This paper introduces two information-theoretically secure protocols that achieve quantum secure direct communication between Alice and Bob in the first case, and among Alice, Bod and Charlie in the second case. Both protocols use the same novel method to embed the secret information in the entangled composite system of the players. The way of encoding the information is the main novelty of this paper and the distinguishing feature compared to previous works in the field. The advantage of this method is that it is easily extensible and can be generalized to a setting involving three, or even more, players, as demonstrated with the second protocol. This trait can be beneficial when two spatially separated players posses only part the secret information that must be combined and transmitted to Alice in order for her to reveal the complete secret. Using the three player protocol, this task can be achieved in one go, without the need to apply a typical QSDC protocol twice, where Alice first receives Bob's information and afterwards Charlie's information. Another characteristic of both protocols their simplicity and uniformity. The two player protocol relies on EPR pairs, and the three player protocol on $\ket{ GHZ_{ 3 } }$ triples, which can be easily prepared with our current technology. In the same vein, the local quantum circuits are similar or identical, and are easily constructible as they employ only Hadamard and CNOT gates.

Keywords:: Quantum Secure Direct Communication, quantum entanglement, Bell states, EPR pairs, GHZ states, quantum games.

§ INTRODUCTION

Nowadays, it is hardly necessary to advocate the importance of privacy and security for every aspect of our life as individuals. Indeed, privacy is a constitutional right that must be respected and protected under all circumstances. This, in turn, has advanced the design and implementation of technical tools that ensure the security of out digital data. Devising bulletproof algorithms and protocols that protect our privacy from unauthorized access is a major trend in current research. This, however, may not be as easy as it sounds. The reason is that we have just entered a new scientific era, the quantum era, which brings the promise of unprecedented computational power. This, unharnessed so far, power offers new algorithms that can, potentially compromise the security offered by established classical methods. Two iconic examples that help drive this point home, are Shor’s [1] and Grover's [2] algorithms. Shor’s algorithm can factorize large numbers in polynomial time and its practical implementation is bound to threaten public key cryptosystems. Grover’s algorithm speeds up unordered search and may also be used to attack symmetric key cryptosystems like AES.

Up to this day, there exist no quantum computers powerful enough to threaten the classical status quo. However, this will probably change sooner than initially anticipated, if one judges by the impressive progress that has been achieved lately. A case in mind is IBM's $127$ qubit Eagle processor [3] and the more recent $433$ qubit Osprey [4] processor. It seems prudent, if not imperative, to find ways to seriously upgrade our algorithms and protocols, before they become a liability to our security infrastructure. The enormous effort to come up with a robust solution, has led to the creation of two new scientific fields, the field of post-quantum or quantum-resistant cryptography and the field of quantum cryptography. The former, is actually an incremental evolution of the current state of affairs [5, 6, 7, 8], reducing security issues to carefully chosen computationally hard problems, an approach that has been vindicated so far. The latter, quantum cryptography, relies on the laws of nature, such as entanglement, monogamy of entanglement, the no-cloning theorem, and nonlocality to ensure ironclad security. Quantum cryptography advocates exploiting the unique and powerful quantum phenomena to design new secure protocols for a plethora of critical applications, such as key distribution [9, 10, 11, 12, 13, 14, 15], secret sharing [16, 17, 18], quantum teleportation [19], cloud storage [20, 21] and blockchain [22, 23, 24].

In the seminal paper [25], the authors proposed a protocol for Quantum Secure Direct Communication (QSDC for short). The characteristic trait of QSDC, which distinguishes it from key distribution that establishes a common random key between two parties, is that QSDC transmits information directly and without using an existing key. Furthermore, the classical channel is employed only for detection purposes and not for transmitting information necessary to decipher the secret message. The intended recipient must be able to uncover the secret information after receiving the quantum states via the quantum channel. Finally, any eavesdropper must be detected, without being allowed to compromise the secret. Almost immediately, in 2003 the researchers in [26] introduced the influential two-step QSDC protocol. Later, [27] presented a QSDC protocol using single photons, [28] gave a protocol based on superdense coding, and [29] proposed the first QSDC protocol with multipartite entanglement. Since then, progress in this are has been non-stop. For a though and comprehensive review of the current state of the field, we refer the reader to the recent [30].

In this work, we initially introduce a new protocol, called $2$PSQDC, for quantum secure direct communication between two entities. Subsequently, the $2$PSQDC is generalized in a intuitive and straightforward manner, so as to provide for quantum secure direct communication among three entities. The resulting protocol, which is called $3$PSQDC, can be seamlessly generalized to an arbitrary number of entities. We present our protocols as games, involving the usual cast of Alice, Bob and Charlie. Hopefully, the pedagogical nature of games will make the presentation of the technical concepts easier to follow. Quantum games, from their inception in 1999 [31, 32], have known great acceptance since quantum strategies are sometimes superior to classical ones [33, 34, 35]. The famous prisoners' dilemma game provides such the most prominent example [32, 36], which also applies to other abstract quantum games [37]. The quantization of many classical systems can even apply to political structures, as was shown in [38]. While on the subject of games on unconventional environments, let us mention that games in biological systems have attracted significant attention [39, 40, 41]. It is interesting to observe that biosystems may give rise to biostrategies superior compared to the classical ones, even in the Prisoners' Dilemma iconic game [42, 43, 44, 45, 46].

Contribution. This paper presents two protocols that achieve quantum secure direct communication between Alice and Bob in the first case, and among Alice, Bod and Charlie in the second case. Both protocols, which are proven to be information-theoretically secure, use the same idea, i.e., embedding the secret information via an oracle that uses the inner product modulo $2$ operation. This way of encoding the information is the main novelty of this paper that distinguishes it from the many previous works in the field. The advantage of this method is that it is seamlessly extensible and can be generalized to a setting involving three, or even more, players, as demonstrated with the $3$PSQDC protocol. This last case, is not only useful, but often necessary, when two spatially separated players posses only part the secret information that must be combined and transmitted to Alice in order for her to reveal the complete secret. Using the $3$PSQDC protocol, this task can be achieved in one go, without the need to apply a typical QSDC protocol twice, where Alice first receives Bob's information and afterwards Charlie's information. Assuming sufficient resources, the $3$PSQDC protocol can be extended in the obvious manner to allow for $n - 1$ players to simultaneously send information to Alice. It is also worth mentioning that both protocols are practically accessible, since they relay on EPR pairs, in the two player case, and $\ket{ GHZ_{ 3 } }$ triples, in the three player case, that can be generated with our current technology. Moreover, the local quantum circuits are characterized by uniformity and symmetry, being similar or identical, and are easily constructible as they employ only Hadamard and CNOT gates.

§.§ Organization

The paper is organized as follows. Section <ref> contains an introduction to the subject along with bibliographic pointers to related works. Section <ref> explains the underlying theory required for the understanding of the protocols. Section <ref> provides a detailed presentation of the $2$PSQDC protocol that tackles information transmission from Alice to Bob. Section <ref> contains a formal presentation of the $3$PSQDC protocol that, in addition to Bob and Alice, also involves Charlie. Finally, Section <ref> gives a brief summary of this work, and outlines directions for future research.

§ PRELIMINARIES

§ BACKGROUND & NOTATION

§.§ $\ket{ \Phi^{ + } }$ EPR pairs

There are certain properties of quantum physics that are quite strange, in the sense that they have no analogue in classical physics and even contradict our everyday intuition. Undoubtedly, entanglement falls into this category. This strange phenomenon is also a source of great potential, as it seems to be one of the keys for achieving things that are difficult or impossible in the classical world. Technically, entanglement appears in composite quantum systems, consisting of at least two subsystems, which can be, and usually are, spatially separated. In mathematical terms, a composite system is entangled, if its state must be described as a linear combination of two or more product states of its subsystems. Bell states, also referred to as EPR pairs, provide us with most well-known example of maximal entanglement for a two-qubit system. There are four Bell states expressed as shown below (see [47]). We use the subscripts $A$ and $B$ to make explicitly clear that the first qubit belongs to Alice and the second to Bob.

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\begin{align} \label{eq:Bell State Phi +}
\ket{ \Phi^{ + } }
%			=
%			\ket{ \beta_{ 00 } }
\frac { \ket{ 0 }_{ A } \ket{ 0 }_{ B } + \ket{ 1 }_{ A } \ket{ 1 }_{ B } } { \sqrt{ 2 } }
\end{align}

\begin{align} \label{eq:Bell State Phi -}
\ket{ \Phi^{ - } }
%			=
%			\ket{ \beta_{ 10 } }
\frac { \ket{ 0 }_{ A } \ket{ 0 }_{ B } - \ket{ 1 }_{ A } \ket{ 1 }_{ B } } { \sqrt{ 2 } }
\end{align}

\begin{align} \label{eq:Bell State Psi +}
\ket{ \Psi^{ + } }
%			=
%			\ket{ \beta_{ 01 } }
\frac { \ket{ 0 }_{ A } \ket{ 1 }_{ B } + \ket{ 1 }_{ A } \ket{ 0 }_{ B } } { \sqrt{ 2 } }
\end{align}

\begin{align} \label{eq:Bell State Psi -}
\ket{ \Psi^{ - } }
%			=
%			\ket{ \beta_{ 11 } }
\frac { \ket{ 0 }_{ A } \ket{ 1 }_{ B } - \ket{ 1 }_{ A } \ket{ 0 }_{ B } } { \sqrt{ 2 } }
\end{align}

One of the critical advantages of quantum entanglement is that when one qubit of the pair gets measured, then the other will immediately collapse to its corresponding basis state, irrespective of the distance between them. It is precisely this celebrated trait of quantum entanglement that is utilized in quantum cryptographic protocols, e.g., for key distribution, secret sharing, etc. Obviously, to achieve the intended result, a sequence of such EPR pairs is required. In the $2$PSQDC protocol, we shall be using $\ket{ \Phi^{ + } }$ pairs. The mathematical description of $m$ pairs in the $\ket{ \Phi^{ + } }$ state is given next.

\begin{align} \label{eq:N Bell States Phi +}
\ket{ \Phi^{ + } }^{ \otimes m }
\frac { 1 } { \sqrt{ 2^m } }
\sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } }
\ket{ \mathbf{ x } }_{ A }
\ket{ \mathbf{ x } }_{ B }
\ ,
\end{align}

where $\mathbb{ B } = \{ 0, 1 \}$.

