1. Introduction
Quantum mechanics is one of the most important inventions in the field of physics in the 20th century [1, 2]. As a crucial research object of quantum mechanics, quantum information has very great significance. In quantum information theory, the evolution in a closed quantum system is unitary and linear [3]. Therefore, unlike the classical world, there are some no-go theorems in the quantum world, such as the no-deleting theorem [4–6], the no-cloning theorem [7–9], the no-broadcasting theorem [10–12] and so on. As one of the no-go theorems, the no-masking theorem [13] plays a vital role in the development of quantum information. Quantum information masking requires that the information contained in subsystems can be encoded into their correlation by some unitary operations, such that the marginal states are identical [14]. In other words, one can hide the initial information contained in the subsystems [15]. This principle greatly improves the security of quantum communication [16, 17]. As a consequence, the masking of quantum information can be used for quantum secret sharing [18–20]. Besides, it also plays a major role in quantum teleportation [21, 22] and key distribution [23]. Therefore, it is necessary to study the quantum states whether can be masked. At present, the research method for this problem is mainly based on the definition of quantum information masking.
In recent works, many scholars have come up with a lot of new discoveries about quantum masking. Modi et al first proposed the no-masking theorem, namely, a unitary operator cannot mask any pure state [13]. Based on this theorem, Ghosh et al considered two special kinds of quantum states to analyze the masking condition [24]. Li and Wang showed that tripartite quantum systems can mask arbitrary quantum pure states [25]. In addition, the relationship between the maskable state set and the hyperdisk was studied by Ding et al [26]. Li and Modi certified that probabilistic universal masking is impossible and derived a necessary condition to make the ε-approximate universal masking possible [27]. Lie et al demonstrated that the maximum number of qubits that can be masked is related to the entropy of the quantum state [28]. Moreover, using time-correlated photons, Liu et al designed a quantum information masking machine [29]. These above papers are important for the development of quantum information masking.
Here, we mainly study the masking of quantum information in two-dimensional Hilbert space. We express the masking conditions by the coefficients of quantum states. As a result, we obtain a system of equations as the masking conditions. Afterward, by observing some concrete maskable non-orthogonal quantum states, we find a common property of these states, i.e. they have the same number of terms and the same basis. Eventually, we analyze two special kinds of quantum states with an orthogonal basis and draw some images for an intuitive description of these examples.
2. No-masking theorem conditions
In this section, we review the definition of masking [13] and discuss the general masking condition of quantum information in ${{\mathbb{C}}}^{2}$.
Let ${{ \mathcal H }}_{X}$ be an n-dimensional Hilbert space associated with the system X. Suppose the states ∣ak⟩A in ${{ \mathcal H }}_{{\rm{A}}}$ contain the quantum information. We say that the quantum information contained in states ∣ak⟩A can be masked, if and only if there is an operator F that maps the states to ∣ψk⟩AB which belong to ${{ \mathcal H }}_{{\rm{A}}}\otimes {{ \mathcal H }}_{{\rm{B}}}$, such that the marginal states of ∣ψk⟩AB satisfy the following two conditions
$\begin{eqnarray*}\begin{array}{rcl}{\rho }_{{\rm{A}}} & = & {\mathrm{Tr}}_{{\rm{B}}}(| {{\rm{\Psi }}}_{k}{\rangle }_{\mathrm{AB}}\langle {{\rm{\Psi }}}_{k}| ),\\ {\rho }_{{\rm{B}}} & = & {\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{k}{\rangle }_{\mathrm{AB}}\langle {{\rm{\Psi }}}_{k}| ).\end{array}\end{eqnarray*}$
That is to say, we cannot tell what the value of k is or what information it carries by looking at the states. Since this is a physical process, we can also write it as F: ∣ak⟩A ⨂ ∣bk⟩B → ∣ψk⟩AB, where the operator F is unitary and is called the masker, $| {b}_{k}{\rangle }_{{\rm{B}}}\in {{ \mathcal H }}_{{\rm{B}}}$. Moreover, in order to complete the difference between the dimensionality of two systems, there is a unitary operator U acting on systems A and B.
Assume that the quantum information in the state ∣b⟩ = α0∣0⟩ + α1∣1⟩ can be masked. Then there exist two quantum states ∣ψ0⟩ and ∣ψ1⟩ which belong to ${{ \mathcal H }}_{{\rm{A}}}\otimes {{ \mathcal H }}_{{\rm{B}}}$, such that
$\begin{eqnarray*}\begin{array}{rcl}| b\rangle & = & {\alpha }_{0}| 0\rangle +{\alpha }_{1}| 1\rangle \to | {\rm{\Psi }}\rangle \\ & = & {\alpha }_{0}| {{\rm{\Psi }}}_{0}\rangle +{\alpha }_{1}| {{\rm{\Psi }}}_{1}\rangle ,\end{array}\end{eqnarray*}$
where ∣α0∣2 + ∣α1∣2 = 1.
For the subsystems A and B, we take the partial trace respectively, therefore, we get
$\begin{eqnarray}\begin{array}{rcl}{\rho }_{{\rm{B}}} & = & {\mathrm{Tr}}_{{\rm{A}}}(| {\rm{\Psi }}\rangle \langle {\rm{\Psi }}| )=| {\alpha }_{0}{| }^{2}{\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{0}| )\\ & & +| {\alpha }_{1}{| }^{2}{\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{1}| )\\ & & +{\alpha }_{0}{\alpha }_{1}^{* }{\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{1}| )\\ & & +{\alpha }_{0}^{* }{\alpha }_{1}{\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{0}| ),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{\rho }_{{\rm{A}}} & = & {\mathrm{Tr}}_{{\rm{B}}}(| {\rm{\Psi }}\rangle \langle {\rm{\Psi }}| )=| {\alpha }_{0}{| }^{2}{\mathrm{Tr}}_{{\rm{B}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{0}| )\\ & & +| {\alpha }_{1}{| }^{2}{\mathrm{Tr}}_{{\rm{B}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{1}| )\\ & & +{\alpha }_{0}{\alpha }_{1}^{* }{\mathrm{Tr}}_{{\rm{B}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{1}| )\\ & & +{\alpha }_{0}^{* }{\alpha }_{1}{\mathrm{Tr}}_{{\rm{B}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{0}| ).\end{array}\end{eqnarray}$
By the definition of masking, the masking conditions are
$\begin{eqnarray*}\begin{array}{rcl}{\rho }_{{\rm{A}}} & = & {\mathrm{Tr}}_{{\rm{B}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{0}| )={\mathrm{Tr}}_{{\rm{B}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{1}| )\\ & = & {\mathrm{Tr}}_{{\rm{B}}}(| {\rm{\Psi }}\rangle \langle {\rm{\Psi }}| ),\\ {\rho }_{{\rm{B}}} & = & {\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{0}| )={\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{1}| )\\ & = & {\mathrm{Tr}}_{{\rm{A}}}(| {\rm{\Psi }}\rangle \langle {\rm{\Psi }}| ).\end{array}\end{eqnarray*}$
In order to fulfill the above masking conditions, the cross-terms in equations (1 ) and (2 ) must vanish, i.e.
