##  Appendix E Matrix Rank Inequalities

Let us introduce some additional theorems regarding symmetric real matrices,
which are the ones resultant from our problem. We start by the following
theorem, which allows us to pick one and only one subspace in the set of
states we can choose to construct a private state:

######  Theorem E.1.

Let $A\in\mathbb{R}^{n\times n}$ be a symmetric matrix given by
$A=\Lambda-\vec{v}\vec{v}^{T}$, where $\Lambda$ is a diagonal matrix with all
entries real and non-negative summing to 1 and $\vec{v}$ is a vector in
$\mathbb{R}^{n}$ such that $\lVert\vec{v}\rVert\leq 1$ and in particular
$\forall i\in\\{1,\dots,n\\}:\lambda_{i}\geq|v_{i}|$. $A$ is positive semi-
definite and the rank of $A$ is at least

$\text{rank}\ \Lambda\geq\text{rank}\ A\geq\text{rank}\ \Lambda-1,$

and $\text{rank}\ A=\text{rank}\ \Lambda-1$ iff. $\lambda_{j}=|v_{j}|$,
$\forall j$.

###### Proof.

We start by reducing the problem into considering only the non-zero entries of
$\Lambda$ and $\vec{v}$, as all the zero entries of $\Lambda$ do not
contribute for the rank and the zero entries of $\vec{v}$ have corresponding
values in $\Lambda$ which are already eigenvalues of $A$. Also, we can never
have a non-zero entry in $\vec{v}$ associated to a zero value in $\Lambda$,
from the conditions stated. Denote the vector $\vec{\lambda}$ as the vector of
all the positive diagonal values of $\Lambda$. The condition imposed on
$\vec{v}$ means that $\forall i\in\\{1,\dots,n\\}:\lambda_{i}\geq|v_{i}|$,
which allows us to say:

$\displaystyle\forall\vec{y}\neq\vec{0}\ ,\exists\
\vec{x}\neq\vec{0}:\vec{y}^{T}\left(\Lambda-\vec{v}\vec{v}^{T}\right)\vec{y}$
$\displaystyle\geq\vec{x}^{T}\left(\Lambda-\vec{\lambda}\vec{\lambda}^{T}\right)\vec{x}$
(70) $\displaystyle\vec{y}^{T}A\vec{y}$
$\displaystyle\geq\vec{x}^{T}B\vec{x}.$

In particular, one could consider the vector $\vec{x}$ where
$(\vec{x})_{j}=|(\vec{y})_{j}|$. By doing this, we can find bound the
eigenvalues of $A$.

$\displaystyle\vec{x}^{T}\left(\text{diag}(\vec{\lambda})-\vec{\lambda}\vec{\lambda}^{T}\right)\vec{x}$
$\displaystyle=\sum_{i}x_{i}^{2}\lambda_{i}-(\sum_{i}x_{i}\lambda_{i})^{2}$
(71)
$\displaystyle=\sum_{i}x_{i}^{2}\lambda_{i}-(\sum_{i}x_{i}\lambda_{i})(\sum_{j}x_{j}\lambda_{j})$
$\displaystyle=\sum_{i}x_{i}^{2}\lambda_{i}(1-\lambda_{i})-\sum_{i,j\neq
i}x_{i}\lambda_{i}x_{j}\lambda_{j}$
$\displaystyle=\sum_{i}x_{i}^{2}\lambda_{i}\sum_{j\neq
i}\lambda_{j}-2\sum_{i,j>i}x_{i}\lambda_{i}x_{j}\lambda_{j}$
$\displaystyle=\sum_{i}\sum_{j>i}\lambda_{i}\lambda_{j}(x_{i}^{2}-2x_{i}x_{j}+x_{j}^{2})$
$\displaystyle=\sum_{i}\sum_{j>i}\lambda_{i}\lambda_{j}(x_{i}-x_{j})^{2}.$

Since each $\lambda_{i}$ is by construction positive and non-zero, this means
this expression is always positive, and is only zero when
$(x_{i}-x_{j})^{2}=0\ \forall i,j$, which only happens for the null vector
$\vec{0}$ and the $\vec{1}/\sqrt{n}$.

Suppose $\exists j:\lambda_{j}>|v_{j}|$. In this case, Eq. 70 is never
saturated, unless $(\vec{y})_{j}$ is zero, which implies
$\vec{x}^{T}B\vec{x}>0$. This means $A$ is definite positive, and consequently
that it has the rank of $\Lambda$. The other possible case is if $\forall\
j:\lambda_{j}=|v_{j}|$. Consider the matrix
$D=diag((\vec{v})_{j}/|(\vec{v})_{j}|)$ which transforms
$\vec{v}\rightarrow\vec{\lambda}$, $D\vec{v}=\vec{\lambda}$. Notice this
matrix is a change of basis as $DD^{T}=\mathbb{1}$ and moreover that
$DAD^{T}=B$ in this case. This allows us to map the eigenvectors of $B$ into
the eigenvectors of $A$, proving the lower bound on the rank, as
$\vec{1}/\sqrt{n}$ is the only non-zero vector than spans the null space of
$B$. This means the rank of $A$ is bounded between the rank of $\Lambda$ and
rank $\Lambda$ -1, where the latest equality only happens iff. $\forall\
j:\lambda_{j}=|v_{j}|$. ∎

######  Proposition E.1.

Let $A\in\mathbb{R}^{n\times n}$ be symmetric and $B\in\mathbb{R}^{k\times
n}$. If $A$ has full rank, then $BAB^{T}$ has rank equal to the rank of $B$.

$\text{rank}\ A=n\quad\implies\quad\text{rank}\ BAB^{T}=\text{rank}\ B$

###### Proof.

Since $A$ is symmetric, consider its diagonalization $A=Q\Lambda Q^{T}$.

$BAB^{T}=BQ\Lambda
Q^{T}B^{T}=BQ\sqrt{\Lambda}Q^{T}Q\sqrt{\Lambda}Q^{T}B^{T}=\tilde{B}\tilde{B}^{T}\implies\text{rank}\
\tilde{B}\tilde{B}^{T}=\text{rank}\tilde{B}=\text{rank}B,$

where used properties of the rank, namely,
$\text{rank}BB^{T}=\text{rank}B^{T}B=\text{rank}B=\text{rank}B^{T}$ and if $A$
is a $l\times m$ matrix with rank $m$ and $B$ is a $m\times n$ matrix,
$\text{rank}AB=\text{rank}B$. ∎

######  Proposition E.2.

Let $A\in\mathbb{R}^{n\times n}$ be symmetric and $B\in\mathbb{R}^{k\times
n}$. If $A$ has rank $n-1$, then:

$\text{rank}\ A=n-1\quad\implies\quad\text{rank}\ B\geq\text{rank}\
BAB^{T}\geq\text{rank}\ B-1.$

###### Proof.

Consider the same as before, but in the last step apply the Sylvester
inequality for the rank of $\sqrt{A}B$:

$\text{rank}\ B\geq\min\\{\text{rank}\ \sqrt{A},\text{rank}\
B\\}\geq\text{rank}\ \tilde{B}\geq\text{rank}\ B+\text{rank}\
\sqrt{A}-n=\text{rank}\ B-1.$

∎

######  Proposition E.3.

Let $V,W$ be two vector spaces. Let $V_{1}$ and $V_{2}$ be two $m\times n$
matrices with vectors in $V$. If $\mathsf{span}(V_{1})=\mathsf{span}(V_{2})$
then $\mathsf{span}(V_{1}W_{1})=\mathsf{span}(V_{2}W_{1})$ for any $n\times p$
matrix $W_{1}$ with vectors in $W$.

