# Special core tensors of multi-qubit states and the concurrency of three
lines

Choong Pak<EMAIL_ADDRESS>Hishamuddin Zainuddin  Chan Kar
Tim222Corresponding author<EMAIL_ADDRESS>Sh. K. Said Husain

(Institute for Mathematical Research, Universiti Putra Malaysia, 43400
Serdang, Selangor, Malaysia.
Date: )

###### Abstract

Classification of multipartite states aims to obtain a set of operationally
useful and finite entanglement classes under the action of either local
unitary (LU) or stochastic local operation and classical communication
(SLOCC). In this work, we propose a computationally simple approach to find
these classes by using higher order singular value decomposition (HOSVD) and
the concurrency of three lines. Since HOSVD simultaneously diagonalizes the
one-body reduced density matrices (RDM) of multipartite states, the core
tensor of multipartite states is the pure-state representation of such
simultaneously diagonalized one-body RDM. We identified the special core
tensors of three and four qubits, which are also genuinely entangled by
default. The special core tensors are further categorized into families of
states based on their first $n$-mode singular values, $\sigma_{1}^{(i)2}$. The
current proposal is limited to multi-qubit system, but it scales well with
large multi-qubit systems and produces a finite number of families of states.

##  1 Introduction

Being a quantum resource under the local operation and classical communication
(LOCC) paradigm [1, 2], numerous efforts have been dedicated to understand
entanglement from various perspectives and mathematical tools [3, 4, 5, 6, 7,
8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27,
28]. To date, even though there is no single unified approach to describe
multipartite entanglement, most discussions focus around the operational
aspects of entanglement in quantum information processing tasks. Since the
local unitary (LU) or stochastic local operation and classical communication
(SLOCC) entanglement classes of multipartite states are claimed to be infinite
[29, 30], the current challenge in the classification of multipartite states
is to find a computationally simple approach that gives operationally
meaningful and finite classification results [31, 32].

Previously [33], we showed that higher order singular value decomposition
(HOSVD) [34, 35] simultaneously diagonalizes the one-body reduced density
matrices of three qubits. Furthermore, by finding all the solutions to the
all-orthogonality conditions of three qubits, we recovered all the special
states of three qubits [4]. The first $n$-mode singular values,
$\sigma_{1}^{(n)2}$, where $n=1,\,2,\,3$, can be used to plot a LU
entanglement polytope similar to that in [25]. However, as the number of
variables grows exponentially with the increase in the number of subsystems,
solving the all-orthogonality conditions of multipartite states is not a
feasible approach in generalizing the methodology to multipartite systems.

Before we proceed further, we would like to point out that from our previous
results, some special states of three qubits are specific cases to a more
generic setting. For example, the bi-separable states $C|AB$ with the first
$n$-mode singular values
$(\sigma_{1}^{(1)2},\,\sigma_{1}^{(2)2},\,\sigma_{1}^{(3)2})=(\sigma_{1}^{(1)2},\,\sigma_{1}^{(1)2},\,1)$

$\displaystyle\left|\text{Bi-
Sep}_{C|AB}\right\rangle=t_{111}\left|111\right\rangle+t_{221}\left|221\right\rangle$

and the three-qubit states with
$(\sigma_{1}^{(1)2},\,\sigma_{1}^{(2)2},\,\sigma_{1}^{(3)2})=(\frac{1}{2},\,\frac{1}{2},\,\sigma_{1}^{(3)2})$
are specific cases to the following Slice states,

$\displaystyle\left|\text{S}_{1}\right\rangle=t_{111}\left|111\right\rangle+t_{112}\left|112\right\rangle+t_{221}\left|221\right\rangle+t_{222}\left|222\right\rangle,$
$\displaystyle\bar{t}_{111}t_{112}+\bar{t}_{221}t_{222}=0,$

where $\sigma_{1}^{(3)2}>\sigma_{1}^{(1)2}=\sigma_{1}^{(2)2}$. Therefore, our
current work focuses on identifying the generic special states of a
multipartite system since the specific cases are inclusive to the generic
special states that we identified.

In this work, we propose a computationally simple approach to identify the
special states of multi-qubit core tensors. This is important because core
tensors are also the pure-state representation of multi-qubit states when
their one-body reduced density matrices (RDM) are simultaneously diagonalized.
Based on the concurrency of three lines [36], we convert the problem of
finding solutions to the set of all-orthogonality conditions into the problem
of satisfying a set of determinants to be zero. This conversion has the added
computational advantage in that satisfying the requirements for a set of
determinants to be zero is easier than finding the solutions to a set of
polynomial equations. To do so, we define a pair of conjugate concurrent
variables (CCV) so that the one-to-one correspondence between the algebraic
manipulations of a set of simultaneous equations and the geometrical idea
based on the concurrency of three lines is preserved. Then, we describe a
general algorithm of this approach and demonstrate it with the case of four
qubits. Even though our approach is unable to identify the generalized GHZ
states, these states have a very recognizable form.

We structure our paper as follows. In Section 2, we provide the original
definitions of matrix unfolding and HOSVD. We show that matrix unfolding is
related to the RDM of multipartite states and HOSVD simultaneously
diagonalizes the one-body RDM of multipartite states. In Section 3, we
summarize our previous results on three qubits, and show that the derivation
from our previous work is equivalent to a geometrical concept in projective
geometry, called the concurrency of three lines. By solving the concurrency of
three lines for three qubits, we identify all the special three-qubit core
tensors using this new approach. Finally, we state a general algorithm for
this approach on multi-qubit core tensors in Section 4, and demonstrate it
with the case of four qubits.

##  2 Matrix unfolding and higher order singular value decomposition

###  2.1 Matrix unfolding

The Hilbert space of a composite quantum system is given by the tensor product
of its subsystems’ Hilbert spaces. Because of this, the probability amplitudes
of multipartite states are elements of higher order tensors, allowing us to
make use of tensor decomposition in the classification of multipartite states
[20, 23, 26, 33]. In order to write down higher order tensors in a way that
obeys the matrix-tensor multiplication rules, a formalism called matrix
unfolding [34] or matricization [35] of tensors was previously introduced.

######  Definition 1 (Matrix unfolding [34]).

Let
$\Psi\in\mathbb{C}^{I_{1}}\otimes\ldots\otimes\mathbb{C}^{I_{n}}\otimes\ldots\otimes\mathbb{C}^{I_{N}}$
be an $N$th-order complex tensor. The $n$-th matrix unfolding, $\Psi_{(n)}$,
is a matrix of size $I_{n}\times(I_{n+1}\times I_{n+2}\times\ldots\times
I_{N}\times I_{1}\times I_{2}\times\ldots\times I_{n-1})$, whereby the tensor
element $\psi_{i_{1}i_{2}\ldots i_{n}\ldots i_{N}}$ will be at the position
with row index $i_{n}$ and column index

$\displaystyle(i_{n+1}-1)I_{n+2}I_{n+3}\ldots I_{N}I_{1}I_{2}\ldots
I_{n-1}+(i_{n+2}-1)I_{n+3}I_{n+4}\ldots I_{N}I_{1}I_{2}\ldots I_{n-1}$
$\displaystyle+\ldots+(i_{N}-1)I_{1}I_{2}\ldots
I_{n-1}+(i_{1}-1)I_{2}I_{3}\ldots I_{n-1}+(i_{2}-1)I_{3}I_{4}\ldots I_{n-1}$
$\displaystyle+\ldots+i_{n-1}.$ (1)

We redefine matrix unfolding by making use of the bra-ket notation.

######  Definition 2 (Matrix unfolding in bra-ket notation).

Let
$\Psi\in\mathbb{C}^{I_{1}}\otimes\ldots\otimes\mathbb{C}^{I_{n}}\otimes\ldots\otimes\mathbb{C}^{I_{N}}$
be an $N$th-order complex tensor. In the bra-ket notation, the $n$-th matrix
unfolding, $\Psi_{(n)}$, rewrites $\Psi$ into the following matrix form,

$\displaystyle\Psi_{(n)}=\sum_{i_{1}\ldots i_{N}}\psi_{i_{1}\ldots
i_{N}}\left|i_{n}\right\rangle\left\langle i_{n+1}\ldots i_{N}i_{1}\ldots
i_{n-1}\right|.$

From Definition 2, we propose the following. The proof can be found in
Appendix A.1.

######  Proposition 1 (Matrix unfolding and reduced density matrices).

The $n$-th matrix unfolding $\Psi_{(n)}$ of an $N$-th order tensor
$\Psi\in\mathbb{C}^{I_{1}}\otimes\ldots\otimes\mathbb{C}^{I_{n}}\otimes\ldots\otimes\mathbb{C}^{I_{N}}$
is related to its one-body and $(n-1)$-body reduced density matrices,
$\rho_{n}$ and $\rho_{n+1\,\ldots N\,1\,\ldots n-1}$ respectively, through the
following relations,

$\displaystyle\Psi_{(n)}\Psi_{(n)}^{\dagger}$ $\displaystyle=\rho_{n},$ (2)
$\displaystyle\Psi_{(n)}^{\text{T}}\bar{\Psi}_{(n)}$
$\displaystyle=\rho_{n+1\,\ldots N\,1\,\ldots n-1}.$ (3)

###  2.2 Higher order singular value decomposition

Next, we introduce higher order singular value decomposition (HOSVD) [34] and
its matrix unfolding variant [34, 23].

