**8. Double entanglement**

Let us introduce a register consisting of two obscure qubits (*L* ¼ 2) in the computational basis ^*i j*<sup>0</sup> � � � <sup>¼</sup> ^*<sup>i</sup>* � � � ⊗ ^*j* 0 � � � E as follows

$$\left| \Psi\_{ab}^{(n=2)}(L=2) \right\rangle = \left| \Psi\_{ab}(2) \right\rangle = \frac{\hat{B}\_{00'} \left| \hat{\mathbf{0}} \mathbf{0}' \right\rangle + \hat{B}\_{10'} \left| \hat{\mathbf{1}} \mathbf{0}' \right\rangle + \hat{B}\_{01'} \left| \hat{\mathbf{0}} \mathbf{1}' \right\rangle + \hat{B}\_{11'} \left| \hat{\mathbf{1}} \mathbf{1}' \right\rangle}{\sqrt{2}}, \tag{84}$$

determined by two-dimensional "vectors" (encoding obscure-quantum amplitudes)

$$
\hat{B}\_{ij'} = \begin{bmatrix} b\_{ij'} \\ \beta\_{ij'} \end{bmatrix}, \qquad i, j = \mathbf{0}, \mathbf{1}, \qquad j' = \mathbf{0}', \mathbf{1}', \tag{85}
$$

where *bi j*<sup>0</sup> ∈ are probability amplitudes for a set of pure states and *βi j*<sup>0</sup> ∈ ℝ are the corresponding membership amplitudes. By analogy with (43) and (46) the normalization factor in (84) is chosen so that

$$
\langle \Psi\_{ob}(\mathbf{2}) | \Psi\_{ob}(\mathbf{2}) \rangle = \mathbf{1}, \tag{86}
$$

if (cf. (14)–(15))

$$\left|\boldsymbol{b}\_{00'}\right|^2 + \left|\boldsymbol{b}\_{10'}\right|^2 + \left|\boldsymbol{b}\_{01'}\right|^2 + \left|\boldsymbol{b}\_{11'}\right|^2 = \mathbf{1},\tag{87}$$

$$
\rho\_{00'}^2 + \rho\_{10'}^2 + \rho\_{01'}^2 + \rho\_{11'}^2 = \mathbf{1}.\tag{88}
$$

A state of two qubits is "entangled", if it cannot be decomposed as a product of two one-qubit states, and otherwise it is "separable" (see, e.g. [1]). We define a product of two obscure qubits (43) as

$$|\Psi\_{ob}\rangle \otimes |\Psi\_{ob}'\rangle = \frac{\hat{A}\_0 \otimes\_H \hat{A}\_0'|\hat{0}\,\mathbf{0}'\rangle + \hat{A}\_1 \otimes\_H \hat{A}\_0'|\hat{1}\,\mathbf{0}'\rangle + \hat{A}\_0 \otimes\_H \hat{A}\_1'|\hat{0}\,\mathbf{1}'\rangle + \hat{A}\_1 \otimes\_H \hat{A}\_1'|\hat{1}\mathbf{1}'\rangle}{2},\tag{89}$$

where ⊗ *<sup>H</sup>* is the Hadamard product (53). Comparing (84) and (89) we obtain two sets of relations, for probability amplitudes and for membership amplitudes

$$b\_{ij'} = \frac{1}{\sqrt{2}} a\_i a\_{j'},\tag{90}$$

$$
\beta\_{ij'} = \frac{1}{\sqrt{2}} a\_i a\_{j'}, \qquad i, j = 0, 1, \qquad j' = 0', 1'. \tag{91}
$$

In this case, the relations (14)–(15) give (87)–(88).