§.§ $\ket{ GHZ_{ 3 } }$ triplets

Obviously, entanglement is a phenomenon that appears in general multipartite system. For a composite system consisting of three or more qubits, one of the most well-known and studied types of maximal entanglement is the so-called GHZ state. In the $3$PSQDC protocol, we shall employ triplets of qubits in the $\ket{ GHZ_{ 3 } }$ state, which is expressed mathematically by the equation (<ref>). As in the case of the $\ket{ \Phi^{ + } }$ pairs, subscripts $A$, $B$ and $C$ are used to make clear that the first qubit belongs to Alice, the second to Bob and the third to Charlie.

\begin{align} \label{eq:GHZ 3 State}
\ket{ GHZ_{ 3 } }
\frac
\ket{ 0 }_{ A }
\ket{ 0 }_{ B }
\ket{ 0 }_{ C }
\ket{ 1 }_{ A }
\ket{ 1 }_{ B }
\ket{ 1 }_{ C }
{ \sqrt{ 2 } }
\ .
\end{align}

A single $\ket{ GHZ_{ 3 } }$ triplet will not suffice for the execution of the $3$PSQDC protocol; $m$ such triplets will be required. A system comprised of $m$ $\ket{ GHZ_{ 3 } }$ triplets is described by the next formula (for its detailed derivation we refer to [17] and [18]).

\begin{align} \label{eq:m GHZ 3 State}
\ket{ GHZ_{ 3 } }^{ \otimes m }
\frac { 1 } { \sqrt{ 2^{ m } } }
\sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } }
\ket{ \mathbf{ x } }_{ A }
\ket{ \mathbf{ x } }_{ B }
\ket{ \mathbf{ x } }_{ C }
\ .
\end{align}

In formula (<ref>), the notation $\mathbf{ x } \in \mathbb{ B }^{ m }$ means that the bit vector $\mathbf{ x }$ ranges through all the $2^{ m }$ bit vector representations of the basis kets. In accordance to what we mentioned before, $\ket{ \mathbf{ x } }_{ A }$, $\ket{ \mathbf{ x } }_{ B }$ and $\ket{ \mathbf{ x } }_{ C }$ correspond to the basis states of Alice, Bob and Charlie's quantum registers, respectively.

Existing quantum computers based on the circuit model can trivially produce the four Bell states. Figure <ref> depicts a quantum circuit that generates $\ket{ \Phi^{ + } }$ pairs. This particular circuit is designed using the IBM Quantum Composer [48]. Similarly, it is easy in principle to construct quantum circuits that produce general $\ket{ GHZ_{ n } }$ states. As a matter of fact there is a methodology for constructing efficient general GHZ circuits [49], in the sense that it require $\lg n$ steps to produce given state. Although, for large $n$ there are practical difficulties in preparing and maintaining a $\ket{ GHZ_{ n } }$ state, for $\ket{ GHZ_{ 3 } }$ that are states necessary for the implementation of the $3$PSQDC protocol, things are quite manageable. Figure <ref> shows a quantum circuit, also designed using the IBM Quantum Composer [48], that prepares $\ket{ GHZ_{ 3 } }$ triples. Figures <ref> and <ref> give the state vector descriptions of the $\ket{ \Phi^{ + } }$ and $\ket{ GHZ_{ 3 } }$ states respectively.

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The above quantum circuit can be used in Qiskit to produce a $\ket{ \Phi^{ + } } = \frac { \ket{ 0 }_{ A } \ket{ 0 }_{ B } + \ket{ 1 }_{ A } \ket{ 1 }_{ B } } { \sqrt{ 2 } }$ pair.

The above quantum circuit can entangle 3 qubits in the $\ket{ GHZ_3 } = \frac { \ket{ 0 }_{ A } \ket{ 0 }_{ B } \ket{ 0 }_{ C } + \ket{ 1 }_{ A } \ket{ 1 }_{ B } \ket{ 1 }_{ C } } { \sqrt{ 2 } }$ state.

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The state vector description of a $\ket{ \Phi^{ + } }$ pair.

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The state vector description of a $\ket{ GHZ_{ 3 } }$ triplet.

Apart from $\ket{ \Phi^{ + } }$ pairs and $\ket{ GHZ_{3} }$ triplets, we use the well-known states $\ket{+}$ and $\ket{-}$. For completeness, we provide their definitions.

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\begin{align} \label{eq:Ket +}
\ket{ + } = H \ket{ 0 } = \frac { \ket{ 0 } + \ket{ 1 } } { \sqrt{ 2 } }
\end{align}

\begin{align} \label{eq:Ket -}
\ket{ - } = H \ket{ 1 } = \frac { \ket{ 0 } - \ket{ 1 } } { \sqrt{ 2 } }
\end{align}

Another useful formula proved in textbooks such as [47] and [50], which will be applied in the explanation of the protocol, is the following

\begin{align} \label{eq:Hadamard m-Fold Ket x}
H^{ \otimes m } \ket{ \mathbf{ x } }
\frac { 1 } { \sqrt{ 2^{ n } } }
\sum_{ \mathbf{ z } \in \mathbb{ B }^{ m } }
( - 1 )^{ \mathbf{ z \cdot x } } \ket{ \mathbf{ z } }
\ .
\end{align}

Finally, we mention that quantum measurements are performed as a rule with respect to the computational basis $\{ \ket{ 0 }, \ket{ 1 } \}$, and, occasionally, with respect to the Hadamard basis $\{ \ket{ + }, \ket{ - } \}$, in which case it is explicitly clarified.

§.§ Inner product modulo $2$

In this work, we follow the typical convention of writing bit vectors $\mathbf{ x } \in \mathbb{ B }^{ m }$ in boldface. That is, a bit vector $\mathbf{ x }$ of length $m$ is a sequence of $m$ bits $\mathbf{ x } = x_{ m - 1 } \dots x_{ 0 }$. The zero bit vector is designated by $\mathbf{ 0 } = 0 \dots 0$.

Given two bit vectors $\mathbf{ x }, \mathbf{ y } \in \mathbb{ B }^{ m }$, where $\mathbf{ x } = x_{ m - 1 } \dots x_{ 0 }$ and $\mathbf{ y } = y_{ m - 1 } \dots y_{ 0 }$, we define the inner product modulo $2$, denoted by $\mathbf{ x \cdot y }$, as

\begin{align} \label{eq:Inner Product Modulo $2$}
\mathbf{ x \cdot y }
x_{ n - 1 } y_{ n - 1 }
\oplus \dots \oplus
x_{ 0 } y_{ 0 }
\ .
\end{align}

In the above formula, $\oplus$ stands for addition modulo $2$. The operation inner product modulo $2$ exhibits a very useful property. If $\mathbf{ c } \in \mathbb{ B }^{ m }$ is different from $\mathbf{ 0 }$, then for half of the elements $\mathbf{ x } \in \mathbb{ B }^{ m }$, $\mathbf{c} \cdot \mathbf{ x }$ is $0$ and for the other half, $\mathbf{ c } \cdot \mathbf{ x }$ is $1$. Obviously, if $\mathbf{ c } = \mathbf{ 0 }$, then for all $\mathbf{ x } \in  \mathbb{ B }^{ m }$, $\mathbf{ c } \cdot \mathbf{ x } = 0$ (a more detailed analysis can be found in [18]). For easy reference, we state this as property (IP).

\begin{align} \label{eq:Inner Product Modulo $2$ Property}
\text{ If } \mathbf{ c } \neq \mathbf{ 0 },
\text{ there are }
&2^{ m - 1 } \text{ bit vectors } \mathbf{ x } \in \mathbb{ B }^{ m },
\text{ such that } \mathbf{ c } \cdot \mathbf{ x } = 0, \text{ and }
\nonumber \\
&2^{ m - 1 } \text{ bit vectors } \mathbf{ x } \in \mathbb{ B }^{ m },
\text{ such that } \mathbf{ c } \cdot \mathbf{ x } = 1 \ .
\tag{IP}
\end{align}

It will also be expedient to extend the operation of addition modulo $2$ to bitwise addition modulo $2$ between bit vectors. Given two bit vectors $\mathbf{ x }, \mathbf{ y } \in \mathbb{ B }^{ m }$, where $\mathbf{ x } = x_{ m - 1 } \dots x_{ 0 }$ and $\mathbf{ y } = y_{ m - 1 } \dots y_{ 0 }$, we define their bitwise addition modulo $2$, denoted by $\mathbf{ x \oplus y }$, as

\begin{align} \label{eq:Bitwise Addition Modulo $2$}
\mathbf{ x \oplus y }
( x_{ n - 1 } \oplus y_{ n - 1 } )
\dots
( x_{ 0 } y_{ 0 } )
\ .
\end{align}

We use the same symbol $\oplus$ for the operation of addition modulo $2$ between bits, and for the operation of bitwise addition modulo $2$ between bit vectors, since the context will help avoid any confusion.

§ THE $2$PSQDC PROTOCOL

The $2$ player scheme for quantum secure direct communication, $2$PSQDC protocol from now on, is designed to allow one party to communicate with a spatially separated second party securely and directly using only the quantum channel. To emphasize that is about the secure direct communication of two players, the protocol is called $2$PSQDC. To enhance its game-like presentation, we call the two parties Alice and Bob. Alice is the one having the initiative and intending to send some information to Bob. The setting is completed by the notorious Eve, a cunning adversary that attempts to steal any information possible. The major advantage that quantum protocols exhibit over classical ones is that communication through the quantum channel involves an array of unique features, such as the no-cloning theorem [51], the monogamy of entanglement [52], and nonlocality [53], that can be used to inhibit Eve.