$\begin{eqnarray}\begin{array}{l}{\alpha }_{0}{\alpha }_{1}^{* }{\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{1}| )\\ \quad +\,{\alpha }_{0}^{* }{\alpha }_{1}{\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{0}| )=0,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{\alpha }_{0}{\alpha }_{1}^{* }{\mathrm{Tr}}_{{\rm{B}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{1}| )\\ \quad +\,{\alpha }_{0}^{* }{\alpha }_{1}{\mathrm{Tr}}_{{\rm{B}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{0}| )=0.\end{array}\end{eqnarray}$
In this paper, we mainly discuss the masking of quantum states in two-dimensional Hilbert space. Without loss of generality, suppose that
$\begin{eqnarray*}\begin{array}{rcl}| {{\rm{\Psi }}}_{0}\rangle & = & {a}_{0}| 00\rangle +{a}_{1}| 01\rangle +{a}_{2}| 10\rangle +{a}_{3}| 11\rangle ,\\ | {{\rm{\Psi }}}_{1}\rangle & = & {b}_{0}| 00\rangle +{b}_{1}| 01\rangle +{b}_{2}| 10\rangle +{b}_{3}| 11\rangle ,\end{array}\end{eqnarray*}$
where ${\sum }_{i=0}^{3}| {a}_{i}{| }^{2}={\sum }_{i=0}^{3}| {b}_{i}{| }^{2}=1$.
When ${\mathrm{Tr}}_{x}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{0}| )={\mathrm{Tr}}_{x}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{1}| )$, where x ∈ {A, B}, the coefficients of ∣ψ0⟩ and ∣ψ1⟩ should meet the following system of equations
$\begin{eqnarray}\left\{\begin{array}{l}| {a}_{0}{| }^{2}+| {a}_{1}{| }^{2}-| {b}_{0}{| }^{2}-| {b}_{1}{| }^{2}=0,\\ | {a}_{0}{| }^{2}+| {a}_{2}{| }^{2}-| {b}_{0}{| }^{2}-| {b}_{2}{| }^{2}=0,\\ | {a}_{1}{| }^{2}+| {a}_{3}{| }^{2}-| {b}_{1}{| }^{2}-| {b}_{3}{| }^{2}=0,\\ | {a}_{2}{| }^{2}+| {a}_{3}{| }^{2}-| {b}_{2}{| }^{2}-| {b}_{3}{| }^{2}=0,\\ {a}_{0}{a}_{1}^{* }+{a}_{2}{a}_{3}^{* }-{b}_{0}{b}_{1}^{* }-{b}_{2}{b}_{3}^{* }=0,\\ {a}_{0}{a}_{2}^{* }+{a}_{1}{a}_{3}^{* }-{b}_{0}{b}_{2}^{* }-{b}_{1}{b}_{3}^{* }=0.\end{array}\right.\end{eqnarray}$
In addition, we take the partial trace with respect to A for ∣ψ0⟩⟨ψ1∣ and ∣ψ1⟩⟨ψ0∣, then we gain
$\begin{eqnarray}\begin{array}{l}{\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{1}| )=({a}_{0}{b}_{0}^{* }+{a}_{2}{b}_{2}^{* })\\ | 0\rangle \langle 0| +({a}_{0}{b}_{1}^{* }+{a}_{2}{b}_{3}^{* })| 0\rangle \langle 1| \\ +\,({a}_{1}{b}_{0}^{* }+{a}_{3}{b}_{2}^{* })| 1\rangle \langle 0| \\ +\,({a}_{1}{b}_{1}^{* }+{a}_{3}{b}_{3}^{* })| 1\rangle \langle 1| ,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{0}| )=({a}_{0}^{* }{b}_{0}+{a}_{2}^{* }{b}_{2})\\ | 0\rangle \langle 0| +({a}_{1}^{* }{b}_{0}+{a}_{3}^{* }{b}_{2})| 0\rangle \langle 1| \\ +\,({a}_{0}^{* }{b}_{1}+{a}_{2}^{* }{b}_{3})| 1\rangle \langle 0| \\ +\,({a}_{1}^{* }{b}_{1}+{a}_{3}^{* }{b}_{3})| 1\rangle \langle 1| .\end{array}\end{eqnarray}$
For the convenience of discussion, we denote
$\begin{eqnarray*}\begin{array}{rcl}M & = & {a}_{0}{b}_{0}^{* }+{a}_{2}{b}_{2}^{* },\\ N & = & {a}_{0}{b}_{1}^{* }+{a}_{2}{b}_{3}^{* },\\ P & = & {a}_{1}{b}_{0}^{* }+{a}_{3}{b}_{2}^{* },\\ Q & = & {a}_{1}^{* }{b}_{1}+{a}_{3}^{* }{b}_{3}.\end{array}\end{eqnarray*}$
Substituting equations (6 ) and (7 ) into equation (3 ), we obtain
$\begin{eqnarray*}\begin{array}{l}({\alpha }_{1}{\alpha }_{2}^{* }M+{\alpha }_{1}^{* }{\alpha }_{2}{M}^{* })| 0\rangle \langle 0| \\ +\,({\alpha }_{1}{\alpha }_{2}^{* }N+{\alpha }_{1}^{* }{\alpha }_{2}{P}^{* })| 0\rangle \langle 1| \\ +\,({\alpha }_{1}{\alpha }_{2}^{* }P+{\alpha }_{1}^{* }{\alpha }_{2}{N}^{* })| 1\rangle \langle 0| \\ +\,({\alpha }_{1}{\alpha }_{2}^{* }Q+{\alpha }_{1}^{* }{\alpha }_{2}{Q}^{* })| 1\rangle \langle 1| =0.\end{array}\end{eqnarray*}$
As a result, the conditions for the establishment of the above formula can be expressed as
$\begin{eqnarray*}\left\{\begin{array}{l}{\alpha }_{1}{\alpha }_{2}^{* }M+{\alpha }_{1}^{* }{\alpha }_{2}{M}^{* }=0,\\ {\alpha }_{1}{\alpha }_{2}^{* }N+{\alpha }_{1}^{* }{\alpha }_{2}{P}^{* }=0,\\ {\alpha }_{1}{\alpha }_{2}^{* }P+{\alpha }_{1}^{* }{\alpha }_{2}{N}^{* }=0,\\ {\alpha }_{1}{\alpha }_{2}^{* }Q+{\alpha }_{1}^{* }{\alpha }_{2}{Q}^{* }=0.\end{array}\right.\end{eqnarray*}$
