1. Introduction
In the theory of quantum information, the problem of quantum state discrimination is receiving increasing attention. While a set of mutually orthogonal quantum states can always be perfectly distinguished through some global measurements, the situation is quite different when only local operations and classical communication (LOCC) are allowed. If a set of orthogonal quantum states cannot be distinguished via LOCC, then this set is termed locally indistinguishable. Locally indistinguishable quantum states have practical applications in designing quantum protocols and play significant roles in areas such as quantum voting [1], quantum data hiding [2, 3] and quantum secret sharing [4–6].
In 1999, Bennett et al [7] presented a set of mutually orthogonal product states in ${{\mathbb{C}}}^{3}\otimes {{\mathbb{C}}}^{3}$ that are indistinguishable under LOCC, a phenomenon they termed ‘nonlocality without entanglement'. Starting from Bennett et al, many scholars studied the nonlocality without entanglement and constructed many locally indistinguishable sets in both bipartite [8–13] and multipartite systems [14–20]. For example, Xu et al gave a novel method to construct a nonlocal set with only 2(m + n) − 4 orthogonal product states in a ${{\mathbb{C}}}^{m}\otimes {{\mathbb{C}}}^{n}$ quantum system for n ≥ m ≥ 3 [13]. Wang et al gave an explicit construction of locally indistinguishable multipartite orthogonal product states for multipartite quantum systems by using a set of locally indistinguishable bipartite orthogonal product states [18]. In addition, Halder showed several sets of LOCC indistinguishable multipartite orthogonal product states [16].
In 2019, Halder et al [21] introduced locally irreducible sets by incorporating the property of local eliminability. Building upon this notion, they introduced strong nonlocality,where a set of orthogonal product states is considered strongly nonlocal if it remains locally irreducible in every bipartition. Strong nonlocality requires irreducibility in every bipartition, a condition that is challenging to satisfy. Naturally, the concept of genuine nonlocality without entanglement was proposed. A set of orthogonal product states in an n-partite quantum system is called genuinely nonlocal if it cannot be perfectly distinguished by any n − 1 parties collaborating through LOCC. Genuine nonlocality represents a notion stronger than nonlocality but weaker than strong nonlocality. In 2023, Xiong et al [22] proved that in the case of subsystem dimension d = 2t there exists a genuinely nonlocal set in system ${{\mathbb{C}}}^{d}\otimes {{\mathbb{C}}}^{d}\otimes {{\mathbb{C}}}^{d}$ with a cardinality of d + 3. Furthermore, in 2024, Xiong et al [23] demonstrated that, as long as condition d ≥ 4 is satisfied, there always exist d + 1 mutually orthogonal generalized GHZ states in ${{\mathbb{C}}}^{d}\otimes {{\mathbb{C}}}^{d}\otimes {{\mathbb{C}}}^{d}$, possessing genuine nonlocality.
Walgate et al [24] showed that two orthogonal pure states, entangled or otherwise, and distributed between any number of parties, can be perfectly distinguished by LOCC. Chen et al [25] showed that two nonorthogonal entangled states with equal prior probabilities can be optimally conclusively distinguished by LOCC. As widely recognized, entanglement is a valuable resource that enables communication between distant parties, as seen in applications such as quantum teleportation [26]. Locally indistinguishable quantum states may become locally distinguishable with a small amount of entanglement resources. Moreover, sets of states exhibiting such ‘nonlocality without entanglement' are connected to unextendible product bases [10]. In 2008, Cohen [27] first proposed a quantum state discrimination protocol utilizing entanglement, and the protocol is more efficient than quantum teleportation. He pointed out that certain types of unextendible product base in ${{\mathbb{C}}}^{m}\otimes {{\mathbb{C}}}^{n}$ systems can be perfectly distinguished using maximum entangled states with ⌈m/2⌉ ⨂ ⌈n/2⌉ dimensions. Subsequently, Zhang et al [28] demonstrated that certain types of orthogonal product state, which are not unextendible product bases, can be perfectly distinguished in ${{\mathbb{C}}}^{m}\otimes {{\mathbb{C}}}^{n}$ systems using entanglement as a resource. Their protocol is also more efficient than quantum teleportation. Li et al [29] introduced a method to distinguish a set of orthogonal product states in ${{\mathbb{C}}}^{d}\otimes {{\mathbb{C}}}^{d}$ systems (where d is odd) solely utilizing a single auxiliary two-qubit maximum entangled state. Then many interesting results sprang up [30–34].
Rout et al [35] classified genuinely nonlocal product bases based on the local eliminability. They perfectly distinguished some genuinely nonlocal product bases in ${{\mathbb{C}}}^{3}\otimes {{\mathbb{C}}}^{3}\otimes {{\mathbb{C}}}^{3}$ and ${{\mathbb{C}}}^{4}\otimes {{\mathbb{C}}}^{4}\otimes {{\mathbb{C}}}^{4}$ using maximally entangled states. Their protocols consumed less entanglement compared to teleportation. In this work, our construction of genuinely nonlocal product basis is based on Rout's classification. We first give a set of nonlocal product states in ${{\mathbb{C}}}^{3}\otimes {{\mathbb{C}}}^{4}$, and let the states in that set make tensor products with ∣4⟩ respectively. Then, combined with the completeness of the basis, we construct a genuinely nonlocal product basis for type I in ${{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}$. Later, we briefly illustrate the genuine nonlocality of the constructed basis. Additionally, we obtain a genuinely nonlocal product basis for type II by replacing some states in type I with some superposition states. In the end, we separately employ two EPR states and one GHZ state to perfectly distinguish the genuinely nonlocal product basis for type II. In distinguishing genuinely nonlocal product basis in ${{\mathbb{C}}}^{d}\otimes {{\mathbb{C}}}^{d}\otimes {{\mathbb{C}}}^{d}$, protocols based on quantum teleportation consume 2 log2 d ebits of entanglements. Our protocols consume 2 or 1 ($\lt 2{\mathrm{log}}_{2}5$) ebits of entanglements, so ours are more efficient than quantum teleportation.
The organization of this paper is as follows. In section 2 , we introduce some coherent concepts and preparatory knowledge. In section 3 , we construct two types of genuinely nonlocal product basis in ${{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}$. In section 4 , we perfectly distinguish the genuinely nonlocal product basis that we constructed by two EPR states and one GHZ state, respectively. In section 5 , we briefly summarize this paper.
2. Preliminaries
Definition 1. Suppose that $D\equiv \{| \psi {\rangle }_{j}={\displaystyle \bigotimes }_{i=1}^{n}| \alpha {\rangle }_{j}^{i}:j\,=1,\ldots ,{{\rm{\Pi }}}_{i=1}^{n}{d}_{i}\}$ is an orthogonal product basis in an n-partite quantum system ${\displaystyle \bigotimes }_{i=1}^{n}{{\mathbb{C}}}^{{d}_{i}}$.
| i It is called nonlocal if states in D cannot be perfectly distinguished by LOCC. | |
| ii It is called genuinely nonlocal if states in D cannot be perfectly distinguished by LOCC even if any $n-1$ parties are allowed to come together. |
In the subsequent discussion, we suppose post-measurement states are mutually orthogonal. A measurement performed to distinguish a set of mutually orthogonal quantum states is called an orthogonality-preserving measurement if the post-measurement states remain mutually orthogonal. Furthermore, such a measurement is called trivial if all the measurement operators are proportional to the identity operator; otherwise it is nontrivial. For a nontrivial orthogonality-preserving measurement, we have the concept of locally irreducible.
