Constructing a maximally multi-qubit entangled state via expanded recurrence relation

Quantum entanglement is a valuable resource for the implementation of quantum computation, and quantum communication protocols, like quantum teleportation [1], quantum secure direct communication [2], remote state preparation [3] and so on. The exploration of quantum entanglement in maximally multipartite quantum systems in a composite system is of great significance to the study of quantum entanglement and maximally multi-qubit entangled states [4, 5]. Due to its great relevance, from both the theoretical and practical points of view, it is imperative to explore and characterize all aspects of the quantum entanglement of maximally multipartite quantum systems [4, 6]. A considerable amount of research has been devoted to the study of multi-qubit entanglement measures defined as the sum of bipartite entanglement measures over all the possible bi-partitions of the full system [7–11]. The multipartite entanglement of system of qubits can be characterized in terms of distribution function of bipartite over all possible bipartitions of the qubits [6, 9, 10]. This led us to formulate the notion of maximally multipartite entangled states (MMES). In 2013, Zha et al presented to identified and generalized criterion for maximally entangled states of 2, 3, 4, 5, 6 and 8 qubits and gave some new maximally multipartite entangled state [11, 12]. In 2019, MMES via recurrence relation was introduced. Later, the five qutrits and seven ququarts MMES have been established using recurrence relation [13, 14]. Therefore, the construct of maximally multipartite state through recurrence relationship bears a great potential.

Recently, Che et al [15] presented a recurrence relation of multi-qubit states. According to this inspiration, we present an expanded recurrence relation of maximally entangled states between the maximally n-qubit entangled state and maximally n + 2-qubit entangled state. Furthermore, some new maximally four-, six-, eight-qubits states are also presented.

Recently, Che et al [15] gave a recurrence relation of multi-qubit states. According to this inspiration, we present an expand recurrence relation of maximally multi-qubit entangled states.

Let ${\left| {{\varphi _M}} \right\rangle _n}$ repress a maximally n-qubit entangled state, then maximally n + 2-qubit entangled state can be written as

$$\begin{align} {\left| {{\varphi _m}} \right\rangle _{12 \cdots n,n + 1,n + 2}} & = a\left({\lambda _1}{\left| {\varphi _m^1} \right\rangle _{12 \cdots n}}{\left| {{i^1}{j^1}} \right\rangle _{n + 1,n + 2}}\right.\nonumber\\ & \quad + {\lambda _2}{\left| {\varphi _m^2} \right\rangle _{12 \cdots n}}{\left| {{i^2}{j^2}} \right\rangle _{n + 1,n + 2}} \nonumber \\ & \quad + {\lambda _3}{\left| {\varphi _m^3} \right\rangle _{12 \cdots n}}{\left| {{i^3}{j^3}} \right\rangle _{n + 1,n + 2}}\nonumber\\ & \left.\quad + {\lambda _4}{\left| {\varphi _m^4} \right\rangle _{12 \cdots n}}{\left| {{i^4}{j^4}} \right\rangle _{n + 1,n + 2}}\right).\end{align} \tag{ 1 }$$

Where ${\left| {\varphi _m^\alpha } \right\rangle _{12 \cdots n}},\alpha = 1,2,3,4$ are all maximally n-qubit entangled state and satisfy

$$\begin{align} {}_n\left\langle {\varphi _m^\alpha } {\left| {\varphi _m^{\alpha {^{^{\prime}}}}} \right\rangle _n}\right. & = {\delta _{\alpha \beta }},\,{\;_{n + 1,n + 2}}\left\langle {{i^\beta }{j^\gamma }} \left| {{i^{\beta {^{^{\prime}}}}}{j^{\gamma {^{^{\prime}}}}}}\right.\right\rangle _{n + 1,n + 2}\nonumber\\ & = {\delta _{\beta \beta {^{^{\prime}}}}}{\delta _{\gamma \gamma {^{^{\prime}}}}}\end{align} \tag{ 2 }$$

${\lambda _{\text{j}}} = 0,{e^{i\theta }};j = 1,2,3,4$; $a$ is coefficients satisfy the normalization condition.

