# Controlled State Reconstruction and Quantum Secret Sharing

Pahulpreet Singh<EMAIL_ADDRESS>Centre for Quantum
Science and Technology
International Institute of Information Technology Hyderabad, Gachibowli,
Hyderabad – 500032, Telangana, India  Center for Computational Natural Science
and Bioinformatics
International Institute of Information Technology Hyderabad, Gachibowli,
Hyderabad – 500032, Telangana, India  Indranil Chakrabarty
<EMAIL_ADDRESS>Centre for Quantum Science and Technology
International Institute of Information Technology Hyderabad, Gachibowli,
Hyderabad – 500032, Telangana, India  Center for Security, Theory and
Algorithmic Research
International Institute of Information Technology Hyderabad, Gachibowli,
Hyderabad – 500032, Telangana, India

###### Abstract

In this article, we present a benchmark for resource characterization in the
process of controlled quantum state reconstruction and secret sharing for
general three-qubit states. This is achieved by providing a closed expression
for the reconstruction fidelity, which relies on the genuine tripartite
correlation and the bipartite channel between the dealer and the reconstructor
characterized by the respective correlation parameters. We formulate the idea
of quantum advantage in approximate state reconstruction as surpassing the
classical limit set at 2/3. This article brings out new interoperability
between teleportation and state reconstruction. This is detailed through a
case-by-case analysis of relevant correlation matrices. We are reformulating
the idea of quantum secret sharing by setting up additional constraints on the
teleportation capacity of the bipartite channels between dealer and
shareholders by ensuring that individually, the shareholders cannot
reconstruct the secret. We believe that this will give us the ideal picture of
how quantum secret sharing should be.

##  I Introduction

With the advent of quantum information theory, sending and sharing quantum 1 ;
2 ; 3 ; 4 ; 5 ; 6 ; 7  as well as classical 8 ; 9 ; 10 ; 11 ; 12 ; 13
information using quantum resources has become a significant area of study.
During the transmission, security of the message becomes a key consideration
and has to be taken care of to set up quantum networks 14 ; 15 ; 16 ; 17 ; 18
; 19 ; 20 ; 21  in the longer run. Secret sharing is a procedure that allows
secure distribution of a secret message into multiple ($n$) parts so that a
certain number ($k$) of parts can be used to get back the original message,
and forbids any subset smaller than $k$ from obtaining useful information
about it. Quantum secret sharing (QSS) 22 ; 23 ; 24 ; 25 ; 26 ; 27 ; 28 ; 29 ;
30 ; 31 ; 32 ; 33 ; 34 ; 35 ; 36  is the extension of this idea where we use a
quantum resource (usually a multi-party entangled state) to distribute the
secret 26 ; 27 ; 28 . The secret itself can be classical or quantum in nature,
and of any size. In this article, we are interested in sharing of a qubit with
two parties such that when both the parties co-operate, they can reconstruct
it at one of the locations. Here, we name this process as controlled state
reconstruction (CSR). CSR is a slightly broader term than the secret sharing
in the sense that it need not account for the security aspects that come with
secret sharing. The security here is defined by the condition that no share
holders can reveal the secret without the help of the other share holders.
This process has also been termed as controlled quantum teleportation (CQT)
since it is analogous to teleporting a qubit from an initial location (dealer)
to a final location (reconstructor) with the help of a third party (assistant)
acting as the control qubit 37 ; 38 ; 39 . Recently, it has been shown that a
witness operator can be constructed to estimate the power of the controller in
CQT 40 .
In this paper, we differentiate between the two scenarios by using the term
"quantum secret sharing" (QSS) specifically when security is a requirement
(i.e. individual shareholders do not have enough information about the secret)
and "controlled state reconstruction" (CSR) as the general term for
transmitting quantum information from one location to another with the help of
an assistant. Most of our work focuses on CSR, however at the end, we
investigate some intricacies that come with QSS.
We consider the simple setting where a dealer (say, Alice) aims to share an
unknown qubit with two shareholders (say, Bob and Charlie). At the start of
the protocol, one of the shareholders is decided to be the re-constructor
(Charlie, for example) which makes the other one (Bob, in this case) the
assistant. Prefect reconstruction of the state after sharing has already been
shown using the GHZ state as the resource 22 . In this paper, we study
approximate controlled reconstruction of the state using general tripartite
entangled states. Hence, if we are using a resource which does not allow
perfect reconstruction, we want the final qubit to be as close to the original
qubit as possible. By quantum advantage, we refer to a situation where the
reconstruction fidelity is better than what can be achieved classically
without having any shared quantum resource.
Our first finding is the classical limit of controlled reconstruction which
happens to be $2/3$. This matches with the classical limit of teleportation
fidelity of a bi partite two qubit channel 2 . Subsequently, we find an
expression for the reconstruction fidelity in terms of the Bloch parameters of
the resource state. It then enables us to find conditions under which a three-
qubit states will have quantum advantage in the process of CSR/CQT. It turns
out that the reconstruction fidelity is dependent on two siginificant
parameters – one is the correlation tensor between three parties, and the
other one is correlation matrix of the dealer - reconstructor pair. It is
intriguing to note that this fidelity does not depend on the correlation
matrix of the assistant - reconstructor pair. In a sense, this fidelity just
quantifies how much quantum information from the initial state is transferred
to the final location. There can be situations with quantum advantage despite
the absence of tripartite correlation. This leads us to believe that this
fidelity does not originate solely from CSR. Rather, there is also
contribution of the teleportation capacity of the bipartite channel between
the dealer and the reconstructor. We not only prove this but consider several
cases based on the correlation matrices that appear in our expression. This
helps us analyze the interoperability between the reconstruction fidelity and
the teleportation fidelity for three-qubit resources in a holistic manner. In
the later part of our article we develop this idea further to introduce QSS in
a different but meaningful way. This paves the way to distinguish QSS from
CSR/CQT. For a QSS protocol to be successful, we make sure that neither of the
share holders have useful information about the secret by enforcing additional
constraints on the teleportation capacities of the dealer to share holder’s
channels. We also specify the states that can be used as resources for
successful QSS.
In section II, we obtain the classical limit of this fidelity without using
any quantum channel to share the state. In section III, we give an expression
for the maximum possible reconstruction fidelity using the parameters of the
resource state. This lets us quantify the quantum advantage an arbitrary
three-qubit resource can provide in reconstruction. In the same section, we
discuss a potential relationship between (2-party) quantum teleportation and
(3-party) state reconstruction. At the end, in section IV, we give criteria to
discern QSS from CSR by adding additional constraints that need to be
satisfied for QSS.