Definition 2. A set of orthogonal quantum states in ${ \mathcal H }={\displaystyle \bigotimes }_{i=1}^{n}{{\mathbb{C}}}^{{d}_{i}}$ with $n\geqslant \,2$ and di $\geqslant 2$, i = 1,…,n, is locally irreducible if it is not possible to eliminate one or more states from the set by nontrivial orthogonality-preserving local measurements.
Figure 1. The relationship between several types of genuinely nonlocal product basis. |
Type I: The basis of this type is locally reducible, meaning that one or more states can be eliminated under nontrivial orthogonality-preserving local measurement.
Type II: The basis of this type is locally irreducible, indicating that no states can be eliminated via nontrivial orthogonality-preserving local measurement.
Type II(a): Such a basis meets the following conditions.
(1) It is locally irreducible when participants are in different locations.
(2) Some states in it can be eliminated under nontrivial orthogonality-preserving local measurement when two of the parties come together.
Type II(b): Such a basis remains locally irreducible regardless of whether participants are in different locations or any two parties come together.
3. Genuinely nonlocal product basis in ${{\mathbb{C}}}^{5}\,\otimes \,{{\mathbb{C}}}^{5}\,\otimes \,{{\mathbb{C}}}^{5}$
In this section, we construct a genuinely nonlocal product basis BI(5, 3) from a nonlocal product basis in ${{\mathbb{C}}}^{3}\otimes {{\mathbb{C}}}^{4}$. Furthermore, we obtain genuinely nonlocal product basis BII(5, 3) by replacing certain states in BI(5, 3). In the appeal statement, BI(5, 3) represents a genuinely nonlocal product basis for type I in ${{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}$ and BII(5, 3) denotes a genuinely nonlocal product basis for type II in ${{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}$.
3.1. Genuinely nonlocal product basis for type I
We present a genuinely nonlocal product basis for type I in quantum system ${{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}$, where the computational basis for ${{\mathbb{C}}}^{5}$ will be denoted by $\{| i\rangle \}{}_{i=0}^{4}$.
First, we give a set of nonlocal quantum states in ${{\mathbb{C}}}^{3}\otimes {{\mathbb{C}}}^{4}$, denoted as
$\begin{eqnarray}{ \mathcal B }=\{| {\phi }_{1,{t}_{1}}\rangle ,| {\phi }_{2,{t}_{2}}\rangle ,| {\phi }_{3,{t}_{1}}\rangle ,| {\phi }_{4,{t}_{2}}\rangle ,| {\psi }_{\mathrm{1,1}}\rangle ,| {\psi }_{\mathrm{1,2}}\rangle \},\end{eqnarray}$
where $\begin{eqnarray*}\begin{array}{rcl}| {\phi }_{1,{t}_{1}}\rangle & = & | 0\rangle (\displaystyle \sum _{j=0}^{2}{\omega }^{{t}_{1}j}| j\rangle );\\ | {\phi }_{2,{t}_{2}}\rangle & = & (\displaystyle \sum _{j=0}^{1}{\left(-1\right)}^{{t}_{2}j}| j\rangle )| 3\rangle ;\\ | {\phi }_{3,{t}_{1}}\rangle & = & | 2\rangle (\displaystyle \sum _{j=0}^{2}{\omega }^{{t}_{1}j}| (j+1)\rangle );\\ | {\phi }_{4,{t}_{2}}\rangle & = & (\displaystyle \sum _{j=0}^{1}{\left(-1\right)}^{{t}_{2}j}| (j+1)\rangle )| 0\rangle ;| {\psi }_{{j}_{1},{j}_{2}}\rangle =| {j}_{1}\rangle | {j}_{2}\rangle ,\\ {j}_{1},{j}_{2} & = & 0,1,2,3,4\end{array}\end{eqnarray*}$
and $\omega ={{\rm{e}}}^{\tfrac{2\pi {\rm{i}}}{3}};$ t1 = 0, 1, 2; t2 = 0, 1.Based on the completeness of the basis, we construct
$\begin{eqnarray}\begin{array}{rcl}{{\bf{B}}}_{{\rm{I}}}(5,3) & = & \{| {\psi }_{h,i,j}\rangle ,| {\psi }_{h,4,k}\rangle ,| {\psi }_{3,k,{k}^{{\prime} }}\rangle ,| {\psi }_{4,h,4}\rangle ,| {\psi }_{4,e,k}\rangle ,\\ & & | 4\rangle | {\phi }_{1,{t}_{1}}\rangle ,| {\phi }_{1,{t}_{1}}\rangle | 4\rangle ,| 4\rangle | {\phi }_{2,{t}_{2}}\rangle ,| {\phi }_{2,{t}_{2}}\rangle | 4\rangle ,\\ & & | 4\rangle | {\phi }_{3,{t}_{1}}\rangle ,| {\phi }_{3,{t}_{1}}\rangle | 4\rangle ,| 4\rangle | {\phi }_{4,{t}_{2}}\rangle ,\\ & & | {\phi }_{4,{t}_{2}}\rangle | 4\rangle ,| 4\rangle | {\psi }_{\mathrm{1,1}}\rangle ,\\ & & | {\psi }_{\mathrm{1,1}}\rangle | 4\rangle ,| 4\rangle | {\psi }_{\mathrm{1,2}}\rangle ,| {\psi }_{\mathrm{1,2}}\rangle | 4\rangle \},\end{array}\end{eqnarray}$
where $| {\psi }_{{i}_{1},{i}_{2},{i}_{3}}\rangle =| {i}_{1}\rangle | {i}_{2}\rangle | {i}_{3}\rangle ;$ ${i}_{1},{i}_{2},{i}_{3},k,k^{\prime} =0,1,2,3,4;$ h, t1 = 0, 1, 2; i, j = 0, 1, 2, 3; e = 3, 4; t2 = 0, 1.The set BI(5, 3) is a genuinely nonlocal product basis for type I in ${{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}$, where BI(5, 3) satisfies the following conditions.
(1) The set ${ \mathcal B }$ is a nonlocal product basis in ${{\mathbb{C}}}^{3}\otimes {{\mathbb{C}}}^{4}$, then the set of tripartite states
$\begin{eqnarray}\{| 4{\rangle }_{1}| \beta {\rangle }_{23}:| \beta \rangle \in { \mathcal B }\}\end{eqnarray}$
is locally indistinguishable in bipartite division 2∣13 as well as 3∣12, where the subscripts 1, 2 and 3 represent the first, second and third quantum systems respectively. In the same way, the set of states $\begin{eqnarray}\{| \beta {\rangle }_{12}| 4{\rangle }_{3}:| \beta \rangle \in { \mathcal B }\}\end{eqnarray}$
is locally indistinguishable in bipartite division 1∣23 and 2∣13.(2) Any participant can eliminate some states by performing a nontrivial orthogonality-preserving measurement. For instance, the first participant can perform a local measurement
$\begin{eqnarray}M\equiv \{| 4\rangle \langle 4| ,{{\mathbb{I}}}_{5}-| 4\rangle \langle 4| \}.\end{eqnarray}$
If the measurement outcome corresponds to ∣4⟩⟨4∣ , then the given state must be one of the following quantum states: $\begin{eqnarray}\{| 4\rangle | \beta \rangle ,| {\psi }_{\mathrm{4,4},k}\rangle ,| {\psi }_{4,h,4}\rangle ,| {\psi }_{4,e,k}\rangle \}.\end{eqnarray}$
Otherwise, it is one of the remaining states.3.2. Genuinely nonlocal product basis for type II
In 3.1, we constructed genuinely nonlocal product basis BI(5, 3). In order to obtain BII(5, 3), we will modify BI(5, 3) through superpositions of ∣2⟩, ∣3⟩ and ∣4⟩.