Without any loss of generality, we first consider maximally two-qubit state.

It is well know that Bell state ${\left| {{\varphi ^ \pm }} \right\rangle _{12}} = \frac{1}{{\sqrt 2 }}{\left( {\left| {00} \right\rangle \pm \left| {11} \right\rangle } \right)_{12}}$, $\left| {{\psi ^ \pm }} \right\rangle = \frac{1}{{\sqrt 2 }}\left( {\left| {01} \right\rangle \pm \left| {10} \right\rangle } \right)$ are maximally two-qubit entangled state,

Now let us construct a maximally four-qubit state. Using expand recurrence relation equation (1)

Let ${\left| {\varphi _m^1} \right\rangle _{12}} = \frac{1}{{\sqrt 2 }}{\left( {\left| {00} \right\rangle + \left| {11} \right\rangle } \right)_{12}}$; ${\left| {\varphi _m^2} \right\rangle _{12}}= \frac{1}{{\sqrt 2 }}$ ${\left( { - \left| {01} \right\rangle + \left| {10} \right\rangle } \right)_{12}}$;

$$\begin{equation}\left| {\varphi _m^3{\rangle _{12}} = \frac{1}{{\sqrt 2 }}{{\left( {\left| {01\rangle + } \right|10\rangle } \right)}_{12}};\,\,} \right|\varphi _m^4{\rangle _{12}} = \frac{1}{{\sqrt 2 }}{\left( { - \left| {00\rangle + } \right|11\rangle } \right)_{12}}.\end{equation} \tag{ 3 }$$

Then, we have

$$\begin{align} |\varphi _m^1{\rangle _{1234}} & = \frac{1}{2}( | \varphi _m^1{\rangle _{12}}\left| {00{\rangle _{34}} + } \right|\varphi _m^2{\rangle _{12}}\left| {01{\rangle _{34}} + } \right|\varphi _m^3{\rangle _{12}}| 10{\rangle _{34}}\nonumber\\ & \quad + |\varphi _m^4{\rangle _{12}} |11{\rangle _{34}}) \nonumber\\ & = \frac{1}{{2\sqrt 2 }}(\left| {0000\rangle + } \right|1100\rangle + | 0110\rangle + |1010\rangle \nonumber\\ & \quad - \left| {0011\rangle + } \right|1111\rangle - \left| {0101\rangle + } \right|1001\rangle )_{1234}.\end{align} \tag{ 4 }$$

This state was presented by Yeo and Chua [16].

Also, let

$$\begin{equation}\left| {\varphi _m^1{\rangle _{13}} = \frac{1}{{\sqrt 2 }}{{\left( {\left| {00\rangle + } \right|11\rangle } \right)}_{13}},\,\,} \right|\varphi _m^4{\rangle _{13}} = \frac{1}{{\sqrt 2 }}{\left( {\left| {00\rangle - } \right|11\rangle } \right)_{13}}.\end{equation} \tag{ 5 }$$

Then, we have

$$\begin{align} |\varphi _m^2{\rangle _{1234}} & = \frac{1}{{\sqrt 2 }}( \left| \varphi _m^1{\rangle _{13}}\left| {00{\rangle _{24}} + } \right|\varphi _m^4{\rangle _{13}} \right|11{\rangle _{24}} ) \nonumber\\ & \quad = \frac{1}{2}{{\left( {\left| {0000\rangle + } \right|0101\rangle + \left| {1010\rangle - } \right|1111\rangle } \right)}_{1234}}.\end{align} \tag{ 6 }$$

This state was just the cluster state given in [17–19]