###### Proof.

If $\mathsf{span}(V_{1})=\mathsf{span}(V_{2})$ then $\exists Q:V_{2}=QV_{1}$
and is invertible. The proof concludes by realising that
$\mathsf{span}(QA)=\mathsf{span}(A)$ for any invertible $Q$, given the
definition of a span of a set of vectors. ∎

##  Appendix F Private States - Proofs

See 3.4

###### Proof.

The proof in the indirect way comes trivially from calculating the privacy for
the family of states $F_{GHZ}$, together with Corol. 2.1, we get that
difference up to a LU operation. Doing the same as before, now we have a
different $\mathfrak{Q}$ and $C$ matrices, as $\vec{n}=\vec{a}$. When looking
at the matrix $\mathfrak{Q}$ note that the first coefficient $\lambda_{0}$ and
$v_{0}$ are associated with the span of the family of states $F_{GHZ}$, as by
varying them, we get all the states in $F_{GHZ}$. Doing the same analysis as
before, we get that now $\vec{a}\in C$ and $dim_{\vec{a}}(C)=1$, $i.e.$ it
only appears once in the lines of $C$. Constructing the same table to analyze
privacy:

rank $\operatorname{supp}\mathfrak{Q}$ |  rank $C_{\operatorname{supp}\mathfrak{Q}}$ |  rank $C\mathfrak{Q}C^{T}$ | Reasoning | Privacy
---|---|---|---|---
any | 1 | $\leq 1$ | Trivial |  Iff. $C_{\operatorname{supp}\mathfrak{Q}}=\vec{a}$
dim $\operatorname{supp}\mathfrak{Q}$ | $\geq 2$ | $\geq 2$ |  Prop. E.1 | Never
dim $\operatorname{supp}\mathfrak{Q}$-1 | 2 | 1 or 2 |  Prop. E.2 | Check
dim $\operatorname{supp}\mathfrak{Q}$-1 | $>2$ | $>1$ |  Prop. E.2 | Never

Checking again the case of rank $\operatorname{supp}\mathfrak{Q}$ = dim
$\operatorname{supp}\mathfrak{Q}-1$ with rank
$C_{\operatorname{supp}\mathfrak{Q}}$ = 2. Using the same reasoning we find a
vector $D\vec{w}$ in the orthogonal subspace of $D\vec{1}$,
$\mathsf{span}P^{\perp}$, acting on two different lines of
$C_{\operatorname{supp}\mathfrak{Q}}$:

$\displaystyle D\vec{w}C_{\operatorname{supp}\mathfrak{Q}}$
$\displaystyle=1/\sqrt{2}\left(\vec{h}_{\boldsymbol{\mathcal{N}}}^{*}(i)\pm\vec{h}_{\boldsymbol{\mathcal{N}}}^{*}(j)\right)$
(72)
$\displaystyle=\begin{cases}2/\sqrt{2}\left[\vec{n}-\left(\vec{h}_{\boldsymbol{\mathcal{N}}}(i)+\vec{h}_{\boldsymbol{\mathcal{N}}}(j)\right)\right]\\\
2/\sqrt{2}\left(\vec{h}_{\boldsymbol{\mathcal{N}}}(i)-\vec{h}_{\boldsymbol{\mathcal{N}}}(j)\right)\end{cases}.$

From here there are two options:

  1. (i)

$i$ or $j$ correspond to the GHZ state, and in that case we have that
$\vec{h}_{\boldsymbol{\mathcal{N}}}(i)\pm\vec{h}_{\boldsymbol{\mathcal{N}}}(j)=\vec{0}\equiv\vec{h}_{\boldsymbol{\mathcal{N}}}(j)=\vec{0}$,
meaning that $i=j$ and contradicting rank-$2$ of
$C_{\operatorname{supp}\mathfrak{Q}}$;

  2. (ii)

neither $i$ nor $j$ correspond to the GHZ state, and we recover that
$C_{\operatorname{supp}\mathfrak{Q}}$ only has vectors
$\vec{c}_{i}\prec\vec{a}$, which is the case for the non-existence of a
private state.

This means also this case is never private. Consequently, the only private
state is the one spanned by a matrix with a seul non-zero entry correspondent
to $\lambda_{0}-v_{0}^{2}$. Moreover, this is maximized for $v_{0}^{2}=0$,
meaning a perfect GHZ state, up to a LU operation.

∎

See 3.7

###### Proof.

This is a direct consequence of the invariance of the Hamming-weight
introduced in Props. 3.1,3.2. Looking at the terms that appear in the QFI, fix
the encoding dynamics basis at $Z$, as we have chosen our distributed
$s$-states (Def. 3.5) in the computational basis. Notice this does not affect
the generality of our proof, as again we are always an LU operation from any
other basis (Thm. 2.1). Using this fact, and the fact that the vectorial
Hamming-weight is invariant under the distributed permutations, we get that:

$\displaystyle\sum_{j\in\mu}Z_{j}\ket{\psi}$
$\displaystyle=h_{\mathcal{N}_{\mu}}(\ket{s({b_{1}}_{\mu},a_{\mu}+{b_{0}}_{\mu})})\ket{\psi}=\left(a_{\mu}+{b_{0}}_{\mu}-{b_{1}}_{\mu}\right)\ket{\psi},$
(73) $\displaystyle\sum_{j\in\mu}Z_{j}\ket{\tilde{\psi}}$
$\displaystyle=h_{\mathcal{N}_{\mu}}(\ket{s(a_{\mu}+{b_{1}}_{\mu},{b_{0}}_{\mu})})\ket{\tilde{\psi}}=\left(-a_{\mu}+{b_{0}}_{\mu}-{b_{1}}_{\mu}\right)\ket{\tilde{\psi}},$

Where we used that $\vec{b}=\vec{b}_{0}+\vec{b}_{1}$ which correspond to the
amount of 0’s and 1’s respectively. Note that $\vec{d}=\vec{b}_{1}$ as well.
After assembling in the QFI matrix we get:

$\displaystyle\boldsymbol{\mathcal{Q}}$
$\displaystyle=\vec{a}\vec{a}^{T}\left[|\alpha^{2}|+|\beta|^{2}-(|\alpha|^{2}-|\beta|^{2})^{2}\right],$
(74)

which means private. Moreover, it is maximised for $|\alpha|=|\beta|$. ∎

See 3.5

###### Proof.

The inverse direction is a direct consequence from Thm. 3.7. The direct proof
comes again from the QFI. Construct the same decomposition into
$C\mathfrak{Q}C^{T}$ as before, where the basis of $\mathfrak{Q}$ are the
families correspondent to $\boldsymbol{F}_{GHZ}$.

rank $\operatorname{supp}\mathfrak{Q}$ |  rank $C_{\operatorname{supp}\mathfrak{Q}}$ |  rank $C\mathfrak{Q}C^{T}$ | Reasoning | Privacy
---|---|---|---|---
any | 1 | $\leq 1$ | Trivial |  Iff. $C_{\operatorname{supp}\mathfrak{Q}}=\vec{a}$
dim $\operatorname{supp}\mathfrak{Q}$ | $\geq 2$ | $\geq 2$ |  Prop. E.1 | Never
dim $\operatorname{supp}\mathfrak{Q}$-1 | 2 | 1 or 2 |  Prop. E.2 | Check
dim $\operatorname{supp}\mathfrak{Q}$-1 | $>2$ | $>1$ |  Prop. E.2 | Never

Now, for the first time we have that $\dim\\{\vec{c}_{j}\in
C:\vec{c}_{j}=\vec{a}\\}\equiv\mathfrak{Q}_{\vec{a}}$ is either 0 or larger
than 1. It is larger than 1 if $\vec{b}=\vec{n}-\vec{a}$ has all entries
divisible by two, and one can find the family of states such that
$\vec{b}_{0}=\vec{b}_{1}$. This is a consequence of the symmetric Hamming
weight that remains the same when adding the same amount of 0s and 1s to a
bitstring.