This set of conditions can be further simplified as
$\begin{eqnarray}\left\{\begin{array}{l}\mathrm{Re}({\alpha }_{1}{\alpha }_{2}^{* }M)=0,\\ \mathrm{Re}({\alpha }_{1}{\alpha }_{2}^{* }Q)=0,\\ {\alpha }_{1}{\alpha }_{2}^{* }N+{\alpha }_{1}^{* }{\alpha }_{2}{P}^{* }=0.\end{array}\right.\end{eqnarray}$
Similar to the above method, for the subsystem B, we acquire that
$\begin{eqnarray}\left\{\begin{array}{l}\mathrm{Re}({\alpha }_{1}{\alpha }_{2}^{* }{M}^{{\prime} })=0,\\ \mathrm{Re}({\alpha }_{1}{\alpha }_{2}^{* }{Q}^{{\prime} })=0,\\ {\alpha }_{1}{\alpha }_{2}^{* }{N}^{{\prime} }+{\alpha }_{1}^{* }{\alpha }_{2}{P}^{{\prime} * }=0,\end{array}\right.\end{eqnarray}$
where
$\begin{eqnarray*}\begin{array}{rcl}{M}^{{\prime} } & = & {a}_{0}{b}_{0}^{* }+{a}_{1}{b}_{1}^{* },\\ {N}^{{\prime} } & = & {a}_{0}{b}_{2}^{* }+{a}_{1}{b}_{3}^{* },\\ {P}^{{\prime} } & = & {a}_{2}{b}_{0}^{* }+{a}_{3}{b}_{1}^{* },\\ {Q}^{{\prime} } & = & {a}_{2}{b}_{2}^{* }+{a}_{3}{b}_{3}^{* }.\end{array}\end{eqnarray*}$
In summary, the masking conditions can be represented by the coefficients of ∣ψ⟩, ∣ψ0⟩ and ∣ψ1⟩, i.e. when their coefficients meet the requirements of equations (5 ), (8 ) and (9 ) at the same time, a single qubit state ∣b⟩ = α0∣0⟩ + α1∣1⟩ can be masked.
Modi et al have shown that masking the quantum information of the two systems is generally impossible, but they have also stated that there are a large number of special quantum states which can be masked [13]. These quantum states which can be masked and the corresponding masker have potential applications in quantum secret sharing and quantum communication protocols. The conditions equations (5 ), (8 ) and (9 ) obtained by us can be used to judge given quantum states in ${{\mathbb{C}}}^{2}$ whether can be masked. It is also useful for subsequent applications in quantum secret sharing to find masking quantum states more quickly by the masking conditions.
3. Masking of two-qubit quantum states
Here, we consider some non-orthogonal quantum states and orthogonal quantum states respectively, then we get some corresponding conclusions as follows.
3.1. Masking of non-orthogonal quantum states
By observing some maskable non-orthogonal quantum states in ${{\mathbb{C}}}^{2}$, we obtain the result that if two non-orthogonal quantum states can mask quantum information contained in a single qubit state, they have the same number of terms and the same basis. Below, we present the analysis process of this result.
Assume that ∣ψ0⟩ and ∣ψ1⟩ are two non-orthogonal quantum states in ${{\mathbb{C}}}^{2}\otimes {{\mathbb{C}}}^{2}$, and ∣ψ⟩ = α0∣ψ0⟩ + α1∣ψ1⟩ can mask the quantum information in the state ∣b⟩ = α0∣0⟩ + α1∣1⟩, where ${\sum }_{i=0}^{1}| {\alpha }_{i}{| }^{2}=1$. Hence, we have
$\begin{eqnarray*}\begin{array}{l}{\mathrm{Tr}}_{{\rm{A}}}(| {\rm{\Psi }}\rangle \langle {\rm{\Psi }}| )={\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{0}| )\\ =\,{\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{1}| ),\\ {\mathrm{Tr}}_{{\rm{B}}}(| {\rm{\Psi }}\rangle \langle {\rm{\Psi }}| )={\mathrm{Tr}}_{{\rm{B}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{0}| )\\ =\,{\mathrm{Tr}}_{{\rm{B}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{1}| ).\end{array}\end{eqnarray*}$
We discuss four possible cases of ∣ψ0⟩ respectively to make an analysis as follows.
Case 1. When ∣ψ0⟩ consists of one term, i.e. ∣ψ0⟩ = ∣i1i2⟩, where i1, i2 ∈ {0, 1}. Assume that ∣ψ1⟩ = b∣i1i2⟩ + c∣j1j2⟩, where b ≠ 0, c ≠ 0, ∣b∣2 + ∣c∣2 = 1 and i1, i2, j1, j2 ∈ {0, 1}. The specific classification of ∣ψ0⟩, ∣ψ1⟩ and the corresponding masking situation are shown in table 1. Considering the compactness, the table lists only situations that can be masked.