Let

$$\begin{equation*}\left| {\varphi _m^1{\rangle _{12}} = \frac{1}{{\sqrt 2 }}{{\left( {\left| {00\rangle + } \right|11\rangle } \right)}_{12}},\,} \right|\varphi _m^4{\rangle _{12}} = \frac{1}{{\sqrt 2 }}{\left( { - \left| {01\rangle + } \right|10\rangle } \right)_{12}}.\end{equation*}$$

Then, we have

$$\begin{align} |\varphi _m^3{\rangle _{1234}} & = \frac{1}{{\sqrt 2 }}\left( {\left| {\varphi _m^1{\rangle _{12}}\left| {00{\rangle _{34}} + } \right|\varphi _m^4{\rangle _{12}}} \right|11{\rangle _{34}}} \right) \nonumber\\ & \quad = \frac{1}{2}{{\left( {\left| {0000\rangle - } \right|0111\rangle + \left| {1011\rangle + } \right|1100\rangle } \right)}_{1234}} .\end{align} \tag{ 7 }$$

By careful calculation, we can obtain

$$\begin{equation*}{\pi _{12}} = Tr\rho _{12}^2 = \frac{1}{2},\,{\pi _{13}} = Tr\rho _{13}^2 = \frac{1}{4},\,\,{\pi _{14}} = Tr\rho _{14}^2 = \frac{1}{4},\end{equation*}$$

$$\begin{equation}{\pi _{23}} = Tr\rho _{23}^2 = \frac{1}{4},\,{\pi _{24}} = Tr\rho _{24}^2 = \frac{1}{4},\,\,{\pi _{34}} = Tr\rho _{34}^2 = \frac{1}{2}.\end{equation} \tag{ 8 }$$

Therefore, we have [11]

$$\begin{equation}{\pi _{{\text{ME}}}} = \frac{1}{6}\left( {{\pi _{12}} + {\pi _{13}} + {\pi _{14}} + {\pi _{23}} + {\pi _{24}} + {\pi _{34}}} \right) = \frac{1}{3}.\end{equation} \tag{ 9 }$$

According to [11], this state is a maximally entangled four-qubit states.

Let

$$\begin{align} {\left| {\varphi _m^1} \right\rangle _{\!12}} & = \frac{1}{{\sqrt 2 }}{\left( {\left| {00} \right\rangle {+} \left| {11} \right\rangle }\! \right)_{12}},\,{\left| {\varphi _m^2} \right\rangle _{\!12}} = \frac{1}{{\sqrt 2 }}{\left( {\left| {01} \right\rangle {+} \left| {10} \right\rangle }\! \right)_{\!12}},\nonumber\\ {\left| {\varphi _m^3} \right\rangle _{12}} & = \frac{1}{{\sqrt 2 }}{\left( {\left| {01} \right\rangle - \left| {10} \right\rangle } \right)_{\!12}},\,{\left| {\varphi _m^4} \right\rangle _{\!12}} = \frac{1}{{\sqrt 2 }}{\left( {\left| {00} \right\rangle - \left| {11} \right\rangle }\! \right)_{12}}\nonumber\\ {\left| {\varphi _m^4} \right\rangle _{1234}} & = \frac{1}{2}( {{\left| {\varphi _m^1} \right\rangle }_{\!12}}{{\left| {+} 0 \right\rangle }_{34}} {+} {{\left| {\varphi _m^2} \right\rangle }_{\!12}}{{\left| { - 1} \right\rangle }_{\!34}}{+}{{\left| {\varphi _m^3} \right\rangle }_{\!12}}{{\left| { + 1} \right\rangle }_{\!34}}\nonumber\\ &\quad + {{\left| {\varphi _m^4} \right\rangle }_{12}}{{\left| { - 0} \right\rangle }_{34}} ) \nonumber\\ & = \frac{1}{2}( \left| {0000} \right\rangle + \left| {0101} \right\rangle - \left| {1011} \right\rangle + \left| {1110} \right\rangle)_{1234} \end{align} \tag{ 10 }$$

where $\left| + \right\rangle = \frac{1}{{\sqrt 2 }}\left( {\left| 0 \right\rangle + \left| 1 \right\rangle } \right),\,\left| - \right\rangle = \frac{1}{{\sqrt 2 }}\left( {\left| 0 \right\rangle - \left| 1 \right\rangle } \right)$