######  Theorem 1 (Higher order singular value decomposition [34]).

Let
$\Psi\in\mathbb{C}^{I_{1}}\otimes\ldots\otimes\mathbb{C}^{I_{n}}\otimes\ldots\otimes\mathbb{C}^{I_{N}}$
be an $N$th-order complex tensor. There exists a core tensor $\mathcal{T}$ of
$\Psi$ and a set of unitary matrices
$U^{(1)},\,\ldots,\,U^{(n)},\,\ldots,\,U^{(N)}$ such that

$\displaystyle\Psi=U^{(1)}\otimes U^{(2)}\otimes\ldots\otimes
U^{(n)}\otimes\ldots\otimes U^{(N)}\mathcal{T}.$ (4)

The core tensor $\mathcal{T}$ is also an $N$th-order complex tensor of which
the subtensors $\mathcal{T}_{i_{n}=\alpha}$, obtained by fixing the $n$-th
index to $\alpha$, have the properties of

  1. 1.

_All-orthogonality_ : Two subtensors $\mathcal{T}_{i_{n}=\alpha}$ and
$\mathcal{T}_{i_{n}=\beta}$ are orthogonal for all possible values of $n$,
$\alpha$ and $\beta$, subject to $\alpha\neq\beta$:

$\displaystyle\left\langle\mathcal{T}_{i_{n}=\alpha},\mathcal{T}_{i_{n}=\beta}\right\rangle$
$\displaystyle=\sum_{i_{1}i_{2}\ldots i_{n-1}i_{n+1}\ldots
i_{N}}\bar{t}_{i_{1}i_{2}\ldots i_{n-1}\alpha i_{n+1}\ldots
i_{N}}t_{i_{1}i_{2}\ldots i_{n-1}\beta i_{n+1}\ldots i_{N}}$
$\displaystyle=0\;\;\text{when}\;\;\alpha\neq\beta;$ (5)

  2. 2.

_Ordering_ :

$\displaystyle\left|\mathcal{T}_{i_{n}=1}\right|\geq\left|\mathcal{T}_{i_{n}=2}\right|\geq\ldots\geq\left|\mathcal{T}_{i_{n}=I_{n}}\right|\geq
0$ (6)

for all possible values of $n$,

where $t_{i_{1}i_{2}\ldots i_{N}}$ is the element of the tensor $\mathcal{T}$.
The Frobenius norm of the subtensors $\left|\mathcal{T}_{i_{n}=i}\right|$ is
given as

$\displaystyle\left|\mathcal{T}_{i_{n}=i}\right|$
$\displaystyle=\sqrt{\langle\mathcal{T}_{i_{n}=i},\,\mathcal{T}_{i_{n}=i}\rangle}$
$\displaystyle=\sqrt{\sum_{i_{1}=1}^{I_{1}}\ldots\sum_{i_{n-1}=1}^{I_{n-1}}\sum_{i_{n+1}=1}^{I_{n+1}}\ldots\sum_{i_{N}=1}^{I_{N}}\bar{t}_{i_{1}\ldots
i_{n-1}ii_{n+1}\ldots i_{N}}t_{i_{1}\ldots i_{n-1}ii_{n+1}\ldots i_{N}}}$
$\displaystyle=\sqrt{\sum_{i_{1}=1}^{I_{1}}\ldots\sum_{i_{n-1}=1}^{I_{n-1}}\sum_{i_{n+1}=1}^{I_{n+1}}\ldots\sum_{i_{N}=1}^{I_{N}}\left|t_{i_{1}\ldots
i_{n-1}ii_{n+1}\ldots i_{N}}\right|^{2}}.$ (7)

and is called the $n$-mode singular value of $\Psi$, $\sigma_{i}^{(n)}$.

######  Theorem 2 (Matrix unfolding of HOSVD [34, 23]).

Let
$\Psi\in\mathbb{C}^{I_{1}}\otimes\ldots\otimes\mathbb{C}^{I_{n}}\otimes\ldots\otimes\mathbb{C}^{I_{N}}$
be an $N$th-order complex tensor and $\mathcal{T}$ be its core tensor. The
matrix unfolding of $\Psi$ and $\mathcal{T}$ can be obtained as

$\displaystyle\Psi_{(n)}=U^{(n)}T_{(n)}(U^{(n+1)}\otimes
U^{(n+2)}\otimes\ldots\otimes U^{(N)}\otimes U^{(1)}\otimes
U^{(2)}\otimes\ldots\otimes U^{(n-1)})^{T},$ (8)

where $\Psi_{(n)}$ and $T_{(n)}$ are complex matrices of size
$I_{n}\times(I_{n+1}\times I_{n+2}\times\ldots\times I_{N}\times I_{1}\times
I_{2}\times\ldots\times I_{n-1})$, and $U^{(n)}$ are unitary matrices of size
$I_{n}\times I_{n}$.

Due to Proposition 1, we state the following. The proof can be found in
Appendix A.2.

######  Theorem 3 (HOSVD and one-body reduced density matrices).

Let
$\Psi\in\mathbb{C}^{I_{1}}\otimes\ldots\otimes\mathbb{C}^{I_{n}}\otimes\ldots\otimes\mathbb{C}^{I_{N}}$
be an $N$th-order complex tensor and $\mathcal{T}$ be its core tensor. HOSVD
simultaneously diagonalizes the set of one-body reduced density matrices of
multipartite states in such a way that the $n$-mode singular values are
ordered. The all-orthogonality conditions are the off-diagonal terms of the
set of one-body reduced density matrices.

##  3 Concurrency of three lines and three qubits

###  3.1 Classification of three qubits

In this section, we briefly discuss the methodology that we have used
previously in [33]. The all-orthogonality conditions of three qubits are given
as

$\displaystyle\bar{t}_{111}t_{211}+\bar{t}_{121}t_{221}+\bar{t}_{112}t_{212}+\bar{t}_{122}t_{222}$
$\displaystyle=0,$ (9)
$\displaystyle\bar{t}_{111}t_{121}+\bar{t}_{211}t_{221}+\bar{t}_{112}t_{122}+\bar{t}_{212}t_{222}$
$\displaystyle=0,$ (10)
$\displaystyle\bar{t}_{111}t_{112}+\bar{t}_{211}t_{212}+\bar{t}_{121}t_{122}+\bar{t}_{221}t_{222}$
$\displaystyle=0.$ (11)

By writing $\bar{t}_{111}$ and $t_{222}$ in terms of other variables,

$\displaystyle
t_{111}=-\frac{\bar{t}_{221}(t_{121}t_{212}-t_{122}t_{211})+t_{112}(\left|t_{212}\right|^{2}-\left|t_{122}\right|^{2})}{t_{212}\bar{t}_{211}-t_{122}\bar{t}_{121}},$
(12) $\displaystyle
t_{222}=\frac{\bar{t}_{112}(t_{121}t_{212}-t_{122}t_{211})+t_{221}(\left|t_{121}\right|^{2}-\left|t_{211}\right|^{2})}{\bar{t}_{212}t_{211}-\bar{t}_{122}t_{121}},$
(13)

we obtain

$\displaystyle(\bar{t}_{221}t_{121}-\bar{t}_{212}t_{112})(\bar{t}_{112}t_{212}+\bar{t}_{121}t_{221})$
$\displaystyle\qquad+(\bar{t}_{122}t_{112}-\bar{t}_{221}t_{211})(\bar{t}_{211}t_{221}+\bar{t}_{112}t_{122})$
$\displaystyle\qquad+(\bar{t}_{212}t_{211}-\bar{t}_{122}t_{121})(\bar{t}_{211}t_{212}+\bar{t}_{121}t_{122})=0.$
(14)

After expanding equation (14), it is possible to separate the real and
imaginary parts,

$\displaystyle\left|t_{112}\right|^{2}(\left|t_{122}\right|^{2}-\left|t_{212}\right|^{2})+\left|t_{121}\right|^{2}(\left|t_{221}\right|^{2}-\left|t_{122}\right|^{2})$
$\displaystyle\qquad+\left|t_{211}\right|^{2}(\left|t_{212}\right|^{2}-\left|t_{221}\right|^{2})=0,$
(15)
$\displaystyle\bar{t}_{112}\bar{t}_{221}(t_{122}t_{211}-t_{121}t_{212})+\bar{t}_{121}\bar{t}_{212}(t_{112}t_{221}-t_{122}t_{211})$
$\displaystyle\qquad+\bar{t}_{122}\bar{t}_{211}(t_{121}t_{212}-t_{112}t_{221})=0.$
(16)

Equation (15) is the basis to our previous work since it provides explicit
relationship between the first $n$-mode singular values, $\sigma_{1}^{(i)2}$,
i.e.

$\displaystyle\left|t_{112}\right|^{2}\left[\sigma_{1}^{(1)2}-\sigma_{1}^{(2)2}\right]+\left|t_{211}\right|^{2}\left[\sigma_{1}^{(2)2}-\sigma_{1}^{(3)2}\right]+\left|t_{121}\right|^{2}\left[\sigma_{1}^{(3)2}-\sigma_{1}^{(1)2}\right]$
$\displaystyle\qquad=0.$ (17)

On the other hand, equation (16) fixes a relative phase of the three-qubit
states. Since our results are based on the first $n$-mode singular values
$\sigma_{1}^{(i)2}$, the relative phase does not affect our results.
Example: Consider the following state $\left|\psi_{1}\right\rangle$,

$\displaystyle\left|\psi_{1}\right\rangle$
$\displaystyle=t_{111}\left|111\right\rangle+t_{112}\left|112\right\rangle+t_{112}\left|121\right\rangle+t_{122}\left|122\right\rangle$
$\displaystyle\quad+t_{211}\left|211\right\rangle+t_{212}\left|212\right\rangle+t_{212}\left|221\right\rangle+t_{222}\left|222\right\rangle,$

where it satisfies one of the bi-separable conditions $A|BC$,
$t_{121}t_{212}=t_{112}t_{221}$. Hence, equation (16) is satisfied. The all-
orthogonality conditions are

$\displaystyle\bar{t}_{111}t_{211}+2\bar{t}_{112}t_{212}+\bar{t}_{122}t_{222}$
$\displaystyle=0,$
$\displaystyle\bar{t}_{111}t_{112}+\bar{t}_{211}t_{212}+\bar{t}_{112}t_{122}+\bar{t}_{212}t_{222}$
$\displaystyle=0.$

The state $\left|\psi_{1}\right\rangle$ has the same property
($\sigma_{1}^{(2)2}=\sigma_{1}^{(3)2}\neq\sigma_{1}^{(1)2}$) as the Slice
state $\left|\text{S}_{3}\right\rangle$,

$\displaystyle\left|\text{S}_{3}\right\rangle$
$\displaystyle=t_{111}\left|111\right\rangle+t_{122}\left|122\right\rangle+t_{211}\left|211\right\rangle+t_{222}\left|222\right\rangle$

with all-orthogonality condition

$\displaystyle\bar{t}_{111}t_{211}+\bar{t}_{122}t_{222}=0.$

Under a coarser classification procedure provided by equation (17), they
belong to the same family of states.