In BI(5, 3), each participant can locally eliminate some quantum states by performing a nontrivial orthogonality-preserving measurement, to distinguish the subspace spanned by ∣4⟩ from the subspace spanned by {∣0⟩, ∣1⟩, ∣2⟩, ∣3⟩}. If we prevent the local elimination of these states, we will obtain BII(5, 3).
We introduce new orthogonal product states
$\begin{eqnarray}| \tau {\rangle }_{{t}_{1}}| 4\rangle | 0\rangle ,| 3\rangle | \tau {\rangle }_{{t}_{1}}| 2\rangle ,| 1\rangle | 4\rangle | \tau {\rangle }_{{t}_{1}},\end{eqnarray}$
and use them to separately replace these states $\begin{eqnarray}\begin{array}{rcl} & & \{| 2\rangle | 4\rangle | 0\rangle ,| 3\rangle | 4\rangle | 0\rangle ,| 4\rangle | 4\rangle | 0\rangle \},\\ & & \{| 3\rangle | 2\rangle | 2\rangle ,| 3\rangle | 3\rangle | 2\rangle ,| 3\rangle | 4\rangle | 2\rangle \},\\ & & \{| 1\rangle | 4\rangle | 2\rangle ,| 1\rangle | 4\rangle | 3\rangle ,| 1\rangle | 4\rangle | 4\rangle \},\end{array}\end{eqnarray}$
in BI(5, 3), where $| \tau {\rangle }_{{t}_{1}}\,=\,{\sum }_{j=0}^{2}{\omega }^{{t}_{1}j}| (j+2)\rangle ,{t}_{1}\,=\,0,1,2$.In this way, any participant cannot discriminate the subspaces spanned by ∣4⟩ and {∣0⟩, ∣1⟩, ∣2⟩, ∣3⟩}, thus he or she cannot eliminate any state via nontrivial orthogonality-preserving measurement.
Therefore, we obtain that
$\begin{eqnarray}\begin{array}{l}{{\bf{B}}}_{\mathrm{II}}(5,3)={\overline{{\bf{B}}}}_{{\rm{I}}}(5,3)\,\cup \{| \tau {\rangle }_{{t}_{1}}| 4\rangle | 0\rangle ,| 3\rangle | \tau {\rangle }_{{t}_{1}}| 2\rangle ,\\ | 1\rangle | 4\rangle | \tau {\rangle }_{{t}_{1}}\}\end{array}\end{eqnarray}$
is a genuinely nonlocal product basis for type II in ${{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}$, where t1 = 0, 1, 2, and ${\overline{{\bf{B}}}}_{{\rm{I}}}(5,3)$ is denoted by $\begin{eqnarray}\begin{array}{rcl}{\overline{{\bf{B}}}}_{{\rm{I}}}(5,3) & = & {{\bf{B}}}_{{\rm{I}}}(5,3)\setminus \{| 2\rangle | 4\rangle | 0\rangle ,| 3\rangle | 4\rangle | 0\rangle ,| 4\rangle | 4\rangle | 0\rangle ,\\ & & | 3\rangle | 2\rangle | 2\rangle ,| 3\rangle | 3\rangle | 2\rangle ,| 3\rangle | 4\rangle | 2\rangle ,\\ & & | 1\rangle | 4\rangle | 2\rangle ,| 1\rangle | 4\rangle | 3\rangle ,| 1\rangle | 4\rangle | 4\rangle \}.\end{array}\end{eqnarray}$
4. Entanglement-assisted discrimination
In this section, we distinguish the genuinely nonlocal product basis in ${{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}$ using entangled states as a resource. Suppose the states in space ${{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}$ are shared among Alice, Bob and Charlie.
First, we consider a genuinely nonlocal product basis for type I. The basis of this type can eliminate some states under nontrivial orthogonality-preserving measurement, and the entangled resource is determined after the elimination. As mentioned in the previous section, if the measurement outcome corresponds to ∣4⟩⟨4∣ we can obtain a subset of BI(5, 3) containing {∣4⟩A∣β⟩BC}. Then entanglement needs to be shared between Bob and Charlie. Otherwise, the entanglement should be shared between Alice and Bob. A maximally entangled state with 3 ⨂ 3 dimensions shared between AB or BC is sufficient to perfectly distinguish BI(5, 3) (see [27]).
In order to achieve perfect discrimination, BII(5, 3) requires entanglement in each bipartition or three parties sharing of a three-qubit GHZ state. We now demonstrate that perfect discrimination of BII(5, 3) requires two additional EPR states or one GHZ state. In ${{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}$, the quantum teleportation protocol consumes 2 log2 5 ebits of entanglement in distinguishing BII(5, 3). We use a smaller amount of entanglement than this, so our protocol is more efficient than quantum teleportation.
Theorem 1. Suppose Alice and Bob share the maximally entangled state $\tfrac{1}{\sqrt{2}}{\left(| 00\rangle +| 11\rangle \right)}_{{a}_{1}{b}_{1}}$ and Alice and Charlie share the maximally entangled state $\tfrac{1}{\sqrt{2}}{\left(| 00\rangle +| 11\rangle \right)}_{{a}_{2}{c}_{1}}$, the set ${{\bf{B}}}_{\mathrm{II}}(5,3)$ in (9) can be perfectly distinguished by LOCC.
Proof. For the entangled state $\tfrac{1}{\sqrt{2}}{\left(| 00\rangle +| 11\rangle \right)}_{{a}_{1}{b}_{1}}$, assume that Alice takes the a1 particle and Bob takes the b1 particle, and for the entangled state $\tfrac{1}{\sqrt{2}}{\left(| 00\rangle +| 11\rangle \right)}_{{a}_{2}{c}_{1}}$, assume that Alice takes the a2 particle and Bob takes the c1 particle.
Accordingly, the initial state is
$\begin{eqnarray}\begin{array}{rcl}| \zeta \rangle & = & | \psi {\rangle }_{\mathrm{ABC}}\otimes \displaystyle \frac{1}{\sqrt{2}}{\left(| 00\rangle +| 11\rangle \right)}_{{a}_{1}{b}_{1}}\\ & & \otimes \,\displaystyle \frac{1}{\sqrt{2}}{\left(| 00\rangle +| 11\rangle \right)}_{{a}_{2}{c}_{1}},\end{array}\end{eqnarray}$
where $| \psi {\rangle }_{\mathrm{ABC}}$ is one of the states from the set ${{\bf{B}}}_{\mathrm{II}}(5,3)$. Next, Alice, Bob and Charlie perform nontrivial orthogonality-preserving measurement separately, and the specific procedure as follows.