Using equation (10), we can obtain

$$\begin{align*}{\pi _{12}} & = Tr\rho _{12}^2 = \frac{1}{4},\,{\pi _{13}} = Tr\rho _{13}^2 = \frac{1}{2},\,\,{\pi _{14}} = Tr\rho _{14}^2 = \frac{1}{4},\nonumber\\ {\pi _{23}} & = Tr\rho _{23}^2 = \frac{1}{4},\,\,{\pi _{24}} = Tr\rho _{24}^2 = \frac{1}{2},\,\,{\pi _{34}} = Tr\rho _{34}^2 = \frac{1}{4}.\end{align*}$$

Then, we have

$$\begin{equation}{\pi _{{\text{ME}}}} = \frac{1}{6}\left( {{\pi _{12}} + {\pi _{13}} + {\pi _{14}} + {\pi _{23}} + {\pi _{24}} + {\pi _{34}}} \right) = \frac{1}{3}.\end{equation} \tag{ 11 }$$

This state is also a maximally entangled four-qubit states.

Using equation (7), let

$$\begin{align}{\left| {\varphi _m^1} \right\rangle _{1234}} &= \frac{1}{2}{\left( {\left| {0000} \right\rangle - \left| {0111} \right\rangle + \left| {1011} \right\rangle + \left| {1100} \right\rangle } \right)_{1234}}\nonumber\\ {\left| {\varphi _m^2} \right\rangle _{1234}} &= \frac{1}{2}{\left( {\left| {0001} \right\rangle - \left| {0110} \right\rangle - \left| {1010} \right\rangle - \left| {1101} \right\rangle } \right)_{1234}}\nonumber\\ {\left| {\varphi _m^3} \right\rangle _{1234}} &= \frac{1}{2}{\left( {\left| {0101} \right\rangle - \left| {0010} \right\rangle + \left| {1110} \right\rangle + \left| {1001} \right\rangle } \right)_{1234}}\nonumber\\ |\varphi _m^4{\rangle _{1234}} &= \frac{1}{2}{\left( {\left| {0100\rangle - } \right|0011\rangle - \left| {1111\rangle - } \right|1000\rangle } \right)_{1234}}.\end{align} \tag{ 12 }$$

According to expand recurrence relation equation (1), we have

$$\begin{align} {\left| {\psi _m^1} \right\rangle _{123456}} & = \frac{1}{2}( {{\left| {\varphi _m^1} \right\rangle }_{1234}}{{\left| {00} \right\rangle }_{56}} - {{\left| {\varphi _m^2} \right\rangle }_{1234}}{{\left| {01} \right\rangle }_{56}}\nonumber\\ &\quad + {{\left| {\varphi _m^3} \right\rangle }_{1234}}{{\left| {10} \right\rangle }_{56}} + {{\left| {\varphi _m^4} \right\rangle }_{1234}}{{\left| {11} \right\rangle }_{56}} ) \nonumber \\ & = \frac{1}{4}[\left| {000000} \right\rangle - \left| {000101} \right\rangle - \left| {001010} \right\rangle - \left| {001111} \right\rangle \nonumber \\ & \quad+ \left| {010011} \right\rangle + \left| {010110} \right\rangle + \left| {011001} \right\rangle - \left| {011100} \right\rangle \nonumber \\ &\quad - \left| {100011} \right\rangle + \left| {100110} \right\rangle + \left| {101001} \right\rangle + \left| {101100} \right\rangle \nonumber \\ & \quad+ \left| {110000} \right\rangle + \left| {110101} \right\rangle + \left| {111010} \right\rangle\nonumber \\ & \quad - \left| {111111} \right\rangle {]_{123456}}.\end{align} \tag{ 13 }$$

Using equation (13), we can obtain

$$\begin{equation}{\pi _{ijk}} = Tr\rho _{ijk}^2 = \frac{1}{8},ijk = 123,124, \cdots ,456.\end{equation} \tag{ 14 }$$

Therefore, according to [14], this state is a maximally entangled six-qubit state.