Two obscure-quantum qubits are entangled, if their joint state (84) cannot be presented as a product of one qubit states (89), and in the opposite case the states are called totally separable. It follows from (90)–(91), that there are two general conditions for obscure qubits to be entangled

$$b\_{00'}b\_{11'} \neq b\_{10'}b\_{01'}, \qquad \text{or } \det \mathbf{b} \neq \mathbf{0}, \ \mathbf{b} = \begin{pmatrix} b\_{00'} & b\_{01'} \\ b\_{10'} & b\_{11'} \end{pmatrix}, \tag{92}$$

$$
\beta\_{00'}\beta\_{11'} \neq \beta\_{10'}\beta\_{01'}, \qquad \text{or } \det \mathfrak{J} \neq \mathbf{0}, \ \mathfrak{J} = \begin{pmatrix} \beta\_{00'} & \beta\_{01'} \\ \beta\_{10'} & \beta\_{11'} \end{pmatrix}. \tag{93}
$$

The first Eq. (92) is the entanglement relation for the standard qubit, while the second condition (93) is for the membership amplitudes of the two obscure qubit joint state (84). The presence of two different conditions (92)–(93) leads to new additional possibilities (which do not exist for ordinary qubits) for "partial" entanglement (or "partial" separability), when only one of them is fulfilled. In this case, the states can be entangled in one subspace (quantum or membership) but not in the other.

The measure of entanglement is numerically characterized by the concurrence. Taking into account the two conditions (92)–(93), we propose to generalize the notion of concurrence for two obscure qubits in two ways. First, we introduce the "vector obscure concurrence"

$$
\hat{\mathbf{C}}\_{\text{vect}} = \begin{bmatrix} \mathbf{C}\_q \\ \mathbf{C}^{(\mu)} \end{bmatrix} = \mathbf{2} \begin{bmatrix} |\det \mathbf{b}| \\ |\det \boldsymbol{\beta}| \end{bmatrix}, \tag{94}
$$

where *b* and *β* are defined in (92)–(93), and 0 ≤*Cq* ≤ 1, 0 ≤*C*ð Þ *<sup>μ</sup>* ≤1. The corresponding "scalar obscure concurrence" can be defined as

$$C\_{scal} = \sqrt{\frac{|\det b|^2 + |\det \beta|^2}{2}},\tag{95}$$

such that 0≤*Cscal* ≤1. Thus, two obscure qubits are totally separable, if *Cscal* ¼ 0. For instance, for an obscure analog of the (maximally entangled) Bell state

$$|\Psi\_{ab}(2)\rangle = \frac{1}{\sqrt{2}} \left( \begin{bmatrix} \mathbf{1} \\ \frac{1}{\sqrt{2}} \\ \mathbf{1} \\ \frac{1}{\sqrt{2}} \end{bmatrix} |\hat{\mathbf{0}}\mathbf{0}'\rangle + \begin{bmatrix} \mathbf{1} \\ \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix} |\hat{\mathbf{1}}\mathbf{1}'\rangle \right) \tag{96}$$

we obtain

$$
\hat{\mathbf{C}}\_{\text{vect}} = \begin{bmatrix} \mathbf{1} \\ \mathbf{1} \end{bmatrix}, \qquad \mathbf{C}\_{\text{scal}} = \mathbf{1}. \tag{97}
$$

A more interesting example is the "intermediately entangled" two obscure qubit state, e.g.

*Obscure Qubits and Membership Amplitudes DOI: http://dx.doi.org/10.5772/intechopen.98685*

$$|\Psi\_{ab}(2)\rangle = \frac{1}{\sqrt{2}} \left( \begin{bmatrix} 1 \\ 2 \\ 1 \\ \frac{1}{\sqrt{2}} \end{bmatrix} |\hat{\mathbf{0}}\mathbf{0}'\rangle + \begin{bmatrix} 1 \\ 4 \\ \frac{\sqrt{5}}{4} \end{bmatrix} |\hat{\mathbf{1}}\mathbf{0}'\rangle + \begin{bmatrix} \frac{\sqrt{3}}{4} \\ \frac{1}{4} \\ \frac{1}{2\sqrt{2}} \end{bmatrix} |\hat{\mathbf{0}}\mathbf{1}'\rangle + \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{4} \\ \frac{1}{4} \end{bmatrix} |\hat{\mathbf{1}}\mathbf{1}'\rangle \right),\tag{98}$$

where the amplitudes satisfy (87)–(88). If the Born-like rule (as in (13)) holds for the membership amplitudes, then the probabilities and membership functions of the states in (98) are

$$p\_{00'} = \frac{1}{4}, \quad p\_{10'} = \frac{1}{16}, \quad p\_{01} = \frac{3}{16}, \quad p\_{11'} = \frac{1}{2},\tag{99}$$

$$
\mu\_{00'} = \frac{1}{2}, \quad \mu\_{10'} = \frac{5}{16}, \quad \mu\_{01'} = \frac{1}{8}, \quad \mu\_{11'} = \frac{1}{16}. \tag{100}
$$