Looking at the case of rank $\operatorname{supp}\mathfrak{Q}$ = dim
$\operatorname{supp}\mathfrak{Q}-1$ with rank
$C_{\operatorname{supp}\mathfrak{Q}}$ = 2, we can find a way to construct all
the others families. Using the same reasoning as before, let $D$ be the matrix
such that $D\vec{1}$ spans the null space of $\boldsymbol{\mathcal{Q}}$.
Denote the orthogonal space of this vector by $P^{\perp}$. Let $W$ be the set
of vectors given by
$W^{\pm}=\\{w|w=\vec{e}_{i}\pm\vec{e}_{j},\text{sign}(D_{ii})=\mp\text{sign}(D_{jj})\\}$,
and Prop. E.3 allows us to use this vectors to continue our proof, as
$\textsf{span}(P^{\perp})=\mathsf{span}(W^{+}\cup W^{-})$. Looking first at
$W^{-}$,the options let us check which vectors $\vec{c}_{i}-\vec{c}_{j}$ are
proportional to $\vec{a}$:

$\displaystyle\vec{h}_{\boldsymbol{\mathcal{N}}}^{*}(i)-\vec{h}_{\boldsymbol{\mathcal{N}}}^{*}(j)$
$\displaystyle=\delta\vec{a},\qquad\delta=\\{\pm 1,\pm 2\\}$ (75)
$\displaystyle\vec{n}-2\vec{h}_{\boldsymbol{\mathcal{N}}}(i)-\left(\vec{n}-2\vec{h}_{\boldsymbol{\mathcal{N}}}(j)\right)$
$\displaystyle=\delta\vec{a},\qquad\delta=\\{\pm 1,\pm 2\\}$
$\displaystyle\vec{h}_{\boldsymbol{\mathcal{N}}}(i)-\vec{h}_{\boldsymbol{\mathcal{N}}}(j)$
$\displaystyle=\tilde{\delta}\vec{a},\qquad\tilde{\delta}=\\{\pm 1\\}$
$\displaystyle\implies\vec{h}_{\boldsymbol{\mathcal{N}}}(i)=\vec{a}+\vec{b}_{j}$
$\displaystyle\wedge\vec{h}_{\boldsymbol{\mathcal{N}}}(j)=\vec{b}_{j},$

where we used the fact that $\vec{h}_{\boldsymbol{\mathcal{N}}}(\cdot)$, is an
positive integer vector and $gcd(\vec{a})=1$. Moreover, we introduce the
notation $\vec{b}_{i}$ as a vector between
$\vec{0}\preceq\vec{b}_{i}\prec\vec{b}$. Substituting for
$\vec{h}_{\boldsymbol{\mathcal{N}}}^{*}(i)$ and
$\vec{h}_{\boldsymbol{\mathcal{N}}}^{*}(j)$ and using
$\vec{b}_{i}=\vec{b}-\vec{b}_{j}$ we observe:

$\displaystyle\vec{h}_{\boldsymbol{\mathcal{N}}}^{*}(i)$
$\displaystyle=\vec{n}-2(\vec{a}+\vec{b}_{j})=\vec{a}+\vec{b}-2(\vec{a}+\vec{b}_{j})=-\vec{a}+\vec{b}-2\vec{b}_{j}=-(\vec{a}+\vec{b}-2\vec{b}_{i})=-(\vec{n}-2\vec{b}_{i})$
(76) $\displaystyle\vec{h}_{\boldsymbol{\mathcal{N}}}^{*}(j)$
$\displaystyle=\vec{n}-2\vec{b}_{j}=\vec{a}+\vec{b}-2\vec{b}_{j}.$

The equations for the subspace $W^{+}$ are:

$\displaystyle\vec{h}_{\boldsymbol{\mathcal{N}}}^{*}(i)+\vec{h}_{\boldsymbol{\mathcal{N}}}^{*}(j)$
$\displaystyle=\delta\vec{a},\qquad\delta=\\{0,\pm 1,\pm 2\\}$ (77)
$\displaystyle\vec{n}-2\vec{h}_{\boldsymbol{\mathcal{N}}}(i)+\left(\vec{n}-2\vec{h}_{\boldsymbol{\mathcal{N}}}(j)\right)$
$\displaystyle=\delta\vec{a},\qquad\delta=\\{0,\pm 1,\pm 2\\}$
$\displaystyle\vec{n}-\left(\vec{h}_{\boldsymbol{\mathcal{N}}}(i)+\vec{h}_{\boldsymbol{\mathcal{N}}}(j)\right)$
$\displaystyle=\tilde{\delta}\vec{a},\qquad\tilde{\delta}=\\{0,\pm 1\\}$
$\displaystyle\implies\left(\vec{h}_{\boldsymbol{\mathcal{N}}}(i)+\vec{h}_{\boldsymbol{\mathcal{N}}}(j)=\vec{a}+\vec{b}\right)$
$\displaystyle\vee\left(\vec{h}_{\boldsymbol{\mathcal{N}}}(i)+\vec{h}_{\boldsymbol{\mathcal{N}}}(j)=\vec{b}\right)\vee\left(\vec{h}_{\boldsymbol{\mathcal{N}}}(i)+\vec{h}_{\boldsymbol{\mathcal{N}}}(j)=2\vec{a}+\vec{b}\right).$

This means the overall solution are the vectors inside
$C_{\operatorname{supp}\mathfrak{Q}}$, such that:

$\displaystyle\begin{cases}\vec{c}_{i}=\vec{n}-2\left(\vec{a}+\vec{b}_{i}\right)\vee\vec{c}_{i}=\vec{n}-2\vec{b}_{i},\\\
\vec{c}_{j}=\vec{n}-2\left(\vec{a}+\vec{b}_{j}\right)\vee\vec{c}_{j}=\vec{n}-2\vec{b}_{j},\end{cases},\qquad\vec{b}_{i}+\vec{b}_{j}=\vec{b}.$
(78)

Substituting we actually see that
$\vec{n}-2\left(\vec{a}+\vec{b}_{i}\right)=-(\vec{n}-2\vec{b}_{j})$, meaning
they are in fact spanned by a state in the same family (see Prop. F.2). This
means choosing only the combination
$\vec{c}_{i}=\vec{n}-2\vec{b}_{i}\wedge\vec{c}_{j}=\vec{n}-2\vec{b}_{j}$ is
sufficient. The set of states with the same vectorial Hamming-weight, are
exactly the states belonging to the equivalence class, by definition. From the
construction of matrix $\mathfrak{Q}$, we get that each line is associated
with the following family of states:

$\displaystyle\overline{F}_{GHZ_{m}}=\left\\{\alpha\ket{m}+\beta\ket{\overline{m}},\alpha,\beta\in\mathbb{C},|\alpha|^{2}+|\beta|^{2}=1\right\\}$
(79)
$\displaystyle\lambda_{i}=|\alpha_{i}|^{2}+|\alpha_{i+1}|^{2},v_{i}=2\text{Re}\left\\{\alpha_{i}\alpha_{i+1}^{*}\right\\},\alpha=\frac{\alpha_{i}+\alpha_{i+1}}{\sqrt{2}},\beta=\frac{\alpha_{i}-\alpha_{i+1}}{\sqrt{2}}$
$\displaystyle\lambda_{i}=|v_{i}|\implies\begin{cases}\alpha=1,\beta=0,v_{i}/|v_{i}|=1\\\
\alpha=0,\beta=1,v_{i}/|v_{i}|=-1\end{cases}.$