Table 1. ∣ψ0⟩ consists of one term (The duplicate parts have been deleted). |
| ∣ψ0⟩ = ∣i1i2⟩ | ∣ψ1⟩ = b∣i1i2⟩ + c∣j1j2⟩ | |
|---|---|---|
| ∣i1i2⟩ | ∣i1i2⟩ | ∣j1j2⟩ |
| ∣00⟩ | ∣00⟩ | ∣00⟩ |
| ∣01⟩ | ∣01⟩ | ∣01⟩ |
| ∣10⟩ | ∣10⟩ | ∣10⟩ |
| ∣11⟩ | ∣11⟩ | ∣11⟩ |
For the cases that cannot be masked and are not listed in table 1, we take the case of i1 = j1 and i2 ≠ j2 as an example. It can be seen that
$\begin{eqnarray*}\begin{array}{l}{\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{0}| )=| {i}_{2}\rangle \langle {i}_{2}| ,\\ {\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{1}| )=| b{| }^{2}| {i}_{2}\rangle \langle {i}_{2}| +{{bc}}^{* }| {i}_{2}\rangle \langle {j}_{2}| \\ +\,{b}^{* }c| {j}_{2}\rangle \langle {i}_{2}| +| c{| }^{2}| {j}_{2}\rangle \langle {j}_{2}| .\end{array}\end{eqnarray*}$
Since b, c ≠ 0, then ${\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{0}| )\ne {\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{1}| )$, which contradicts the masking condition.
Case 2. Suppose ∣ψ0⟩ = a0∣i1i2⟩ + a1∣j1j2⟩, where a0, a1 ≠ 0, ∣a0∣2 + ∣a1∣2 = 1 and i1, i2, j1, j2 ∈ {0, 1}.
Assume ∣ψ1⟩ = b0∣i1i2⟩ + b1∣k1k2⟩ + b2∣l1l2⟩, where ${\sum }_{i=0}^{2}| {b}_{i}{| }^{2}=1$, b0, b1, b2 ≠ 0 and i1, i2, k1, k2, l1, l2 ∈ {0, 1}. By classifying and discussing the relationship among i1, i2, j1, j2, k1, k2, l1, l2, we take the partial trace of ∣ψ0⟩ and ∣ψ1⟩ respect to both system A and system B, through which we can know whether ∣ψ0⟩ and ∣ψ1⟩ can mask the single qubit state. The situations that can be masked are shown in table 2.
Table 2. ∣ψ0⟩ consists of two terms (The duplicate parts have been deleted). |
| ∣ψ0⟩ = a0∣i1i2⟩ + a1∣j1j2⟩ | ∣ψ1⟩ = b0∣i1i2⟩ + b1∣k1k2⟩ + b2∣l1l2⟩ | |||
|---|---|---|---|---|
| ∣i1i2⟩ | ∣j1j2⟩ | ∣i1i2⟩ | ∣k1k2⟩ | ∣l1l2⟩ |
| ∣00⟩ | ∣01⟩ | ∣00⟩ | ∣00⟩ | ∣01⟩ |
| ∣00⟩ | ∣10⟩ | ∣00⟩ | ∣00⟩ | ∣10⟩ |
| ∣00⟩ | ∣11⟩ | ∣00⟩ | ∣00⟩ | ∣11⟩ |
| ∣01⟩ | ∣10⟩ | ∣01⟩ | ∣01⟩ | ∣10⟩ |
| ∣01⟩ | ∣11⟩ | ∣01⟩ | ∣01⟩ | ∣11⟩ |
| ∣10⟩ | ∣11⟩ | ∣10⟩ | ∣10⟩ | ∣11⟩ |
Choose the case where ∣ψ0⟩ = a0∣00⟩ + a1∣01⟩ and ∣ψ1⟩ = b0∣00⟩ + b1∣01⟩ + b2∣10⟩ to prove that it cannot be masked.
We obtain that
$\begin{eqnarray*}\begin{array}{l}\qquad {\mathrm{Tr}}_{{\rm{B}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{0}| )=| 0\rangle \langle 0| ,\\ \qquad {\mathrm{Tr}}_{{\rm{B}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{1}| )=(| {b}_{0}{| }^{2}+| {b}_{1}{| }^{2})| 0\rangle \langle 0| \\ \,+\,{b}_{0}{b}_{2}^{* }| 0\rangle \langle 1| +{b}_{0}^{* }{b}_{2}| 1\rangle \langle 0| +| {b}_{1}{| }^{2}| 1\rangle \langle 1| ,\end{array}\end{eqnarray*}$
where b0, b1, b2 ≠ 0. Then ${\mathrm{Tr}}_{{\rm{B}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{0}| )\ne {\mathrm{Tr}}_{{\rm{B}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{1}| )$, which contradicts the condition of masking.
Case 3. When ∣ψ0⟩ is constitutive of three terms, namely, ∣ψ0⟩ = a0∣i1i2⟩ + a1∣j1j2⟩ + a2∣k1k2⟩, where a0, a1, a2 ≠ 0, ${\sum }_{i=0}^{2}| {a}_{i}{| }^{2}=1$ and i1, i2, j1, j2, k1, k2 ∈ {0, 1}.
Suppose that ∣ψ1⟩ = b0∣i1i2⟩ + b1∣l1l2⟩ + b2∣m1m2⟩ + b3∣n1n2⟩, where b0, b1, b2, b3 ≠ 0, ${\sum }_{i=0}^{3}| {b}_{i}{| }^{2}=1$ and i1, i2, l1, l2, m1, m2, n1, n2 ∈ {0, 1}. We get the table 3, similar to the above.