§.§ Entanglement distribution phase

The $2$PSQDC protocol evolves in phases. Initially, during the entanglement distribution phase, Alice, or a third trusted source, prepares $\ket{ \Phi^{ + } }$ pairs. As customary, we assume the existence of a trusted quantum source, which may not necessarily be Alice, that is responsible for this task. In any event, the produced $\ket{ \Phi^{ + } }$ pairs are shared between Alice and Bob, according to the following pattern.

* Alice sends to Bob a sequence of $m$ qubits $b_{ 0 }, \dots, b_{ m - 1 }$, called the transmission sequence, which Bob stores in his register.

* At the same time, Alice stores in her register a corresponding sequence of $m$ qubits $a_{ 0 }, \dots, a_{ m - 1 }$, called the embedding sequence.

* The $k^{ th }$ qubit in the transmission sequence, $0 \leq k \leq m - 1$, is either

* the first qubit of a $\ket{ \Phi^{ + } }$ pair, in which case the $k^{ th }$ qubit in the embedding sequence is the second qubit of the same $\ket{ \Phi^{ + } }$ pair, or

* a nonentangled qubit in a state randomly chosen, with equal probability, from $\{ \ket{ 0 }, \ket{ 1 }, \ket{ + }, \ket{ - } \}$, in which case the $k^{ th }$ qubit in the embedding sequence is also a nonentangled qubit in the same state. A qubit of this type is called a decoy.

* In all, Alice sends to Bob $d$ decoys. It is imperative that Alice insert the decoys randomly within the  transmission sequence. Obviously, Alice keeps track of the position and the state of all her decoys.

The integer $m$ should be sufficiently large to ensure the secure execution of the protocol, and the integer $d$ should be significantly smaller than $m$. In our forthcoming mathematical analysis in Section <ref>, we shall be more precise.

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(Alice) – (Bob) node [ above, midway, text = black ] Spatially Separated;

Alice shares the $\ket{ \Phi^{ + } }$ pairs with Bob. She keeps the first qubit of each pair (drawn in red), and sends the second to Bob through the quantum channel (drawn in green).

The situation before the entanglement distribution is visualized in Figure <ref>. At the end of the distribution phase, both Alice and Bob have in their own quantum registers $m$ qubits each. Their registers are correlated because Alice and Bob's corresponding qubits are entangled in the state $\ket{ \Phi^{ + } }$. The setup at this point is described in Figure <ref>. To complete this phase, Alice and Bob conduct the first consistency check on their registers. If the test is successful, they proceed to the secret embedding phase. If not, then their communication has been compromised by Eve, and, so, they abort the protocol. We defer the detailed explanation of the actions taking place during the first consistency check until Section <ref>.

§.§ Secret embedding phase

During this phase, Alice encodes the secret she intends to transmit to Bob. To achieve this, she acts locally upon her register using the local quantum circuit outlined in Figure <ref>. Although Alice and Bob are spatially separated and they both operate via their local quantum circuits, the entanglement correlating their registers results in one composite system, consisting of Alice and Bob's subsystems.

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[ scale = 1.00 ]

nobit AUX_B_0;
[ name = IBREG ] qubits  $BR$   BR;
nobit AUX_B_1;
nobit AUX_B_2;
nobit AUX_B_3;

[ name = wave, register/minimum height=5mm ] nobit wave;

nobit AUX_A_0;
nobit AUX_A_1;
nobit AUX_A_2;
[ name = IAREG ] qubits  $AR$   AR;
nobit AUX_A_3;
qubit  $AQ$:  $\ket{ - }$  AQ;
nobit AUX_A_4;
nobit AUX_A_5;

[ name = InitialState, WordBlueVeryLight, line width = 0.50 mm, label = Initial State ]
barrier ( - ) ;

[ draw = GreenLighter2, fill = GreenLighter2, radius = 0.7 cm ] box    U$_{ A }$  ( AR - AQ );

[ name = SecretEmbedding, WordBlueVeryLight, line width = 0.50 mm, label = Secret Embedding ]
barrier ( - ) ;

output   BR;

output   AR;

[on background layer]
[ wave, fill = WordDarkTealLighter80, fit = (wave) (current bounding box.east |- wave), inner ysep = 3.00 mm, inner xsep = 0pt ] Spatially Separated;
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(AliceInput.west) – (BobInput.west)
node [ MagentaLighter, midway, rotate = 90, yshift = 5.00 mm ]
$\frac { 1 } { \sqrt{ 2^{ m } } } \sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } } \ket{ \mathbf{ x } }_{ A } \ket{ \mathbf{ x } }_{ B }$ ;
MagentaLighter, line width = 0.50 mm, decorate, decoration =  brace, mirror, amplitude = 5pt , xshift = 0.0 mm, yshift = 0.0 mm
(AliceOutput.west) – (BobOutput.west)
node [ MagentaLighter, midway, rotate = 270, yshift = 5.00 mm ]
$\frac { 1 } { \sqrt{ 2^{ m } } } \sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } } ( - 1 )^{ \mathbf{ s } \cdot \mathbf{ x } } \ket{ \mathbf{ x } }_{ A } \ket{ \mathbf{ x } }_{ B }$ ;
[ below = 3.00 cm ] at (InitialState)  $\ket{ \psi_{ 0 } }$ ;
[ below = 3.00 cm ] at (SecretEmbedding)  $\ket{ \psi_{ 1 } }$ ;

The above figure is an abstract visualization of the quantum circuits employed by Alice and Bob in the secret embedding phase. Although the local quantum circuits are spatially separated, they constitute a composite system, going from state $\ket{\psi_{0}}$ to state $\ket{\psi_{1}}$.

In the above Figure <ref>, $AR$ and $BR$ designate Alice and Bob's entangled registers, respectively, and $AQ$ stands for Alice's qubit, initialized in state $\ket{ - }$. The subscripts $A$ and $B$ are used to distinguish between Alice and Bob's qubits and registers.

The information that Alice aims to send to Bob is represented by the secret bit vector $\mathbf{ s }$, where the bits corresponding to the positions of the decoys are set to $0$.

\begin{align} \label{eq:Secret Bit Vector}
\mathbf{ s }
s_{ m - 1 }
\dots
s_{ 0 }
\ .
\end{align}

Alice embeds the secret into the state of the composite quantum system by using the unitary transform $U_{ A }$ on her quantum register. $U_{ A }$ is based on the function

\begin{align} \label{eq:Alice's Function}
f_{ A } ( \mathbf{ x } )
\mathbf{ s }
\cdot
\mathbf{ x }
\ .
\end{align}

The complete definition of $U_{ A }$ follows the typical rule given below

\begin{align} \label{eq:Alice's Transform}
U_{ A }
\ket{ y }_{ A }
\ket{ \mathbf{ x } }_{ A }
\rightarrow
\ket{ y \oplus f_{ A } ( \mathbf{ x } ) }_{ A }
\ket{ \mathbf{ x } }_{ A }
\ ,
\end{align}

where $\ket{ y }_{ A }$ and $\ket{ \mathbf{ x } }_{ A }$ represent the state of Alice's qubit $AQ$ and register $AR$, respectively. Taking into account equation (<ref>) and the fact that $\ket{ y }_{ A } = \ket{ - }_{ A }$, equation (<ref>) becomes

\begin{align} \label{eq:Explicit Alice's Transform}
U_{ A }
\ket{ - }_{ A }
\ket{ \mathbf{ x } }_{ A }
\rightarrow
( - 1 )^{ \mathbf{ s } \cdot \mathbf{ x } }
\ket{ - }_{ A }
\ket{ \mathbf{ x } }_{ A }
\ .
\end{align}

The quantum circuit of Figure <ref> begins its operation in the initial state $\ket{ \psi_0 }$. By invoking (<ref>), $\ket{ \psi_{ 0 } }$ can be expressed as

\begin{align} \label{eq:$2$PSQDC Part 1 Initial State}
\ket{ \psi_{ 0 } }
\frac { 1 } { \sqrt{ 2^{ m } } }
\sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } }
\ket{ - }_{ A }
\ket{ \mathbf{ x } }_{ A }
\ket{ \mathbf{ x } }_{ B }
\ .
\end{align}

Alice acts on her quantum register with the unitary transform (<ref>) driving the composite system into the next state $\ket{ \psi_{ 1 } }$:

\begin{align} \label{eq:$2$PSQDC Part 1 Secret Embedding State}
\ket{ \psi_{ 1 } }
&\overset { (\ref{eq:Explicit Alice's Transform}) } { = }
\frac { 1 } { \sqrt{ 2^{ m } } }
\sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } }
( - 1 )^{ \mathbf{ s } \cdot \mathbf{ x } }
\ket{ - }_{ A }
\ket{ \mathbf{ x } }_{ A }
\ket{ \mathbf{ x } }_{ B }
\ .
\end{align}

At the end of this phase, Alice sends to Bob the $m$ qubits in her register through the quantum channel. Bob receives and places these qubits in a second register denoted by $BR_{ A }$.

§.§ Secret decryption phase

During this phase, Bob will complete the protocol and decipher Alice's secret locally, using the quantum circuit outlined in Figure <ref>. In addition to the quantum register $BR$, utilized in the circuit of Figure <ref>, Bob now also uses register $BR_{ A }$. These two registers are entangled and, by acting on both of them, Bob will uncover the secret bit vector $\mathbf{ s }$. To avoid any confusion, henceforth we shall use subscript $1$ to designate the state and contents of $BR_{ A }$ and subscript $0$ to designate the state and contents of register $BR$.