This means we can only take states $i$ such that, up to permutation of qubits
(see Prop. F.1), that have
$\vec{h}_{\boldsymbol{\mathcal{N}}}(i)=\vec{b}_{i}$, together with states $j$
that $\vec{h}_{\boldsymbol{\mathcal{N}}}(\overline{j})=\vec{a}+\vec{b}_{j}$.
Given that we can only choose $\vec{b}_{i},\vec{b}_{j}$ such that
$\vec{b}_{i}+\vec{b}_{j}=\vec{b}$ (Eq. 78) and
$\vec{h}_{\boldsymbol{\mathcal{N}}}(\overline{j})=\vec{n}-\vec{h}_{\boldsymbol{\mathcal{N}}}(j)=\vec{n}-(\vec{a}+\vec{b}_{j})=\vec{b}-\vec{b}_{j}$,
then this are exactly the family of states given by
$\mathcal{F}(\boldsymbol{\mathcal{N}},\vec{a},\vec{b}_{i})$. By doing the same
for each possible set of $\vec{b}_{i},\vec{b}_{j}$, one generates each of the
possible families of private states. Note however that one cannot create a
superposition of states of different families, as that means the rank of
$C_{\operatorname{supp}\mathfrak{Q}}$ would be larger than 2, which would
always mean a non-private state (check table). Moreover, one recovers the fact
that if $\vec{b}$ is divisible by 2, then for
$\vec{b}_{i}=\vec{b}_{j}=\vec{b}/2$ is the case of a rank-1
$C_{\operatorname{supp}\mathfrak{Q}}$. Given that all other possibilities
create either not rank-1 QFI matrices, or QFI matrices which are not
proportional to $\vec{a}\vec{a}^{T}$, this are the only sets of private
states. ∎

Below we provide two additional propositions that allowed some simplifications
on the theorems above.

######  Proposition F.1.

Let $\sigma\in S_{\boldsymbol{\mathcal{N}}}$. Let
$\ket{GHZ_{m}}=\alpha\ket{m}+\beta\ket{\overline{m}}$, where $m$ is a binary
string with length $\lVert\vec{n}\rVert_{1}$ and $\overline{(\cdot)}$ is the
binary conjugation of $(\cdot)$.

$\displaystyle
h_{\boldsymbol{\mathcal{N}}}(m)=h_{\boldsymbol{\mathcal{N}}}\left(\sigma(m)\right)\Longleftrightarrow\mathcal{Q}(\ket{GHZ_{m}})=\mathcal{Q}\left(\sigma(\ket{GHZ_{m}})\right)$
(80)

######  Proposition F.2.

Let $\ket{GHZ_{m}}=\alpha\ket{m}+\beta\ket{\overline{m}}$ and
$\ket{GHZ_{\overline{m}}}=\alpha\ket{\overline{m}}+\beta\ket{m}$, where $m$ is
a binary string with length $\lVert\vec{n}\rVert_{1}$ and $\overline{(\cdot)}$
is the binary conjugation of $(\cdot)$.

$\displaystyle\mathcal{Q}(\ket{GHZ_{m}})=\mathcal{Q}\left(\ket{GHZ_{\overline{m}}}\right)$
(81)

Finally, the statement for when we have multiples of the vector of resources:

See 3.6

###### Proof.

From Thm. 2.1 we get that we can fix dynamics as $Z$ dynamics, without loosing
any generality, and work with the regular computational basis for the qubits.
This comprises the statement up to LU operations. Then, starting from:

$\displaystyle\sum_{j\in
l}Z_{j}\ket{0_{L}}_{l}\equiv\boldsymbol{Z}_{\mu}^{l}\ket{0_{L}}_{l}$
$\displaystyle=\left(a_{\mu}+{b_{0}}_{\mu}^{l}-{b_{1}}_{\mu}^{l}\right)\ket{0_{L}}=\left(a_{\mu}+\Delta
b_{\mu}^{l}\right)\ket{0_{L}},$ (82) $\displaystyle\sum_{j\in
l}Z_{j}\ket{1_{L}}_{l}\equiv\boldsymbol{Z}_{\mu}^{l}\ket{1_{L}}_{l}$
$\displaystyle=\left(-a_{\mu}+{b_{0}}_{\mu}^{l}-{b_{1}}_{\mu}^{l}\right)\ket{1_{L}}=\left(-a_{\mu}+\Delta
b_{\mu}^{l}\right)\ket{1_{L}},$

define $\ket{j}=\ket{j_{1}j_{2}\cdots j_{d}}$, such that
$\boldsymbol{Z}_{\mu}^{l}\ket{j_{L}}_{l}=\left((-1)^{j_{l}}a_{\mu}+\Delta
b_{\mu}^{l}\right)\ket{j_{L}}_{l}$ and note that:

$\displaystyle\boldsymbol{Z}_{\mu}\ket{j}=\sum_{l=1}^{d}\boldsymbol{Z}_{\mu}^{l}\ket{j}$
$\displaystyle=\left(h^{*}(j)a_{\mu}+\sum_{l=1}^{d}\Delta
b_{\mu}^{l}\right)\ket{j}.$ (83)

This means that, for $\ket{\psi}\in\mathfrak{F}$, the QFI will have the
following form:

$\displaystyle\mathcal{Q}_{\mu\nu}$
$\displaystyle\propto\bra{\psi}\boldsymbol{Z}_{\mu}\boldsymbol{Z}_{\nu}\ket{\psi}-\bra{\psi}\boldsymbol{Z}_{\mu}\ket{\psi}\bra{\psi}\boldsymbol{Z}_{\nu}\ket{\psi}$
(84)
$\displaystyle=\sum_{j,m=0}^{2^{d}-1}\alpha_{j}^{*}\alpha_{m}\bra{j}\boldsymbol{Z}_{\mu}\boldsymbol{Z}_{\nu}\ket{m}-\sum_{j,m=0}^{2^{d}-1}\alpha_{j}^{*}\alpha_{m}\bra{j}\boldsymbol{Z}_{\mu}\ket{m}\sum_{p,q=0}^{2^{d}-1}\alpha_{p}^{*}\alpha_{q}\bra{p}\boldsymbol{Z}_{\nu}\ket{k}$
$\displaystyle=\sum_{j=0}^{2^{d}-1}|\alpha_{j}|^{2}\left(h^{*}(j)a_{\mu}+\sum_{l=1}^{d}\Delta
b_{\mu}^{l}\right)\left(h^{*}(j)a_{\nu}+\sum_{l=1}^{d}\Delta
b_{\nu}^{l}\right)$
$\displaystyle\qquad\qquad-\sum_{j,p=0}^{2^{d}-1}|\alpha_{j}|^{2}|\alpha_{p}|^{2}\left(h^{*}(j)a_{\mu}+\sum_{l=1}^{d}\Delta
b_{\mu}^{l}\right)\left(h^{*}(p)a_{\nu}+\sum_{l=1}^{d}\Delta
b_{\nu}^{l}\right)$
$\displaystyle=\left[\sum_{j=0}^{2^{d}-1}|\alpha_{j}|^{2}h^{*}(j)^{2}-\sum_{j,p=0}^{2^{d}-1}|\alpha_{j}|^{2}|\alpha_{p}|^{2}h^{*}(j)h^{*}(p)\right]a_{\mu}a_{\nu},$

which makes the QFI matrix always proportional to $\vec{a}\vec{a}^{T}$,
resulting always in a private state.