###  3.2 Concurrency of three lines

Now, let $L_{1},\,L_{2},\,L_{3}$ to be three lines intersecting at one point
$(x,\,y)$,

$\displaystyle L_{1}$ $\displaystyle\equiv a_{1}x+b_{1}y+c_{1}=0,$ (18)
$\displaystyle L_{2}$ $\displaystyle\equiv a_{2}x+b_{2}y+c_{2}=0,$ (19)
$\displaystyle L_{3}$ $\displaystyle\equiv a_{3}x+b_{3}y+c_{3}=0,$ (20)

where $a_{i},\,b_{i},\,c_{i}$ for $i=1,\,2,\,3$ are some coefficients and
$x,\,y$ are indeterminates. In order to find the solution $(x,\,y)$ to the set
of lines, we can substitute $x$ from $L_{1}$ and $y$ from $L_{2}$ into $L_{3}$
to get

$\displaystyle
a_{3}(b_{1}c_{2}-b_{2}c_{1})+b_{3}(a_{2}c_{1}-a_{1}c_{2})+c_{3}(a_{1}b_{2}-a_{2}b_{1})=0.$
(21)

Equation (21) can be written into a more concise form as

$\displaystyle\begin{vmatrix}a_{1}&b_{1}&c_{1}\\\ a_{2}&b_{2}&c_{2}\\\
a_{3}&b_{3}&c_{3}\end{vmatrix}=0,$ (22)

which is called the concurrency of three lines [36]. By comparison, it is
obvious that the derivation in Section 3.1 is the same as the concurrency of
three lines, with $(x,\,y)=(\bar{t}_{111},\,t_{222})$. Since we did not
specify the underlying field when deriving equation (21), the concurrency of
three lines can be applied to complex field as long as the inherent properties
of $x$ and $y$ (i.e. complex conjugate of $x$ and $y$) are not being used
while solving the set of equations algebraically [37].

The biggest advantage in using the concurrency of three lines is that it is
easier to find the solutions in the determinant form (22) in contrary to the
polynomial form (21). There are two ways for a determinant to be zero,

  1. 1.

At least one row (column) of the determinant is zero.

  2. 2.

At least one row (column) of the determinant is linearly dependent to the
other row (column).

However, the linear dependence between rows (columns) of a determinant can
always be decomposed into a combination of the former scenario, i.e. one row
(column) of the determinant is zero. For instance, the linear dependence
$L_{1}=k^{\prime}_{2}L_{2}+k^{\prime}_{3}L_{3}$, where
$k^{\prime}_{i}=-\frac{k_{i}}{k_{1}}$ and $i=2,\,3$ can be written as

$\displaystyle\begin{vmatrix}a_{1}&b_{1}&c_{1}\\\ a_{2}&b_{2}&c_{2}\\\
a_{3}&b_{3}&c_{3}\end{vmatrix}$
$\displaystyle=\begin{vmatrix}k^{\prime}_{2}a_{2}&k^{\prime}_{2}b_{2}&k^{\prime}_{2}c_{2}\\\
a_{2}&b_{2}&c_{2}\\\
a_{3}&b_{3}&c_{3}\end{vmatrix}+\begin{vmatrix}k^{\prime}_{3}a_{3}&k^{\prime}_{3}b_{3}&k^{\prime}_{3}c_{3}\\\
a_{2}&b_{2}&c_{2}\\\ a_{3}&b_{3}&c_{3}\end{vmatrix}$
$\displaystyle=\begin{vmatrix}0&0&0\\\ a_{2}&b_{2}&c_{2}\\\
a_{3}&b_{3}&c_{3}\end{vmatrix}+\begin{vmatrix}0&0&0\\\ a_{2}&b_{2}&c_{2}\\\
a_{3}&b_{3}&c_{3}\end{vmatrix}$
$\displaystyle=\begin{vmatrix}k^{\prime}_{2}a_{2}&k^{\prime}_{2}b_{2}&k^{\prime}_{2}c_{2}\\\
0&0&0\\\
a_{3}&b_{3}&c_{3}\end{vmatrix}+\begin{vmatrix}k^{\prime}_{3}a_{3}&k^{\prime}_{3}b_{3}&k^{\prime}_{3}c_{3}\\\
a_{2}&b_{2}&c_{2}\\\ 0&0&0\end{vmatrix}=\ldots$

Therefore, we will be able to identify all unique solutions to equation (22)
by studying only the former scenario.

Furthermore, in order to simplify the computational process, we focus only on
the minimum requirement for a determinant to be zero, i.e. when one row
(column) of the determinant is zero. This consideration does not generate
generalized GHZ states, however it can be recognized as

$\displaystyle\left|\text{GHZ}\right\rangle=t_{1\ldots 1}\left|1\ldots
1\right\rangle+t_{2\ldots 2}\left|2\ldots 2\right\rangle.$ (23)

###  3.3 Conjugate concurrent variables

Since the inherent properties of $x$ and $y$ cannot be used while solving the
all-orthogonality conditions, we can make use of the concurrency of three
lines. We formalize this idea with the following definition.

######  Definition 3 (Conjugate concurrent variables).

Let $\\{L_{i}\\}$ be a set of all-orthogonality conditions. A pair of
conjugate concurrent variables $(x,\,y)$ satisfies the following two
criteria:-

  1. 1.

The relative phase between the conjugate concurrent variables is preserved
throughout the all-orthogonality conditions;

  2. 2.

The pair of conjugate concurrent variables must exist in every all-
orthogonality conditions.

The first criterion is stated so that we do not make use of the inherent
properties of the conjugate concurrent variables (CCV). From equations (9) to
(11), there are four pairs of CCV: $(\bar{t}_{111},\,t_{222})$,
$(\bar{t}_{112},\,t_{221})$, $(\bar{t}_{121},\,t_{212})$ and
$(\bar{t}_{122},\,t_{211})$. Meanwhile, $(\bar{t}_{111},\,\bar{t}_{112})$ is
not a pair of CCV because the relative phase between $\bar{t}_{111}$ and
$\bar{t}_{112}$ changes in equation (11). One has to make use of the inherent
property of $\bar{t}_{112}$ as a complex variable to be able to solve the all-
orthogonality conditions.

The second criterion is required so that the solutions that we found will
satisfy every all-orthogonality conditions. This implies that that the current
approach is limited to multi-qubit systems. For instance, if we consider the
all-orthogonality conditions of a $(2\times 2\times 3)$-system,

$\displaystyle\bar{t}_{111}t_{211}+\bar{t}_{112}t_{212}+\bar{t}_{113}t_{213}+\bar{t}_{121}t_{221}+\bar{t}_{122}t_{222}+\bar{t}_{123}t_{223}$
$\displaystyle=0,$
$\displaystyle\bar{t}_{111}t_{121}+\bar{t}_{211}t_{221}+\bar{t}_{112}t_{122}+\bar{t}_{212}t_{222}+\bar{t}_{113}t_{123}+\bar{t}_{213}t_{223}$
$\displaystyle=0,$
$\displaystyle\bar{t}_{111}t_{112}+\bar{t}_{121}t_{122}+\bar{t}_{211}t_{212}+\bar{t}_{221}t_{222}$
$\displaystyle=0,$
$\displaystyle\bar{t}_{111}t_{113}+\bar{t}_{121}t_{123}+\bar{t}_{211}t_{213}+\bar{t}_{221}t_{223}$
$\displaystyle=0,$
$\displaystyle\bar{t}_{112}t_{113}+\bar{t}_{122}t_{123}+\bar{t}_{212}t_{213}+\bar{t}_{222}t_{223}$
$\displaystyle=0,$

we can see that some of the variables do not exist in every all-orthogonality
conditions. Therefore, we say that a pair of CCV does not exist in this
$(2\times 2\times 3)$-system.

###  3.4 Special three-qubit core tensors by concurrency of three lines

From equations (9) to (11), the concurrency of three lines for all-
orthogonality conditions of three qubits is given by

$\displaystyle\begin{vmatrix}t_{211}&\bar{t}_{122}&\bar{t}_{121}t_{221}+\bar{t}_{112}t_{212}\\\
t_{121}&\bar{t}_{212}&\bar{t}_{211}t_{221}+\bar{t}_{112}t_{122}\\\
t_{112}&\bar{t}_{221}&\bar{t}_{211}t_{212}+\bar{t}_{121}t_{122}\end{vmatrix}=0,$
(24)

where $(x,\,y)=(\bar{t}_{111},t_{222})$. The details of our calculations will
be shown in Appendix B. The results are summarized in Table
LABEL:TableConcurrencyThreeQubits.