##  II Classical Limit of Controlled state Reconstruction

We define the classical limit as the expected fidelity score obtained if only
classical channels are used to share a qubit. This will allow us to define the
threshold above which a quantum advantage can be claimed. Here, we consider a
three-party scheme where Alice is the dealer, Bob is the assistant, and
Charlie is the reconstructor. Let $\ket{q}$ denote Alice’s qubit which is to
be shared. She can measure in some basis (say $\ket{\uparrow}$ ,
$\ket{\downarrow}$), which would agreed upon by the parties beforehand. Then
this measurement result can be encoded into a single classical bit s (say 0
for $\ket{\uparrow}$ and 1 for $\ket{\downarrow}$). This can be split into two
shares $s_{1}$ and $s_{2}$ such that $s_{1}\oplus s_{2}=s$ It can be shown
that (for a given s), it would be optimal for Alice to choose the appropriate
$s_{1}$, $s_{2}$ from a uniform distribution (see Appendix A). Alice transmits
the respective bits to Bob and Charlie through classical channels. During the
reconstruction phase, Bob and Charlie cooperate to find s, using which they
can construct the corresponding quantum state $\ket{s}$ (again,
$\ket{\uparrow}$ for s=0 and $\ket{\downarrow}$ for s=1) which is an
approximation of Alice’s original quantum state $\ket{q}$, which we represent
as:

$\ket{q}=\cos\frac{\theta}{2}\ket{\uparrow}+e^{i\phi}\sin\frac{\theta}{2}\ket{\downarrow}$
(1)

Since the final state $\ket{s}$ is eventually dependent only on Alice’s
measurement, we can say $\ket{s}$ is $\ket{\uparrow}$ with probability
$\cos^{2}\frac{\theta}{2}$ and $\ket{\downarrow}$ with probability
$\sin^{2}\frac{\theta}{2}$. Now, calculating the fidelity between the two
qubit states,

$\displaystyle F(q,s)$ $\displaystyle=\|\braket{q}{s}\|^{2}{}$ (2)
$\displaystyle=Pr[s=0]\cdot\|\braket{q}{\uparrow}\|^{2}+Pr[s=1]\cdot\|\braket{q}{\downarrow}\|^{2}{}$
$\displaystyle=1-\frac{1}{2}\sin^{2}\theta.$

Taking the expectation fidelity over all states $|q\rangle$ on the Bloch
sphere,

$\displaystyle\mathcal{F}$ $\displaystyle=\langle F(q,s)\rangle{}$ (3)
$\displaystyle=\frac{1}{4\pi}\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi}\sin\theta\,d\theta\,d\phi(1-\frac{1}{2}\sin^{2}\theta){}$
$\displaystyle=\frac{1}{2}(2)-\frac{1}{4}(\frac{4}{3})=\boxed{\frac{2}{3}}.$

Hence the Fidelity for reconstruction is $\mathcal{F}_{c}=\dfrac{2}{3}$.

This is the value of the classical limit of the reconstruction fidelity of the
state shared by the dealer. If one is able to achieve a fidelity more than
this with the help of a shared quantum resource (in this case, a tripartite
entangled state), we say that there is a quantum advantage.

Note: We recently became aware of 41  who have done this calculation for a
similar scenario and arrived at the same value, i.e. 2/3.

##  III Approximate Controlled Reconstruction And Quantum Advantage

We start with a three-qubit resource state $\rho_{ABC}$ in the space
$\mathcal{H}_{A}\otimes\mathcal{H}_{B}\otimes\mathcal{H}_{C}$. We can write it
in parametric form as

$\displaystyle\rho_{ABC}=\frac{1}{8}[I^{\otimes
3}+\sum_{i=1}^{3}a_{i}.\sigma_{i}\otimes I^{\otimes 2}+{}$
$\displaystyle\sum_{j=1}^{3}I\otimes b_{j}.\sigma_{j}\otimes
I+\sum_{k=1}^{3}I^{\otimes 2}\otimes c_{k}.\sigma_{k}+{}$
$\displaystyle\sum_{i,j=1}^{3}q_{ij}\sigma_{i}\otimes\sigma_{j}\otimes
I+\sum_{i,k=1}^{3}r_{ik}\sigma_{i}\otimes I\otimes\sigma_{k}+{}$
$\displaystyle\sum_{j,k=1}^{3}s_{jk}I\otimes\sigma_{j}\otimes\sigma_{k}+\sum_{i,j,k=1}^{3}t_{ijk}\sigma_{i}\otimes\sigma_{j}\otimes\sigma_{k}].$
(4)

Here $a_{i},b_{j},c_{k}$ are local Bloch vectors and the correlation matrices
are given by, $Q=\\{q_{ij}\\}=Tr(\rho_{ABC}(\sigma_{i}\otimes\sigma_{j}\otimes
I)),R=\\{r_{ik}\\}=Tr(\rho_{ABC}(\sigma_{i}\otimes I\otimes\sigma_{k}))$ and
$S=\\{s_{jk}\\}=Tr(\rho_{ABC}(I\otimes\sigma_{j}\otimes\sigma_{k}))$ are the
correlation matrices of order $3\times 3$. Here
$\tau=t_{ijk}=Tr(\rho_{ABC}(\sigma_{i}\otimes\sigma_{j}\otimes\sigma_{k}))$ is
the correlation tensor.