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[ scale = 1.00 ]

nobit AUX_B_04;
[ name = IBREG ] qubits   $BR$   BR;
nobit AUX_B_02;
nobit AUX_B_01;
[ name = BOB ] nobit AUX_B_00;
nobit AUX_B_11;
nobit AUX_B_12;
[ name = IBREGA ] qubits  $BR_{ A }$   BRA;
nobit AUX_B_14;
nobit AUX_B_15;

[ name = Input, WordBlueVeryLight, line width = 0.50 mm, label = Input ]
barrier ( - ) ;

[ draw = RedPurple!50, fill = RedPurple!50, radius = 0.5 cm ] box H$^{\otimes m}$ BR;
[ draw = RedPurple!50, fill = RedPurple!50, radius = 0.5 cm ] box H$^{\otimes m}$ BRA;

[ name = Expansion, WordBlueVeryLight, line width = 0.50 mm, label = Expansion ]
barrier ( - ) ;

cnot BR | BRA;

[ name = Decryption, WordBlueVeryLight, line width = 0.50 mm, label = Decryption ]
barrier ( - ) ;

[ fill = yellow!25, radius = 0.5 cm ] measure BR;

hspace 0.5 cm BR;
output $\ket{ \mathbf{ s } }$ BR;

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left = 1.50 cm of BOB,
label =  [ label distance = 0.10 cm ] west: Bob
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(BobInputA)  ;
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$\frac { 1 } { \sqrt{ 2^{ m } } } \sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } } ( - 1 )^{ \mathbf{ s } \cdot \mathbf{ x } } \ket{ \mathbf{ x } }_{ 1 } \ket{ \mathbf{ x } }_{ 0 }$ ;
[ below = 3.00 cm ] at (Input)  $\ket{ \psi_{ 1 } }$ ;
[ below = 3.00 cm ] at (Expansion)  $\ket{ \psi_{ 2 } }$ ;
[ below = 3.00 cm ] at (Decryption)  $\ket{ \psi_{ 3 } }$ ;

This figure shows the quantum circuit used locally by Bob during the secret decryption phase to extract Alice's secret $\mathbf{ s }$.

The circuit depicted in Figure <ref> starts its operation in the $\ket{ \psi_{ 1 } }$ state. Bob applies to both his registers $BR$ and $BR_{ A }$ the $m$-fold Hadamard transform, driving the system to the next state $\ket{ \psi_{ 2 } }$. Note that in the rest of the computations we ignore Alice's local qubit $AQ$ in state $\ket{ - }$, as it has served its intended purpose to introduce the relative phase $( - 1 )^{ \mathbf{ s } \cdot \mathbf{ x } }$.

\begin{align} \label{eq:$2$PSQDC Part 2 Expansion 1}
\ket{ \psi_{ 2 } }
\frac { 1 } { \sqrt{ 2^{ m } } }
\sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } }
( - 1 )^{ \mathbf{ s } \cdot \mathbf{ x } }
H^{ \otimes m } \ket{ \mathbf{ x } }_{ 1 }
H^{ \otimes m } \ket{ \mathbf{ x } }_{ 0 }
\nonumber \\
&\overset { ( \ref{eq:Hadamard m-Fold Ket x} ) } { = }
\frac { 1 } { \sqrt{ 2^{ m } } }
\sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } }
( - 1 )^{ \mathbf{ s } \cdot \mathbf{ x } }
\left(
\frac { 1 } { \sqrt{ 2^{ m } } }
\sum_{ \mathbf{ a } \in \mathbb{ B }^{ m } }
( - 1 )^{ \mathbf{ a } \cdot \mathbf{ x } }
\ket{ \mathbf{ a } }_{ 1 }
\right)
\left(
\frac { 1 } { \sqrt{ 2^{ m } } }
\sum_{ \mathbf{ b } \in  \mathbb{ B }^{ m } }
( - 1 )^{ \mathbf{ b } \cdot \mathbf{ x } }
\ket{ \mathbf{ b } }_{ 0 }
\right)
\nonumber \\
\frac { 1 } { 2^{ m } \sqrt{ 2^{ m } } }
\sum_{ \mathbf{ a } \in \mathbb{ B }^{ m } }
\sum_{ \mathbf{ b } \in \mathbb{ B }^{ m } }
\sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } }
( - 1 )^{ ( \mathbf{ s } \oplus \mathbf{ a } \oplus \mathbf{ b } ) \cdot \mathbf{ x } }
\ket{ \mathbf{ a } }_{ 1 }
\ket{ \mathbf{ b } }_{ 0 }
\ .
\end{align}

The above expression (<ref>) can be simplified via the use of property (<ref>) of the inner product modulo $2$, which asserts that if

\begin{align} \label{eq:Fundamental Correlation Property 1}
\mathbf{ a } \oplus \mathbf{ b } \oplus \mathbf{ s } = \mathbf{ 0 }
\Leftrightarrow
\mathbf{ a } \oplus \mathbf{ b } = \mathbf{ s }
\ ,
\end{align}

the sum $\sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } } ( - 1 )^{ ( \mathbf{ s } \oplus \mathbf{ a } \oplus \mathbf{ b } ) \cdot \mathbf{ x } } \ket{ \mathbf{ a } }_{ 1 } \ket{ \mathbf{ b } }_{ 0 }$ is equal to $2^{ m }$ $\ket{ \mathbf{ a } }_{ 1 }$ $\ket{ \mathbf{ b } }_{ 0 }$, whereas if $\mathbf{ a } \oplus \mathbf{ b } \oplus \mathbf{ s } \neq \mathbf{ 0 }$, the sum reduces to $0$. This allows the rewriting of $\ket{ \psi_{ 2 } }$ as

\begin{align} \label{eq:$2$PSQDC Part 2 Expansion 2}
\ket{ \psi_{ 2 } }
\frac { 1 } { \sqrt{ 2^{ m } } }
\sum_{ \mathbf{ a } \in \mathbb{ B }^{ m } }
\sum_{ \mathbf{ b } \in \mathbb{ B }^{ m } }
\ket{ \mathbf{ a } }_{ 1 }
\ket{ \mathbf{ b } }_{ 0 }
\ , \quad \text{where} \quad
\mathbf{ a } \oplus \mathbf{ b } = \mathbf{ s }
\ .
\end{align}

The above formula is the mathematical expression of the correlation of the contents of the registers $BR$ and $BR_{ A }$. This is a result of the entanglement between Alice and Bob's registers in the initial state of the quantum circuit of Figure <ref>. The practical significance of this fact is that the contents of the two registers do not vary independently of each other, as they must always obey (<ref>).

Let us now recall that a CNOT (controlled-NOT) gate acts by negating the second qubit, called target qubit, if and only the first qubit, called control qubit, is in state $\ket{ 1 }$. One way to express its operation is by writing

\begin{align} \label{eq:CNOT Gate}
{ \rm CNOT }
\ket{ x }
\ket{ y }
\ket{ x }
\ket{ x \oplus y }
\ .
\end{align}

Moreover, it will convenient to express $\ket{ \mathbf{ a } }_{ 1 }$ and $\ket{ \mathbf{ b } }_{ 0 }$ as

\begin{align}
\ket{ \mathbf{ a } }_{ 1 }
\ket{ a_{ m - 1 } }
\dots
\ket{ a_{ 0 } }
\label{eq:Ket a Expansion 1}
\\
\ket{ \mathbf{ b } }_{ 0 }
\ket{ b_{ m - 1 } }
\dots
\ket{ b_{ 0 } }
\label{eq:Ket b Expansion 1}
\ .
\end{align}

Having established notation, we now proceed to explain in detail Bob's actions to decipher Alice's secret. Bob applies $m$ CNOT gates to his registers $BR_{ A }$ and $BR$. Each of the $m$ qubits in $BR_{ A }$ will serve as a control qubit targeting the corresponding qubit in $BR$. The formal mathematical description of this process is given below

\begin{align} \label{eq:$2$PSQDC Part 2 Decryption 1}
&{ \rm CNOT }
\ket{ a_{ m - 1 } }
\ket{ b_{ m - 1 } }
\dots
{ \rm CNOT }
\ket{ a_{ 0 } }
\ket{ b_{ 0 } }
\nonumber \\
\ket{ a_{ m - 1 } }
\ket{ a_{ m - 1 } \oplus b_{ m - 1 } }
\dots
\ket{ a_{ 0 } }
\ket{ a_{ 0 } \oplus b_{ 0 } }
\nonumber \\
&\overset { ( \ref{eq:Fundamental Correlation Property 1} ) } { = }
\ket{ \mathbf{ a } }_{ 1 }
\ket{ \mathbf{ s } }_{ 0 }
\ .
\end{align}

Hence, in view of equations (<ref>) and (<ref>), state $\ket{ \psi_{ 3 } }$ can be expressed as

\begin{align} \label{eq:$2$PSQDC Part 2 Decryption 2}
\ket{ \psi_{ 3 } }
\frac { 1 } { \sqrt{ 2^{ m } } }
\sum_{ \mathbf{ a } \in \mathbb{ B }^{ m } }
\ket{ \mathbf{ a } }_{ 1 }
\ket{ \mathbf{ s } }_{ 0 }
\ .
\end{align}

Now, Bob measures (in the computational basis) the contents of the register $BR$ and obtains the secret bit vector $\mathbf{ s }$.

To complete the protocol, Alice and Bob perform the second consistency check. If the test is successful, then Bob accepts the obtained secret bit vector $\mathbf{ s }$. If not, then their second communication through the quantum channel was compromised by Eve, and, so, Bob rejects $\mathbf{ s }$ and the protocol starts all over again. We defer the detailed explanation of the actions taking place during the second consistency check until Section <ref>. If the secret bit vector $\mathbf{ s }$ is accepted, then Alice may further apply standard techniques, like quantum error correction and privacy amplification [54, 55].

§ THE $3$PSQDC PROTOCOL

The $3$ player scheme for quantum secure direct communication, $3$PSQDC protocol from now on, is a generalization of the $2$PSQDC, where three spatially separated players interact in order to enable two of them to communicate with the third player directly and securely using only the quantum channel. Many real-life situations involve more than two agents. Therefore, it is advantageous to possess algorithms and techniques facilitating secure communication and information exchange among an arbitrary number of players.