From Table LABEL:TableConcurrencyThreeQubits, we can see that by considering
only the minimum requirements to satisfy equation (24), it is enough to
recover all the generic special states of three qubits besides the generalized
GHZ states,

$\displaystyle\left|\text{GHZ}\right\rangle=t_{111}\left|111\right\rangle+t_{222}\left|222\right\rangle.$
(25)

Some of the special states that we have identified are not generic because of
the ordering property of higher order singular value decomposition (HOSVD). As
an example, if we study the following three-qubit state,

$\displaystyle\left|\psi\right\rangle=t_{121}\left|121\right\rangle+t_{122}\left|122\right\rangle+t_{211}\left|211\right\rangle+t_{212}\left|212\right\rangle,$
$\displaystyle\bar{t}_{211}t_{212}+\bar{t}_{121}t_{122}=0,$

the first $n$-mode singular values are given as

$\displaystyle\sigma_{1}^{(1)2}$
$\displaystyle=\left|t_{121}\right|^{2}+\left|t_{122}\right|^{2}=\sigma_{2}^{(2)2},$
$\displaystyle\sigma_{1}^{(2)2}$
$\displaystyle=\left|t_{211}\right|^{2}+\left|t_{212}\right|^{2}=\sigma_{2}^{(1)2},$
$\displaystyle\sigma_{1}^{(3)2}$
$\displaystyle=\left|t_{121}\right|^{2}+\left|t_{211}\right|^{2}.$

Due to the ordering property, equality is possible only when
$(\sigma_{1}^{(1)2},\,\sigma_{1}^{(2)2},\,\sigma_{1}^{(3)2})=(\frac{1}{2},\,\frac{1}{2},\,\sigma_{1}^{(3)2})$.
Therefore, it is not a generic special state of three qubits.

Table 1: Special three-qubit core tensors due to the concurrency of three lines for all-orthogonality conditions of three qubits Row (Column) checking |  States
---|---
1\. Column 1 = 0 |  $\left|\text{B}_{1}\right\rangle=t_{111}\left|111\right\rangle+t_{122}\left|122\right\rangle+t_{212}\left|212\right\rangle+t_{221}\left|221\right\rangle$
2\. Column 2 = 0 |  $\left|\text{B}_{2}\right\rangle=t_{112}\left|112\right\rangle+t_{121}\left|121\right\rangle+t_{211}\left|211\right\rangle+t_{222}\left|222\right\rangle$
3\. Column 3 = 0 (Non-generic) |  (a) $t_{112}=t_{221}=0$
|  (a)
$\left|\psi\right\rangle=t_{121}\left|121\right\rangle+t_{122}\left|122\right\rangle+t_{211}\left|211\right\rangle+t_{212}\left|212\right\rangle,$
|  (a) $\bar{t}_{211}t_{212}+\bar{t}_{121}t_{122}=0$
|  (b) $t_{121}=t_{212}=0$
|  (b)
$\left|\psi\right\rangle=t_{112}\left|112\right\rangle+t_{122}\left|122\right\rangle+t_{211}\left|211\right\rangle+t_{221}\left|221\right\rangle,$
|  (b) $\bar{t}_{211}t_{221}+\bar{t}_{112}t_{122}=0$
|  (c) $t_{122}=t_{211}=0$
|  (c)
$\left|\psi\right\rangle=t_{112}\left|112\right\rangle+t_{121}\left|121\right\rangle+t_{212}\left|212\right\rangle+t_{221}\left|221\right\rangle,$
|  (c) $\bar{t}_{121}t_{221}+\bar{t}_{112}t_{212}=0$
4\. Row 1 = 0 |  (a) $t_{111}=t_{222}=0$
|  (a) Same as 3(c)
|  (b) $t_{112}=t_{221}=0$
|  (b)
$\left|\text{S}_{2}\right\rangle=t_{111}\left|111\right\rangle+t_{121}\left|121\right\rangle+t_{212}\left|212\right\rangle+t_{222}\left|222\right\rangle,$
|  (b) $\bar{t}_{111}t_{121}+\bar{t}_{212}t_{222}=0$
|  (c) $t_{121}=t_{212}=0$
|  (c)
$\left|\text{S}_{1}\right\rangle=t_{111}\left|111\right\rangle+t_{112}\left|112\right\rangle+t_{221}\left|221\right\rangle+t_{222}\left|222\right\rangle,$
|  (c) $\bar{t}_{111}t_{112}+\bar{t}_{221}t_{222}=0$
5\. Row 2 = 0 |  (a) $t_{111}=t_{222}=0$
|  (a) Same as 3(b)
|  (b) $t_{112}=t_{221}=0$
|  (b)
$\left|\text{S}_{3}\right\rangle=t_{111}\left|111\right\rangle+t_{122}\left|122\right\rangle+t_{211}\left|211\right\rangle+t_{222}\left|222\right\rangle,$
|  (b) $\bar{t}_{111}t_{211}+\bar{t}_{122}t_{222}=0$
|  (c) $t_{121}=t_{212}=0$
|  (c) Same as 4 (c)
6\. Row 3 = 0 |  (a) $t_{111}=t_{222}=0$
|  (a) Same as 3(a)
|  (b) $t_{122}=t_{211}=0$
|  (b) Same as 4 (b)
|  (c) $t_{121}=t_{212}=0$
|  (c) Same as 5 (c)

##  4 Four qubits and beyond

###  4.1 Generalization to multi-qubit states

For multi-qubit states, we can generalize our approach by the following
algorithm.

  1. 1.

Select a pair of conjugate concurrent variables (CCV) and formulate the
concurrency of three lines accordingly;

  2. 2.

Perform row (column) checking on the concurrency of three lines;

  3. 3.

For a system of all-orthogonality conditions without a pair of CCV, find its
family of states;

  4. 4.

For a system of all-orthogonality conditions with a pair of CCV, select
another pair of CCV and formulate the next iteration of concurrency of three
lines accordingly;

  5. 5.

The process stops when at most two all-orthogonality conditions are left.

There are $n$ number of all-orthogonality conditions for $n$-qubit states. In
order to formulate the concurrency of three lines for the set of all-
orthogonality conditions, we need to exhaust all the possible combinations
between the $n$ number of all-orthogonality conditions. This is a
combinatorial problem of selecting three out of $n$-th all-orthogonality
conditions, therefore the number of simultaneous concurrency of three lines
that we can form is given by $\frac{n!}{3!(n-3)!}$.

In order to explore all the minimum requirements for the set of concurrency of
three lines to be true, we need to have at least $n-2$ number of rows to be
zero during the row checking. This is another combinatorial problem of
selecting $n-2$ out of $n$ rows, which requires $\frac{n!}{2!(n-2)!}$ of row
checking in total for one iteration. For column checking, we always need three
checks regardless of the number of simultaneous concurrency of three lines
that we have.

pt$n$-qubit systemptAll-orthogonality conditionsMatrix unfoldingsConjugate
concurrent variables (CCV)ptCCVptFamily of statesptCCVptConcurrency of three
linesRow iterationsColumn iterationsptRow/Column iterationFalseTrue Figure 1:
Flowchart for the identification of special multi-qubit states

Since we explore every requirements to satisfy the concurrency of three lines
for the set of all-orthogonality conditions by going through several
iterations, the choice of CCV does not matter.

###  4.2 Special four-qubit core tensors by concurrency of three lines

The all-orthogonality conditions for four qubits are given as

$\displaystyle\bar{t}_{1111}t_{2111}+\bar{t}_{1112}t_{2112}+\bar{t}_{1121}t_{2121}+\bar{t}_{1122}t_{2122}+\bar{t}_{1211}t_{2211}+\bar{t}_{1212}t_{2212}$
$\displaystyle\qquad+\bar{t}_{1221}t_{2221}+\bar{t}_{1222}t_{2222}=0,$ (26)
$\displaystyle\bar{t}_{1111}t_{1211}+\bar{t}_{2111}t_{2211}+\bar{t}_{1112}t_{1212}+\bar{t}_{2112}t_{2212}+\bar{t}_{1121}t_{1221}+\bar{t}_{2121}t_{2221}$
$\displaystyle\qquad+\bar{t}_{1122}t_{1222}+\bar{t}_{2122}t_{2222}=0,$ (27)
$\displaystyle\bar{t}_{1111}t_{1121}+\bar{t}_{1211}t_{1221}+\bar{t}_{2111}t_{2121}+\bar{t}_{2211}t_{2221}+\bar{t}_{1112}t_{1122}+\bar{t}_{1212}t_{1222}$
$\displaystyle\qquad+\bar{t}_{2112}t_{2122}+\bar{t}_{2212}t_{2222}=0,$ (28)
$\displaystyle\bar{t}_{1111}t_{1112}+\bar{t}_{1121}t_{1122}+\bar{t}_{1211}t_{1212}+\bar{t}_{1221}t_{1222}+\bar{t}_{2111}t_{2112}+\bar{t}_{2121}t_{2122}$
$\displaystyle\qquad+\bar{t}_{2211}t_{2212}+\bar{t}_{2221}t_{2222}=0.$ (29)

We can formulate four concurrency of three lines from equations (26) to (29).
By selecting $\bar{t}_{1111}$ and $t_{2222}$ as the pair of conjugate
concurrent variables (CCV), the first iteration is given by