Now we find the reconstruction fidelity of the state shared in terms of the
Bloch parameters of the three-qubit resource state $\rho_{ABC}$. The qubit
$\S$ on Alice’s side, parameterized by the Bloch vector $\phi$, is given by

$\rho_{\S}=\frac{1}{2}(I+\sum_{i}\phi_{i}.\sigma_{i}).$ (5)

In the standard scheme, measurement takes place at two phases of the protocol.
First, at Alice’s side on the to be shared qubit ($\S$) along with Alice’s
share of the resource ($A$), with projectors
$P_{l}=|\Psi_{l}\rangle\langle\Psi_{l}|$ ($l=0,1,2,3$) and second, on Bob’s
qubit, with projectors $P_{x}=|x\rangle\langle x|$ ($x=+,-$). Here, the Bell
states and Hadamard states are given as,
$|\Psi_{(0)}^{3}\rangle=\frac{1}{\sqrt{2}}(|01\rangle\pm|10\rangle),|\Psi_{(1)}^{2}\rangle=\frac{1}{\sqrt{2}}(|00\rangle\pm|11\rangle)$
and $|x_{\pm}\rangle=\frac{1}{\sqrt{2}}(|0\rangle\pm|1\rangle)$ respectively.
The Bell state projectors can be written in the form

$P_{l}=\frac{1}{4}(I^{\otimes 2}+\sum_{ij}t_{ij}\sigma_{i}\otimes\sigma_{j}).$
(6)

The coefficients $t_{ij}$ form a correlation matrix $T_{l}$ ($l$ = 0,1,2,3 for
different projectors). These are given by $T_{0}$ = diag(-1, -1, -1), $T_{1}$
= diag(-1, +1, +1), $T_{2}$ = diag(+1, -1, +1) and $T_{3}$ = diag(+1, +1, -1).
The set of Hadamard projectors on Bob’s side is given by
$P_{x}=\frac{1}{2}(I+\mathbf{x}\cdot\sigma)$. Here, $\mathbf{x}=(\pm 1,0,0)$.

Now we find the output state of Charlie’s qubit, after the two measurements
followed by applying appropriate unitaries.

$\displaystyle p_{\alpha}\varrho_{\alpha}=$ (7)
$\displaystyle\mathrm{Tr}_{123}\left[(P_{l}\otimes P_{x}\otimes
U_{\alpha})(\rho_{\S}\otimes\rho_{ABC})(P_{l}\otimes P_{x}\otimes
U_{\alpha}^{\dagger})\right].$

The trace is taken over the original qubit, Alice’s share, and Bob’s share.
$\alpha$ acts as multi-index for the pair $(l,x)$. Here,
$p_{\alpha}=\mathrm{Tr}\left((P_{l}\otimes P_{x}\otimes
I)(\rho_{\S}\otimes\rho_{ABC})\right)$ is the probability of getting the
measurement corresponding to the combination $(P_{l},P_{x})$. Finally,
$U_{\alpha}$ is the unitary operator chosen to reconstruct (a close
approximation of) the state at Charlie’s side. Substituting the expressions
for the states and the projection operators, we obtain:

$\displaystyle\varrho_{\alpha}=\frac{1}{16p_{\alpha}}\Bigg{(}\bigg{[}1+\frac{1}{2}\sum_{i}(T_{l})_{ii}A_{i}\phi_{i}+\sum_{i}B_{i}\phi_{i}+$
$\displaystyle\frac{1}{2}\sum_{i,j}(T_{l})_{ii}Q_{ij}\phi_{i}x_{j}\bigg{]}I+\sum_{jk}\Omega_{jk}\bigg{[}\sum_{j}C_{j}+\sum_{ij}S_{ij}x_{i}$
$\displaystyle+\frac{1}{2}\sum_{ij}(T_{l})_{ii}R_{ij}\phi_{i}+\frac{1}{2}\sum_{ijm}(T_{l})_{mm}t_{mij}x_{i}\phi_{m}\bigg{]}\sigma_{k}\Bigg{)}.$
(8)

Here, $\\{\Omega_{\alpha}\\}$ are rotations in $\mathbb{R}^{3}$ obtained from
the unitaries $\\{U_{\alpha}\\}$, given by the relation:

$U_{\alpha}\hat{n}\cdot\sigma
U_{\alpha}^{\dagger}=(\Omega^{\dagger}\hat{n})\cdot\sigma=\sum_{ij}\Omega_{ij}n_{i}\sigma_{j}.$
(9)

Now, the expected fidelity of reconstruction, i.e. the "closeness" of
Charlie’s qubit to the original state, is given by the following integral over
the Bloch sphere with uniform distribution $M$:

$\mathcal{F}=\oint
dM(\phi)\sum_{\alpha}p_{\alpha}\mathrm{Tr}\left(\varrho_{\alpha}\rho_{\S}\right).$
(10)

After substituting expressions from eq (8) and (5), followed by omitting the
terms that do not contribute to the integral, and using the relation,

$\oint\langle\phi,Y\phi\rangle dM(\phi)=\frac{1}{3}\mathrm{Tr}\left(Y\right),$
(11)

the integral in (10) reduces to

$\displaystyle\mathcal{F}=\frac{1}{16}\sum_{\alpha}\bigg{[}1+\mathbf{B\cdot
x}+\frac{1}{3}\mathrm{Tr}\left(\Omega_{\alpha}^{\dagger}R^{\dagger}T_{l}\right)$
$\displaystyle+\frac{1}{3}\mathrm{Tr}\left(\Omega_{\alpha}^{\dagger}(\tau_{\lambda\mu\nu}x^{\mu})^{\dagger}T_{l}\right)\bigg{]}.$
(12)