The $3$PSQDC protocol evolves as a game among Alice, Bob and Charlie. To make the game more interesting this time, we assume that Bob and Charlie, being loyal agents of Alice, each have come up with some information, which is incomplete by itself. Only by combining the two pieces of information can the complete secret be revealed. Therefore, Bob and Charlie must send their information to Alice, so that she may uncover the secret. A similar setting in which an arbitrary number of agents send information to Alice in order for her to compose the complete secret, was analyzed in [18]. The major difference compared to the present work, is that in [18] the protocol at the final stage requires the agents to communicate to Alice the information necessary to unlock the secret through the classical channel. Eve, as usual, undertakes the role of the adversary aiming to sabotage the protocol and steal the secret. The advantage that quantum protocols exhibit over classical ones is that communication through the quantum channel involves an array of unique features, such as the no-cloning theorem [51], the monogamy of entanglement [52], and nonlocality [53], that can be used to inhibit Eve.

§.§ Entanglement distribution phase

We may conceptually divide the $3$PSQDC protocol in phases. The first phase is the entanglement distribution phase, during which Alice, or a third trusted source, prepares $\ket{ GHZ_{ 3 } }$ triplets. For the execution of the protocol, it is immaterial whether it is Alice or another trusted source that undertakes this task. What matters is that using a quantum computer running the quantum circuit depicted in Figure <ref>, or some equivalent apparatus, entangled triplets can be generated in order to be shared among Alice, Bob, and Charlie, according to the following scheme.

* Alice sends to Bob a sequence of $m$ qubits $b_{ 0 }, \dots, b_{ m - 1 }$, called Bob's transmission sequence, which Bob stores in his register. Symmetrically, Alice also sends to Charlie a sequence of $m$ qubits $c_{ 0 }, \dots, c_{ m - 1 }$, called Charlie's transmission sequence, which Charlie stores in his register.

* At the same time, Alice stores in her register a corresponding sequence of $m$ qubits $a_{ 0 }, \dots, a_{ m - 1 }$, called Alice's sequence.

* The $k^{ th }$ qubit, $0 \leq k \leq m - 1$, in all the above sequences is either

* one qubit from the same $\ket{ GHZ_{ 3 } }$ triplet, or

* a nonentangled qubit in a state randomly chosen, with equal probability, from $\{ \ket{ 0 }, \ket{ 1 }, \ket{ + }, \ket{ - } \}$, called a decoy.

* In all, Alice sends to each of Bob and Charlie $d$ decoys. It is imperative that Alice insert the decoys randomly within the  transmission sequence. Obviously, Alice keeps track of the position and the state of all her decoys.

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$\ket{ GHZ_{ 3 } }$ Triplet Preparation ;
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(GHZ3-C-m-1)  ;
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(GHZ3-A-0)  ;
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(GHZ3-B-0)  ;
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(GHZ3-C-0)  ;

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(Alice) – (Charlie) node [ above, midway, text = black ]  ;

Alice creates the $m$ $\ket{ GHZ_{ 3 } }$ triplets that she will share with Bob and Charlie, who are in different spatial locations.

[ scale = 0.50 ]

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(GHZ3-C-0)  ;
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(GHZ3-B-0) – (GHZ3-A-0) – (GHZ3-C-0);

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(Alice) – (Bob) node [ above, midway, text = black ]  ;
[ WordBlueDarker25, <->, >=stealth, line width = 1.75 mm, shade, left color = WordBlueDarker25, right color = white ]
(Alice) – (Charlie) node [ above, midway, text = black ]  ;

Alice shares the $m$ $\ket{ GHZ_{ 3 } }$ triplets with Bob and Charlie. She keeps the first qubit of each triplet (drawn in red), and sends the second to Bob (drawn in green) and the third to Charlie (drawn in blue).

We point out that $m$ should be sufficiently large to ensure the secure implementation of the protocol, and the integer $d$ should be significantly smaller than $m$. In our forthcoming mathematical analysis in Section <ref>, we shall be more precise.

The situation before Alice distributes Bob and Charlie's sequences is visualized in Figure <ref>. At the end of the distribution phase, Alice, Bob and Charlie have in their local quantum registers $m$ qubits each. All three registers are correlated because Alice, Bob and Charlie's corresponding qubits are entangled in the $\ket{ GHZ_{ 3 } }$ state, and the whole setup is shown in Figure <ref>. To complete this phase, Alice and Bob and, simultaneously, Alice and Charlie conduct the first consistency check on their registers. If the test is successful, they proceed to the secret embedding phase. If at least one consistency check fails, then their communication has been compromised by Eve, and, so, they abort the protocol. We defer the detailed explanation of the actions taking place during the first consistency check until Section <ref>.

§.§ Secret embedding phase

The idea behind this generalization is that the two agents Bob and Charlie have in their possession part of a secret that they must transmit to Alice in order for her to obtain the complete secret $\mathbf{ s }$. To succeed in this task, the three players use their local circuits outlined in Figure <ref>. Despite spatial separation between the three, the correlations among their registers, due to entanglement, effectively create one quantum distributed system.

grow to left by = 0.00 cm,
grow to right by = 0.00 cm,
colback = MagentaVeryLight!03,
enhanced jigsaw,

sharp corners,

toprule = 1.0 pt,
bottomrule = 1.0 pt,
leftrule = 0.1 pt,
rightrule = 0.1 pt,
sharp corners,
center title,
fonttitle = ]

[ scale = 1.00 ]

nobit AUX_C_0;
nobit AUX_C_1;
[ name = CREG ] qubits  $CR$   CR;
nobit AUX_C_2;
qubit  $CQ$: $\ket{ - }$   CQ;
nobit AUX_C_3;
nobit AUX_C_4;

[ name = waveC, register/minimum height=5mm ] nobit waveC;

nobit AUX_B_0;
nobit AUX_B_1;
[ name = BREG ] qubits  $BR$   BR;
nobit AUX_B_2;
qubit  $BQ$: $\ket{ - }$   BQ;
nobit AUX_B_3;
nobit AUX_B_4;

[ name = waveB, register/minimum height=5mm ] nobit waveB;

nobit AUX_A_0;
nobit AUX_A_1;
[ name = AREG ] qubits  $AR$   AR;
nobit AUX_A_3;
nobit AUX_A_4;

[ name = InitialState, WordBlueVeryLight, line width = 0.50 mm, label = Initial State ]
barrier ( - ) ;

[ draw = GreenLighter2, fill = GreenLighter2, radius = 0.7 cm ] box    U$_{ C }$  ( CR - CQ );

[ draw = GreenLighter2, fill = GreenLighter2, radius = 0.7 cm ] box    U$_{ B }$  ( BR - BQ );

[ name = SecretEmbedding, WordBlueVeryLight, line width = 0.50 mm, label = Secret Embedding ]
barrier ( - ) ;

output   CR;

output   BR;

output   AR;

[on background layer]
[ wave, fill = WordDarkTealLighter80, fit = (waveC) (current bounding box.east |- waveC), inner ysep = 3.00 mm, inner xsep = 0pt ] Spatially Separated;
[ wave, fill = WordDarkTealLighter80, fit = (waveB) (current bounding box.east |- waveB), inner ysep = 3.00 mm, inner xsep = 0pt ] Spatially Separated;
shirt = WordBlueVeryLight,
scale = 1.50,
anchor = center,
left = 0.50 cm of CREG,
label =  [ label distance = 0.00 cm ] north: Charlie
(Charlie)  ;
anchor = center,
right = -0.40 cm of CREG,
(CharlieInput)  ;
anchor = center,
right = 7.25 cm of CREG,
(CharlieOutput)  ;
shirt = GreenLighter2,
scale = 1.50,
anchor = center,
left = 0.50 cm of BREG,
label =  [ label distance = 0.10 cm ] west: Bob
( )  ;
shirt = RedPurple,
scale = 1.50,
anchor = center,
left = 0.50 cm of AREG,
label =  [ label distance = 0.10 cm ] west: Alice
( )  ;
anchor = center,
right = -0.40 cm of AREG,
(AliceInput)  ;
anchor = center,
right = 7.25 cm of AREG,
(AliceOutput)  ;
MagentaLighter, line width = 0.50 mm, decorate, decoration =  brace, amplitude = 5pt , xshift = 0.0 mm, yshift = 0.0 mm
(AliceInput.west) – (CharlieInput.west)
node [ MagentaLighter, midway, rotate = 90, yshift = 5.00 mm ]
$\frac { 1 } { \sqrt{ 2^{ m } } } \sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } } \ket{ \mathbf{ x } }_{ A } \ket{ \mathbf{ x } }_{ B } \ket{ \mathbf{ x } }_{ C }$ ;
MagentaLighter, line width = 0.50 mm, decorate, decoration =  brace, mirror, amplitude = 5pt , xshift = 0.0 mm, yshift = 0.0 mm
(AliceOutput.west) – (CharlieOutput.west)
node [ MagentaLighter, midway, rotate = 270, yshift = 5.00 mm ]
$\frac { 1 } { \sqrt{ 2^{ m } } } \sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } } ( - 1 )^{ \mathbf{ s } \cdot \mathbf{ x } } \ket{ \mathbf{ x } }_{ A } \ket{ \mathbf{ x } }_{ B } \ket{ \mathbf{ x } }_{ C }$ ;
[ below = 4.50 cm ] at (InitialState)  $\ket{ \psi_{ 0 } }$ ;
[ below = 4.50 cm ] at (SecretEmbedding)  $\ket{ \psi_{ 1 } }$ ;

This figure presents an outline of the quantum circuits employed by Alice, Bob and Charlie in the secret embedding phase. This distributed system evolves from the initial state $\ket{\psi_{0}}$ to the state $\ket{\psi_{1}}$.

In Figure <ref>, $AR$, $BR$ and $CR$ stand for Alice, Bob and Charlie's entangled registers, respectively, while $BQ$ and $CQ$ denote Bob and Charlie's qubits, both initialized in state $\ket{ - }$. To avoid any confusion, we use subscripts $A$, $B$ and $C$ to distinguish among Alice, Bob and Charlie's qubits and registers.