${\bf{Step}}\,{\bf{1}}$ Bob selects the measurement operator in ${\bf{M}}$ to measure the initial state $| \zeta \rangle $, where
$\begin{eqnarray}\begin{array}{l}{\bf{M}}\equiv \{{M}_{1}:= | 0\rangle \langle 0{| }_{B}\otimes | 0\rangle \langle 0{| }_{{b}_{1}}+(| 1\rangle \langle 1| +| 2\rangle \langle 2| \\ \,{+| 3\rangle \langle 3| +| 4\rangle \langle 4| )}_{{\rm{B}}}\otimes | 1\rangle \langle 1{| }_{{b}_{1}},\,{\bar{M}}_{1}:= {\unicode{120128}}-{M}_{1}\}.\end{array}\end{eqnarray}$
Then Charlie selects the measurement operator in ${\bf{N}}$ to measure Bob's post-measurement states, where $\begin{eqnarray}\begin{array}{l}{\bf{N}}\equiv \{{N}_{1}:= | 0\rangle \langle 0{| }_{C}\otimes | 0\rangle \langle 0{| }_{{c}_{1}}+\left(| 1\rangle \langle 1| +| 2\rangle \langle 2| \right.\\ \,+\,| 3\rangle \langle 3| {\left.+| 4\rangle \langle 4| \right)}_{C}\otimes | 1\rangle \langle 1{| }_{{c}_{1}},\,{\overline{N}}_{1}:= {\mathbb{I}}-{N}_{1}\}.\end{array}\end{eqnarray}$
Suppose Bob selects M1 and Charlie selects N1; the post-measurement states are shown in table 1. ${\bf{Step}}\,{\bf{2}}$ Alice selects the measurement operator in ${\bf{K}}$ to measure the post-measurement states in table 1, where
$\begin{eqnarray}\begin{array}{rcl}{\bf{K}}\equiv \{{K}_{1} & := & {\left(| 1\rangle \langle 1| +| 2\rangle \langle 2| \right)}_{A}\otimes | 0\rangle \langle 0{| }_{{a}_{1}}\otimes {{\mathbb{I}}}_{{a}_{2}},\\ {K}_{2} & := & | 1\rangle \langle 1{| }_{A}\otimes | 1\rangle \langle 1{| }_{{a}_{1}}\otimes | 0\rangle \langle 0{| }_{{a}_{2}},\\ {K}_{3} & := & | 3\rangle \langle 3{| }_{A}\otimes | 0\rangle \langle 0{| }_{{a}_{1}}\otimes {{\mathbb{I}}}_{{a}_{2}},\\ {K}_{4} & := & | 2\rangle \langle 2{| }_{A}\otimes | 1\rangle \langle 1{| }_{{a}_{1}}\otimes | 1\rangle \langle 1{| }_{{a}_{2}},\\ {K}_{5} & := & | 3\rangle \langle 3{| }_{A}\otimes | 1\rangle \langle 1{| }_{{a}_{1}}\otimes | 1\rangle \langle 1{| }_{{a}_{2}},\\ {K}_{6} & := & {\mathbb{I}}-{K}_{1}-{K}_{2}-{K}_{3}-{K}_{4}-{K}_{5}\}.\end{array}\end{eqnarray}$
States corresponding to the measurement operators K1, K2, K3, K4 and K5 separately are $\begin{eqnarray*}\begin{array}{rcl}{K}_{1} & \Rightarrow & \{| 1\rangle | 0\rangle | 0\rangle ,| 2\rangle | 0\rangle | 0\rangle ,| 1\rangle | 0\rangle | f\rangle ,| 2\rangle | 0\rangle | f\rangle ,| {\phi }_{4,{t}_{2}}\rangle | 4\rangle \},\\ f & = & 1,2,3;\,{t}_{2}=0,1.\\ {K}_{2} & \Rightarrow & \{| 1\rangle | m\rangle | 0\rangle \},m=1,2,3,4.\\ {K}_{3} & \Rightarrow & \{| 3\rangle | 0\rangle | k\rangle \},k=0,1,2,3,4.\\ {K}_{4} & \Rightarrow & \{| 2\rangle | m\rangle | f\rangle ,| 2\rangle | 4\rangle | 4\rangle ,| {\phi }_{3,{t}_{1}}\rangle | 4\rangle \},\\ m & = & 1,2,3,4;\,f=1,2,3;\,{t}_{1}=0,1,2.\\ {K}_{5} & \Rightarrow & \{| 3\rangle | 1\rangle | m\rangle ,| 3\rangle | 2\rangle | l\rangle ,| 3\rangle | 3\rangle | l\rangle ,| 3\rangle | 4\rangle | l\rangle ,| 3\rangle | \tau {\rangle }_{{t}_{1}}| 2\rangle \},\\ m & = & 1,2,3,4;\,l=1,3,4;\,{t}_{1}=0,1,2.\end{array}\end{eqnarray*}$
The states in each of these sets are LOCC distinguishable. If Alice selects the measurement operator K6, the state is one of the remaining states. After Alice's measurements, we are able to screen out some locally distinguishable sets of quantum states. The remaining quantum states are locally indistinguishable and remain unchanged. Further measurements can be performed on these remaining quantum states. This is true for every subsequent step.
${\bf{Step}}\,{\bf{3}}$ Bob selects the measurement operator in ${\bf{M}}^{\prime} $ to measure the remaining quantum states in table 1, where
$\begin{eqnarray}\begin{array}{l}{\bf{M}}^{\prime} \equiv \{{M}_{12}:= | 4\rangle \langle 4{| }_{B}\otimes {{\mathbb{I}}}_{{b}_{1}},\,{M}_{22}:= | 3\rangle \langle 3{| }_{B}\\ \,\otimes \,{{\mathbb{I}}}_{{b}_{1}},\,{M}_{32}:= {\mathbb{I}}-{M}_{12}-{M}_{22}\}.\end{array}\end{eqnarray}$
States corresponding to the measurement operators M12 and M22 separately are $\begin{eqnarray*}\begin{array}{l}{M}_{12}\Rightarrow \{| 0\rangle | 4\rangle | k\rangle ,| 1\rangle | 4\rangle | 1\rangle ,| 4\rangle | 4\rangle | m\rangle ,\\ \,\ \ \,| 1\rangle | 4\rangle | \tau {\rangle }_{{t}_{1}},| \tau {\rangle }_{{t}_{1}}| 4\rangle | 0\rangle \},\\ k=0,1,2,3,4;\,m=1,2,3,4;\,{t}_{1}=0,1,2.\\ {M}_{22}\Rightarrow \{| 0\rangle | 3\rangle | 0\rangle ,| 2\rangle | 3\rangle | 0\rangle ,| 3\rangle | 3\rangle | 0\rangle ,| 4\rangle | 3\rangle | 0\rangle ,\\ \,\ \ \,| g\rangle | 3\rangle | f\rangle ,| 4\rangle | 3\rangle | m\rangle ,| {\phi }_{2,{t}_{2}}\rangle | 4\rangle \},\\ g=0,1;f=1,2,3;\,m=1,2,3,4;\,{t}_{2}=0,1.\end{array}\end{eqnarray*}$
The states in each of these sets are LOCC distinguishable. If Bob selects the measurement operator M32, the state is one of the remaining states. ${\bf{Step}}\,{\bf{4}}$ Alice selects the measurement operator in ${\bf{K}}^{\prime} $ to measure the remaining quantum states in table 1, where
$\begin{eqnarray}\begin{array}{rcl}{\bf{K}}^{\prime} & \equiv & \{{K}_{12}:= | 1\rangle \langle 1{| }_{A}\otimes {{\mathbb{I}}}_{{a}_{1}}\otimes {{\mathbb{I}}}_{{a}_{2}},\\ {K}_{22} & := & | 4\rangle \langle 4{| }_{A}\otimes | 1\rangle \langle 1{| }_{{a}_{1}}\otimes | 0\rangle \langle 0{| }_{{a}_{2}},\\ {K}_{32} & := & | 0\rangle \langle 0{| }_{A}\otimes {{\mathbb{I}}}_{{a}_{1}}\otimes | 1\rangle \langle 1{| }_{{a}_{2}},\\ {K}_{42} & := & {\left(| 0\rangle \langle 0| +| 2\rangle \langle 2| +| 3\rangle \langle 3| \right)}_{A}\otimes | 1\rangle \langle 1{| }_{{a}_{1}}\otimes | 0\rangle \langle 0{| }_{{a}_{2}},\\ {K}_{52} & := & {\mathbb{I}}-{K}_{12}-{K}_{22}-{K}_{32}-{K}_{42}\}.\end{array}\end{eqnarray}$