This is being summed up over all $\alpha$, i.e. all the $(l,x)$ possibilities
of the two measurements. Note that $\sum_{\alpha}\mathbf{B\cdot
x}=\sum_{l}\sum_{x}\mathbf{B\cdot x}=0$. Let $T$ be the matrix formed by the
elements $\\{\sum_{j}t_{ijk}x_{j}\\}$, or in tensor notation,

$T=\tau_{\lambda\mu\nu}x^{\mu},$ (13)

for $\mathbf{x}=(+1,0,0)$. Then, for $\mathbf{x}=(-1,0,0)$, we have
$\tau_{\lambda\mu\nu}x^{\mu}=-T$. Thus, the summation can be split into two,
based on $x$,

$\displaystyle\mathcal{F}=\frac{1}{2}+\frac{1}{16}.\frac{1}{3}\sum_{l}\mathrm{Tr}\left(T_{l}^{\dagger}(R+T)\Omega_{(l,+)}\right)$
$\displaystyle+\frac{1}{16}.\frac{1}{3}\sum_{l}\mathrm{Tr}\left(T_{l}^{\dagger}(R-T)\Omega_{(l,-)}\right).$
(14)

Here we have expanded the multi-index $\alpha$ back in the pair form $(l,x)$.
Now, we want to choose the rotations to maximise $\mathcal{F}$. As
$-T_{l}^{\dagger}$ is also a rotation, the $\Omega_{\alpha}$’s can be chosen
independently to maximize each term. We take $\Omega^{\prime}$ to be the the
rotation that maximises the terms corresponding to $T_{l}^{\dagger}(R-T)$ and
$\Omega^{\prime\prime}$ for the terms corresponding to $T_{l}^{\dagger}(R+T)$.
Now, this expression is independent of $l$.

$\mathcal{F}_{\max}=\max_{\Omega^{\prime},\Omega^{\prime\prime}}\frac{1}{2}\big{(}1-\frac{1}{6}\mathrm{Tr}\left[(R+T)\Omega\right]-\frac{1}{6}\mathrm{Tr}\left[(R-T)\Omega^{\prime}\right]\big{)},$
(15)

where the maximum is taken over all rotations
$\Omega^{\prime},\Omega^{\prime\prime}$. Since $\Omega^{\prime}$ and
$\Omega^{\prime\prime}$ can be independent of each other, we get the maximum
as

$\displaystyle\mathcal{F}_{\max}=\frac{1}{2}\bigg{(}1+\frac{1}{6}\mathrm{Tr}\left(\sqrt{(R+T)^{\dagger}(R+T)}\right)+$
$\displaystyle\frac{1}{6}\mathrm{Tr}\left(\sqrt{(R-T)^{\dagger}(R-T)}\right)\bigg{)}.$
(16)

A tripartite resource state $\varrho$ is useful for reconstruction of the
state only when $\mathcal{F}_{\max}>2/3$, or when $\vartheta(\varrho)>1$,
where we define $\vartheta(\varrho)$ as:

$\vartheta(\varrho):=\frac{1}{2}\bigg{(}\|R+T\|_{1}+\|R-T\|_{1}\bigg{)},$ (17)

such that $\mathcal{F}_{\max}=\dfrac{1}{2}(1+\dfrac{1}{3}\vartheta(\varrho))$,
and $\|\cdot\|_{1}$ denotes the trace norm of a matrix, given by
$\|Z\|_{1}=\mathrm{Tr}\sqrt{Z^{\dagger}Z}$. Since $R$ shows up in the
expression, we conclude that the reconstruction of the state is not entirely
because of the controlled reconstruction capability. There can be a situation
when $T=O$ ($O$ denotes the null matrix), but the value of $\mathcal{F}$ is
greater than $2/3$. Here, as there is no tripartite correlation 42note , Bob’s
involvement seems inconsequential. In a sense, this fidelity quantifies the
information that can be retrieved as a result of this process. Hence, the CSR
fidelity has contributions from both, the reconstruction capacity of the
three-qubit resource and the teleportation capacity of the two-qubit channel
between the dealer and receiver.

Here, the dealer-reconstructor subsystem of the resource $\rho_{ABC}$ (from eq
(4)) is given by:

$\displaystyle\rho_{AC}=\mathrm{Tr}_{B}(\rho_{ABC})=$ (18)
$\displaystyle\frac{1}{4}[I^{\otimes 2}+\sum_{i}a_{i}.\sigma_{i}\otimes
I+\sum_{k}I\otimes
c_{k}.\sigma_{k}+\sum_{ik}r_{ik}\sigma_{i}\otimes\sigma_{k}].$

Using the result from 2 , we can write the teleportation fidelity of
$\rho_{AC}$ as:

$\mathcal{F}^{\prime}_{AC}=\frac{1}{2}\big{(}1+\frac{1}{3}\mathrm{Tr}\sqrt{R^{\dagger}R}\big{)},$
(19)

This analysis is for the case when Alice is the dealer, and the final qubit is
being reconstructed at Charlie’s end with the assistance of Bob. However,
there can be other cases with the same resource state when the roles are
interchanged. This gives us an ordered triplet of (dealer, assistant,
reconstructor), which we call the setting.