$\displaystyle\begin{vmatrix}t_{2111}&\bar{t}_{1222}&c_{1}\\\
t_{1211}&\bar{t}_{2122}&c_{2}\\\
t_{1121}&\bar{t}_{2212}&c_{3}\end{vmatrix}=0,$ (30)
$\displaystyle\begin{vmatrix}t_{2111}&\bar{t}_{1222}&c_{1}\\\
t_{1211}&\bar{t}_{2122}&c_{2}\\\
t_{1112}&\bar{t}_{2221}&c_{4}\end{vmatrix}=0,$ (31)
$\displaystyle\begin{vmatrix}t_{2111}&\bar{t}_{1222}&c_{1}\\\
t_{1121}&\bar{t}_{2212}&c_{3}\\\
t_{1112}&\bar{t}_{2221}&c_{4}\end{vmatrix}=0,$ (32)
$\displaystyle\begin{vmatrix}t_{1211}&\bar{t}_{2122}&c_{2}\\\
t_{1121}&\bar{t}_{2212}&c_{3}\\\
t_{1112}&\bar{t}_{2221}&c_{4}\end{vmatrix}=0,$ (33)

where

$\displaystyle c_{1}$
$\displaystyle=\bar{t}_{1112}t_{2112}+\bar{t}_{1121}t_{2121}+\bar{t}_{1122}t_{2122}+\bar{t}_{1211}t_{2211}+\bar{t}_{1212}t_{2212}+\bar{t}_{1221}t_{2221},$
(34) $\displaystyle c_{2}$
$\displaystyle=\bar{t}_{2111}t_{2211}+\bar{t}_{1112}t_{1212}+\bar{t}_{2112}t_{2212}+\bar{t}_{1121}t_{1221}+\bar{t}_{2121}t_{2221}+\bar{t}_{1122}t_{1222},$
(35) $\displaystyle c_{3}$
$\displaystyle=\bar{t}_{1211}t_{1221}+\bar{t}_{2111}t_{2121}+\bar{t}_{2211}t_{2221}+\bar{t}_{1112}t_{1122}+\bar{t}_{1212}t_{1222}+\bar{t}_{2112}t_{2122},$
(36) $\displaystyle c_{4}$
$\displaystyle=\bar{t}_{1121}t_{1122}+\bar{t}_{1211}t_{1212}+\bar{t}_{1221}t_{1222}+\bar{t}_{2111}t_{2112}+\bar{t}_{2121}t_{2122}+\bar{t}_{2211}t_{2212}.$
(37)

As mentioned in Section 4.1, we need to allow two rows to be zero in order to
minimally satisfy the simultaneous concurrency of three lines from equations
(30) to (33). We need to perform six row-checkings, i.e. rows $(1\text{-}2)$,
$(1\text{-}3)$, $(1\text{-}4)$, $(2\text{-}3)$, $(2\text{-}4)$ and
$(3\text{-}4)$. The number of column checking we need to perform is 3. To
summarize, we performed a total of 73 of row and column checking across 3
iterations for four qubits. We used Mathematica to perform all the
computational tasks.

Our results can be summarized as follows:-

Table 2: Special four-qubit core tensors due to the concurrency of three lines for all-orthogonality conditions Cases |  States
---|---
1\. $\sigma_{1}^{(1)2}\neq\sigma_{1}^{(2)2}\neq\sigma_{1}^{(3)2}\neq\sigma_{1}^{(4)2}$ |  $\left|\psi\right\rangle=t_{1111}\left|1111\right\rangle+t_{1122}\left|1122\right\rangle+t_{1212}\left|1212\right\rangle+t_{1221}\left|1221\right\rangle+t_{2112}\left|2112\right\rangle$
|  $\left|\psi\right\rangle=$
$+t_{2121}\left|2121\right\rangle+t_{2211}\left|2211\right\rangle+t_{2222}\left|2222\right\rangle$
|
$\left|\psi\right\rangle=t_{1112}\left|1112\right\rangle+t_{1121}\left|1121\right\rangle+t_{1211}\left|1211\right\rangle+t_{1222}\left|1222\right\rangle+t_{2111}\left|2111\right\rangle$
|  $\left|\psi\right\rangle=$
$+t_{2122}\left|2122\right\rangle+t_{2212}\left|2212\right\rangle+t_{2221}\left|2221\right\rangle$
|
$\left|\psi\right\rangle=t_{1122}\left|1122\right\rangle+t_{1212}\left|1212\right\rangle+t_{1221}\left|1221\right\rangle+t_{2112}\left|2112\right\rangle+t_{2121}\left|2121\right\rangle$
|  $\left|\psi\right\rangle=$ $+t_{2211}\left|2211\right\rangle$
|
$\left|\psi\right\rangle=t_{1111}\left|1111\right\rangle+t_{1122}\left|1122\right\rangle+t_{1212}\left|1212\right\rangle+t_{2112}\left|2112\right\rangle+t_{2221}\left|2221\right\rangle$
|
$\left|\psi\right\rangle=t_{1111}\left|1111\right\rangle+t_{1122}\left|1122\right\rangle+t_{1221}\left|1221\right\rangle+t_{2121}\left|2121\right\rangle+t_{2212}\left|2212\right\rangle$
|
$\left|\psi\right\rangle=t_{1111}\left|1111\right\rangle+t_{1212}\left|1212\right\rangle+t_{1221}\left|1221\right\rangle+t_{2122}\left|2122\right\rangle+t_{2211}\left|2211\right\rangle$
|
$\left|\psi\right\rangle=t_{1112}\left|1112\right\rangle+t_{1211}\left|1211\right\rangle+t_{1222}\left|1222\right\rangle+t_{2121}\left|2121\right\rangle+t_{2212}\left|2212\right\rangle$
|
$\left|\psi\right\rangle=t_{1121}\left|1121\right\rangle+t_{1212}\left|1212\right\rangle+t_{2111}\left|2111\right\rangle+t_{2122}\left|2122\right\rangle+t_{2221}\left|2221\right\rangle$
|
$\left|\psi\right\rangle=t_{1112}\left|1112\right\rangle+t_{1221}\left|1221\right\rangle+t_{2121}\left|2121\right\rangle+t_{2211}\left|2211\right\rangle+t_{2222}\left|2222\right\rangle$
|
$\left|\psi\right\rangle=t_{1121}\left|1121\right\rangle+t_{1212}\left|1212\right\rangle+t_{2112}\left|2112\right\rangle+t_{2211}\left|2211\right\rangle+t_{2222}\left|2222\right\rangle$
|
$\left|\psi\right\rangle=t_{1122}\left|1122\right\rangle+t_{1211}\left|1211\right\rangle+t_{2112}\left|2112\right\rangle+t_{2121}\left|2121\right\rangle+t_{2222}\left|2222\right\rangle$
2\. $\sigma_{1}^{(i)2}=\sigma_{1}^{(j)2},\,i\neq j$ |  (a) $\sigma_{1}^{(1)2}=\sigma_{1}^{(2)2}$
|  (a)
$\left|\psi\right\rangle=t_{1111}\left|1111\right\rangle+t_{1112}\left|1112\right\rangle+t_{1121}\left|1121\right\rangle+t_{1122}\left|1122\right\rangle$
|  (a) $\left|\psi\right\rangle=$
$+t_{2211}\left|2211\right\rangle+t_{2212}\left|2212\right\rangle+t_{2221}\left|2221\right\rangle+t_{2222}\left|2222\right\rangle,$
|  (a)
$\bar{t}_{1111}t_{1121}+\bar{t}_{1112}t_{1122}+\bar{t}_{2211}t_{2221}+\bar{t}_{2212}t_{2222}=0,$
|  (a)
$\bar{t}_{1111}t_{1112}+\bar{t}_{1121}t_{1122}+\bar{t}_{2211}t_{2212}+\bar{t}_{2221}t_{2222}=0$
|  (b) $\sigma_{1}^{(1)2}=\sigma_{1}^{(3)2}$
|  (b)
$\left|\psi\right\rangle=t_{1111}\left|1111\right\rangle+t_{1112}\left|1112\right\rangle+t_{1211}\left|1211\right\rangle+t_{1212}\left|1212\right\rangle$
|  (b) $\left|\psi\right\rangle=$
$+t_{2121}\left|2121\right\rangle+t_{2122}\left|2122\right\rangle+t_{2221}\left|2221\right\rangle+t_{2222}\left|2222\right\rangle,$
|  (b)
$\bar{t}_{1111}t_{1211}+\bar{t}_{1112}t_{1212}+\bar{t}_{2121}t_{2221}+\bar{t}_{2122}t_{2222}=0,$
|  (b)
$\bar{t}_{1111}t_{1112}+\bar{t}_{1211}t_{1212}+\bar{t}_{2121}t_{2122}+\bar{t}_{2221}t_{2222}=0$
|  (c) $\sigma_{1}^{(1)2}=\sigma_{1}^{(4)2}$
|  (c)
$\left|\psi\right\rangle=t_{1111}\left|1111\right\rangle+t_{1121}\left|1121\right\rangle+t_{1211}\left|1211\right\rangle+t_{1221}\left|1221\right\rangle$
|  (c) $\left|\psi\right\rangle=$
$+t_{2112}\left|2112\right\rangle+t_{2122}\left|2122\right\rangle+t_{2212}\left|2212\right\rangle+t_{2222}\left|2222\right\rangle,$
|  (c)
$\bar{t}_{1111}t_{1211}+\bar{t}_{1121}t_{1221}+\bar{t}_{2112}t_{2212}+\bar{t}_{2122}t_{2222}=0,$
|  (c)
$\bar{t}_{1111}t_{1121}+\bar{t}_{1211}t_{1221}+\bar{t}_{2112}t_{2122}+\bar{t}_{2212}t_{2222}=0$
|  (d) $\sigma_{1}^{(2)2}=\sigma_{1}^{(3)2}$
|  (d)
$\left|\psi\right\rangle=t_{1111}\left|1111\right\rangle+t_{1112}\left|1112\right\rangle+t_{1221}\left|1221\right\rangle+t_{1222}\left|1222\right\rangle$
|  (d) $\left|\psi\right\rangle=$
$+t_{2111}\left|2111\right\rangle+t_{2112}\left|2112\right\rangle+t_{2221}\left|2221\right\rangle+t_{2222}\left|2222\right\rangle,$
|  (d)
$\bar{t}_{1111}t_{2111}+\bar{t}_{1112}t_{2112}+\bar{t}_{1221}t_{2221}+\bar{t}_{1222}t_{2222}=0,$
|  (d)
$\bar{t}_{1111}t_{1112}+\bar{t}_{1221}t_{1222}+\bar{t}_{2111}t_{2112}+\bar{t}_{2221}t_{2222}=0$
|  (e) $\sigma_{1}^{(2)2}=\sigma_{1}^{(4)2}$
|  (e)
$\left|\psi\right\rangle=t_{1111}\left|1111\right\rangle+t_{1121}\left|1121\right\rangle+t_{1212}\left|1212\right\rangle+t_{1222}\left|1222\right\rangle$
|  (e) $\left|\psi\right\rangle=$
$+t_{2111}\left|2111\right\rangle+t_{2121}\left|2121\right\rangle+t_{2212}\left|2212\right\rangle+t_{2222}\left|2222\right\rangle,$
|  (e)
$\bar{t}_{1111}t_{2111}+\bar{t}_{1121}t_{2121}+\bar{t}_{1212}t_{2212}+\bar{t}_{1222}t_{2222}=0,$
|  (e)
$\bar{t}_{1111}t_{1121}+\bar{t}_{1212}t_{1222}+\bar{t}_{2111}t_{2121}+\bar{t}_{2212}t_{2222}=0$
|  (f) $\sigma_{1}^{(3)2}=\sigma_{1}^{(4)2}$
|  (f)
$\left|\psi\right\rangle=t_{1111}\left|1111\right\rangle+t_{1122}\left|1122\right\rangle+t_{1211}\left|1211\right\rangle+t_{1222}\left|1222\right\rangle$
|  (f) $\left|\psi\right\rangle=$
$+t_{2111}\left|2111\right\rangle+t_{2122}\left|2122\right\rangle+t_{2211}\left|2211\right\rangle+t_{2222}\left|2222\right\rangle,$
|  (f)
$\bar{t}_{1111}t_{2111}+\bar{t}_{1122}t_{2122}+\bar{t}_{1211}t_{2211}+\bar{t}_{1222}t_{2222}=0,$
|  (f)
$\bar{t}_{1111}t_{1211}+\bar{t}_{1122}t_{1222}+\bar{t}_{2111}t_{2211}+\bar{t}_{2122}t_{2222}=0$
3\. $\sigma_{1}^{(i)2}=\sigma_{1}^{(j)2},\,\sigma_{1}^{(k)2}=\sigma_{1}^{(l)2},$ |  (a) $\sigma_{1}^{(1)2}=\sigma_{1}^{(2)2},\,\sigma_{1}^{(3)2}=\sigma_{1}^{(4)2}$
3. $i\neq j\neq k\neq l$ |  (a) $\left|\psi\right\rangle=t_{1111}\left|1111\right\rangle+t_{1122}\left|1122\right\rangle+t_{2211}\left|2211\right\rangle+t_{2222}\left|2222\right\rangle$
|  (b)
$\sigma_{1}^{(1)2}=\sigma_{1}^{(3)2},\,\sigma_{1}^{(2)2}=\sigma_{1}^{(4)2}$
|  (b)
$\left|\psi\right\rangle=t_{1111}\left|1111\right\rangle+t_{1212}\left|1212\right\rangle+t_{2121}\left|2121\right\rangle+t_{2222}\left|2222\right\rangle$
|  (c)
$\sigma_{1}^{(1)2}=\sigma_{1}^{(4)2},\,\sigma_{1}^{(2)2}=\sigma_{1}^{(3)2}$
|  (c)
$\left|\psi\right\rangle=t_{1111}\left|1111\right\rangle+t_{1221}\left|1221\right\rangle+t_{2112}\left|2112\right\rangle+t_{2222}\left|2222\right\rangle$
4\. $\sigma_{1}^{(i)2}=\sigma_{1}^{(j)2}=\sigma_{1}^{(k)2},$ |  (a) $\sigma_{1}^{(1)2}=\sigma_{1}^{(2)2}=\sigma_{1}^{(3)2}$
4. $i\neq j\neq k$ |  (a) $\left|\psi\right\rangle=t_{1111}\left|1111\right\rangle+t_{1112}\left|1112\right\rangle+t_{2221}\left|2221\right\rangle+t_{2222}\left|2222\right\rangle,$
|  (a) $\bar{t}_{1112}t_{1111}+\bar{t}_{2222}t_{2221}=0$
|  (b) $\sigma_{1}^{(1)2}=\sigma_{1}^{(2)2}=\sigma_{1}^{(4)2}$
|  (b)
$\left|\psi\right\rangle=t_{1111}\left|1111\right\rangle+t_{1121}\left|1121\right\rangle+t_{2212}\left|2212\right\rangle+t_{2222}\left|2222\right\rangle,$
|  (b) $\bar{t}_{1111}t_{1121}+\bar{t}_{2212}t_{2222}=0$
|  (c) $\sigma_{1}^{(1)2}=\sigma_{1}^{(3)2}=\sigma_{1}^{(4)2}$
|  (c)
$\left|\psi\right\rangle=t_{1111}\left|1111\right\rangle+t_{1211}\left|1211\right\rangle+t_{2122}\left|2122\right\rangle+t_{2222}\left|2222\right\rangle,$
|  (c) $\bar{t}_{1111}t_{1211}+\bar{t}_{2122}t_{2222}=0$
|  (d) $\sigma_{1}^{(2)2}=\sigma_{1}^{(3)2}=\sigma_{1}^{(4)2}$
|  (d)
$\left|\psi\right\rangle=t_{1111}\left|1111\right\rangle+t_{1222}\left|1222\right\rangle+t_{2111}\left|2111\right\rangle+t_{2222}\left|2222\right\rangle,$
|  (d) $\bar{t}_{1111}t_{2111}+\bar{t}_{1222}t_{2222}=0$