Using equation (10), let

$$\begin{align} {\left| {\varphi _m^1} \right\rangle _{1234}} & = \frac{1}{2}{\left( {\left| {0000} \right\rangle + \left| {0101} \right\rangle - \left| {1011} \right\rangle + \left| {1110} \right\rangle } \right)_{1234}}\nonumber\\ {\left| {\varphi _m^2} \right\rangle _{1234}} & = {\sigma _{3z}}{\sigma _{3x}}{\sigma _{4z}}{\left| {\varphi _m^1} \right\rangle _{1234}} \nonumber \\ & = \frac{1}{2}{\left( { - \left| {0010} \right\rangle + \left| {0111} \right\rangle + \left| {1001} \right\rangle + \left| {1100} \right\rangle } \right)_{1234}} \nonumber \\ {\left| {\varphi _m^3} \right\rangle _{1234}} & = {\sigma _{3x}}{\sigma _{4x}}{\left| {\varphi _m^1} \right\rangle _{1234}} \nonumber \\ & = \frac{1}{2}{\left( {\left| {0011} \right\rangle + \left| {0110} \right\rangle - \left| {1000} \right\rangle + \left| {1101} \right\rangle } \right)_{1234}} \nonumber \\ | \varphi _m^4{\rangle _{1234}} & = {\sigma _{1z}}{\sigma _{2z}}{\sigma _{3z}}{\sigma _{4x}} |\varphi _m^1{\rangle _{1234}} \nonumber \\ & = \frac{1}{2}{{\left( {\left| {0001\rangle - } \right|0100\rangle - \left| {1010\rangle - } \right|1111\rangle } \right)}_{1234}}.\end{align} \tag{ 15 }$$

According to expand recurrence relation equation (1), we have

$$\begin{align} {\left| {\psi _m^2} \right\rangle _{123456}} & = \frac{1}{2}( {{\left| {\varphi _m^1} \right\rangle }_{1234}}{{\left| {00} \right\rangle }_{56}} + {{\left| {\varphi _m^2} \right\rangle }_{1234}}{{\left| {01} \right\rangle }_{56}}\nonumber\\ &\quad + {{\left| {\varphi _m^3} \right\rangle }_{1234}}{{\left| {10} \right\rangle }_{56}} + {{\left| {\varphi _m^4} \right\rangle }_{1234}}{{\left| {11} \right\rangle }_{56}} ) \nonumber \\ & = \frac{1}{4}[\left| {000000} \right\rangle + \left| {000111} \right\rangle - \left| {001001} \right\rangle + \left| {001110} \right\rangle \nonumber \\ &\quad - \left| {010011} \right\rangle + \left| {010100} \right\rangle + \left| {011010} \right\rangle + \left| {011101} \right\rangle \nonumber \\ &\quad - \left| {100010} \right\rangle + \left| {100101} \right\rangle - \left| {101011} \right\rangle - \left| {101100} \right\rangle \nonumber \\ &\quad + \left| {110001} \right\rangle + \left| {110110} \right\rangle + \left| {111000} \right\rangle\nonumber \\ &\quad - \left| {111111} \right\rangle {]_{123456}}.\end{align} \tag{ 16 }$$

Similarly, we can show

$$\begin{equation}{\pi _{ijk}} = Tr\rho _{ijk}^2 = \frac{1}{8},ijk = 123,124, \cdots ,456.\end{equation} \tag{ 17 }$$

This six-qubit state have discovered by Borras et al [20]