Table 3. ∣ψ0⟩ consists of three terms (The duplicate parts have been deleted). |
| ∣ψ0⟩ = a0∣i1i2⟩ + a1∣j1j2⟩ + a2∣k1k2⟩ | ∣ψ1⟩ = b0∣i1i2⟩ + b1∣l1l2⟩ + b2∣m1m2⟩ + b3∣n1n2⟩ | |||||
|---|---|---|---|---|---|---|
| ∣i1i2⟩ | ∣j1j2⟩ | ∣k1k2⟩ | ∣i1i2⟩ | ∣l1l2⟩ | ∣m1m2⟩ | ∣n1n2⟩ |
| ∣00⟩ | ∣01⟩ | ∣10⟩ | ∣00⟩ | ∣00⟩ | ∣01⟩ | ∣10⟩ |
| ∣00⟩ | ∣01⟩ | ∣11⟩ | ∣00⟩ | ∣00⟩ | ∣01⟩ | ∣11⟩ |
| ∣00⟩ | ∣10⟩ | ∣11⟩ | ∣00⟩ | ∣00⟩ | ∣10⟩ | ∣11⟩ |
| ∣01⟩ | ∣10⟩ | ∣11⟩ | ∣01⟩ | ∣01⟩ | ∣10⟩ | ∣11⟩ |
We choose the case where ∣ψ0⟩ = a0∣00⟩ + a1∣01⟩ + a2∣10⟩ and ∣ψ1⟩ = b0∣00⟩ + b1∣01⟩ + b2∣10⟩ + b3∣11⟩ to analyze the situation that cannot be masked. Then we have
$\begin{eqnarray*}\begin{array}{l}{\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{0}| )=(| {a}_{0}{| }^{2}+| {a}_{2}{| }^{2})| 0\rangle \langle 0| \\ +\,| {a}_{1}{| }^{2}| 1\rangle \langle 1| +{a}_{0}{a}_{1}^{* }| 0\rangle \langle 1| +{a}_{0}^{* }{a}_{1}| 1\rangle \langle 0| ,\\ {\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{1}| )=(| {b}_{0}{| }^{2}+| {b}_{2}{| }^{2})| 0\rangle \langle 0| \\ +\,(| {b}_{1}{| }^{2}+| {b}_{3}{| }^{2})| 1\rangle \langle 1| \\ +\,({b}_{0}{b}_{1}^{* }+{b}_{2}{b}_{3}^{* })| 0\rangle \langle 1| +{b}_{0}^{* }{b}_{1}| 1\rangle \langle 0| .\end{array}\end{eqnarray*}$
Since ${\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{0}| )={\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{1}| )$, it can be seen that
$\begin{eqnarray*}\left\{\begin{array}{l}| {a}_{0}{| }^{2}+| {a}_{2}{| }^{2}-| {b}_{0}{| }^{2}-| {b}_{2}{| }^{2}=0,\\ | {a}_{1}{| }^{2}-| {b}_{1}{| }^{2}-| {b}_{3}{| }^{2}=0,\\ {a}_{0}{a}_{1}^{* }-{b}_{0}{b}_{1}^{* }-{b}_{2}{b}_{3}^{* }=0,\\ {a}_{0}^{* }{a}_{1}-{b}_{0}^{* }{b}_{1}=0.\end{array}\right.\end{eqnarray*}$
After the calculation, we get ${b}_{2}{b}_{3}^{* }=0$, i.e. b2 = 0 or b3 = 0. This is in contradiction with the assumption of b2, b3 ≠ 0.
Case 4. ∣ψ0⟩ consists of four terms, that is, ∣ψ0⟩ = a0∣00⟩ + a1∣01⟩ + a2∣10⟩ + a3∣11⟩, where a0, a1, a2, a3 ≠ 0 and ${\sum }_{i=0}^{3}| {a}_{i}{| }^{2}=1$. So we acquire
$\begin{eqnarray*}\begin{array}{l}{\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{0}| )=(| {a}_{0}{| }^{2}+| {a}_{2}{| }^{2})| 0\rangle \langle 0| \\ +\,(| {a}_{1}{| }^{2}+| {a}_{3}{| }^{2})| 1\rangle \langle 1| \\ +\,({a}_{0}{a}_{1}^{* }+{a}_{2}{a}_{3}^{* })| 0\rangle \langle 1| +{a}_{0}^{* }{a}_{1}| 1\rangle \langle 0| ,\\ {\mathrm{Tr}}_{{\rm{B}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{0}| )=(| {a}_{0}{| }^{2}+| {a}_{1}{| }^{2})| 0\rangle \langle 0| \\ +\,(| {a}_{2}{| }^{2}+| {a}_{3}{| }^{2})| 1\rangle \langle 1| \\ +\,({a}_{0}{a}_{2}^{* }+{a}_{1}{a}_{3}^{* })| 0\rangle \langle 1| \\ +\,({a}_{0}^{* }{a}_{2}+{a}_{1}^{* }{a}_{3})| 1\rangle \langle 0| .\end{array}\end{eqnarray*}$
Suppose that ∣ψ1⟩ = b0∣i1i2⟩ + b1∣j1j2⟩ + b2∣k1k2⟩ + b3∣l1l2⟩, where ${\sum }_{i=0}^{3}| {b}_{i}{| }^{2}=1$, b0, b1, b2, b3 ≠ 0 and i1, i2, j1, j2, k1, k2, l1, l2 ∈ {0, 1}. We make a similar argument as the proof of Case 3 and find that ∣ψ⟩ = α0∣ψ0⟩ + α1∣ψ1⟩ can mask the quantum information in the state ∣b⟩ = α0∣0⟩ + α1∣1⟩ only if ∣ψ1⟩ = b0∣00⟩ + b1∣01⟩ + b2∣10⟩ + b3∣11⟩.
If not, considering the case of ∣ψ1⟩ = b0∣00⟩ + b1∣00⟩ + b2∣10⟩ + b3∣11⟩, we obtain
$\begin{eqnarray*}\begin{array}{l}{\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{1}| )=(| {b}_{0}{| }^{2}\\ +\,{b}_{0}{b}_{1}^{* }+{b}_{0}^{* }{b}_{1}+| {b}_{1}{| }^{2}+| {b}_{2}{| }^{2})| 0\rangle \langle 0| \\ +\,| {b}_{3}{| }^{2}| 1\rangle \langle 1| +{b}_{2}{b}_{3}^{* }| 0\rangle \langle 1| +{b}_{2}^{* }{b}_{3}| 1\rangle \langle 0| .\end{array}\end{eqnarray*}$
Due to ${\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{0}| )={\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{1}| )$, we know that
$\begin{eqnarray*}\left\{\begin{array}{l}| {a}_{0}{| }^{2}+| {a}_{2}{| }^{2}-| {b}_{0}{| }^{2}-{b}_{0}{b}_{1}^{* }-{b}_{0}^{* }{b}_{1}-| {b}_{1}{| }^{2}| {b}_{2}{| }^{2}=0,\\ | {a}_{1}{| }^{2}+| {a}_{3}{| }^{2}-| {b}_{3}{| }^{2}=0,\\ {a}_{0}{a}_{1}^{* }+{a}_{2}{a}_{3}^{* }-{b}_{2}{b}_{3}^{* }=0,\\ {a}_{0}^{* }{a}_{1}-{b}_{2}^{* }{b}_{3}=0.\end{array}\right.\end{eqnarray*}$
As a result, we acquire that a2 = 0 or a3 = 0, which contradicts the condition a2, a3 ≠ 0.