The partial secrets that Bob and Charlie intend to send to Alice are encoded in the secret bit vectors $\mathbf{ s }_{ B }$ and  $\mathbf{ s }_{ C }$, where the bits corresponding to the positions of the decoys are set to $0$. Alice must combine both of them via bitwise addition in order to uncover the complete secret $\mathbf{ s }$:

\begin{align} \label{eq:Alice Bob and Charlie's Secret Bit Vectors}
\mathbf{ s } = \mathbf{ s }_{ B } \oplus \mathbf{ s }_{ B }
\ .
\end{align}

Bob and Charlie encode their secrets into the state of the composite system by using the unitary transforms $U_{ B }$ and $U_{ C }$ on their registers, where

\begin{align} \label{eq:Bob and Charlie's Explicit Transforms}
U_{ B }
\ket{ - }_{ B }
\ket{ \mathbf{ x } }_{ B }
\rightarrow
( - 1 )^{ \mathbf{ s }_{ B } \cdot \mathbf{ x } }
\ket{ - }_{ B }
\ket{ \mathbf{ x } }_{ B }
\ , \text{ and }
\nonumber \\
U_{ C }
\ket{ - }_{ C }
\ket{ \mathbf{ x } }_{ C }
\rightarrow
( - 1 )^{ \mathbf{ s }_{ C } \cdot \mathbf{ x } }
\ket{ - }_{ C }
\ket{ \mathbf{ x } }_{ C }
\ .
\end{align}

Using (<ref>), we can express the initial state $\ket{ \psi_0 }$ of the circuit of Figure <ref> as

\begin{align} \label{eq:$3$PSQDC Part 1 Initial State}
\ket{ \psi_{ 0 } }
\frac { 1 } { \sqrt{ 2^{ m } } }
\sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } }
\ket{ \mathbf{ x } }_{ A }
\ket{ - }_{ B }
\ket{ \mathbf{ x } }_{ B }
\ket{ - }_{ C }
\ket{ \mathbf{ x } }_{ C }
\ .
\end{align}

The application of the unitary transforms (<ref>) sends the system to the next state $\ket{ \psi_{ 1 } }$:

\begin{align} \label{eq:$3$PSQDC Part 1 Secret Embedding State}
\ket{ \psi_{ 1 } }
&\overset { (\ref{eq:Bob and Charlie's Explicit Transforms}) } { = }
\frac { 1 } { \sqrt{ 2^{ m } } }
\sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } }
\ket{ \mathbf{ x } }_{ A }
( - 1 )^{ \mathbf{ s }_{ B } \cdot \mathbf{ x } }
\ket{ - }_{ B }
\ket{ \mathbf{ x } }_{ B }
( - 1 )^{ \mathbf{ s }_{ C } \cdot \mathbf{ x } }
\ket{ - }_{ C }
\ket{ \mathbf{ x } }_{ C }
\nonumber \\
&\overset { (\ref{eq:Alice Bob and Charlie's Secret Bit Vectors}) } { = }
\frac { 1 } { \sqrt{ 2^{ m } } }
\sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } }
( - 1 )^{ \mathbf{ s } \cdot \mathbf{ x } }
\ket{ \mathbf{ x } }_{ A }
\ket{ - }_{ B }
\ket{ \mathbf{ x } }_{ B }
\ket{ - }_{ C }
\ket{ \mathbf{ x } }_{ C }
\ .
\end{align}

At the end of this phase, Bob sends to Alice the $m$ qubits in his register through the quantum channel. Alice organizes these qubits in an additional register denoted by $AR_{ B }$. Similarly, Alice receives from Charlie his $m$ qubits and places them into a third register, designated by $AR_{ C }$.

§.§ Secret decryption phase

grow to left by = 1.00 cm,
grow to right by = 1.00 cm,
colback = MagentaVeryLight!03,
enhanced jigsaw,

sharp corners,

toprule = 1.0 pt,
bottomrule = 1.0 pt,
leftrule = 0.1 pt,
rightrule = 0.1 pt,
sharp corners,
center title,
fonttitle = ]

[ scale = 1.00 ]

nobit AUX_C_0;
[ name = AREGC ] qubits   $AR_{ C }$   ARC;
nobit AUX_B_0;
[ name = AREGB ] qubits   $AR_{ B }$   ARB;
nobit AUX_A_0;
[ name = AREG ] qubits  $AR$   AR;
nobit AUX_A_1;

[ name = Input, WordBlueVeryLight, line width = 0.50 mm, label = Input ]
barrier ( - ) ;

[ draw = RedPurple!50, fill = RedPurple!50, radius = 0.5 cm ] box H$^{\otimes m}$ ARC;
[ draw = RedPurple!50, fill = RedPurple!50, radius = 0.5 cm ] box H$^{\otimes m}$ ARB;
[ draw = RedPurple!50, fill = RedPurple!50, radius = 0.5 cm ] box H$^{\otimes m}$ AR;

[ name = Expansion, WordBlueVeryLight, line width = 0.50 mm, label = Expansion ]
barrier ( - ) ;

cnot ARB | ARC;
hspace  0.5 cm  -;
cnot AR | ARB;

[ name = Decryption, WordBlueVeryLight, line width = 0.50 mm, label = Decryption ]
barrier ( - ) ;

[ fill = yellow!25, radius = 0.5 cm ] measure AR;

hspace 0.5 cm AR;
output $\ket{ \mathbf{ s } }$ AR;

shirt = RedPurple,
scale = 1.50,
anchor = center,
left = 0.50 cm of AREGB,
label =  [ label distance = 0.10 cm ] west: Alice
( )  ;
anchor = center,
right = -0.50 cm of AREG,
(AliceInput)  ;
anchor = center,
right = -0.50 cm of AREGC,
(AliceInputC)  ;
MagentaLighter, line width = 0.50 mm, decorate, decoration =  brace, amplitude = 5pt , xshift = 0.0 mm, yshift = 0.0 mm
(AliceInput.east) – (AliceInputC.east)
node [ MagentaLighter, midway, rotate = 90, yshift = 5.00 mm ]
$\frac { 1 } { \sqrt{ 2^{ m } } } \sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } } ( - 1 )^{ \mathbf{ s } \cdot \mathbf{ x } } \ket{ \mathbf{ x } }_{ A } \ket{ \mathbf{ x } }_{ B } \ket{ \mathbf{ x } }_{ C }$ ;
[ below = 3.00 cm ] at (Input)  $\ket{ \psi_{ 1 } }$ ;
[ below = 3.00 cm ] at (Expansion)  $\ket{ \psi_{ 2 } }$ ;
[ below = 3.00 cm ] at (Decryption)  $\ket{ \psi_{ 3 } }$ ;

This figure shows the quantum circuit used locally by Bob during the secret decryption phase to extract Alice's secret $\mathbf{ s }$.

During this phase, Alice will use the quantum circuit outlined in Figure <ref> to decrypt the complete secret bit vector $\mathbf{ s }$. In addition to the quantum register $AR$, shown in the circuit of Figure <ref>, Alice now uses registers $AR_{ B }$ and $AR_{ C }$. All these registers are entangled and, by using all of them, Alice will uncover the secret bit vector $\mathbf{ s }$. To avoid any confusion, in the formulas below we employ subscripts $0$, $1$, and $2$ to designate the state and contents of registers $AR_{ C }$, $AR_{ B }$ and $AR$, respectively.

The circuit depicted in Figure <ref> is initialized in state $\ket{ \psi_{ 1 } }$. Alice applies to all her registers $m$-fold Hadamard transforms, driving the system to the next state $\ket{ \psi_{ 2 } }$. Note that in the rest of the computations we ignore the qubits $BQ$ and $CQ$ because they have completed their purpose, i.e., the introduction of the relative phase $( - 1 )^{ \mathbf{ s } \cdot \mathbf{ x } }$.

\begin{align} \label{eq:$3$PSQDC Part 2 Expansion 1}
\ket{ \psi_{ 2 } }
\frac { 1 } { \sqrt{ 2^{ m } } }
\sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } }
( - 1 )^{ \mathbf{ s } \cdot \mathbf{ x } }
H^{ \otimes m } \ket{ \mathbf{ x } }_{ 2 }
H^{ \otimes m } \ket{ \mathbf{ x } }_{ 1 }
H^{ \otimes m } \ket{ \mathbf{ x } }_{ 0 }
\nonumber \\
&\overset { ( \ref{eq:Hadamard m-Fold Ket x} ) } { = }
\frac { 1 } { \sqrt{ 2^{ m } } }
\sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } }
( - 1 )^{ \mathbf{ s } \cdot \mathbf{ x } }
\left(
\frac { 1 } { \sqrt{ 2^{ m } } }
\sum_{ \mathbf{ a } \in \mathbb{ B }^{ m } }
( - 1 )^{ \mathbf{ a } \cdot \mathbf{ x } }
\ket{ \mathbf{ a } }_{ 2 }
\right)
\left(
\frac { 1 } { \sqrt{ 2^{ m } } }
\sum_{ \mathbf{ b } \in  \mathbb{ B }^{ m } }
( - 1 )^{ \mathbf{ b } \cdot \mathbf{ x } }
\ket{ \mathbf{ b } }_{ 1 }
\right)
\left(
\frac { 1 } { \sqrt{ 2^{ m } } }
\sum_{ \mathbf{ c } \in  \mathbb{ B }^{ m } }
( - 1 )^{ \mathbf{ c } \cdot \mathbf{ x } }
\ket{ \mathbf{ c } }_{ 0 }
\right)
\nonumber \\
\frac { 1 } { 2^{ m } 2^{ m } }
\sum_{ \mathbf{ a } \in \mathbb{ B }^{ m } }
\sum_{ \mathbf{ b } \in \mathbb{ B }^{ m } }
\sum_{ \mathbf{ c } \in \mathbb{ B }^{ m } }
\sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } }
( - 1 )^{ ( \mathbf{ s } \oplus \mathbf{ a } \oplus \mathbf{ b } \oplus \mathbf{ c } ) \cdot \mathbf{ x } }
\ket{ \mathbf{ a } }_{ 2 }
\ket{ \mathbf{ b } }_{ 1 }
\ket{ \mathbf{ c } }_{ 0 }
\ .
\end{align}