States corresponding to the measurement operators K12, K22, K32 and K42 separately are $\begin{eqnarray*}\begin{array}{rcl}{K}_{12} & \Rightarrow & \{| 1\rangle | 1\rangle | f\rangle ,| 1\rangle | 2\rangle | f\rangle ,| {\psi }_{\mathrm{1,1}}\rangle | 4\rangle ,| {\psi }_{\mathrm{1,2}}\rangle | 4\rangle \},\\ f & = & 1,2,3.\\ {K}_{22} & \Rightarrow & \{| 4\rangle | {\phi }_{4,{t}_{2}}\rangle \},{t}_{2}=0,1.\\ {K}_{32} & \Rightarrow & \{| 0{\rangle }_{{\rm{A}}}[| 0{\rangle }_{{\rm{B}}}\otimes | 00{\rangle }_{{a}_{1}{b}_{1}}+{\omega }^{{t}_{1}}| 1\rangle \\ & & {+\,{\omega }^{2{t}_{1}}| 2\rangle }_{{\rm{B}}}\otimes | 11\rangle {a}_{1}{b}_{1}]| 4{\rangle }_{{\rm{C}}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},| 0\rangle | h\rangle | f\rangle \},\\ h & = & 0,1,2;\,f\,=\,1,2,3;\,{t}_{1}\,=\,0,1,2;\,\omega ={{\rm{e}}}^{\tfrac{2\pi {\rm{i}}}{3}}.\\ {K}_{42} & \Rightarrow & \{| 0\rangle | 1\rangle | 0\rangle ,| 0\rangle | 2\rangle | 0\rangle ,\\ & & | 2\rangle | 1\rangle | 0\rangle ,| 2\rangle | 2\rangle | 0\rangle ,| 3\rangle | 1\rangle | 0\rangle ,| 3\rangle | 2\rangle | 0\rangle \}.\end{array}\end{eqnarray*}$
The states in each of these sets are LOCC distinguishable. If Alice selects the measurement operator K52, the state is one of the remaining states. ${\bf{Step}}\,{\bf{5}}$ Bob selects the measurement operator in ${\bf{M}}^{\prime\prime} $ to measure the remaining quantum states in table 1, where
$\begin{eqnarray}\begin{array}{rcl}{\bf{M}}^{\prime\prime} & \equiv & \{{M}_{13}:= | 2\rangle \langle 2{| }_{{\rm{B}}}\otimes {{\unicode{120128}}}_{{b}_{1}},\\ \,{M}_{23} & := & {\left(| 0\rangle \langle 0| +| 1\rangle \langle 1| \right)}_{{\rm{B}}}\otimes {{\unicode{120128}}}_{{b}_{1}},\,{M}_{33}:= {\unicode{120128}}-{M}_{13}-{M}_{23}\}.\end{array}\end{eqnarray}$
States corresponding to the measurement operators M13 and M23 are $\begin{eqnarray*}\begin{array}{rcl}{M}_{13} & \Rightarrow & \{| 4\rangle | 2\rangle | 4\rangle ,| 4\rangle | {\phi }_{3,{t}_{1}}\rangle \},\,{t}_{1}=0,1,2.\\ {M}_{23} & \Rightarrow & \{| 0\rangle | 0\rangle | 0\rangle ,| 4\rangle | 0\rangle | 4\rangle ,| 4\rangle | 1\rangle | 4\rangle ,| 4\rangle | {\psi }_{\mathrm{1,1}}\rangle ,| 4\rangle | {\psi }_{\mathrm{1,2}}\rangle ,\\ & & | 4{\rangle }_{\rm{A}}| 0{\rangle }_{{\rm{B}}}[| 0{\rangle }_{{\rm{C}}}\otimes | 00{\rangle }_{{a}_{2}{c}_{1}}{\left({\omega }^{{t}_{1}}| 1\rangle +{\omega }^{2{t}_{1}}| 2\rangle \right)}_{{\rm{C}}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}}]\\ & & \otimes | 00{\rangle }_{{a}_{1}{b}_{1}},\\ & & | 4{\rangle }_{{\rm{A}}}[| 0{\rangle }_{{\rm{B}}}\otimes | 00{\rangle }_{{a}_{1}{b}_{1}}\pm | 1{\rangle }_{{\rm{B}}}\otimes | 11{\rangle }_{{a}_{1}{b}_{1}}]| 3{\rangle }_{{\rm{C}}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}}\},\\ & & \omega ={{\rm{e}}}^{\tfrac{2\pi {\rm{i}}}{3}};{t}_{1}\,=\,0,1,2.\end{array}\end{eqnarray*}$
The states in each of these sets are LOCC distinguishable. In ${\bf{Step}}\,{\bf{1}}$, if Bob selects M1 and Charlie selects ${\overline{N}}_{1}$, or other situations, there will also be a similar protocol. □
Table 1. Post-measurement states after Step 1. |
| Post-measurement states | Range |
|---|---|
| $| i\rangle | 0\rangle | 0\rangle \otimes | 00{\rangle }_{{a}_{1}{b}_{1}}\otimes | 00{\rangle }_{{a}_{2}{c}_{1}}.$ | i = 0, 1, 2, 3. |
| | |
| $| h\rangle | 0\rangle | f\rangle \otimes | 00{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},\,\,\,| 3\rangle | 0\rangle | m\rangle \otimes | 00{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},$ | h = 0, 1, 2; f = 1, 2, 3; |
| $| 4\rangle | 0\rangle | 4\rangle \otimes | 00{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},\,\,\,\,| {\phi }_{4,{t}_{2}}\rangle | 4\rangle \otimes | 00{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}}.$ | m = 1, 2, 3, 4; t2 = 0, 1. |
| | |
| $| g\rangle | m\rangle | 0\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 00{\rangle }_{{a}_{2}{c}_{1}},\,\,| 2\rangle | f\rangle | 0\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 00{\rangle }_{{a}_{2}{c}_{1}},$ | g, t2 = 0, 1; |
| $| 3\rangle | f\rangle | 0\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 00{\rangle }_{{a}_{2}{c}_{1}},\,\,\,| 4\rangle | 3\rangle | 0\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 00{\rangle }_{{a}_{2}{c}_{1}},$ | m = 1, 2, 3, 4; |
| $| 4\rangle | {\phi }_{4,{t}_{2}}\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 00{\rangle }_{{a}_{2}{c}_{1}},\,\,\,| \tau {\rangle }_{{t}_{1}}| 4\rangle | 0\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 00{\rangle }_{{a}_{2}{c}_{1}}.$ | f = 1, 2, 3; t1 = 0, 1, 2. |
| | |
| $| h\rangle | f\rangle | f^{\prime} \rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},\,\,| 0\rangle | 4\rangle | m\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},$ | |
| $| 1\rangle | 4\rangle | 1\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},\,\,\,\,| 1\rangle | 4\rangle | \tau {\rangle }_{{t}_{1}}\otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},$ | |
| $| 2\rangle | 4\rangle | m\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},\,\,\,| 3\rangle | 1\rangle | m\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},$ | |
| $| 3\rangle | 2\rangle | 1\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},\,\,\,\,| 3\rangle | \tau {\rangle }_{{t}_{1}}| 2\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},$ | h, t1 = 0, 1, 2; $f,f^{\prime} =1,2,3;$ |