For some resource states like the GHZ state, all six settings are equivalent
due to its symmetry. However, this cannot be generalized, as $\mathcal{F}$ for
all the settings need not be the same. We, thus, need to define three
different $T$ matrices:
$T_{AB}=\\{Tr(\rho_{ABC}(\sigma_{i}\otimes\sigma_{j}\otimes\sigma_{x}))\\}_{ij}$,
$T_{AC}=\\{Tr(\rho_{ABC}(\sigma_{i}\otimes\sigma_{x}\otimes\sigma_{j}))\\}_{ij}$
and
$T_{BC}=\\{Tr(\rho_{ABC}(\sigma_{x}\otimes\sigma_{i}\otimes\sigma_{j}))\\}_{ij}$.
Here, the subscripts denote the subsystems that contribute to the matrix. For
example, in $T_{AB}$, the matrix indices correspond to the first and second
subsystems of the three-qubit resource, as seen above. We can, hence, re-write
eq (17) for the six settings, as shown in table 1. As the table shows, the
expression varies only based on the assistant and is symmetric in swapping the
dealer and reconstructor. This symmetry is shared with the expression for
fidelity of teleportation 2 .

S. No. | Setting | Expression for $\vartheta(\rho)$
---|---|---
1 | (Alice, Bob, Charlie) | $\vartheta_{AC}(\rho)=\displaystyle\frac{1}{2}\big{(}\|R+T_{AC}\|_{1}$
2 | (Charlie, Bob, Alice) | $+\|R-T_{AC}\|_{1}\big{)}$
3 | (Alice, Charlie, Bob) | $\vartheta_{AB}(\rho)=\displaystyle\frac{1}{2}\big{(}\|Q+T_{AB}\|_{1}$
4 | (Bob, Charlie, Alice) | $+\|Q-T_{AB}\|_{1}\big{)}$
5 | (Bob, Alice, Charlie) | $\vartheta_{BC}(\rho)=\displaystyle\frac{1}{2}\big{(}\|S+T_{BC}\|_{1}$
6 | (Charlie, Alice, Bob) | $+\|S-T_{BC}\|_{1}\big{)}$
Table 1: A table showing the expression of $\vartheta(\rho)$ for different
settings of (dealer, assistant, reconstructor).

For simplicity, we will henceforth use the setting (Alice, Bob, Charlie) by
default, and follow the representation in eq (17), i.e. $T=T_{AC}$ and
$\vartheta(\varrho)=\vartheta_{AC}(\varrho)$, unless specified otherwise. We
have seen that the correlation matrix $S$ is not present in eq (17). It is
important to note that this does not rule out the role of Bob in the
reconstruction of the state. It only tells that the prior correlation between
Bob and Charlie does not affect reconstruction fidelity. The only factors in
determining it are the genuine correlation between Alice, Bob and Charlie,
captured by $T$, and the correlation matrix between the dealer and the
reconstructor, denoted by $R$. Hence, we study different cases based on $R,T$
which give an overview of how this score captures both state reconstructing
fidelity and the teleportation fidelity of the channel between the source and
the reconstructor. These also present us with some conditions on quantum
advantage in terms of $R,T$.

Case 1: R $\neq$ O, T $\neq$ O This is the most general scenario when both $R$
and $T$ can take any value. In this case, it is not evident whether the
fidelity score is entirely because of the state reconstruction capacity of the
entire three-qubit state or, due to the teleportation capacity of the dealer-
reconstructor channel, or both. Consequently, we can not pinpoint the main
reason behind the quantum advantage. One way to address this situation is to
look into the teleportation fidelity of the dealer-reconstructor subsystem. If
the teleportation fidelity is at most $2/3$ and the reconstruction fidelity is
greater than $2/3$, then we can conjecture that there is a quantum advantage
because of the state reconstruction resource. We discuss different cases with
the help of the following examples:

Example 1: As the first simple case we consider the GHZ state.
$\ket{GHZ}=\dfrac{1}{\sqrt{2}}(\ket{000}+\ket{111})$.

The matrices $R$ and $T$ for the GHZ state can be found after writing the
corresponding density state in its Bloch form:

$R=\left(\begin{array}[]{ccc}0&0&0\\\ 0&0&0\\\
0&0&1\end{array}\right),T=\left(\begin{array}[]{ccc}1&0&0\\\ 0&-1&0\\\
0&0&0\end{array}\right).$

This gives

$\displaystyle\vartheta(\rho_{W})=3,$ $\displaystyle\mathcal{F}_{\max}=1,$
(20)

which is expected since it is already known that the GHZ is used for perfect
reconstruction. Note that the teleportation fidelity (from eq (19)) for the
dealer-reconstructor subsystem, of this state is
$\mathcal{F}^{\prime}=\frac{1}{2}(1+\frac{1}{3})=\dfrac{2}{3}$. This clearly
gives us a case with a quantum advantage arising from the state reconstruction
ability of three-qubit resource states.

Example 2: In a case where both of the fidelities $\mathcal{F}^{\prime}$ and
$\mathcal{F}$ are greater than $2/3$, we cannot be sure whether the quantum
advantage in the reconstruction fidelity $\mathcal{F}$ is entirely because of
the state reconstruction resource. The W state, known to exhibit a different
nature of entanglement from the GHZ state 43 , is given by
$\ket{W}=\frac{1}{\sqrt{3}}(\ket{001}+\ket{010}+\ket{100})$. $R$ and $T$ for
the W state, from its Bloch representation of $\rho_{W}=\ket{W}\bra{W}$, are:

$R=\left(\begin{array}[]{ccc}2/3&0&0\\\ 0&2/3&0\\\
0&0&-1/3\end{array}\right),T=\left(\begin{array}[]{ccc}0&0&2/3\\\ 0&0&0\\\
2/3&0&0\end{array}\right),$

which gives

$\displaystyle\vartheta(\rho_{W})=\frac{7}{3},$
$\displaystyle\mathcal{F}_{\max}=\frac{8}{9}\approx 0.89.$ (21)

We see that $\mathcal{F}_{\max}\neq 1$, which is expected since it is already
known that W states cannot be used for perfect controlled reconstruction 44 ,
but $\vartheta(\rho_{W})>1$ (equivalently, $\mathcal{F}_{\max}>2/3$). In this
case, the subsystem-teleportation fidelity for the dealer-reconstructor
channel is found out to be
$\mathcal{F}^{\prime}=\frac{1}{2}(1+\frac{5}{9})=\dfrac{7}{9}$. Since in this
case $\mathcal{F}^{\prime}>2/3$, we can not claim that the quantum advantage
here is due to genuine tripartite entanglement.