Applying the property (<ref>) of the inner product modulo $2$,
which says that if

\begin{align} \label{eq:Fundamental Correlation Property 2}
\mathbf{ a } \oplus \mathbf{ b } \oplus \mathbf{ c } \oplus \mathbf{ s } = \mathbf{ 0 }
\Leftrightarrow
\mathbf{ a } \oplus \mathbf{ b } \oplus \mathbf{ c } = \mathbf{ s }
\ ,
\end{align}

the sum $\sum_{ \mathbf{ x } \in \mathbb{ B }^{ m } } ( - 1 )^{ ( \mathbf{ s } \oplus \mathbf{ a } \oplus \mathbf{ b } \oplus \mathbf{ c } ) \cdot \mathbf{ x } } \ket{ \mathbf{ a } }_{ 2 } \ket{ \mathbf{ b } }_{ 1 } \ket{ \mathbf{ c } }_{ 0 }$ is equal to $2^{ m }$ $\ket{ \mathbf{ a } }_{ 2 }$ $\ket{ \mathbf{ b } }_{ 1 } \ket{ \mathbf{ c } }_{ 0 }$, whereas if $\mathbf{ a } \oplus \mathbf{ b } \oplus \mathbf{ c } \oplus \mathbf{ s } \neq \mathbf{ 0 }$, the sum reduces to $0$. Hence, expression (<ref>) can be simplified as

\begin{align} \label{eq:$3$PSQDC Part 2 Expansion 2}
\ket{ \psi_{ 2 } }
\frac { 1 } { 2^{ m } }
\sum_{ \mathbf{ a } \in \mathbb{ B }^{ m } }
\sum_{ \mathbf{ b } \in \mathbb{ B }^{ m } }
\sum_{ \mathbf{ c } \in \mathbb{ B }^{ m } }
\ket{ \mathbf{ a } }_{ 2 }
\ket{ \mathbf{ b } }_{ 1 }
\ket{ \mathbf{ c } }_{ 0 }
\ , \quad \text{where} \quad
\mathbf{ a } \oplus \mathbf{ b } \oplus \mathbf{ c } = \mathbf{ s }
\ .
\end{align}

The above formula conveys the correlation of the contents of the registers $AR$, $AR_{ B }$, and $AR_{ C }$. This is a result of the entanglement between Alice, Bob and Charlie's registers in the initial state of the quantum circuit of Figure <ref>. Conceptually, we may consider this situation as follows: THE contents of any two of the three registers may vary of each other, but then the contents of the remaining third register are completed defined by (<ref>).

If we expand $\ket{ \mathbf{ a } }_{ 2 }$, $\ket{ \mathbf{ b } }_{ 1 }$ and $\ket{ \mathbf{ c } }_{ 0 }$ as

\begin{align}
\ket{ \mathbf{ a } }_{ 2 }
\ket{ a_{ m - 1 } }
\dots
\ket{ a_{ 0 } }
\label{eq:Ket a Expansion 2}
\\
\ket{ \mathbf{ b } }_{ 1 }
\ket{ b_{ m - 1 } }
\dots
\ket{ b_{ 0 } }
\label{eq:Ket b Expansion 2}
\\
\ket{ \mathbf{ c } }_{ 0 }
\ket{ c_{ m - 1 } }
\dots
\ket{ b_{ 0 } }
\label{eq:Ket c Expansion 2}
\ ,
\end{align}

then the action of the first group of $m$ CNOT gates, where each of the $m$ qubits in $AR_{ C }$ serves as a control qubit targeting the corresponding qubit in $AR_{ B }$, results in

\begin{align} \label{eq:$3$PSQDC Part 2 Decryption 1}
&{ \rm CNOT }
\ket{ c_{ m - 1 } }
\ket{ b_{ m - 1 } }
\dots
{ \rm CNOT }
\ket{ c_{ 0 } }
\ket{ b_{ 0 } }
\nonumber \\
\ket{ c_{ m - 1 } }
\ket{ b_{ m - 1 } \oplus c_{ m - 1 } }
\dots
\ket{ c_{ 0 } }
\ket{ b_{ 0 } \oplus c_{ 0 } }
\ .
\end{align}

The subsequent action of the second group of $m$ CNOT gates, where each of the $m$ qubits in $AR_{ B }$ serves as a control qubit targeting the corresponding qubit in $AR$, drives the circuit into the next state

\begin{align} \label{eq:$3$PSQDC Part 2 Decryption 2}
&{ \rm CNOT }
\ket{ b_{ m - 1 } \oplus c_{ m - 1 } }
\ket{ a_{ m - 1 } }
\dots
{ \rm CNOT }
\ket{ b_{ 0 } \oplus c_{ 0 } }
\ket{ a_{ 0 } }
\nonumber \\
\ket{ b_{ m - 1 } \oplus c_{ m - 1 } }
\ket{ a_{ m - 1 } \oplus b_{ m - 1 } \oplus c_{ m - 1 } }
\dots
\ket{ b_{ 0 } \oplus c_{ 0 } }
\ket{ a_{ 0 } \oplus b_{ 0 } \oplus c_{ 0 } }
\nonumber \\
&\overset { ( \ref{eq:Fundamental Correlation Property 2} ) } { = }
\ket{ \mathbf{ s } }_{ 2 }
\ket{ \mathbf{ b } \oplus \mathbf{ c } }_{ 1 }
\ket{ \mathbf{ c } }_{ 0 }
\ .
\end{align}

Thus, in view of equations (<ref>) and (<ref>), state $\ket{ \psi_{ 3 } }$ can be written as

\begin{align} \label{eq:$3$PSQDC Part 2 Decryption 3}
\ket{ \psi_{ 3 } }
\frac { 1 } { 2^{ m } }
\sum_{ \mathbf{ b } \in \mathbb{ B }^{ m } }
\sum_{ \mathbf{ c } \in \mathbb{ B }^{ m } }
\ket{ \mathbf{ s } }_{ 2 }
\ket{ \mathbf{ b } \oplus \mathbf{ c } }_{ 1 }
\ket{ \mathbf{ c } }_{ 0 }
\ .
\end{align}

Now, Alice measures (in the computational basis) the contents of the register $AR$ and obtains the secret bit vector $\mathbf{ s }$.

To complete the protocol, Alice performs the second consistency check. If the test is successful, then Alice accepts the obtained secret bit vector $\mathbf{ s }$. If not, then her second communication with either Bob or Charlie through the quantum channel was compromised by Eve. In such an eventuality, Alice rejects $\mathbf{ s }$ and the protocol starts all over again. We defer the detailed explanation of the actions taking place during the second consistency check until Section <ref>. As we have pointed out before, if the secret bit vector $\mathbf{ s }$ is accepted, then Alice may further apply standard techniques, like quantum error correction and privacy amplification [54, 55].

§ SECURITY ANALYSIS

The current section presents a unified security analysis of the $2$PSQDC and $3$PSQDC protocols. The ubiquitous Eve, as usual, undertakes the role of the adversary aiming to sabotage the protocol and steal the secret. We shall also make use of a classical authenticated channel, in order to detect the presence of the eavesdropper Eve and perform error correction, and not for the transition of information pertaining to the secret bit vector. For a recent comprehensive text analyzing security issues of quantum protocols in general we refer to [56] and the more recent [57]. Extensive security analysis specifically for QSDC can be found in the thorough and very recent [30].

We analyze the $3$ players case, involving Alice, Bob and Charlie, since the $2$ player case can be viewed as a special case. The setting now includes, in addition to our three protagonists, a forth notorious entity, traditionally named Eve, whose sole purpose is to devise and implement attacks against our protocol, aiming to acquire a piece of the secret information, or even, the complete secret information.

Ultimately the security analysis of any quantum protocol rests on certain well-understood assumptions. For the sake of completeness, we briefly mention them at this point. First, we assume that quantum theory is correct, which in turn means that hallmark features such as the no-cloning theorem [51], the monogamy of entanglement [52], and nonlocality [53] are valid. Clearly, if quantum protocols did not exhibit these properties, they would be useless. Secondly, we assume that quantum theory is complete, which implies that Eve is constrained by the laws of physics, and she cannot derive more information beyond what is predicted by quantum mechanics.

§.§ First consistency check

The secret embedding phase begins only after the successful completion of the first consistency check and consists of the following steps.

* Alice communicates to Bob and Charlie the positions and the intended basis of measurement for the decoys.

* Bob and Charlie send back to Alice the results of their measurements.

* Alice analyzes the results received from Bob and Charlie, and decides whether the test was successful or not, according to the following rationale.

$\diamond$	If no or very few inconsistencies are found, then Alice deems that the first consistency check was successful.

$\diamond$	If the number of inconsistencies is $\approx \frac { d } { 4 }$, or above a predefined threshold, then Alice deems that the first consistency check failed, aborts and terminates the protocol.

In the ideal scenario, where there is no eavesdropping and the quantum channel is perfect, there will be $0$ inconsistencies. In a more realistic scenario, we can expect a few inconsistencies ($\approx 0$) due to the channel imperfections. On the other hand, if Eve managed to eavesdrop, there should be $\approx \frac { d } { 4 }$ different measurement outcomes. The probability of Eve using the wrong basis is $\frac { 1 } { 2 }$, and, in such a case, the probability of ultimately getting the wrong result is also $\frac { 1 } { 2 }$. Thus, Eve's eavesdropping would cause Bob or Charlie to obtain the wrong outcome with a probability of $\frac { 1 } { 4 }$.

Let us be clear that even if Eve successfully eavesdrop during the distribution phase, she will get no information whatsoever because no information has been encoded yet. However it is still possible that she may still disrupt the execution of the protocol. The probability of achieving that without being detected is practically zero. Let us consider what Eve may do during the distribution phase.