As expected, we did not recover the generalized GHZ states of four qubits by
using this approach, however it is recognized as

$\displaystyle\left|\text{GHZ}\right\rangle=t_{1111}\left|1111\right\rangle+t_{2222}\left|2222\right\rangle,$
(38)

with the first $n$-mode singular values
$\sigma_{1}^{(1)2}=\sigma_{1}^{(2)2}=\sigma_{1}^{(3)2}=\sigma_{1}^{(4)2}$.

##  5 Conclusion

In this work, we discussed how the matrix unfolding of a multipartite state is
related to its reduced density matrices, and how the higher order singular
value decomposition (HOSVD) is related to the simultaneous diagonalization of
one-body reduced density matrices (RDM). While these results are not new and
have been discussed in the past (for example, the $A\text{-}BC$, $B\text{-}AC$
and $C\text{-}AB$ bipartite decomposition of three qubits [38]; tensor
flattening [32]; trace decomposition [15, 16]; simultaneous diagonalization of
one-body RDM due to the momentum map and Cartan subalgebra [18]), we showed
these results from the perspectives of matrix unfolding and HOSVD.

From our previous work [33], we solved for the solutions to the set of all-
orthogonality conditions for three qubits and obtained some results equivalent
to the local unitary (LU) classification of three qubits [4]. Stemming from
the same methodology, we proposed a simpler coarse-grained method to identify
special multi-qubit core tensors by using the concurrency of three lines. A
detailed study on the special core tensors based on their entanglement and
geometrical properties is an interesting future direction we wish to pursue.

##  6 Acknowledgement

This research was supported by Fundamental Research Grant Scheme (FRGS) funded
by Ministry of Higher Education of Malaysia with reference code
FRGS/1/2019/STG02/UPM/02/3. The first author is sponsored by Ministry of
Education (MOE) under MyBrainSc.

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##  Appendix A Proofs to Proposition 1 and Theorem 3

###  A.1 Proposition 1

######  Proposition 1 (Matrix unfolding and reduced density matrices).

The $n$-th matrix unfolding $\Psi_{(n)}$ of an $N$-th order tensor
$\Psi\in\mathbb{C}^{I_{1}}\otimes\ldots\otimes\mathbb{C}^{I_{n}}\otimes\ldots\otimes\mathbb{C}^{I_{N}}$
is related to its one-body and $(n-1)$-body reduced density matrices through
the following relations respectively,

$\displaystyle\Psi_{(n)}\Psi_{(n)}^{\dagger}$ $\displaystyle=\rho_{n},$ (39)
$\displaystyle\Psi_{(n)}^{\text{T}}\bar{\Psi}_{(n)}$
$\displaystyle=\rho_{n+1\,\ldots N\,1\,\ldots n-1}.$ (40)

###### Proof.

First, we generalize the partial trace operation to multipartite states

$\displaystyle\text{Tr}_{n}(\left|i_{1}\ldots i_{N}\right\rangle\left\langle
j_{1}\ldots j_{N}\right|)$ $\displaystyle\quad=\left|i_{1}\ldots
i_{n-1}i_{n+1}\ldots i_{N}\right\rangle\left\langle j_{1}\ldots
j_{n-1}j_{n+1}\ldots
j_{N}\right|\text{Tr}(\left|i_{n}\right\rangle\left\langle j_{n}\right|)$
$\displaystyle\quad=\left|i_{1}\ldots i_{n-1}i_{n+1}\ldots
i_{N}\right\rangle\left\langle j_{1}\ldots j_{n-1}j_{n+1}\ldots
j_{N}\right|\langle i_{n}|i_{n}\rangle\langle j_{n}|i_{n}\rangle$
$\displaystyle\quad=\langle j_{n}|i_{n}\rangle\left|i_{1}\ldots
i_{n-1}i_{n+1}\ldots i_{N}\right\rangle\left\langle j_{1}\ldots
j_{n-1}j_{n+1}\ldots j_{N}\right|$
$\displaystyle\quad=\delta_{i_{n}j_{n}}\left|i_{1}\ldots i_{n-1}i_{n+1}\ldots
i_{N}\right\rangle\left\langle j_{1}\ldots j_{n-1}j_{n+1}\ldots j_{N}\right|$
(41)

such that the $(n-1)$-body reduced density matrix is given by

$\displaystyle\rho_{1\,\ldots n-1\,n+1\ldots N}$
$\displaystyle\quad=\sum_{\mathcal{I}}\sum_{\mathcal{J}}\psi_{i_{1}\ldots
i_{n}\ldots i_{N}}\bar{\psi}_{j_{1}\ldots i_{n}\ldots j_{N}}\left|i_{1}\ldots
i_{n-1}i_{n+1}\ldots i_{N}\right\rangle\left\langle j_{1}\ldots
j_{n-1}j_{n+1}\ldots j_{N}\right|,$

where $\mathcal{I}$ and $\mathcal{J}$ are the index sets.