Example 3: Next, we consider another example where we show the existence of a
state within the paradigm of $R\neq O,T\neq O$, for which the reconstruction
fidelity is greater than $2/3$, whereas the teleportation fidelity of the
dealer-reconstructor subsystem is $\leq 2/3$. This is a clear example of a
state (other than the well-known GHZ state) for which quantum advantage is
because of tripartite controlled reconstruction. In this context, let us
consider a generalized W Class of states. These states can be expressed as 45
; 46 :

$\ket{\psi_{W}}=\lambda_{0}\ket{000}+\lambda_{1}\ket{100}+\lambda_{2}\ket{101}+\lambda_{3}\ket{110}$
(22)

where $\lambda_{i}\in\mathbb{R}$, $\lambda_{i}\geq 0$ and
$\sum_{i}\lambda_{i}^{2}=1$. For
$\rho_{\tilde{W}}=\ket{\psi_{W}}\bra{\psi_{W}}$, the matrices of interest are:

$R_{\tilde{W}}=2\left(\begin{array}[]{ccc}\lambda_{0}\lambda_{2}&0&\lambda_{0}\lambda_{1}\\\
0&-\lambda_{0}\lambda_{2}&0\\\
-\lambda_{1}\lambda_{2}&0&\frac{1}{2}-\lambda_{1}^{2}-\lambda_{3}^{2}\end{array}\right),$
$T_{\tilde{W}}=2\left(\begin{array}[]{ccc}0&0&\lambda_{0}\lambda_{3}\\\
0&0&0\\\ -\lambda_{2}\lambda_{3}&0&-\lambda_{1}\lambda 3\end{array}\right).$

In Fig. 1, we plot the teleportation fidelity of the dealer-reconstructor
subsystem (given by eq (19)) against the reconstruction fidelity (given by eq
(17)) of these states, denoted as $F^{\prime}$ and $F$ respectively in the
figure. The figure represents the entire set of W states (irrespective of
$R,T$), with the orange region depicting states with $F^{\prime}\leq 2/3$,
which is of interest, and correspondingly, the blue region has states with
$F^{\prime}>2/3$. This tells us that states in the orange region are depicting
quantum advantage not because of the teleportation channel, while it is still
inconclusive for the blue region. Since these are all pure states, $F\geq
2/3$. Not all of these states satisfy $R\neq O,T\neq O$, but one state that
falls in this paradigm has been marked in the figure as an example of quantum
advantage, as it lies in the orange region. This state is described by
parameters $\lambda_{0}=\lambda_{1}=0.7,\lambda_{2}\approx
0.09,\lambda_{3}\approx 0.11$. The standard W state has also been marked,
which lies in the blue region, as expected from the discussion in Example 2.

Figure 1: Plot comparing the reconstruction fidelity and subsystem-
teleportation fidelity for $4\times 10^{6}$ pure state resources uniformly
sampled from the entire W class.

Case 2: R = O, T $\neq$ O In this case, since the correlation matrix $R=O$, it
can be said with certainty that there is no direct correlation between the
dealer and the reconstructor. This means there will be no teleportation
capacity of the channel between them, and hence, it will not affect the total
reconstruction fidelity. In other words, if there is a quantum advantage in
this region, we can surely claim that this is because of tripartite’s
controlled reconstruction capacity. However, the converse is not true. There
can be states with $R\neq O$ for which there is no correlation between the
dealer and the reconstructor. In other words, there can be quantum states
whose quantum advantage arises solely from the $T$ component of correlation
even when $R\neq O$.

Theorem 1 : If R = O, T $\neq$ O, the observed quantum advantage is only
because of tripartite state reconstruction.

Proof: Putting $R=O$ in the expression for subsystem-teleportation fidelity
(from eq (19)), gives $\mathcal{F}^{\prime}=\dfrac{1}{2}$. Hence, no quantum
information flow can flow from dealer to reconstructor through teleportation
alone. Then, the contribution to the reconstruction fidelity has to come from
the tripartite channel. $\blacksquare$

Example: Consider the following states:

$\ket{\gamma_{\pm}}=\frac{\ket{000}\pm\ket{100}\pm\ket{110}+\ket{111}}{2},$

Both these states fall into Case 1 but if we take their equal mixture
$\rho_{\gamma}=\frac{1}{2}(\ket{\gamma_{-}}\bra{\gamma_{-}}+\ket{\gamma_{+}}\bra{\gamma_{+}}),$
$R$ is found to be $O$. Moreover, we get $\mathcal{F}=3/4>2/3$, giving us a
quantum advantage. Since $R=O$ and teleportation cannot contribute to the
final fidelity, we infer that this advantage arises purely from the tripartite
state reconstruction.

Case 3: R $\neq$ O, T = O If in equation (17), $R\neq O$, but $T=O$, then it
can be said that there is no tripartite correlation and hence there is no
involvement of Bob. So, in principle, there is no question of controlled
reconstruction in this case. In such a case, if the reconstruction fidelity is
greater than $2/3$, that is purely because of the teleportation capacity of
the subsystem. However, the converse is not true as there can be tripartite
states with $T\neq O$ but not having genuine quantum correlation.

Theorem 2 : If $R\neq O$ and $T=O$, then the quantum advantage in the
reconstruction is entirely because of the teleportation capacity of the
subsystem between the dealer and the reconstructor.