* Measure and Resend. Eve intercepts two qubits from each $\ket{ GHZ_{ 3 } }$ triplet during their transmission from Alice to Bob and Charlie, measures them and resends them back to Bob and Charlie. Eve will fail to discover any information because at this phase the $\ket{ GHZ_{ 3 } }$ triplets do not be carry any information. By the acts of measurement, Eve destroys the entanglement. Assuming that Eve randomly chooses the measurement basis between the computational or the Hadamard basis with equal probability, she will introduce $\approx \frac { d } { 4 }$ wrong outcomes, which will be reveled during the first consistency check.

* Intercept and Resend fake $\ket{ GHZ_{ 3 } }$ triplets. Eve intercepts two qubits from each $\ket{ GHZ_{ 3 } }$ triplet during their transmission from Alice to Bob and Charlie. Although cloning is prohibited by the no-cloning theorem, if Eve has already created a sufficient number of her own $\ket{ GHZ_{ 3 } }$ triplets, she may store the intercepted qubits, and, in their place, forward her own qubits. By employing such a strategy she will gain no information because at this phase the $\ket{ GHZ_{3} }$ triplets carry no information. Moreover, the fact that Eve knows nothing regarding the decoys, will again lead to $\approx \frac { d } { 4 }$ wrong outcomes when Bod and Charlie measure the decoys during the first consistency check.

§.§ Second consistency check

Alice accepts as valid the secret bit vector $\mathbf{ s }$ only after the successful completion of the second consistency check, which consists of the following steps.

* Alice chooses a verification sequence $v_{ 1 }, \dots, v_{ f }$ of $f$ positions within the sequence $0, \dots, m - 1$, excluding of course the decoys. It is critical that Alice chooses the verification sequence randomly using an appropriate probability distribution.

* Alice communicates to Bob and Charlie the the verification sequence.

* Bob and Charlie send back to Alice the bits $s_{ v_{ 1 } }^{ B }, \dots, s_{ v_{ f } }^{ B }$ and $s_{ v_{ 1 } }^{ C }, \dots, s_{ v_{ f } }^{ C }$, respectively, of their secret bit vectors.

* Alice compares the bit values $s_{ v_{ 1 } }, \dots, s_{ v_{ f } }$ of her own $\mathbf{ s }$ with the results received from Bob and Charlie, and decides whether the test was successful or not, according to the following rationale.

$\diamond$	If no or very few inconsistencies are found, then Alice deems that the second consistency check was successful.

$\diamond$	If the number of inconsistencies is $\approx \frac { f } { 2 }$, or above a predefined threshold, then Alice deems that the second consistency check failed, aborts and terminates the protocol.

Obviously, in the ideal scenario, where there is no eavesdropping and the quantum channel is perfect, there will be $0$ inconsistencies. In a more realistic scenario, we can expect a few inconsistencies ($\approx 0$) due to the channel imperfections. On the other hand, if Eve managed to eavesdrop, there should be $\approx \frac { f } { 2 }$ different measurement outcomes.

Let us now consider what attacks might Eve devise during the second quantum transmission, where Bob and Charlie each send $m$ qubits to Alice.

* Measure and Resend. Eve intercepts two qubits from each $\ket{ GHZ_{ 3 } }$ triplet during their transmission from Bob and Charlie to Alice, measures them and resends them back to Alice. Eve will fail to discover any information because she has no access to Alice's registers. By the acts of measurement, Eve destroys the entanglement. This random collapse of entanglement will result to $\approx \frac { f } { 2 }$ wrong outcomes in the verification sequence, which will be reveled during the second consistency check.

* Intercept and Resend fake $\ket{ GHZ_{ 3 } }$ triplets. Eve intercepts two qubits from each $\ket{ GHZ_{ 3 } }$ triplet during their transmission from Bob and Charlie to Alice. Although cloning is prohibited by the no-cloning theorem, if Eve has already created a sufficient number of her own $\ket{ GHZ_{ 3 } }$ triplets, she may store the intercepted qubits, and, in their place, forward her own qubits. The flaw in this scenario is that Eve's entangled qubits do not carry the information embedded in Bob and Charlie's qubits. By employing such a strategy she will gain no information because she lacks the third piece of information existing in Alice's register. Obviously, the newly transmitted qubits will not contain the information Alice requires to reveal the correct $\mathbf{ s }$. In the resulting derivation, approximately half the bits will be wrong, and, thus, the verification sequence will contain $\approx \frac { f } { 2 }$ inconsistencies.

* Entangle and Measure. Once again, Eve intercepts two qubits from each $\ket{ GHZ_{ 3 } }$ triplet during their transmission from Bob and Charlie to Alice. This time Eve does not measure them, but entangles them with her ancilla state, and then sends the corresponding GHZ qubits to Alice. Eve waits until the protocol completes before measuring her qubits, hoping to gain useful information. However, the result of Eve's actions is that, instead of having $\approx m$ $\ket{ GHZ_{ 3 } }$ triplets evenly distributed among Alice, Bob and Charlie, to end up with $\approx m$ $\ket{ GHZ_{ 4 } }$ quadruples evenly distributed among Alice, Bob, Charlie, and Eve. Ultimately, Alice will derive an incorrect $\mathbf{ s }$, a fact that she will realize the second consistency check. Eve will also fail gain information about the correct $\mathbf{ s }$ because that would require the contents of Alice's register. Therefore, in this case too, Eve will fail, whereas Alice will be able to infer that Eve tempered with the protocol.

* PNS. The photon number splitting attack (PNS), introduced in [58] and subsequently analyzed in [59, 60], is regarded as one of the most effective attack strategies that Eve can employ against any quantum protocol. This attack exploits the fact that, due to technological limitations, photon sources occasionally do not emit single-photon signals, but may produce multiple identical photons instead of just one. This allows Eve to intercept pulses emanating from Alice for the distribution of the $\ket{ GHZ_{ 3 } }$ triplets, keep one photon from the multi-photon pulse for herself and send the remaining photons to Bob and Charlie without being detected during the transmission phase. As far as the $3$PSQDC protocol is concerned, this case resembles the Entangle and Measure attack analyzed above. Again, instead of $\ket{ GHZ_{ 3 } }$ triplets evenly distributed among Alice, Bob and Charlie, in reality there are $\ket{ GHZ_{ 4 } }$ quadruples evenly distributed among Alice, Bob, Charlie, and Eve. Eve becomes effectively the fourth player and is unable to gain any information about the other players' measurements for the same reasons as in the previous case.

The above security analysis demonstrates that both the $2$PSQDC and $3$PSQDC protocols information-theoretically secure. Parameters $d$ and $f$ are critical for the correct execution of the first and second consistency checks, respectively. In a perfect or near perfect channel their values need not be large; the probability that Eve eavesdrops undetected converges rapidly to zero, as shown in Table <ref>. Things become significantly more complicated in the case of a noisy channel, where proper values must be selected in order to ensure that the probability that Eve eavesdrops undetected is negligible.

Some numerical values for the $d$ and $f$ parameters.

5c$d, f$ & Probability

$d, f$

$\frac { d } { 4 }$

$\frac { f } { 2 }$

$P( \text{ Eve cheats FCC } )$

$P( \text{ Eve cheats SCC } )$

4 // 4

4 // 2

Probability1 ( m ) := ( 3 / 4 ) ** m;
Probability1 ( 4 // 4 )

Probability2 ( m ) := ( 1 / 2 ) ** m;
Probability2 ( 4 // 2 )

8 // 4

8 // 2

Probability1 ( m ) := ( 3 / 4 ) ** m;
Probability1 ( 8 // 4 )

Probability2 ( m ) := ( 1 / 2 ) ** m;
Probability2 ( 8 // 2 )

16 // 4

16 // 2

Probability1 ( m ) := ( 3 / 4 ) ** m;
Probability1 ( 16 // 4 )

Probability2 ( m ) := ( 1 / 2 ) ** m;
Probability2 ( 16 // 2 )

32 // 4

32 // 2

Probability1 ( m ) := ( 3 / 4 ) ** m;
Probability1 ( 32 // 4 )

Probability2 ( m ) := ( 1 / 2 ) ** m;
Probability2 ( 32 // 2 )

64 // 4

64 // 2

Probability1 ( m ) := ( 3 / 4 ) ** m;
Probability1 ( 64 // 4 )

Probability2 ( m ) := ( 1 / 2 ) ** m;
Probability2 ( 64 // 2 )

128 // 4

128 // 2

Probability1 ( m ) := ( 3 / 4 ) ** m;
Probability1 ( 128 // 4 )

Probability2 ( m ) := ( 1 / 2 ) ** m;
Probability2 ( 128 // 2 )

§ DISCUSSION AND CONCLUSIONS

In this work, we have introduced two new protocols for quantum secure direct communication. The first, called $2$PSQDC, involves communication between two entities, Alice and Bob. Subsequently, the $2$PSQDC was generalized in a intuitive and straightforward manner, so as to provide for quantum secure direct communication among three entities. The resulting protocol, which is called $3$PSQDC, can be seamlessly generalized to an arbitrary number of entities. Both protocols, which are proven to be information-theoretically secure, use the same idea, i.e., embedding the secret information via an oracle that uses the inner product modulo $2$ operation. This way of encoding the information is the main novelty of this paper that distinguishes it from the many previous works in the field. The advantage of this method is that it is seamlessly extensible and can be generalized to a setting involving three, or even more, players, as demonstrated with the $3$PSQDC protocol. This last case, is not only useful, but often necessary, when two spatially separated players posses only part the secret information that must be combined and transmitted to Alice in order for her to reveal the complete secret. Using the $3$PSQDC protocol, this task can be achieved in one go, without the need to apply a typical QSDC protocol twice, where Alice first receives Bob's information and afterwards Charlie's information. Assuming sufficient resources, the $3$PSQDC protocol can be extended in the obvious manner to allow for $n - 1$ players to simultaneously send information to Alice. It is also worth mentioning that both protocols are practically accessible, since they relay on EPR pairs, in the two player case, and $\ket{ GHZ_{ 3 } }$ triples, in the three player case, that can be generated with our current technology. Moreover, the local quantum circuits are characterized by uniformity and symmetry, being similar or identical, and are easily constructible as they employ only Hadamard and CNOT gates.

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