This means that, e.g., the state ^10<sup>0</sup> � � � will be measured with the quantum probability 1*=*16 and the membership function ("truth" value) 5*=*16. For the entangled obscure qubit (98) we obtain the concurrences

$$\begin{aligned} \hat{C}\_{\text{net}} &= \begin{bmatrix} \frac{1}{2}\sqrt{2} - \frac{1}{8}\sqrt{3} \\\\ \frac{1}{8}\sqrt{2}\sqrt{5} - \frac{1}{4}\sqrt{2} \end{bmatrix} = \begin{bmatrix} 0.491 \\ 0.042 \end{bmatrix}, \quad \mathcal{C}\_{\text{sal}} = \sqrt{\frac{53}{128} - \frac{1}{16}\sqrt{5} - \frac{1}{16}\sqrt{2}\sqrt{3}} \\ &= 0.348. \end{aligned} \tag{101}$$

In the vector representation (49)–(52) we have

$$\left|\hat{\mathbf{i}}\,\hat{\mathbf{j}}'\right> = \left|\hat{\mathbf{i}}\right> \otimes \left|\hat{\mathbf{j}}'\right> = \begin{bmatrix} \hat{\mathbf{e}}\_i \otimes\_K \hat{\mathbf{e}}\_{j'}\\ \boldsymbol{\varepsilon}\_i \otimes\_{K} \boldsymbol{\varepsilon}\_{j'} \end{bmatrix}, \qquad i, j = \mathbf{0}, \mathbf{1}, \quad j' = \mathbf{0}', \mathbf{1}', \tag{102}$$

where ⊗ *<sup>K</sup>* is the Kronecker product (48), and ^*ei*, *ε<sup>i</sup>* are defined in (50)–(51). Using (85) and the Kronecker-like product (49), we put (informally, with no summation)

$$
\hat{B}\_{i'}|\hat{i}j'\rangle = \begin{bmatrix} b\_{i'}\hat{e}\_i \otimes\_K \hat{e}\_{j'} \\ \beta\_{i'} e\_i \otimes\_{K^{\mathcal{E}\_{j'}}} \end{bmatrix}, \qquad i, j = \mathbf{0}, \mathbf{1}, \qquad j' = \mathbf{0}', \mathbf{1}'. \tag{103}
$$

To clarify our model, we show here a manifest form of the two obscure qubit state (98) in the vector representation

$$|\Psi\_{ab}(2)\rangle = \frac{1}{\sqrt{2}} \left( \begin{bmatrix} 1 \\ 1 \\ 2 \\ 1 \\ 0 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \\ \end{bmatrix} \right) + \begin{bmatrix} 0 \\ 1 \\ 4 \\ 1 \\ 0 \\ \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \\ \end{bmatrix} + \begin{bmatrix} 1 \\ 0 \\ \frac{\sqrt{3}}{4} \\ 0 \\ 1 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \\ \end{bmatrix} \right) + \begin{bmatrix} 1 \\ 1 \\ \frac{1}{\sqrt{2}} \\ 1 \\ 1 \\ \frac{1}{2\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \\ \end{pmatrix} \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \\ \frac{1}{4} \\ \frac{1}{4} \\ 1 \\ \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \\ \end{bmatrix} \right). \tag{104}$$

The states above may be called "symmetric two obscure qubit states". However, there are more general possibilities, as may be seen from the r.h.s. of (103) and (104), when the indices of the first and second rows do not coincide. This would allow more possible states, which we call "non-symmetric two obscure qubit states". It would be worthwhile to establish their possible physical interpretation.

The above constructions show that quantum computing using Kronecker obscure qubits can involve a rich structure of states, giving a more detailed description with additional variables reflecting vagueness.