| $| 3\rangle | 2\rangle | 3\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},\,\,\,\,| 3\rangle | 2\rangle | 4\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},$ | m = 1, 2, 3, 4; |
| $| 3\rangle | 3\rangle | l\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},\,\,\,\,\,| 3\rangle | 4\rangle | l\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},$ | l = 1, 3, 4; |
| $| 4\rangle | 1\rangle | 4\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},\,\,\,\,| 4\rangle | 2\rangle | 4\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},$ | t2 = 0, 1. |
| $| 4\rangle | 3\rangle | m\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},\,\,\,| 4\rangle | 4\rangle | m\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},$ | |
| $| 4\rangle | {\phi }_{3,{t}_{1}}\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},\,\,\,\,| 4\rangle | {\psi }_{\mathrm{1,1}}\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},$ | |
| $| 4\rangle | {\psi }_{\mathrm{1,2}}\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},\,\,\,\,\,| {\phi }_{2,{t}_{2}}\rangle | 4\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},$ | |
| $| {\phi }_{3,{t}_{1}}\rangle | 4\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},\,\,\,\,| {\psi }_{\mathrm{1,1}}\rangle | 4\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},$ | |
| $| {\psi }_{\mathrm{1,2}}\rangle | 4\rangle \otimes | 11{\rangle }_{{a}_{1}{b}_{1}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}}.$ | |
| | |
| $| 4{\rangle }_{{\rm{A}}}| 0{\rangle }_{{\rm{B}}}[| 0{\rangle }_{{\rm{C}}}\otimes | 00{\rangle }_{{a}_{2}{c}_{1}}+{\left({\omega }^{{t}_{1}}| 1\rangle +{\omega }^{2{t}_{1}}| 2\rangle \right)}_{{\rm{C}}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}}]\otimes | 00{\rangle }_{{a}_{1}{b}_{1}},$ | |
| $| 4{\rangle }_{{\rm{A}}}[| 0{\rangle }_{{\rm{B}}}\otimes | 00{\rangle }_{{a}_{1}{b}_{1}}\pm | 1{\rangle }_{{\rm{B}}}\otimes | 11{\rangle }_{{a}_{1}{b}_{1}}]| 3{\rangle }_{{\rm{C}}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}},$ | $\omega ={{\rm{e}}}^{\tfrac{2\pi {\rm{i}}}{3}}$; |
| $| 0{\rangle }_{{\rm{A}}}[| 0{\rangle }_{{\rm{B}}}\otimes | 00{\rangle }_{{a}_{1}{b}_{1}}+{\left({\omega }^{{t}_{1}}| 1\rangle +{\omega }^{2{t}_{1}}| 2\rangle \right)}_{{\rm{B}}}\otimes | 11{\rangle }_{{a}_{1}{b}_{1}}]| 4{\rangle }_{{\rm{C}}}\otimes | 11{\rangle }_{{a}_{2}{c}_{1}}.$ | t1 = 0, 1, 2. |
Thus, we have demonstrated that two EPR states serving as resources can perfectly distinguish BII(5, 3). Next, we proceed to prove that only one GHZ state is also sufficient for the perfect discrimination of BII(5, 3).
Theorem 2. Suppose Alice, Bob and Charlie share the three-qubit maximally entangled state $\tfrac{1}{\sqrt{2}}{\left(| 000\rangle +| 111\rangle \right)}_{{abc}}$, then the set ${{\bf{B}}}_{\mathrm{II}}(5,3)$ in (9) can be perfectly distinguished by LOCC.
Proof. For the entangled state $\tfrac{1}{\sqrt{2}}{\left(| 000\rangle +| 111\rangle \right)}_{{abc}}$, assume that Alice takes the a particle, Bob takes the b particle and Charlie takes the c particle. The initial state is
$\begin{eqnarray}| \xi \rangle =| \psi {\rangle }_{\mathrm{ABC}}\otimes \displaystyle \frac{1}{\sqrt{2}}{\left(| 000\rangle +| 111\rangle \right)}_{{abc}},\end{eqnarray}$
where $| \psi {\rangle }_{\mathrm{ABC}}$ is one of the states from the set ${{\bf{B}}}_{\mathrm{II}}(5,3)$. Then, Alice, Bob and Charlie perform nontrivial orthogonality-preserving measurement separately; the specific procedure is as follows.
${\bf{Step}}\,{\bf{1}}$ Bob selects the measurement operator in ${\bf{M}}$ to measure the initial state $| \xi \rangle $, where
$\begin{eqnarray}\begin{array}{rcl} & & {\bf{M}}\equiv \{{M}_{1}:= | 0\rangle \langle 0{| }_{{\rm{B}}}\otimes | 0\rangle \langle 0{| }_{b}+(| 1\rangle \langle 1| +\\ & & | {2\rangle \langle 2| +| 3\rangle \langle 3| +| 4\rangle \langle 4| )}_{{\rm{B}}}\\ & & \otimes | 1\rangle \langle 1{| }_{b},\,{\bar{M}}_{1}:= {\unicode{120128}}-{M}_{1}\}.\end{array}\end{eqnarray}$
Suppose Bob selects the measurement operators M1, then the post-measurement states are respectively shown in table 2. ${\bf{Step}}\,{\bf{2}}$ Alice selects the measurement operator in ${\bf{K}}$ to measure the post-measurement states in table 2, where
$\begin{eqnarray}\begin{array}{rcl} & & {\bf{K}}\equiv \{{K}_{1}:= | 3\rangle \langle 3{| }_{{\rm{A}}}\otimes | 0\rangle \langle 0{| }_{a},\,{K}_{2}:= {\left(| 1\rangle \langle 1| +| 2\rangle \langle 2| \right)}_{{\rm{A}}}\\ & & \,\otimes | 0\rangle \langle 0{| }_{a},\,{K}_{3}:= {\unicode{120128}}-{K}_{1}-{K}_{2}\}.\end{array}\end{eqnarray}$
States corresponding to the measurement operators K1 and K2 separately are $\begin{eqnarray*}\begin{array}{rcl}{K}_{1} & \Rightarrow & \{| 3\rangle | 0\rangle | k\rangle \},k=0,1,2,3,4.\\ {K}_{2} & \Rightarrow & \{| 1\rangle | 0\rangle | j\rangle ,| 2\rangle | 0\rangle | j\rangle ,| {\phi }_{4,{t}_{2}}\rangle | 4\rangle \},j=0,1,2,3;{t}_{2}=0,1.\end{array}\end{eqnarray*}$
The states in each of these sets are LOCC distinguishable. If Alice selects the operator K3, the state is one of the remaining states. After Alice's measurement, we are able to filter out certain quantum states that are locally distinguishable. The remaining quantum states are locally indistinguishable and remain unchanged, allowing for further measurements to be conducted on them. This holds true for every subsequent step.