∎

##  Appendix G Private States Existence for General Hamiltonians

Let us start by repeating and proving Thm. 3.7:

See 3.7

###### Proof.

Given the rewriting of the QFI matrix in Eq. 31, we already now from Thm. E.1
that the rank of this matrix is given by the rank
$\operatorname{supp}\mathfrak{Q}-1$, similarly to what we have done before. By
choosing the orthogonal subspace to the null space of
$\operatorname{supp}\mathfrak{Q}$ we realize that the span of this space is
generated by the vectors in $W=\\{w|w=\vec{e}_{\vec{i}}-\vec{e}_{\vec{j}}\\}$.
Note that $WC_{\operatorname{supp}\mathfrak{Q}}$ generates a space of vectors
$\mathcal{C}=\\{\vec{c}|\vec{c}=\vec{c}_{\vec{i}}-\vec{c}_{\vec{j}},\vec{c}_{\vec{i}},\vec{c}_{\vec{j}}\in\mathcal{O}\\}$.
This is exactly the higher order discrete subspace generated by $\mathcal{O}$,
$i.e.$
$\mathcal{C}\mathrel{\begin{subarray}{c}\textstyle\subset\\\\[-0.68887pt]
\textstyle\sim\end{subarray}}\mathcal{O}^{2}_{-}$. This already proves the
first part of the theorem about the existence of private states, by choosing
states with $\alpha_{\vec{i}},\alpha_{\vec{j}}$ to be the only non-zero terms,
associated to vectors
$\vec{c}_{\vec{i}}-\vec{c}_{\vec{j}}=\vec{a}\in\mathcal{O}^{2}_{-}$.

The fact that only vectors in the equivalence class of $\mathcal{O}^{2}_{-}$
have private states is a consequence of
$\mathsf{span}(\mathcal{C})=\\{\alpha\vec{a},\alpha\in\mathbb{R}\\}$ in order
for privacy to be verified. If $\alpha\vec{a}\not\in\mathcal{O}^{2}_{-}$ than
one cannot build $\mathsf{span}(\mathcal{C})$, as every
$\vec{c}_{\vec{i}}-\vec{c}_{\vec{j}}$ has to be proportional to $\vec{a}$ and
$\vec{c}_{\vec{i}},\vec{c}_{\vec{j}}$ can only be chosen from $\mathcal{O}$. ∎

To prove Prop. 3.8 let us first introduce and prove some additional
propositions that will allows us to prove it:

######  Proposition G.1.

Let $X\in\mathcal{P}_{n}$ be a Pauli string and
$\mathcal{H}\subseteq\mathcal{P}_{n}$ a subset of Pauli strings. Define an
Hamiltonian as $H=\sum_{P\in\mathcal{H}}c_{P}P$, where all $c_{P}\neq 0$.
Then:

$[X,H]_{\pm}=0\Leftrightarrow[X,P]_{\pm}=0,\quad\forall\ P\in\mathcal{H}.$

###### Proof.

The inverse implication comes trivially from the linearity of the
(anti)commutator $[\cdot,\cdot]_{\pm}$. The direct implication comes as
follows: divide $\mathcal{H}$ into two disjoint sets
$\mathcal{H}=\mathcal{H}_{+}\cup\mathcal{H}_{-}$ such that $[X,P]_{\pm}=0,\
\forall\ P\in\mathcal{H}_{\pm}$. This is always possible given that two Pauli
strings either commute or anticommute. Then:

± $\displaystyle=\sum_{P\in\mathcal{H}}c_{P}[X,P]_{\pm}$
$\displaystyle=\sum_{P\in\mathcal{H}_{\pm}}c_{P}[X,P]_{\pm}+\sum_{P\in\mathcal{H}_{\mp}}c_{P}[X,P]_{\pm}$
$\displaystyle=0+\sum_{P\in\mathcal{H}_{\mp}}c_{P}[X,P]_{\pm}$
$\displaystyle=2\sum_{P\in\mathcal{H}_{\mp}}c_{P}PX$
$\displaystyle=2\sum_{\tilde{P}\in\mathcal{H}_{\mp}X}c_{\tilde{P}}\tilde{P},$

which can only be zero if all $c_{\tilde{P}}$ are 0, as the set of Pauli
strings generates a basis for complex matrices, leading to a contradiction. ∎

######  Proposition G.2.

Let $\mathcal{H}\subseteq\mathcal{P}_{n}$ be a subset of Pauli strings. Define
an Hamiltonian as $H=\sum_{P\in\mathcal{H}}c_{P}P$, where all $c_{P}>0$. Then:

$\exists
X\in\mathcal{P}_{n}:\\{X,H\\}=0\Leftrightarrow\\{\lambda_{H}\\}=\\{+\lambda_{H}\\}\cup\\{-\lambda_{H}\\}$

Meaning for each eigenvalue of the Hamiltonian $H$, $\lambda_{H}$, its
symmetric counterpart, $-\lambda_{H}$, is also an eigenvalue of $H$.

###### Proof.

The direct proof follows very simply. $X\in\mathcal{P}_{n}\implies
X^{2}=\mathbb{1}$.

$\displaystyle\det(H-\lambda\mathbb{1})$
$\displaystyle=\det(H-\lambda\mathbb{1})\det(X^{2})$
$\displaystyle=\det(X)\det(H-\lambda\mathbb{1})\det(X)$
$\displaystyle=\det(XHX-\lambda X\mathbb{1}X)$
$\displaystyle=\det(-H-\lambda\mathbb{1}),$

which means the eigenvalues admit solutions for $\det(H-\lambda\mathbb{1})$
and $\det(H+\lambda\mathbb{1})$ and therefore if $\lambda_{H}\in$ ev,
$-\lambda_{H}\in$ ev. The inverse proof comes as a consequence of the Pauli
strings forming a basis for the space of complex matrices
$\mathbb{C}^{2^{n}}\times\mathbb{C}^{2^{n}}$. Take $U$ to be a unitary
operation that switches from positive eigenstates to the negative ones:
$U\ket{\lambda_{H}^{\pm}}=\ket{\lambda_{H}^{\mp}}$. Then:

$\displaystyle UH\ket{\lambda_{H}^{\pm}}$
$\displaystyle=\pm\lambda_{H}U\ket{\lambda_{H}^{\pm}}=\pm\lambda_{H}\ket{\lambda_{H}^{\mp}}$
$\displaystyle HU\ket{\lambda_{H}^{\pm}}$
$\displaystyle=H\ket{\lambda_{H}^{\mp}}=\mp\lambda_{H}\ket{\lambda_{H}^{\mp}}$
$\displaystyle\implies$
$\displaystyle(UH+HU)\ket{\lambda_{H}^{\pm}}=0,\quad\forall\
\ket{\lambda_{H}^{\pm}},$

which means there is a unitary satisfying the anti-commutation relation. Given
that $U$ is a complex matrix, one can find a decomposition in terms of Pauli
strings. Using Prop. G.1 and similar arguments for the proof of Prop. G.1,
then one has to choose the decomposition among the set of anti-commuting Pauli
strings only. This consequently means there is at least one Pauli string that
anti-commutes with the Hamiltonian. ∎

This allows us to prove Prop. 3.8: See 3.8

###### Proof.

Follows directly from Prop. G.2. The stated conditions for each local
Hamiltonian imply that the local eigenvalues $\lambda_{\mu}$ belong to a set
such that if $+\lambda$ exists so does $-\lambda$. This means that if the
vector $\vec{c}_{i}$ belongs to the discrete subset $\mathcal{O}$, so does the
vector $-\vec{c}_{i}$. This consequently means that $\mathcal{O}^{2}_{-}$
where the vectors are built from $\vec{c}_{i}-\vec{c}_{j}$ will, in
particular, include the vectors $\vec{c}_{i}-(-\vec{c}_{i})=2\vec{c}_{i}$. So
$\mathcal{O}\sim
2\mathcal{O}\subseteq\mathcal{O}^{2}_{-}\equiv\mathcal{O}\mathrel{\begin{subarray}{c}\textstyle\subset\\\\[-0.68887pt]
\textstyle\sim\end{subarray}}\mathcal{O}^{2}_{-}$. ∎

Finally, we are able to prove Corol. 3.1:

See 3.1

###### Proof.

This is a consequence of the decomposition of $\mathcal{Q}=C\mathfrak{Q}C^{T}$
of Eq. 31 and the fact that $\lVert\vec{a}\rVert_{2}$ is maximum for
$\vec{a}\in V(\overline{\mathcal{O}}^{2}_{-})$. Among only the private states,
$\vec{a}\in\overline{\mathcal{O}}^{2}_{-}$, one can calculate the maximal QFI
as:

$\max\mathcal{Q}(\vec{a})=\max\vec{a}^{T}\mathcal{Q}\vec{a}=\lVert\vec{a}\rVert_{2}^{2}=\lVert\vec{c}_{i}-\vec{c}_{j}\rVert_{2}^{2}.$

The privacy measure already incapsulates the fact that any private state will
maximise information about the target function. This means the maximum
information one can get about the maximum function is necessarily private. ∎

##  Appendix H QFI for Mixed States - Derivations

As said in the main text, to find the expression for the QFI in the mixed
state case, we start by the most elementar expression for the QFI:

$\displaystyle\mathcal{Q}_{\mu\nu}(\rho_{{\boldsymbol{\theta}}})=\operatorname{Tr}\mathcal{R}_{\rho_{{\boldsymbol{\theta}}}}^{-1}(\partial_{\theta_{\mu}}\rho_{{\boldsymbol{\theta}}})\rho_{{\boldsymbol{\theta}}}\mathcal{R}_{\rho_{{\boldsymbol{\theta}}}}^{-1}(\partial_{\theta_{\nu}}\rho_{{\boldsymbol{\theta}}}).$
(85)

Under the assumptions of the main text regarding the superoperator
$\mathcal{R}_{\rho_{{\boldsymbol{\theta}}}}$, which corresponds to the
symmetric logarithmic derivative, we can derive it in terms of the density
matrix eigenvectors:

$\displaystyle\mathcal{R}_{\rho}^{-1}(\partial_{\theta_{\mu}}\rho_{{\boldsymbol{\theta}}})$
$\displaystyle=\sum_{j,k}\frac{2}{\lambda_{j}+\lambda_{k}}\bra{\mathcal{G}_{j}}\partial_{\theta_{\mu}}\rho_{{\boldsymbol{\theta}}}\ket{\mathcal{G}_{k}}\ket{\mathcal{G}_{j}}\bra{\mathcal{G}_{k}}.$
(86)

Using the notation
$\ket{\mathcal{G}_{k}^{\prime}}=U_{\boldsymbol{\theta}}\ket{\mathcal{G}_{k}}$:

$\displaystyle\partial_{\theta_{\mu}}\rho_{{\boldsymbol{\theta}}}$
$\displaystyle=\partial_{\theta_{\mu}}\sum_{k\in\operatorname{supp}}\lambda_{k}\ket{\mathcal{G}_{k}^{\prime}}\bra{\mathcal{G}_{k}^{\prime}}$
(87)
$\displaystyle=\sum_{k\in\operatorname{supp}}\lambda_{k}\left[\ket{\partial_{\theta_{\mu}}\mathcal{G}_{k}^{\prime}}\bra{\mathcal{G}_{k}^{\prime}}+\ket{\mathcal{G}_{k}^{\prime}}\bra{\partial_{\theta_{\mu}}\mathcal{G}_{k}^{\prime}}\right],$

where we assumed that the encoding is unitary and done over the density
matrix.

$\displaystyle
D_{ij}^{\mu}\equiv\bra{\mathcal{G}_{i}^{\prime}}\partial_{\theta_{\mu}}\rho_{{\boldsymbol{\theta}}}\ket{\mathcal{G}_{j}^{\prime}}$
$\displaystyle=\bra{\mathcal{G}_{i}^{\prime}}\sum_{k}\lambda_{k}\left[\ket{\partial_{\theta_{\mu}}\mathcal{G}_{k}^{\prime}}\bra{\mathcal{G}_{k}^{\prime}}+\ket{\mathcal{G}_{k}^{\prime}}\bra{\partial_{\theta_{\mu}}\mathcal{G}_{k}^{\prime}}\right]\ket{\mathcal{G}_{j}^{\prime}}$
(88)
$\displaystyle=\sum_{k}\lambda_{k}\left[\braket{\mathcal{G}_{i}^{\prime}}{\partial_{\theta_{\mu}}\mathcal{G}_{k}^{\prime}}\delta_{kj}+\delta_{ik}\braket{\partial_{\theta_{\mu}}\mathcal{G}_{k}^{\prime}}{\mathcal{G}_{j}^{\prime}}\right]$
$\displaystyle=\text{id}_{\operatorname{supp}}(j)\lambda_{j}\braket{\mathcal{G}_{i}^{\prime}}{\partial_{\theta_{\mu}}\mathcal{G}_{j}^{\prime}}+\text{id}_{\operatorname{supp}}(i)\lambda_{i}\braket{\partial_{\theta_{\mu}}\mathcal{G}_{i}^{\prime}}{\mathcal{G}_{j}^{\prime}}$
$\displaystyle=\left[\text{id}_{\operatorname{supp}}(i)\lambda_{i}-\text{id}_{\operatorname{supp}}(j)\lambda_{j}\right]\braket{\partial_{\theta_{\mu}}\mathcal{G}_{i}^{\prime}}{\mathcal{G}_{j}^{\prime}},$

where the last step was taking into consideration that:

$\displaystyle\partial_{\theta_{\mu}}(\braket{\mathcal{G}_{i}^{\prime}}{\mathcal{G}_{j}^{\prime}})=\partial_{\theta_{\mu}}(\braket{\mathcal{G}_{i}}{\mathcal{G}_{j}})$
$\displaystyle=\partial_{\theta_{\mu}}(\delta_{ij})=0$ (89)
$\displaystyle=\braket{\partial_{\theta_{\mu}}\mathcal{G}_{i}^{\prime}}{\mathcal{G}_{j}^{\prime}}+\braket{\mathcal{G}_{i}^{\prime}}{\partial_{\theta_{\mu}}\mathcal{G}_{j}^{\prime}}.$

Plugging it in the QFI we get that:

$\displaystyle\mathcal{Q}_{\mu\nu}(\rho_{\boldsymbol{\theta}})$
$\displaystyle=\text{Tr}\left[\mathcal{R}_{\rho_{\boldsymbol{\theta}}}^{-1}(\partial_{\theta_{\mu}}\rho_{\boldsymbol{\theta}})\rho_{\boldsymbol{\theta}}\mathcal{R}_{\rho_{\boldsymbol{\theta}}}^{-1}(\partial_{\theta_{\nu}}\rho_{\boldsymbol{\theta}})\right]$
(90)
$\displaystyle=\operatorname{Tr}\left[\sum_{i,j}\frac{2}{\lambda_{i}+\lambda_{j}}D_{ij}^{\mu}\ket{\mathcal{G}_{i}^{\prime}}\bra{\mathcal{G}_{j}^{\prime}}\cdot\sum_{k\in\operatorname{supp}}\lambda_{k}\ket{\mathcal{G}_{k}^{\prime}}\bra{\mathcal{G}_{k}^{\prime}}\cdot\sum_{m,n}\frac{2}{\lambda_{m}+\lambda_{n}}D_{mn}^{\nu}\ket{\mathcal{G}_{m}^{\prime}}\bra{\mathcal{G}_{n}^{\prime}}\right]$
$\displaystyle=\operatorname{Tr}\left[\sum_{i,j,m,n}\frac{2D_{ij}^{\mu}}{\lambda_{i}+\lambda_{j}}\cdot\frac{2D_{mn}^{\nu}}{\lambda_{m}+\lambda_{n}}\sum_{k\in\operatorname{supp}}\lambda_{k}\ket{\mathcal{G}_{i}^{\prime}}\braket{\mathcal{G}_{j}^{\prime}}{\mathcal{G}_{k}^{\prime}}\braket{\mathcal{G}_{k}^{\prime}}{\mathcal{G}_{m}^{\prime}}\bra{\mathcal{G}_{n}^{\prime}}\right]$
$\displaystyle=\operatorname{Tr}\left[\sum_{\begin{subarray}{c}i,n\\\
k\in\operatorname{supp}\end{subarray}}\frac{2D_{ik}^{\mu}}{\lambda_{i}+\lambda_{k}}\cdot\frac{2D_{kn}^{\nu}}{\lambda_{k}+\lambda_{n}}\lambda_{k}\ket{\mathcal{G}_{i}^{\prime}}\bra{\mathcal{G}_{n}^{\prime}}\right]$
$\displaystyle=\sum_{\begin{subarray}{c}n\in\operatorname{supp}\\\
k\in\operatorname{supp}\end{subarray}}\frac{2D_{nk}^{\mu}}{\lambda_{n}+\lambda_{k}}\cdot\frac{2D_{kn}^{\nu}}{\lambda_{k}+\lambda_{n}}\lambda_{k}+\sum_{\begin{subarray}{c}n\in\text{null}\\\
k\in\operatorname{supp}\end{subarray}}\frac{2D_{nk}^{\mu}}{\lambda_{n}+\lambda_{k}}\cdot\frac{2D_{kn}^{\nu}}{\lambda_{k}+\lambda_{n}}\lambda_{k}$
$\displaystyle=\sum_{\begin{subarray}{c}n\in\operatorname{supp}\\\
k\in\operatorname{supp}\end{subarray}}\frac{2D_{nk}^{\mu}D_{kn}^{\nu}}{\lambda_{n}+\lambda_{k}}\cdot\left(\frac{\lambda_{k}}{\lambda_{k}+\lambda_{n}}+\frac{\lambda_{n}}{\lambda_{n}+\lambda_{k}}\right)+\sum_{\begin{subarray}{c}n\in\text{null}\\\
k\in\operatorname{supp}\end{subarray}}\frac{2D_{nk}^{\mu}}{\lambda_{k}}\cdot\frac{2D_{kn}^{\nu}}{\lambda_{k}}\lambda_{k}$
$\displaystyle=\sum_{\begin{subarray}{c}n\in\operatorname{supp}\\\
k\in\operatorname{supp}\end{subarray}}\frac{2D_{nk}^{\mu}D_{kn}^{\nu}}{\lambda_{n}+\lambda_{k}}+\sum_{\begin{subarray}{c}n\in\text{null}\\\
k\in\operatorname{supp}\end{subarray}}\frac{4D_{nk}^{\mu}D_{kn}^{\nu}}{\lambda_{k}}$
$\displaystyle=\sum_{\begin{subarray}{c}n\in\operatorname{supp}\\\
k\in\operatorname{supp}\end{subarray}}2\frac{(\lambda_{n}-\lambda_{k})^{2}}{\lambda_{n}+\lambda_{k}}\braket{\partial_{\theta_{\mu}}\mathcal{G}_{n}^{\prime}}{\mathcal{G}_{k}^{\prime}}\braket{\mathcal{G}_{k}^{\prime}}{\partial_{\theta_{\nu}}\mathcal{G}_{n}^{\prime}}+\sum_{\begin{subarray}{c}n\in\text{null}\\\
k\in\operatorname{supp}\end{subarray}}4\lambda_{k}\braket{\mathcal{G}_{n}^{\prime}}{\partial_{\theta_{\mu}}\mathcal{G}_{k}^{\prime}}\braket{\partial_{\theta_{\nu}}\mathcal{G}_{k}^{\prime}}{\mathcal{G}_{n}^{\prime}}$
$\displaystyle=\sum_{\begin{subarray}{c}n\in\operatorname{supp}\\\
k\in\operatorname{supp}\end{subarray}}2\frac{(\lambda_{n}-\lambda_{k})^{2}}{\lambda_{n}+\lambda_{k}}\text{Re}\left\\{\braket{\mathcal{G}_{n}}{\boldsymbol{G}_{\mu}}{\mathcal{G}_{k}}\braket{\mathcal{G}_{k}}{\boldsymbol{G}_{\nu}}{\mathcal{G}_{n}}\right\\}+\sum_{\begin{subarray}{c}n\in\text{null}\\\
k\in\operatorname{supp}\end{subarray}}4\lambda_{k}\text{Re}\left\\{\braket{\mathcal{G}_{k}}{\boldsymbol{G}_{\mu}}{\mathcal{G}_{n}}\braket{\mathcal{G}_{n}}{\boldsymbol{G}_{\nu}}{\mathcal{G}_{k}}\right\\}.$

Let us retrieve the pure state equation from this. Assume the quantum state
density matrix can be described by that of a pure state
$\rho=\ket{\psi}\bra{\psi}$. This means $\lambda_{1}=1$ and every other
$\lambda_{j}=0$. Moreover, since independent of the $\ket{\psi}$ choice, one
can always find an orthonormal basis such that $\ket{\psi}$ is part of. This
means in particular that
$\exists\mathcal{B}=\\{\ket{\psi_{j}}\\}_{j=1,\cdots,2^{n}}$, such that
$\ket{\psi_{1}}=\ket{\psi}$ and
$\sum_{j=1}^{2^{n}}\ket{\psi_{j}}\bra{\psi_{j}}=\mathbb{1}$. Plugging $\rho$
into Eq. 36 we get:

$\displaystyle\mathcal{Q}_{\mu\nu}(\rho_{\boldsymbol{\theta}})$
$\displaystyle=4\text{Re}\left\\{\bra{\psi_{1}}\boldsymbol{G}_{\mu}\left(\sum_{k=2}^{2^{n}}\ket{\psi_{k}}\bra{\psi_{k}}\right)\boldsymbol{G}_{\nu}\ket{\psi_{1}}\right\\}$
(91)
$\displaystyle=4\text{Re}\left\\{\bra{\psi_{1}}\boldsymbol{G}_{\mu}\left(\mathbb{1}-\ket{\psi_{1}}\bra{\psi_{1}}\right)\boldsymbol{G}_{\nu}\ket{\psi_{1}}\right\\}$
$\displaystyle=4\text{Re}\left\\{\bra{\psi_{1}}\boldsymbol{G}_{\mu}\boldsymbol{G}_{\nu}\ket{\psi_{1}}-\bra{\psi_{1}}\boldsymbol{G}_{\mu}\ket{\psi_{1}}\bra{\psi_{1}}\boldsymbol{G}_{\nu}\ket{\psi_{1}}\right\\}.$

This allows us to get one additional way to calculate the QFI of a mixed
state. Suppose the density matrix is a sum of orthogonal states (even if it is
not, there is always a decomposition into orthogonal states by changing basis,
consequence of a diagonalization of the matrix). Let:

$\displaystyle\rho=\sum_{j\in\operatorname{supp}}\lambda_{j}\ket{\mathcal{G}_{j}}\bra{\mathcal{G}_{j}},\quad\text{such
that }\braket{\mathcal{G}_{j}}{\mathcal{G}_{k}}=\delta_{jk}.$ (92)

Then:

$\displaystyle\mathcal{Q}_{\mu\nu}(\rho_{\boldsymbol{\theta}})$
$\displaystyle=\sum_{\begin{subarray}{c}n\in\operatorname{supp}\\\
k\in\operatorname{supp}\end{subarray}}\\!2\frac{(\lambda_{n}-\lambda_{k})^{2}}{\lambda_{n}+\lambda_{k}}\text{Re}\left\\{\braket{\mathcal{G}_{n}}{\boldsymbol{G}_{\mu}}{\mathcal{G}_{k}}\braket{\mathcal{G}_{k}}{\boldsymbol{G}_{\nu}}{\mathcal{G}_{n}}\right\\}+\sum_{\begin{subarray}{c}n\in\text{null}\\\
k\in\operatorname{supp}\end{subarray}}4\lambda_{k}\text{Re}\left\\{\braket{\mathcal{G}_{k}}{\boldsymbol{G}_{\mu}}{\mathcal{G}_{n}}\braket{\mathcal{G}_{n}}{\boldsymbol{G}_{\nu}}{\mathcal{G}_{k}}\right\\}$
(93) $\displaystyle=\sum_{\begin{subarray}{c}n\in\operatorname{supp}\\\
k\in\operatorname{supp}\end{subarray}}2\frac{(\lambda_{n}-\lambda_{k})^{2}}{\lambda_{n}+\lambda_{k}}\text{Re}\left\\{a_{kn}^{\mu}a_{nk}^{\nu}\right\\}+\sum_{\begin{subarray}{c}k\in\operatorname{supp}\end{subarray}}4\lambda_{k}\text{Re}\left\\{\bra{\mathcal{G}_{k}}\boldsymbol{G}_{\nu}\left[\sum_{\begin{subarray}{c}n\in\text{null}\end{subarray}}\ket{\mathcal{G}_{n}}\bra{\mathcal{G}_{n}}\right]\boldsymbol{G}_{\mu}\ket{\mathcal{G}_{k}}\right\\}$
$\displaystyle=\smashoperator[]{\sum_{\begin{subarray}{c}n\in\operatorname{supp}\\\
k\in\operatorname{supp}\end{subarray}}^{}}2\frac{(\lambda_{n}-\lambda_{k})^{2}}{\lambda_{n}+\lambda_{k}}\text{Re}\left\\{a_{kn}^{\mu}a_{nk}^{\nu}\right\\}+\smashoperator[]{\sum_{\begin{subarray}{c}k\in\operatorname{supp}\end{subarray}}^{}}4\lambda_{k}\text{Re}\left\\{\bra{\mathcal{G}_{k}}\boldsymbol{G}_{\nu}\left[\mathbb{1}-\smashoperator[]{\sum_{\begin{subarray}{c}n\in\operatorname{supp}\end{subarray}}^{}}\ket{\mathcal{G}_{n}}\bra{\mathcal{G}_{n}}\right]\boldsymbol{G}_{\mu}\ket{\mathcal{G}_{k}}\right\\}$
$\displaystyle=\smashoperator[]{\sum_{\begin{subarray}{c}n\in\operatorname{supp}\\\
k\in\operatorname{supp}\end{subarray}}^{}}2\frac{(\lambda_{n}-\lambda_{k})^{2}}{\lambda_{n}+\lambda_{k}}\text{Re}\left\\{a_{kn}^{\mu}a_{nk}^{\nu}\right\\}+\smashoperator[]{\sum_{\begin{subarray}{c}k\in\operatorname{supp}\end{subarray}}^{}}4\lambda_{k}\text{Re}\left\\{\bra{\mathcal{G}_{k}}\boldsymbol{G}_{\nu}\left[\mathbb{1}-\ket{\mathcal{G}_{k}}\bra{\mathcal{G}_{k}}-\smashoperator[]{\sum_{\begin{subarray}{c}n\in\operatorname{supp}\\\
n\neq
k\end{subarray}}^{}}\ket{\mathcal{G}_{n}}\bra{\mathcal{G}_{n}}\right]\boldsymbol{G}_{\mu}\ket{\mathcal{G}_{k}}\right\\}$
$\displaystyle=\smashoperator[]{\sum_{\begin{subarray}{c}n\in\operatorname{supp}\\\
k\in\operatorname{supp}\end{subarray}}^{}}2\frac{(\lambda_{n}-\lambda_{k})^{2}}{\lambda_{n}+\lambda_{k}}\text{Re}\left\\{a_{kn}^{\mu}a_{nk}^{\nu}\right\\}+\smashoperator[]{\sum_{\begin{subarray}{c}k\in\operatorname{supp}\end{subarray}}^{}}\lambda_{k}\mathcal{Q}_{\mu\nu}(\ket{\mathcal{G}_{k}})-\smashoperator[]{\sum_{\begin{subarray}{c}n,k\in\operatorname{supp}\\\
k\neq
n\end{subarray}}^{}}4\lambda_{k}\text{Re}\left\\{a_{kn}^{\mu}a_{nk}^{\nu}\right\\}$
$\displaystyle=\sum_{\begin{subarray}{c}n,k\in\operatorname{supp}\\\ n\neq
k\end{subarray}}\left[2\frac{(\lambda_{n}-\lambda_{k})^{2}}{\lambda_{n}+\lambda_{k}}-4\lambda_{k}\right]\text{Re}\left\\{a_{kn}^{\mu}a_{nk}^{\nu}\right\\}+\sum_{\begin{subarray}{c}k\in\operatorname{supp}\end{subarray}}\lambda_{k}\mathcal{Q}_{\mu\nu}(\ket{\mathcal{G}_{k}})$
$\displaystyle=\sum_{n>k\in\operatorname{supp}}4\left[\frac{(\lambda_{n}-\lambda_{k})^{2}}{\lambda_{n}+\lambda_{k}}-(\lambda_{n}+\lambda_{k})\right]\text{Re}\left\\{a_{nk}^{\nu}a_{kn}^{\mu}\right\\}+\sum_{\begin{subarray}{c}k\in\operatorname{supp}\end{subarray}}\lambda_{k}\mathcal{Q}_{\mu\nu}(\ket{\mathcal{G}_{k}}).$

In the case the state density matrix is described by a sum of non-orthogonal
vectors:

$\displaystyle\rho=\sum_{j\in\operatorname{supp}}\lambda_{j}\ket{\phi_{j}}\bra{\phi_{j}},\quad\text{such
that }\braket{\phi_{j}}{\phi_{k}}=\alpha_{jk}\leq 1,$ (94)

such that $\boldsymbol{G}_{\mu}\ket{\phi_{j}}$ is equal for every
$\ket{\phi_{j}}$, then diagonalizing $\rho$ into vectors that are orthogonal
between themselves, call it $\ket{\tilde{\phi}_{j}}$ still holds the fact that
$\boldsymbol{G}_{\mu}\ket{\tilde{\phi}_{j}}$ is equal for every
$\ket{\tilde{\phi}_{j}}$. If, more than that,
$\text{Re}\left\\{a_{nk}^{\nu}a_{kn}^{\mu}\right\\}\propto\mathcal{Q}_{\nu\mu}(\ket{\mathcal{G}_{n}})$,
then $\lambda_{n}$ should simply be scaled into a $\tilde{\lambda}_{n}$, but
the information is still structurally the same, $i.e.$ it is the same function
of parameters.

This gives some intuition about how to analyze Eq. 93 and when to use it over
Eq. 90. In Eq. 93 the first term corresponds to a distinguishability between
the states under the encoding dynamics. If the generators of the dynamics are
orthogonal to the noise, then this goes to zero. The second part of Eq. 93 can
be seen as the local informations of each states in a perfect scenario
weighted by the corresponding weights. Note that the expression:

$4\left[\frac{(\lambda_{n}-\lambda_{k})^{2}}{\lambda_{n}+\lambda_{k}}-(\lambda_{n}+\lambda_{k})\right],$

is always negative, meaning that if they are completely distinguishable, then
we recover the notion of convexity of the QFI. If they are not completely
distinguishable, then the amount of information is decreased by a factor
proportional to $\text{Re}\left\\{a_{nk}^{\nu}a_{kn}^{\mu}\right\\}$. On the
other hand, when the noise is only in the same direction of the generators of
the encoding, then Eq. 90 becomes simpler, as the second part goes to zero,
and we are left simply with an expression that weights the distinguishability
of the states, under the encoding dynamics.