Permutation matrices can act on the $(n-1)$-body reduced density matrix so
that the labeling of qubits can be rearranged. Particularly, we want a cyclic
permutation in such a way that qubits labeled 1 to $n-1$ are permuted to the
back,

$\displaystyle P_{\pi}\left(\rho_{1\,\ldots n-1\,n+1\ldots
N}\right)P^{\text{T}}_{\pi}$ $\displaystyle\,=\rho_{n+1\,\ldots N\,1\,\ldots
n-1}$ $\displaystyle\,=\sum_{\mathcal{I}}\sum_{\mathcal{J}}\psi_{i_{1}\ldots
i_{n}\ldots i_{N}}\bar{\psi}_{j_{1}\ldots i_{n}\ldots
j_{N}}P_{\pi}\left|i_{1}\ldots i_{n-1}i_{n+1}\ldots
i_{N}\right\rangle\left\langle j_{1}\ldots j_{n-1}j_{n+1}\ldots
j_{N}\right|P^{\text{T}}_{\pi}$
$\displaystyle\,=\sum_{\mathcal{I}}\sum_{\mathcal{J}}\psi_{i_{1}\ldots
i_{n}\ldots i_{N}}\bar{\psi}_{j_{1}\ldots i_{n}\ldots
j_{N}}\left|i_{n+1}\ldots i_{N}i_{1}\ldots i_{n-1}\right\rangle\left\langle
j_{n+1}\ldots j_{N}j_{1}\ldots j_{n-1}\right|$
$\displaystyle\,=\Psi_{(n)}^{\text{T}}\bar{\Psi}_{(n)},$

where $P_{\pi}$ is the matrix for the desired permutation and $\Psi_{(n)}$ is
the $n$-th matrix unfolding of $\Psi$.

In addition, one can perform partial trace operation $N-1$ times on
$N$-partite states to obtain a set of one-body reduced density matrices. From
equation (41), every time an $n$-partial trace operation is performed, a
Kronecker delta $\delta_{i_{n}j_{n}}$ will be produced. Thus, the one-body
reduced density matrices will have the following generic form,

$\displaystyle\rho_{n}=\sum_{\mathcal{I}}\sum_{\mathcal{J}}\psi_{i_{1}\ldots
i_{n}\ldots i_{N}}\bar{\psi}_{i_{1}\ldots j_{n}\ldots
i_{N}}\left|i_{n}\right\rangle\left\langle
j_{n}\right|=\Psi_{(n)}\Psi_{(n)}^{\dagger}.$ (42)

∎

###  A.2 Theorem 3

######  Theorem 3 (HOSVD and one-body reduced density matrices).

Let
$\Psi\in\mathbb{C}^{I_{1}}\otimes\ldots\otimes\mathbb{C}^{I_{n}}\otimes\ldots\otimes\mathbb{C}^{I_{N}}$
be an $N$th-order complex tensor and $\mathcal{T}$ be its core tensor. HOSVD
simultaneously diagonalizes the set of one-body reduced density matrices of
multipartite states in such a way that the $n$-mode singular values are
ordered.

###### Proof.

From equation (42), the summation of the two index sets $\mathcal{I}$ and
$\mathcal{J}$ is between two subtensors $\Psi_{i_{n}}$ and $\Psi_{j_{n}}$. Due
to Theorem 2, we can write

$\displaystyle\Psi_{(n)}\Psi_{(n)}^{\dagger}=\rho_{n}=U^{(n)}T_{(n)}T_{(n)}^{\dagger}U^{(n)\dagger}=U^{(n)}\rho_{n}^{d}U^{(n)\dagger},$
(43)

where $\rho_{n}^{d}=T_{(n)}T_{(n)}^{\dagger}$ is the $n$-th diagonalized one-
body reduced density matrix. The one-body reduced density matrix is
diagonalized because when $i_{n}=j_{n}$, we obtain the square of $n$-mode
singular values, $\sigma_{i}^{(n)2}$, whereas when $i_{n}\neq j_{n}$, we have
the all-orthogonality conditions, which are zero due to HOSVD. ∎

##  Appendix B Special three-qubit core tensors by concurrency of three lines

By definition, the all-orthogonality conditions of three qubits are

$\displaystyle\bar{t}_{111}t_{211}+\bar{t}_{121}t_{221}+\bar{t}_{112}t_{212}+\bar{t}_{122}t_{222}$
$\displaystyle=0,$ (44)
$\displaystyle\bar{t}_{111}t_{121}+\bar{t}_{211}t_{221}+\bar{t}_{112}t_{122}+\bar{t}_{212}t_{222}$
$\displaystyle=0,$ (45)
$\displaystyle\bar{t}_{111}t_{112}+\bar{t}_{211}t_{212}+\bar{t}_{121}t_{122}+\bar{t}_{221}t_{222}$
$\displaystyle=0.$ (46)

By writing $\bar{t}_{111}$ and $t_{222}$ in terms of other unknowns, we can
reformulate the solutions to the above all-orthogonality conditions in the
form of concurrency of three lines,

$\displaystyle\begin{vmatrix}t_{211}&\bar{t}_{122}&\bar{t}_{121}t_{221}+\bar{t}_{112}t_{212}\\\
t_{121}&\bar{t}_{212}&\bar{t}_{211}t_{221}+\bar{t}_{112}t_{122}\\\
t_{112}&\bar{t}_{221}&\bar{t}_{211}t_{212}+\bar{t}_{121}t_{122}\end{vmatrix}=0,$
(47)

where $(x,\,y)=(\bar{t}_{111},t_{222})$. Now, we study all possible solutions
to the above determinant.

### B.1 Column 1 = 0: $t_{112}=t_{121}=t_{211}=0$

We have

$\displaystyle\bar{t}_{122}t_{222}$ $\displaystyle=0,$ (48)
$\displaystyle\bar{t}_{212}t_{222}$ $\displaystyle=0,$ (49)
$\displaystyle\bar{t}_{221}t_{222}$ $\displaystyle=0.$ (50)

Let $t_{222}=0$, we have
$\left|\text{B}_{1}\right\rangle=t_{111}\left|111\right\rangle+t_{122}\left|122\right\rangle+t_{212}\left|212\right\rangle+t_{221}\left|221\right\rangle$.

### B.2 Column 2 = 0: $t_{122}=t_{212}=t_{221}=0$

We have

$\displaystyle\bar{t}_{111}t_{211}$ $\displaystyle=0,$ (51)
$\displaystyle\bar{t}_{111}t_{121}$ $\displaystyle=0,$ (52)
$\displaystyle\bar{t}_{111}t_{112}$ $\displaystyle=0.$ (53)

Let $t_{111}=0$, we have
$\left|\text{B}_{2}\right\rangle=t_{112}\left|112\right\rangle+t_{121}\left|121\right\rangle+t_{211}\left|211\right\rangle+t_{222}\left|222\right\rangle$.

### B.3 Column 3 = 0

We have

$\displaystyle\bar{t}_{121}t_{221}+\bar{t}_{112}t_{212}$ $\displaystyle=0,$
(54) $\displaystyle\bar{t}_{211}t_{221}+\bar{t}_{112}t_{122}$
$\displaystyle=0,$ (55)
$\displaystyle\bar{t}_{211}t_{212}+\bar{t}_{121}t_{122}$ $\displaystyle=0,$
(56)

in addition to the original all-orthogonality conditions that have to be
satisfied, which are reduced to

$\displaystyle\bar{t}_{111}t_{211}+\bar{t}_{122}t_{222}$ $\displaystyle=0,$
(57) $\displaystyle\bar{t}_{111}t_{121}+\bar{t}_{212}t_{222}$
$\displaystyle=0,$ (58)
$\displaystyle\bar{t}_{111}t_{112}+\bar{t}_{221}t_{222}$ $\displaystyle=0.$
(59)

From equations (57) and (59), since we are looking for minimum requirements to
satisfy the set of equations, we have to let $t_{111}=t_{222}=0$. From
equations (54) and (55), we have

$\displaystyle\frac{\bar{t}_{112}}{t_{221}}=-\frac{\bar{t}_{121}}{t_{212}}=-\frac{\bar{t}_{211}}{t_{122}},$
(60)

but

$\displaystyle\frac{\bar{t}_{121}}{t_{212}}=-\frac{\bar{t}_{211}}{t_{122}}$
(61)

from equation (56). In order to resolve this contradiction, we consider the
following possibilities:-

  1. 1.