${\bf{Step}}\,{\bf{3}}$ Charlie selects the measurement operator in ${\bf{N}}$ to measure the remaining quantum states in table 2, where
$\begin{eqnarray}\begin{array}{l}{\bf{N}}\equiv \{{N}_{1}:= {\left(| 0\rangle \langle 0| +| 1\rangle \langle 1| +| 2\rangle \langle 2| \right)}_{{\rm{C}}}\\ \,\otimes \,| 0\rangle \langle 0{| }_{c},\,{N}_{2}:= | 0\rangle \langle 0{| }_{{\rm{C}}}\\ \,\otimes \,| 1\rangle \langle 1{| }_{c},\,{N}_{3}:= {\unicode{120128}}-{N}_{1}-{N}_{2}\}.\end{array}\end{eqnarray}$
States corresponding to the measurement operators N1 and N2 separately are $\begin{eqnarray*}\begin{array}{rcl}{N}_{1} & \Rightarrow & \{| 0\rangle | 0\rangle | h\rangle ,| 4\rangle | {\phi }_{1,{t}_{1}}\rangle \},h=0,1,2;{t}_{1}=0,1,2.\\ {N}_{2} & \Rightarrow & \{| g\rangle | m\rangle | 0\rangle ,| 2\rangle | f\rangle | 0\rangle ,| 3\rangle | f\rangle | 0\rangle ,\\ & & \,| 4\rangle | 3\rangle | 0\rangle ,| 4\rangle | {\phi }_{4,{t}_{2}}\rangle ,| \tau {\rangle }_{{t}_{1}}| 4\rangle | 0\rangle \},\\ g & = & 0,1;m=1,2,3,4;f=1,2,3;\\ {t}_{1} & = & 0,1,2;{t}_{2}=0,1.\end{array}\end{eqnarray*}$
The states in each of these sets are LOCC distinguishable. If Bob selects the operator N3, the state is one of the remaining states. ${\bf{Step}}\,{\bf{4}}$ Alice selects the measurement operator in ${\bf{K}}^{\prime} $ to measure the remaining quantum states in table 2, where
$\begin{eqnarray}\begin{array}{l}{\bf{K}}^{\prime} \equiv \{{K}_{12}:= | 3\rangle \langle 3{| }_{{\rm{A}}}\otimes | 1\rangle \langle 1{| }_{a},\\ {K}_{22}\,:= \,| 2\rangle \langle 2{| }_{{\rm{A}}}\otimes {{\unicode{120128}}}_{a},\,{K}_{32}:= {\unicode{120128}}-{K}_{12}-{K}_{22}\}.\end{array}\end{eqnarray}$
States corresponding to the measurement operators K12 and K22 separately are $\begin{eqnarray*}\begin{array}{rcl}{K}_{12} & \Rightarrow & \{| 3\rangle | 1\rangle | m\rangle ,| 3\rangle | 2\rangle | l\rangle ,| 3\rangle | 3\rangle | l\rangle ,| 3\rangle | 4\rangle | l\rangle ,| 3\rangle | \tau {\rangle }_{{t}_{1}}| 2\rangle \},\\ l & = & 1,3,4;{t}_{1}=0,1,2.\\ {K}_{22} & \Rightarrow & \{| 2\rangle | f\rangle | f^{\prime} \rangle ,| 2\rangle | 4\rangle | m\rangle ,| {\phi }_{3,{t}_{1}}\rangle | 4\rangle \},\\ f,f^{\prime} & = & 1,2,3;m=1,2,3,4;{t}_{1}=0,1,2.\end{array}\end{eqnarray*}$
The states in each of these sets are LOCC distinguishable. If Alice selects the operator K32, the state is one of the remaining states. ${\bf{Step}}\,{\bf{5}}$ Bob selects the measurement operator in ${\bf{M}}^{\prime} $ to measure the remaining quantum states in table 2, where
$\begin{eqnarray}\begin{array}{l}{\bf{M}}^{\prime} \equiv \{{M}_{12}:= | 4\rangle \langle 4{| }_{{\rm{B}}}\otimes {{\unicode{120128}}}_{b},\,{M}_{22}:= | 3\rangle \langle 3{| }_{{\rm{B}}}\\ \,\ \ \otimes {{\unicode{120128}}}_{b},\,{M}_{32}:= {\unicode{120128}}-{M}_{12}-{M}_{22}\}.\end{array}\end{eqnarray}$
States corresponding to the measurement operators M12 and M22 separately are $\begin{eqnarray*}\begin{array}{rcl}{M}_{12} & \Rightarrow & \{| 0\rangle | 4\rangle | m\rangle ,| 1\rangle | 4\rangle | 1\rangle ,| 4\rangle | 4\rangle | m\rangle ,| 1\rangle | 4\rangle | \tau {\rangle }_{{t}_{1}}\},\\ m & = & 1,2,3,4;{t}_{1}=0,1,2.\\ {M}_{22} & \Rightarrow & \{| h\rangle | 3\rangle | f\rangle ,| 4\rangle | 3\rangle | m\rangle ,| {\phi }_{2,{t}_{2}}\rangle | 4\rangle \},\\ h & = & 0,1,2;f=1,2,3;m=1,2,3,4;{t}_{2}=0,1.\end{array}\end{eqnarray*}$
The states in each of these sets are LOCC distinguishable. If Bob selects the operator M32, the state is one of the remaining states. ${\bf{Step}}\,{\bf{6}}$ Alice selects the measurement operator in ${\bf{K}}^{\prime\prime} $ to measure the remaining quantum states in table 2, where
$\begin{eqnarray}{\bf{K}}^{\prime\prime} \equiv \{{K}_{13}:= | 1\rangle \langle 1{| }_{{\rm{A}}}\otimes {{\unicode{120128}}}_{a},\,{K}_{23}:= {\unicode{120128}}-{K}_{13}\}.\end{eqnarray}$
States corresponding to the measurement operators K13 are $\begin{eqnarray*}\begin{array}{rcl}{K}_{13} & \Rightarrow & \{| 1\rangle | 1\rangle | f\rangle ,| 1\rangle | 2\rangle | f\rangle ,| {\psi }_{\mathrm{1,1}}\rangle | 4\rangle ,| {\psi }_{\mathrm{1,2}}\rangle | 4\rangle \},\\ f & = & 1,2,3;\,{t}_{1}=0,1,2.\end{array}\end{eqnarray*}$
The states in each of these sets are LOCC distinguishable. If Alice selects the operator K23, the state is one of the remaining states. ${\bf{Step}}\,{\bf{7}}$ Charlie selects the measurement operator in ${\bf{N}}^{\prime} $ to measure the remaining quantum states in table 2, where
$\begin{eqnarray}\begin{array}{l}{\bf{N}}^{\prime} \equiv \{{N}_{12}:= {\left(| 1\rangle \langle 1+| 2\rangle \langle 2| +| 3\rangle \langle 3| \right)}_{{\rm{C}}}\\ \,\otimes {{\unicode{120128}}}_{c},\,{N}_{22}:= | 4\rangle \langle 4{| }_{{\rm{C}}}\otimes {{\unicode{120128}}}_{c}\}.\end{array}\end{eqnarray}$