Proof : Putting $T=O$ in eq (17), we get the expression

$\vartheta(\rho_{ABC})=\|R\|_{1}.$ (23)

The expression in eq (19) now matches with the expression for reconstruction
fidelity given by

$\mathcal{F}_{\max}=\dfrac{1}{2}(1+\dfrac{1}{3}\vartheta(\rho_{ABC}))=\dfrac{1}{2}(1+\dfrac{1}{3}\|R\|_{1}).$

Hence, it can be concluded that quantum advantage in this case is solely
because of the teleportation channel between the dealer and the reconstructor.
$\blacksquare$

Example: Consider the following states:

$\ket{\delta_{\pm}}=\frac{\ket{000}\pm\ket{100}\pm\ket{110}+\ket{111}}{2},$

Both of these have $R\neq O\neq T$, but when their equal mixture is
considered, then for the mixed state $\rho_{\delta}$, we find $T=O$, where
$\rho_{\delta}=\frac{1}{2}(\ket{\delta_{-}}\bra{\delta_{-}}+\ket{\delta_{+}}\bra{\delta_{+}}),$
In this case as well, we get $\mathcal{F}=3/4>2/3$. However, since $T$ is $O$,
we argue that this fidelity arises solely from the teleportation capacity of
the channel between the sender and the reconstructor and, hence, the quantum
advantage.

Case 4: R = O, T = O Here, $\vartheta(\rho)=0$, giving us
$\mathcal{F}_{\max}=1/2$ which is no better than a random guess. In this case,
there cannot be any quantum advantage because there is no flow of information
from the dealer to the reconstructor.

##  IV Quantum Secret Sharing

For any secret-sharing protocol to be successful, neither of the shareholders
should have useful information about the secret on their own. In other words,
Charlie should not be able to reconstruct the secret without Bob’s
involvement, and vice-versa. In this context, we use the term useful
information as the amount of extra information that can be obtained over the
classical limit of the respective channels. One quantification can be the
teleportation capability of the bipartite channels between the dealer and the
shareholders. If this is more than $2/3$, then there is an information gain
through the bipartite channel compared to what can be achieved classically.
This enforces additional conditions on the resource to ensure the protocol is
secure against dishonest parties. The maximum expected fidelity that Bob can
obtain on his own accord is equivalent to the teleportation capacity of
subsystem $\rho_{AB}$ of the resource since Charlie is not involved. The state
$\rho_{AB}$ is given by,

$\displaystyle\rho_{AB}=\mathrm{Tr}_{C}(\rho_{ABC})=$
$\displaystyle\frac{1}{4}[I^{\otimes 2}+\sum_{i}a_{i}.\sigma_{i}\otimes
I+\sum_{j}I\otimes
b_{j}.\sigma_{j}+\sum_{ij}q_{ij}\sigma_{i}\otimes\sigma_{j}].$

The teleportation capacity of this bipartite resource, i.e. the maximum
expected share of Bob without involving Charlie, is:

$\mathcal{F}^{\prime}_{AB}=\frac{1}{2}\big{(}1+\frac{1}{3}\mathrm{Tr}\sqrt{Q^{\dagger}Q}\big{)},$
(24)

We have already seen the expression of $\mathcal{F}^{\prime}_{AC}$ for
Charlie’s case in eq (19). Since we know that a state is useful for quantum
teleportation when $\mathcal{F}^{\prime}>2/3$ 2 , we want:

$\displaystyle\mathcal{F}^{\prime}_{AB}\leq\frac{2}{3}$
$\displaystyle\mathcal{F}^{\prime}_{AC}\leq\frac{2}{3}$
$\displaystyle\implies\mathrm{Tr}\sqrt{Q^{\dagger}Q}\leq 1$
$\displaystyle\mathrm{Tr}\sqrt{R^{\dagger}R}\leq 1$ (25)

These constraints, along with $\vartheta(\rho)>1$ (or, equivalently,
$\mathcal{F}>2/3$), together form the three conditions for successful QSS (see
Fig. 2). This ensures that the secret is faithfully reconstructed.
Additionally it accounts for dishonest recipients (either Bob or Charlie) as
this ensures that the information they can retrieve from the quantum resource,
without faithfully cooperating does not exceed the information that can be
obtained classically even in absence of quantum channel. Thus, we are able to
provide a way to characterize the states useful for secret sharing more
precisely.

ABC$\leqslant\frac{2}{3}$$\leqslant\frac{2}{3}$$>\frac{2}{3}$

Figure 2: Conditions for secret sharing (A: dealer, B: assistant, C:
reconstructor)

Example: Consider the following states:

$\ket{\beta_{\pm}}=\frac{\ket{000}+\ket{100}+\ket{101}\pm\ket{110}}{2},$

The equal mixture of $\ket{\beta_{+}}\bra{\beta_{+}}$ and
$\ket{\beta_{-}}\bra{\beta_{-}}$ gives
$\mathrm{Tr}\sqrt{Q^{\dagger}Q}=1/2=\mathrm{Tr}\sqrt{R^{\dagger}R}$, and still
gives $\mathcal{F}>2/3$, fulfilling all three conditions. This example
demonstrates successful secret sharing, in addition to the well-known GHZ
state. This is contrary to the belief that only the GHZ state is useful for
secret sharing.

##  V Discussion

In this article, we have given an expression for the maximum expected CSR
fidelity for general three-qubit states in terms of their Bloch parameters.
This fidelity has contributions from both the tripartite correlation tensor
and the correlation matrix between the dealer-reconstructor subsystem. We
report a quantum advantage in the reconstruction of the state over its
classical limit ($\frac{2}{3}$). Different cases, in terms of the involved
correlation matrices, are then investigated with respect to the CSR fidelity.
We provide examples where the advantage is only because of the teleportation
capacity of the subsystem and those where the advantage is mainly because of
the state reconstruction capacity of the tripartite resource. We also discuss
cases which are ambiguous when it comes to the cause of the quantum advantage.
Our results introduce interoperability between quantum teleportation and CSR.
This also opens up new avenues of research related to the interoperability
that can exist in different quantum information processing tasks. We extend
our analysis by outlining additional conditions that must be satisfied to
ensure the protocol is secure by preventing any dishonest participant from
having a quantum advantage. These conditions distinguish CSR from QSS in a
true sense. Further work can be done in this context, analysing the
interference of eavesdroppers, and how they can be detected. From a resource
point of view, we are able to characterize the states that provide quantum
advantage in CSR. In addition, we are able to identify the states that can be
used as resources for QSS.