Using equation (16), let

$$\begin{equation*}\begin{gathered} {\left| {\phi _m^1} \right\rangle _{123456}} \hfill \\ = \frac{1}{4}[\left| {000000} \right\rangle + \left| {000111} \right\rangle - \left| {001001} \right\rangle + \left| {001110} \right\rangle \hfill \\ \quad - \left| {010011} \right\rangle + \left| {010100} \right\rangle + \left| {011010} \right\rangle + \left| {011101} \right\rangle \hfill \\ \quad - \left| {100010} \right\rangle + \left| {100101} \right\rangle - \left| {101011} \right\rangle - \left| {101100} \right\rangle \hfill \\ \quad + \left| {110001} \right\rangle + \left| {110110} \right\rangle + \left| {111000} \right\rangle - \left| {111111} \right\rangle {]_{123456}} \hfill \\ \end{gathered} \end{equation*}$$

$$\begin{equation*}\begin{gathered} {\left| {\phi _m^2} \right\rangle _{123456}} = {\sigma _{1x}}{\sigma _{2x}}{\left| {\phi _m^1} \right\rangle _{123456}} \hfill \\ = \frac{1}{4}[\left| {000001} \right\rangle + \left| {000110} \right\rangle + \left| {001000} \right\rangle - \left| {001111} \right\rangle \hfill \\ \quad - \left| {010010} \right\rangle + \left| {010101} \right\rangle - \left| {011011} \right\rangle - \left| {011100} \right\rangle \hfill \\ \quad - \left| {100011} \right\rangle + \left| {100100} \right\rangle + \left| {101010} \right\rangle + \left| {101101} \right\rangle \hfill \\ \quad + \left| {110000} \right\rangle + \left| {110111} \right\rangle - \left| {111001} \right\rangle + \left| {111110} \right\rangle {]_{123456}} \hfill \\ \end{gathered} \end{equation*}$$

$$\begin{equation*}\begin{gathered} {\left| {\phi _m^3} \right\rangle _{123456}} = {\sigma _{1y}}{\sigma _{2y}}{\left| {\phi _m^1} \right\rangle _{123456}} \hfill \\ = \frac{1}{4}[ - \left| {000001} \right\rangle - \left| {000110} \right\rangle - \left| {001000} \right\rangle + \left| {001111} \right\rangle \hfill \\ \quad - \left| {010010} \right\rangle + \left| {010101} \right\rangle - \left| {011011} \right\rangle - \left| {011100} \right\rangle \hfill \\ \quad - \left| {100011} \right\rangle + \left| {100100} \right\rangle + \left| {101010} \right\rangle + \left| {101101} \right\rangle \hfill \\ \quad - \left| {110000} \right\rangle - \left| {110111} \right\rangle + \left| {111001} \right\rangle - \left| {111110} \right\rangle {]_{123456}} \hfill \\ \end{gathered} \end{equation*}$$

$$\begin{equation}\begin{gathered} {\left| {\phi _m^4} \right\rangle _{123456}} = {\sigma _{1z}}{\sigma _{2z}}{\left| {\phi _m^1} \right\rangle _{123456}} \hfill \\ = \frac{1}{4}[\left| {000000} \right\rangle + \left| {000111} \right\rangle - \left| {001001} \right\rangle + \left| {001110} \right\rangle \hfill \\ \quad + \left| {010011} \right\rangle - \left| {010100} \right\rangle - \left| {011010} \right\rangle - \left| {011101} \right\rangle \hfill \\ \quad + \left| {100010} \right\rangle - \left| {100101} \right\rangle + \left| {101011} \right\rangle + \left| {101100} \right\rangle \hfill \\ \quad + \left| {110001} \right\rangle + \left| {110110} \right\rangle + \left| {111000} \right\rangle - \left| {111111} \right\rangle {]_{123456}} \hfill \\ \end{gathered} .\end{equation} \tag{ 18 }$$