States corresponding to the measurement operators N1 and N2 separately are $\begin{eqnarray*}\begin{array}{l}{N}_{12}\Rightarrow \{| 0\rangle | 1\rangle | f\rangle ,| 0\rangle | 2\rangle | f\rangle ,| 0\rangle | 0\rangle | 3\rangle ,| 4\rangle | {\phi }_{3,{t}_{1}}\rangle ,\\ \,\ \ \,| 4\rangle | {\psi }_{\mathrm{1,1}}\rangle ,| 4\rangle | {\psi }_{\mathrm{1,2}}\rangle ,\\ | 4{\rangle }_{{\rm{A}}}(| 0{\rangle }_{{\rm{B}}}\otimes | 000{\rangle }_{{abc}}\pm | 1{\rangle }_{{\rm{B}}}\otimes | 111{\rangle }_{{abc}})| 3{\rangle }_{{\rm{C}}}\},\\ \,\ \ \,f=1,2,3;{t}_{1}=0,1,2.\\ {N}_{22}\Rightarrow \{| 4\rangle | h\rangle | 4\rangle ,| 0{\rangle }_{{\rm{A}}}[| 0{\rangle }_{{\rm{B}}}\otimes | 000{\rangle }_{{abc}}\\ \,\ \ +\,{\left({\omega }^{{t}_{1}}| 1\rangle +{\omega }^{2{t}_{1}}| 2\rangle \right)}_{{\rm{B}}}\otimes | 111{\rangle }_{{abc}}]| 4{\rangle }_{{\rm{C}}}\},\\ \ \ \ \omega ={{\rm{e}}}^{\tfrac{2\pi {\rm{i}}}{3}};h=0,1,2;{t}_{1}\,=\,0,1,2.\end{array}\end{eqnarray*}$
The states in each of these sets are LOCC distinguishable. In ${\bf{Step}}\,{\bf{1}}$, if Bob selects ${\overline{M}}_{1}$, there will also be a similar protocol. □
Table 2. Post-measurement states after Step 1. |
| Post-measurement States | Range |
|---|---|
| ∣h⟩∣0⟩∣j⟩ ⨂ ∣000⟩abc, ∣3⟩∣0⟩∣k⟩ ⨂ ∣000⟩abc, | h, t1 = 0, 1, 2; j = 0, 1, 2, 3; |
| $| 4\rangle | 0\rangle | 4\rangle \otimes | 000{\rangle }_{{abc}},\,\,\,| 4\rangle | {\phi }_{1,{t}_{1}}\rangle \otimes | 000{\rangle }_{{abc}},$ | k = 0, 1, 2, 3, 4; t2 = 0, 1. |
| $| {\phi }_{4,{t}_{2}}\rangle | 4\rangle \otimes | 000{\rangle }_{{abc}}.$ | |
| | |
| ∣h⟩∣f⟩∣j⟩ ⨂ ∣111⟩abc, ∣0⟩∣4⟩∣k⟩ ⨂ ∣111⟩abc, | |
| $| 1\rangle | 4\rangle | g\rangle \otimes | 111{\rangle }_{{abc}},\,\,\,\,| 1\rangle | 4\rangle | \tau {\rangle }_{{t}_{1}}\otimes | 111{\rangle }_{{abc}},$ | |
| $| \tau {\rangle }_{{t}_{1}}| 4\rangle | 0\rangle \otimes | 111{\rangle }_{{abc}},\,\,| 2\rangle | 4\rangle | m\rangle \otimes | 111{\rangle }_{{abc}},$ | |
| ∣3⟩∣1⟩∣k⟩ ⨂ ∣111⟩abc, ∣3⟩∣2⟩∣n⟩ ⨂ ∣111⟩abc, | h, t1 = 0, 1, 2; |
| ∣3⟩∣3⟩∣n⟩ ⨂ ∣111⟩abc, ∣3⟩∣4⟩∣1⟩ ⨂ ∣111⟩abc, | f = 1, 2, 3; |
| ∣3⟩∣4⟩∣3⟩ ⨂ ∣111⟩abc, ∣3⟩∣4⟩∣4⟩ ⨂ ∣111⟩abc, | j = 0, 1, 2, 3; |
| $| 3\rangle | \tau {\rangle }_{{t}_{1}}| 2\rangle \otimes | 111{\rangle }_{{abc}},\,\,| 4\rangle | 1\rangle | 4\rangle \otimes | 111{\rangle }_{{abc}},$ | k = 0, 1, 2, 3, 4; |
| ∣4⟩∣2⟩∣4⟩ ⨂ ∣111⟩abc, ∣4⟩∣3⟩∣k⟩ ⨂ ∣111⟩abc, | g, t2 = 0, 1; |
| $| 4\rangle | 4\rangle | m\rangle \otimes | 111{\rangle }_{{abc}},\,\,\,| 4\rangle | {\phi }_{3,{t}_{1}}\rangle \otimes | 111{\rangle }_{{abc}},$ | m = 1, 2, 3, 4; |
| $| 4\rangle | {\phi }_{4,{t}_{2}}\rangle \otimes | 111{\rangle }_{{abc}},\,\,\,\,| 4\rangle | {\psi }_{\mathrm{1,1}}\rangle \otimes | 111{\rangle }_{{abc}},$ | n = 0, 1, 3, 4. |
| $| 4\rangle | {\psi }_{\mathrm{1,2}}\rangle \otimes | 111{\rangle }_{{abc}},\,\,\,\,\,| {\phi }_{2,{t}_{2}}\rangle | 4\rangle \otimes | 111{\rangle }_{{abc}},$ | |
| $| {\phi }_{3,{t}_{1}}\rangle | 4\rangle \otimes | 111{\rangle }_{{abc}},\,\,\,\,| {\psi }_{\mathrm{1,1}}\rangle | 4\rangle \otimes | 111{\rangle }_{{abc}},$ | |
| ∣ψ1,2⟩∣4⟩ ⨂ ∣111⟩abc. | |
| | |
| ∣4⟩A(∣0⟩B ⨂ ∣000⟩abc ± ∣1⟩B ⨂ ∣111⟩abc)∣3⟩C, | $\omega ={{\rm{e}}}^{\tfrac{2\pi {\rm{i}}}{3}};$ |
| $| 0{\rangle }_{{\rm{A}}}[| 0{\rangle }_{{\rm{B}}}\otimes | 000{\rangle }_{{abc}}+{\left({\omega }^{{t}_{1}}| 1\rangle +{\omega }^{2{t}_{1}}| 2\rangle \right)}_{{\rm{B}}}\otimes | 111{\rangle }_{{abc}}]| 4{\rangle }_{{\rm{C}}}.$ | t1 = 0, 1, 2. |
5. Conclusion
Distinguishability of quantum states is an interesting question in quantum information processing. In this paper, we study the local discriminability of genuinely nonlocal sets with entanglement. The methodology proposed by Rout et al [35], utilizing nonlocal product states in ${{\mathbb{C}}}^{3}\otimes {{\mathbb{C}}}^{3}$ to construct A genuinely nonlocal product basis in ${{\mathbb{C}}}^{4}\otimes {{\mathbb{C}}}^{4}\otimes {{\mathbb{C}}}^{4}$, inspired us to construct such A basis in ${{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}$ using nonlocal product states in ${{\mathbb{C}}}^{3}\otimes {{\mathbb{C}}}^{4}$.
We firstly give a set of nonlocal product states in ${{\mathbb{C}}}^{3}\otimes {{\mathbb{C}}}^{4}$, and let the states in that set make tensor products with ∣4⟩ respectively. Then based on the completeness of the basis, we construct a genuinely nonlocal product basis for type I in ${{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}$. Besides, we obtain a genuinely nonlocal product basis for type II by replacing some states in type I with some superposition states. In the end, we perfectly distinguished the genuinely nonlocal product basis for type II we constructed with two Bell states and one GHZ state, respectively. In distinguishing a genuinely nonlocal product basis in ${{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}\otimes {{\mathbb{C}}}^{5}$, protocols based on quantum teleportation consume 2 log2 5 ebits of entanglements. Our protocols consume 2 or 1 ($\lt 2{\mathrm{log}}_{2}5$) ebits of entanglements, so our protocols are more efficient than quantum teleportation.