##  VI Acknowledgements

The authors thank Dr Shantanav Chakrabarty (Centre for Quantum Science &
Technology, IIIT Hyderabad) and Dr Nirman Ganguly (Department of Mathematics,
BITS-Pilani Hyderabad) for their valuable inputs.

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##  Appendix A The Case of Dishonest Party in Classical Scenario

Here, we wish to see what happens in the case if one of the receiving parties
(either Bob or Charlie) is dishonest. Since information from both the parties
is required to reconstruct the qubit, the honest one can keep a check on the
dishonest one. But we also have to take a look at what might happen in case
the dishonest person wants to guess the shared qubit based on what information
they have.

Before we move on, let us have a look at the distributions of the values
$s_{1},s_{2}$ can take. Let $s_{i}\leftarrow\\{p_{0},p_{1}\\}$ denote that
$s_{i}$ takes the value 0 with probability $p_{0}$ and 1 with probability
$p_{1}$. Now if we have $s_{1}\leftarrow\\{p,1-p\\}$, then either
$s_{2}\leftarrow\\{p,1-p\\}$ (if $s=0$) or $s_{2}\leftarrow\\{1-p,p\\}$ (if
$s=1$). In either case, both the distributions are dependant on each other,
and once one of the bits is sampled, the other bit is known with certainty.

Since the distributions are symmetric, without the loss of generality, we can
take Bob* as the dishonest party. He knows $s_{1}$ but has no knowledge of
$s_{2}$ without communicating with Charlie. So he has to guess the bit $s$,
and let’s denote his guess of the bit as $s^{\prime}$ which is either same as
his bit ($s_{1}$) or the negation of it.

Case - 1 $s^{\prime}=s_{1}$, or equivalently, Bob* guesses $s_{2}=0$

Let $s_{2}\leftarrow\\{p,1-p\\}$. The fidelity between the original qubit and
the one reconstructed by Bob can be expressed as: ($F^{\prime}$ is used to
denote fidelity with guess).

$\displaystyle F^{\prime}(q,s^{\prime})$
$\displaystyle=Pr[s=s^{\prime}]F(q,s)+Pr[s\neq s^{\prime}]F(q,\neg s)$
$\displaystyle=Pr[s_{2}=0]\big{(}Pr[s=0]\cdot\|\braket{q}{\uparrow}\|^{2}+Pr[s=1]\cdot\|\braket{q}{\downarrow}\|^{2}\big{)}$
$\displaystyle\hskip 40.0pt+Pr[s_{2}=1]\big{(}Pr[\neg
s=0]\cdot\|\braket{q}{\uparrow}\|^{2}+Pr[\neg
s=1]\cdot\|\braket{q}{\downarrow}\|^{2}\big{)}$
$\displaystyle=(p)\big{(}\cos^{2}{\frac{\theta}{2}}\cdot\cos^{2}{\frac{\theta}{2}}+\sin^{2}{\frac{\theta}{2}}\cdot\sin^{2}{\frac{\theta}{2}}\big{)}+(1-p)\big{(}\sin^{2}{\frac{\theta}{2}}\cdot\cos^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}}\cdot\sin^{2}{\frac{\theta}{2}}\big{)}$
$\displaystyle=p\big{(}1-\frac{1}{2}\sin^{2}\theta\big{)}+\frac{1}{2}(1-p)\sin^{2}\theta=p\cdot\cos^{2}\theta+\frac{1}{2}\sin^{2}\theta$

Hence the expected "guess" fidelity in this case:

$\displaystyle\mathcal{F^{\prime}}=\langle F^{\prime}(q,s^{\prime})\rangle$
$\displaystyle=\frac{1}{4\pi}\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi}\sin\theta\,d\theta\,d\phi(p\cdot\cos^{2}\theta+\frac{1}{2}\sin^{2}\theta)$
$\displaystyle=\frac{1}{4\pi}\cdot
2\pi\Big{(}\frac{4}{3}-\frac{2}{3}p\Big{)}=\boxed{\frac{2-p}{3}}$

We want Bob*’s guess to be no better than a random one. Or equivalently, we
want

$\mathcal{F^{\prime}}\leq\frac{1}{2}\implies\frac{2-p}{3}\leq\frac{1}{2}\implies\boxed{p\geq\frac{1}{2}}$
(26)

Case - 2 $s^{\prime}=\neg s_{1}$, or equivalently, Bob* guesses $s_{2}=1$

In this case,

$\displaystyle F^{\prime}(q,s^{\prime})$
$\displaystyle=Pr[s_{2}=1]F(q,s)+Pr[s_{2}=0]F(q,\neg s)$
$\displaystyle=(1-p)\cos^{2}\theta+\frac{1}{2}\sin^{2}\theta$ and
$\displaystyle\mathcal{F^{\prime}}$ $\displaystyle=\langle
F^{\prime}(q,s^{\prime})\rangle=\boxed{\frac{1+p}{3}}$

Enforcing the same condition as last time,

$\mathcal{F^{\prime}}\geq\frac{1}{2}\implies\frac{1+p}{3}\leq\frac{1}{2}\implies\boxed{p\leq\frac{1}{2}}$
(27)

Both the conditions will be satisfied if we set $p=\frac{1}{2}$. Hence, Alice
can prevent either party from cheating if the uniform distribution is used to
sample one of the bits.