In summary, the conclusion is correct, i.e. a single qubit state can be masked by two non-orthogonal quantum states, which have the same number of terms and the same basis.
Based on the general masking conditions of the quantum state obtained in the previous section, we further discuss the masking of non-orthogonal quantum states and derive a conclusion, which provides a more efficient method to judge whether the quantum states can be masked. In other words, for non-orthogonal quantum states in ${{\mathbb{C}}}^{2}$, they cannot be used to mask the quantum information in a single qubit state as long as do not have the same number of terms or the same basis. Therefore, the conclusion is useful for the study of quantum information masking.
3.2. Masking of orthogonal quantum states
In this section, we discuss the further optimization of the masking conditions when ∣ψ0⟩ and ∣ψ1⟩ are orthogonal.
Let ∣ψ0⟩ = a0∣00⟩ + a1∣11⟩, ∣ψ1⟩ = b0∣01⟩ + b1∣10⟩ and ∣ψ⟩ = α0∣ψ0⟩ + α1∣ψ1⟩, where ${\sum }_{i=0}^{1}| {a}_{i}{| }^{2}={\sum }_{i=0}^{1}| {b}_{i}{| }^{2}={\sum }_{i=0}^{1}| {\alpha }_{i}{| }^{2}=1$. Under this assumption, the masking conditions equations (5 ), (8 ) and (9 ) can be further simplified as
$\begin{eqnarray}\left\{\begin{array}{l}| {a}_{0}{| }^{2}=| {a}_{1}{| }^{2}=| {b}_{0}{| }^{2}=| {b}_{1}{| }^{2}=\displaystyle \frac{1}{2},\\ {\alpha }_{0}{\alpha }_{1}^{* }{a}_{0}{b}_{0}^{* }+{\alpha }_{0}^{* }{\alpha }_{1}{a}_{1}^{* }{b}_{1}=0,\\ {\alpha }_{0}{\alpha }_{1}^{* }{a}_{0}{b}_{1}^{* }+{\alpha }_{0}^{* }{\alpha }_{1}{a}_{1}^{* }{b}_{0}=0,\end{array}\right.\end{eqnarray}$
and the quantum states ∣ψ0⟩, ∣ψ1⟩ and ∣ψ⟩ satisfy the masking definition
$\begin{eqnarray*}\begin{array}{l}{\mathrm{Tr}}_{{\rm{A}}}(| {\rm{\Psi }}\rangle \langle {\rm{\Psi }}| )={\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{0}| )\\ =\,{\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{1}| )=\displaystyle \frac{| 0\rangle \langle 0| +| 1\rangle \langle 1| }{2},\\ {\mathrm{Tr}}_{{\rm{B}}}(| {\rm{\Psi }}\rangle \langle {\rm{\Psi }}| )={\mathrm{Tr}}_{{\rm{B}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{0}| )\\ =\,{\mathrm{Tr}}_{{\rm{B}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{1}| )=\displaystyle \frac{| 0\rangle \langle 0| +| 1\rangle \langle 1| }{2}.\end{array}\end{eqnarray*}$
We show that when the coefficients of ∣ψ0⟩, ∣ψ1⟩ and ∣ψ⟩ satisfy equation (10 ), quantum information contained in the quantum state ∣b⟩ = α0∣0⟩ + α1∣1⟩ can be masked. To demonstrate this claim, we consider the following two examples.
Example 1. For equation (10 ), we analyze the case that a0, a1, b0 and b1 are given. Then we discuss what kind of state ∣b⟩ = α0∣0⟩ + α1∣1⟩ can be masked by ∣ψ0⟩ and ∣ψ1⟩.
Given
$\begin{eqnarray*}\begin{array}{rcl}| {{\rm{\Psi }}}_{0}\rangle & = & \displaystyle \frac{1}{\sqrt{2}}| 00\rangle +\displaystyle \frac{{\rm{i}}}{\sqrt{2}}| 11\rangle ,\\ | {{\rm{\Psi }}}_{1}\rangle & = & \displaystyle \frac{1}{\sqrt{2}}| 01\rangle +\displaystyle \frac{1}{\sqrt{2}}| 10\rangle ,\end{array}\end{eqnarray*}$
we obtain$\begin{eqnarray*}\begin{array}{l}| {\rm{\Psi }}\rangle ={\alpha }_{0}| {{\rm{\Psi }}}_{0}\rangle +{\alpha }_{1}| {{\rm{\Psi }}}_{1}\rangle ={\alpha }_{0}\left(\displaystyle \frac{1}{\sqrt{2}}| 00\rangle \right.\\ \left.+\,\displaystyle \frac{{\rm{i}}}{\sqrt{2}}| 11\rangle \right)+{\alpha }_{1}\left(\displaystyle \frac{1}{\sqrt{2}}| 01\rangle +\displaystyle \frac{1}{\sqrt{2}}| 10\rangle \right).\end{array}\end{eqnarray*}$
Therefore, equation (10 ) can be simplified to
$\begin{eqnarray}\displaystyle \frac{1}{2}{\alpha }_{0}{\alpha }_{1}^{* }-\displaystyle \frac{{\rm{i}}}{2}{\alpha }_{0}^{* }{\alpha }_{1}=0.\end{eqnarray}$
Let α0 = x0 + y0i and α1 = x1 + y1i. Due to ${\sum }_{i=0}^{1}| {\alpha }_{i}{| }^{2}=1$, we acquire
$\begin{eqnarray*}{y}_{1}=\pm \sqrt{1-({x}_{0}^{2}+{y}_{0}^{2}+{x}_{1}^{2})},\end{eqnarray*}$
where x0, x1, y0 ∈ [−1, 1] and ${x}_{0}^{2}+{y}_{0}^{2}+{x}_{1}^{2}\in [0,1]$.