According to expand recurrence relation equation (1), we have

$$\begin{align} {\left| {\psi _m^1} \right\rangle _{12345678}} &= \frac{1}{2}\left( {{{\left| {\phi _m^1} \right\rangle }_{123456}}{{\left| {00} \right\rangle }_{78}} + {{\left| {\phi _m^2} \right\rangle }_{123456}}{{\left| {01} \right\rangle }_{78}} + {{\left| {\phi _m^3} \right\rangle }_{123456}}{{\left| {10} \right\rangle }_{78}} + {{\left| {\phi _m^4} \right\rangle }_{123456}}{{\left| {11} \right\rangle }_{78}}} \right) \nonumber \\ &= \frac{1}{8}[{\left( {\left| {0000} \right\rangle - \left| {0011} \right\rangle + \left| {1101} \right\rangle + \left| {1110} \right\rangle } \right)_{1236}} \otimes {\left( {\left| {0000} \right\rangle + \left| {0011} \right\rangle + \left| {1101} \right\rangle - \left| {1110} \right\rangle } \right)_{4578}} \nonumber \\ &\quad+ {\left( {\left| {0001} \right\rangle + \left| {0010} \right\rangle + \left| {1100} \right\rangle - \left| {1111} \right\rangle } \right)_{1236}} \otimes {\left( {\left| {0001} \right\rangle - \left| {0010} \right\rangle + \left| {1100} \right\rangle + \left| {1111} \right\rangle } \right)_{4578}} \nonumber \\ &\quad+ {\left( {\left| {0100} \right\rangle + \left| {0111} \right\rangle + \left| {1001} \right\rangle - \left| {1010} \right\rangle } \right)_{1236}} \otimes {\left( { - \left| {0101} \right\rangle - \left| {0110} \right\rangle + \left| {1000} \right\rangle - \left| {1011} \right\rangle } \right)_{4578}} \nonumber \\ &\quad+ {\left( {\left| {0101} \right\rangle - \left| {0110} \right\rangle + \left| {1000} \right\rangle + \left| {1011} \right\rangle } \right)_{1236}} \otimes {\left( { - \left| {0100} \right\rangle + \left| {0111} \right\rangle + \left| {1001} \right\rangle + \left| {1010} \right\rangle } \right)_{4578}}{\text{ }}] \ .\end{align} \tag{ 19 }$$

By careful calculation, we can obtain,

$$\begin{align*} {\pi _{1236}} &= {\pi _{4578}} = {\pi _{1245}} = {\pi _{3678}} = {\pi _{1278}} = {\pi _{3456}} = \frac{1}{4};\nonumber\\ {\pi _{1348}} & = {\pi _{2567}} = {\pi _{1357}} = {\pi _{2468}} = {\pi _{1467}} = {\pi _{2358}}\nonumber\\ &= {\pi _{1568}} = {\pi _{2347}} = \frac{1}{8};\end{align*}$$

others

$$\begin{equation*}{\pi _{ijkl}} = T{r_{ijkl}}\rho _{ijkl}^2 = \frac{1}{{16}},ijkl = 1234,1235, \cdots ,5678.\end{equation*}$$

That is say, there are 56 of ${\pi _{ijkl}} = \frac{1}{{16}}$. 8 of ${\pi _{ijkl}} = \frac{1}{8}$. 6 of ${\pi _{ijkl}} = \frac{1}{4}$.

Therefore, we have

$$\begin{equation}{\pi _{{\text{ME}}}} = \frac{1}{{70}}\left( {56 \times \frac{1}{{16}} + 8 \times \frac{1}{8} + 6 \times \frac{1}{4}} \right) = \frac{3}{{35}}.\end{equation} \tag{ 20 }$$

According to the criterion for a maximally entangled eight-qubits [11], this state is a maximally entangled states.

In summary, we have presented a expand recurrence relationship between the maximally n-qubit entangled state and maximally n + 2-qubit entangled state. Furthermore, some new maximally four-, six-, eight-qubits states are given. We believe this method can play an important role in constructed maximally entangled states.

This work was supported by the National Natural Science Foundation of China (61705182), National Science Foundation of Shannxi Province (2017JQ6024).