Then equation (11 ) is reformulated as
$\begin{eqnarray}\begin{array}{l}{x}_{0}{x}_{1}+{y}_{0}\sqrt{1-({x}_{0}^{2}+{y}_{0}^{2}+{x}_{1}^{2})}\\ +\,{x}_{0}\sqrt{1-({x}_{0}^{2}+{y}_{0}^{2}+{x}_{1}^{2})}-{x}_{1}{y}_{0}=0,\end{array}\end{eqnarray}$
or
$\begin{eqnarray}\begin{array}{l}{x}_{0}{x}_{1}-{y}_{0}\sqrt{1-({x}_{0}^{2}+{y}_{0}^{2}+{x}_{1}^{2})}\\ -\,{x}_{0}\sqrt{1-({x}_{0}^{2}+{y}_{0}^{2}+{x}_{1}^{2})}-{x}_{1}{y}_{0}=0.\end{array}\end{eqnarray}$
As shown in figure 1, we find a kind of state ∣b⟩ = α0∣0⟩ + α1∣1⟩ which can be masked, i.e. when the coefficients of the state ∣b⟩ satisfy the relationship in figure 1, the state ∣b⟩ = α0∣0⟩ + α1∣1⟩ can be masked by given quantum states $| {{\rm{\Psi }}}_{0}\rangle =\tfrac{1}{\sqrt{2}}| 00\rangle +\tfrac{{\rm{i}}}{\sqrt{2}}| 11\rangle $ and $| {{\rm{\Psi }}}_{1}\rangle =\tfrac{1}{\sqrt{2}}| 01\rangle \,+\tfrac{1}{\sqrt{2}}| 10\rangle $.
Figure 1. (a) $| {{\rm{\Psi }}}_{0}\rangle =\tfrac{1}{\sqrt{2}}| 00\rangle +\tfrac{{\rm{i}}}{\sqrt{2}}| 11\rangle $ and $| {{\rm{\Psi }}}_{1}\rangle =\tfrac{1}{\sqrt{2}}(| 01\rangle +| 10\rangle )$ can mask information in ∣b⟩ = α0∣0⟩ + α1∣1⟩, if and only if α0 = x0 + y0i and ${\alpha }_{1}={x}_{1}+\sqrt{1-({x}_{0}^{2}+{y}_{0}^{2}+{x}_{1}^{2})}{\rm{i}}$ satisfy this figure. (b) $| {{\rm{\Psi }}}_{0}\rangle =\tfrac{1}{\sqrt{2}}| 00\rangle +\tfrac{{\rm{i}}}{\sqrt{2}}| 11\rangle $ and $| {{\rm{\Psi }}}_{1}\rangle =\tfrac{1}{\sqrt{2}}(| 01\rangle +| 10\rangle )$ can mask information in ∣b⟩ = α0∣0⟩ + α1∣1⟩, if and only if α0 = x0 + y0i and ${\alpha }_{1}={x}_{1}-\sqrt{1-({x}_{0}^{2}+{y}_{0}^{2}+{x}_{1}^{2})}{\rm{i}}$ satisfy this figure. |
In particular, according to figure 1(b), we consider the case of ${x}_{0}=-\tfrac{1}{5}$, x1 = 0 and ${y}_{0}=-\tfrac{1}{5}$, namely, ${\alpha }_{0}=\tfrac{1-{i}}{5}$ and ${\alpha }_{1}=\tfrac{\sqrt{23}}{5}$.
Through calculation, we obtain
$\begin{eqnarray*}\begin{array}{l}{\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{0}| )={\mathrm{Tr}}_{{\rm{A}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{1}| )\\ =\,{\mathrm{Tr}}_{{\rm{A}}}(| {\rm{\Psi }}\rangle \langle {\rm{\Psi }}| )=\displaystyle \frac{| 0\rangle \langle 0| +| 1\rangle \langle 1| }{2},\\ {\mathrm{Tr}}_{{\rm{B}}}(| {{\rm{\Psi }}}_{0}\rangle \langle {{\rm{\Psi }}}_{0}| )={\mathrm{Tr}}_{{\rm{B}}}(| {{\rm{\Psi }}}_{1}\rangle \langle {{\rm{\Psi }}}_{1}| )\\ =\,{\mathrm{Tr}}_{{\rm{B}}}(| {\rm{\Psi }}\rangle \langle {\rm{\Psi }}| )=\displaystyle \frac{| 0\rangle \langle 0| +| 1\rangle \langle 1| }{2}.\end{array}\end{eqnarray*}$
In other words, we know that the state $| b\rangle =\tfrac{1-{\rm{i}}}{5}| 0\rangle +\tfrac{\sqrt{23}}{5}| 1\rangle $ can be masked by the given quantum states $| {{\rm{\Psi }}}_{0}\rangle =\tfrac{1}{\sqrt{2}}| 00\rangle +\tfrac{{\rm{i}}}{\sqrt{2}}| 11\rangle $ and $| {{\rm{\Psi }}}_{1}\rangle =\tfrac{1}{\sqrt{2}}| 01\rangle +\tfrac{1}{\sqrt{2}}| 10\rangle $.
Example 2. By equation (10 ), we consider that if α0 and α1 are certain, what kind of quantum states ∣ψ0⟩ and ∣ψ1⟩ can mask the quantum information in ∣b⟩ = α0∣0⟩ + α1∣1⟩.
Let ${\alpha }_{0}=\tfrac{1}{\sqrt{1+{\lambda }^{2}}}$ and ${\alpha }_{1}=\left(\tfrac{\lambda }{\sqrt{1+{\lambda }^{2}}}\right){\rm{i}}$, where $\lambda \in {\mathbb{R}}$, and denote
$\begin{eqnarray*}\begin{array}{l}{a}_{0}={x}_{0}+{y}_{0}{\rm{i}},\qquad {a}_{1}={x}_{1}+{y}_{1}{\rm{i}},\\ {b}_{0}={x}_{2}+{y}_{2}{\rm{i}},\qquad {b}_{1}={x}_{3}+{y}_{3}{\rm{i}}.\end{array}\end{eqnarray*}$
Then, equation (10 ) can be reduced to
$\begin{eqnarray}\left\{\begin{array}{l}{x}_{0}^{2}+{y}_{0}^{2}={x}_{1}^{2}+{y}_{1}^{2}={x}_{2}^{2}+{y}_{2}^{2}={x}_{3}^{2}+{y}_{3}^{2}=\displaystyle \frac{1}{2},\\ -\,{x}_{0}{x}_{2}+{x}_{3}{x}_{1}-{y}_{0}{y}_{2}+{y}_{3}{y}_{1}=0,\\ -\,{x}_{0}{x}_{3}+{x}_{2}{x}_{1}-{y}_{0}{y}_{3}+{y}_{2}{y}_{1}=0,\\ -\,{x}_{0}{y}_{2}+{x}_{2}{y}_{0}+{x}_{3}{y}_{1}-{x}_{1}{y}_{3}=0,\\ {x}_{3}{y}_{0}-{x}_{0}{y}_{3}+{x}_{2}{y}_{1}-{x}_{1}{y}_{2}=0.\end{array}\right.\end{eqnarray}$
Particularly, when x0 = x2 and y0 = y2, equation (14 ) is further changed to
$\begin{eqnarray}\left\{\begin{array}{l}{x}_{3}{x}_{1}+{y}_{3}{y}_{1}-\displaystyle \frac{1}{2}=0,\\ {x}_{3}{y}_{1}-{x}_{1}{y}_{3}=0,\\ ({x}_{0}+{y}_{0})({x}_{3}-{x}_{1})+({y}_{0}-{x}_{0})({y}_{3}-{y}_{1})=0.\end{array}\right.\end{eqnarray}$
Since ${x}_{i}^{2}+{y}_{i}^{2}=\tfrac{1}{2}$, where i ∈ {0, 1, 2, 3}, denoting
$\begin{eqnarray*}\left\{\begin{array}{l}{y}_{0}=\pm \sqrt{\displaystyle \frac{1}{2}-{x}_{0}^{2}},\\ {y}_{1}=\pm \sqrt{\displaystyle \frac{1}{2}-{x}_{1}^{2}},\\ {y}_{2}={y}_{0}=\pm \sqrt{\displaystyle \frac{1}{2}-{x}_{0}^{2}},\\ {y}_{3}=\pm \sqrt{\displaystyle \frac{1}{2}-{x}_{3}^{2}}.\end{array}\right.\end{eqnarray*}$
Through the above equations, we get the solutions to equation (15 ), that is, ${x}_{1}={x}_{3}=\tfrac{\sqrt{2}}{2}$, ${x}_{0}\in {\mathbb{R}}$ or ${x}_{1}={x}_{3}=-\tfrac{\sqrt{2}}{2}$, ${x}_{0}\in {\mathbb{R}}$. Indeed, we obtain that
$\begin{eqnarray}\left\{\begin{array}{l}| {{\rm{\Psi }}}_{0}\rangle =({x}_{0}+{y}_{0}{\rm{i}})| 00\rangle +\displaystyle \frac{\sqrt{2}}{2}| 11\rangle ,\\ | {{\rm{\Psi }}}_{1}\rangle =({x}_{0}+{y}_{0}{\rm{i}})| 01\rangle +\displaystyle \frac{\sqrt{2}}{2}| 10\rangle ,\end{array}\right.\end{eqnarray}$
or $\begin{eqnarray}\left\{\begin{array}{l}| {{\rm{\Psi }}}_{0}\rangle =({x}_{0}+{y}_{0}{\rm{i}})| 00\rangle -\displaystyle \frac{\sqrt{2}}{2}| 11\rangle ,\\ | {{\rm{\Psi }}}_{1}\rangle =({x}_{0}+{y}_{0}{\rm{i}})| 01\rangle -\displaystyle \frac{\sqrt{2}}{2}| 10\rangle .\end{array}\right.\end{eqnarray}$
According to equations (16 ) and (17 ), respectively, equation (10 ) would be simplified as
$\begin{eqnarray}\left(-{x}_{0}^{2}-{y}_{0}^{2}+\displaystyle \frac{1}{2}\right)\displaystyle \frac{\lambda }{1+{\lambda }^{2}}=0.\end{eqnarray}$
Moreover, let λ ∈ [−1, 1], then we draw figure 2 for an intuitive description.
Figure 2. $| b\rangle =(\tfrac{1}{\sqrt{1+{\lambda }^{2}}})| 0\rangle +(\tfrac{\lambda {\rm{i}}}{\sqrt{1+{\lambda }^{2}}})| 1\rangle $ can be masked by $| {{\rm{\Psi }}}_{0}\rangle \,=({x}_{0}+{y}_{0}{\rm{i}})| 00\rangle +\tfrac{\sqrt{2}}{2}| 11\rangle $ and $| {{\rm{\Psi }}}_{1}\rangle =({x}_{0}+{y}_{0}{\rm{i}})| 01\rangle +\tfrac{\sqrt{2}}{2}| 10\rangle $ or $| {{\rm{\Psi }}}_{0}\rangle =({x}_{0}+{y}_{0}{\rm{i}})| 00\rangle -\tfrac{\sqrt{2}}{2}| 11\rangle $ and $| {{\rm{\Psi }}}_{1}\rangle =({x}_{0}+{y}_{0}{\rm{i}})| 01\rangle -\tfrac{\sqrt{2}}{2}| 10\rangle $, if and only if their coefficients λ, x0, y0 satisfy the relationship represented in this image. |
As shown in figure 2, for the state $| b\rangle =(\tfrac{1}{\sqrt{1+{\lambda }^{2}}})| 0\rangle +(\tfrac{\lambda {\rm{i}}}{\sqrt{1+{\lambda }^{2}}})| 1\rangle $, there are quantum states ∣ψ0⟩ and ∣ψ1⟩ in equation (16 ) or equation (17 ), which can mask the state ∣b⟩, if and only if their coefficients satisfy the relationship shown in figure 2.
4. Conclusion
In two-dimensional Hilbert space, we expressed the quantum information masking conditions by a system of equations for the coefficients of quantum states. Moreover, by observing the characteristic of the maskable non-orthogonal two-qubit quantum states, we obtained the conclusion that if two non-orthogonal quantum states can mask a single qubit state, they have the same number of terms and the same basis. Furthermore, we considered two examples of orthogonal quantum states and calculated the masking conditions. Finally, we gave the corresponding images for an intuitive description. Our results would provide a foundation idea for further study on the masking of higher dimensional quantum states.