$t_{112}=t_{221}=0$
$\left|\psi\right\rangle=t_{121}\left|121\right\rangle+t_{122}\left|122\right\rangle+t_{211}\left|211\right\rangle+t_{212}\left|212\right\rangle,\,\bar{t}_{211}t_{212}+\bar{t}_{121}t_{122}=0$;
$\sigma_{1}^{(1)2}=\sigma_{1}^{(3)2}=\left|t_{121}\right|^{2}+\left|t_{122}\right|^{2},\,\sigma_{1}^{(2)2}=\sigma_{2}^{(1)2}=\left|t_{211}\right|^{2}+\left|t_{212}\right|^{2}$.
From the ordering property of higher order singular value decomposition, since
$\sigma_{1}^{(2)2}$ is the largest 2-mode singular value, the only way
$\sigma_{1}^{(2)2}=\sigma_{2}^{(1)2}$ can be satisfied is when
$\sigma_{1}^{(2)2}=\frac{1}{2}$, resulting to
$\sigma_{1}^{(1)2}=\sigma_{1}^{(2)2}=\sigma_{1}^{(3)2}=\frac{1}{2}$. This is
not a generic special state that we are looking for.

  2. 2.

$t_{121}=t_{212}=0$
$\left|\psi\right\rangle=t_{112}\left|112\right\rangle+t_{122}\left|122\right\rangle+t_{211}\left|211\right\rangle+t_{221}\left|221\right\rangle,\,\bar{t}_{211}t_{221}+\bar{t}_{112}t_{122}=0$;
$\sigma_{1}^{(1)2}=\sigma_{1}^{(2)2}=\left|t_{112}\right|^{2}+\left|t_{122}\right|^{2},\,\sigma_{1}^{(3)2}=\sigma_{2}^{(1)2}=\left|t_{211}\right|^{2}+\left|t_{221}\right|^{2}$.
From the ordering property of higher order singular value decomposition, since
$\sigma_{1}^{(3)2}$ is the largest 3-mode singular value, the only way
$\sigma_{1}^{(3)2}=\sigma_{2}^{(1)2}$ can be satisfied is when
$\sigma_{1}^{(3)2}=\frac{1}{2}$, resulting to
$\sigma_{1}^{(1)2}=\sigma_{1}^{(2)2}=\sigma_{1}^{(3)2}=\frac{1}{2}$. This is
not a generic special state that we are looking for.

  3. 3.

$t_{122}=t_{211}=0$
$\left|\psi\right\rangle=t_{112}\left|112\right\rangle+t_{121}\left|121\right\rangle+t_{212}\left|212\right\rangle+t_{221}\left|221\right\rangle,\,\bar{t}_{121}t_{221}+\bar{t}_{112}t_{212}=0$;
$\sigma_{1}^{(1)2}=\sigma_{1}^{(2)2}=\left|t_{112}\right|^{2}+\left|t_{121}\right|^{2},\,\sigma_{1}^{(3)2}=\sigma_{2}^{(2)2}=\left|t_{121}\right|^{2}+\left|t_{221}\right|^{2}$.
From the ordering property of higher order singular value decomposition, since
$\sigma_{1}^{(3)2}$ is the largest 3-mode singular value, the only way
$\sigma_{1}^{(3)2}=\sigma_{2}^{(2)2}$ can be satisfied is when
$\sigma_{1}^{(3)2}=\frac{1}{2}$, resulting to
$\sigma_{1}^{(1)2}=\sigma_{1}^{(2)2}=\sigma_{1}^{(3)2}=\frac{1}{2}$. This is
not a generic special state that we are looking for.

### B.4 Row 1 = 0

We have

$\displaystyle\bar{t}_{112}t_{212}+\bar{t}_{121}t_{221}$ $\displaystyle=0,$
(62) $\displaystyle\bar{t}_{111}t_{121}+\bar{t}_{212}t_{222}$
$\displaystyle=0,$ (63)
$\displaystyle\bar{t}_{111}t_{112}+\bar{t}_{221}t_{222}$ $\displaystyle=0.$
(64)

From equations (63) and (64), we have

$\displaystyle\frac{\bar{t}_{111}}{t_{222}}=-\frac{\bar{t}_{212}}{t_{121}}=-\frac{\bar{t}_{221}}{t_{112}},$
(65)

but

$\displaystyle\frac{\bar{t}_{212}}{t_{121}}=-\frac{\bar{t}_{221}}{t_{112}}$
(66)

from equation (62). In order to resolve this contradiction, we consider the
following possibilities:-

  1. 1.

$t_{111}=t_{222}=0$
$\left|\psi\right\rangle=t_{112}\left|112\right\rangle+t_{121}\left|121\right\rangle+t_{212}\left|212\right\rangle+t_{221}\left|221\right\rangle,\,\bar{t}_{121}t_{221}+\bar{t}_{112}t_{212}=0$;
$\sigma_{1}^{(1)2}=\sigma_{1}^{(2)2}=\left|t_{112}\right|^{2}+\left|t_{121}\right|^{2},\,\sigma_{1}^{(3)2}=\sigma_{2}^{(2)2}=\left|t_{121}\right|^{2}+\left|t_{221}\right|^{2}$.
From the ordering property of higher order singular value decomposition, since
$\sigma_{1}^{(3)2}$ is the largest 3-mode singular value, the only way
$\sigma_{1}^{(3)2}=\sigma_{2}^{(2)2}$ can be satisfied is when
$\sigma_{1}^{(3)2}=\frac{1}{2}$, resulting to
$\sigma_{1}^{(1)2}=\sigma_{1}^{(2)2}=\sigma_{1}^{(3)2}=\frac{1}{2}$. This is
not a generic special state that we are looking for.

  2. 2.

$t_{112}=t_{221}=0$
$\left|\text{S}_{2}\right\rangle=t_{111}\left|111\right\rangle+t_{121}\left|121\right\rangle+t_{212}\left|212\right\rangle+t_{222}\left|222\right\rangle,\,\bar{t}_{111}t_{121}+\bar{t}_{212}t_{222}=0$.

  3. 3.

$t_{121}=t_{212}=0$
$\left|\text{S}_{1}\right\rangle=t_{111}\left|111\right\rangle+t_{112}\left|112\right\rangle+t_{221}\left|221\right\rangle+t_{222}\left|222\right\rangle,\,\bar{t}_{111}t_{112}+\bar{t}_{221}t_{222}=0$.

### B.5 Row 2 = 0

We have

$\displaystyle\bar{t}_{211}t_{221}+\bar{t}_{112}t_{122}$ $\displaystyle=0,$
(67) $\displaystyle\bar{t}_{111}t_{211}+\bar{t}_{122}t_{222}$
$\displaystyle=0,$ (68)
$\displaystyle\bar{t}_{111}t_{112}+\bar{t}_{221}t_{222}$ $\displaystyle=0.$
(69)

From equations (68) and (69), we have

$\displaystyle\frac{\bar{t}_{111}}{t_{222}}=-\frac{\bar{t}_{122}}{t_{211}}=-\frac{\bar{t}_{221}}{t_{112}},$
(70)

but

$\displaystyle\frac{\bar{t}_{122}}{t_{211}}=-\frac{\bar{t}_{221}}{t_{112}}$
(71)

from equation (67). In order to resolve this contradiction, we consider the
following possibilities:-

  1. 1.

$t_{111}=t_{222}=0$
$\left|\psi\right\rangle=t_{121}\left|121\right\rangle+t_{122}\left|122\right\rangle+t_{211}\left|211\right\rangle+t_{212}\left|212\right\rangle,\,\bar{t}_{211}t_{212}+\bar{t}_{121}t_{122}=0$;
$\sigma_{1}^{(1)2}=\sigma_{1}^{(3)2}=\left|t_{121}\right|^{2}+\left|t_{122}\right|^{2},\,\sigma_{1}^{(2)2}=\sigma_{2}^{(1)2}=\left|t_{211}\right|^{2}+\left|t_{212}\right|^{2}$.
From the ordering property of higher order singular value decomposition, since
$\sigma_{1}^{(2)2}$ is the largest 2-mode singular value, the only way
$\sigma_{1}^{(2)2}=\sigma_{2}^{(1)2}$ can be satisfied is when
$\sigma_{1}^{(2)2}=\frac{1}{2}$, resulting to
$\sigma_{1}^{(1)2}=\sigma_{1}^{(2)2}=\sigma_{1}^{(3)2}=\frac{1}{2}$. This is
not a generic special state that we are looking for.

  2. 2.

$t_{122}=t_{211}=0$
$\left|\text{S}_{3}\right\rangle=t_{111}\left|111\right\rangle+t_{122}\left|122\right\rangle+t_{211}\left|211\right\rangle+t_{222}\left|222\right\rangle,\,\bar{t}_{111}t_{211}+\bar{t}_{122}t_{222}=0$.

  3. 3.

$t_{122}=t_{211}=0$
$\left|\text{S}_{1}\right\rangle=t_{111}\left|111\right\rangle+t_{112}\left|112\right\rangle+t_{221}\left|221\right\rangle+t_{222}\left|222\right\rangle,\,\bar{t}_{111}t_{112}+\bar{t}_{221}t_{222}=0$.

### B.6 Row 3 = 0

We have

$\displaystyle\bar{t}_{211}t_{212}+\bar{t}_{121}t_{122}$ $\displaystyle=0,$
(72) $\displaystyle\bar{t}_{111}t_{211}+\bar{t}_{122}t_{222}$
$\displaystyle=0,$ (73)
$\displaystyle\bar{t}_{111}t_{121}+\bar{t}_{212}t_{222}$ $\displaystyle=0.$
(74)

From equations (73) and (74), we have

$\displaystyle\frac{\bar{t}_{111}}{t_{222}}=-\frac{\bar{t}_{122}}{t_{211}}=-\frac{\bar{t}_{212}}{t_{121}},$
(75)

but

$\displaystyle\frac{\bar{t}_{122}}{t_{211}}=-\frac{\bar{t}_{212}}{t_{121}}$
(76)

from equation (72). In order to resolve this contradiction, we consider the
following possibilities:-

  1. 1.

$t_{111}=t_{222}=0$
$\left|\psi\right\rangle=t_{112}\left|112\right\rangle+t_{122}\left|122\right\rangle+t_{211}\left|211\right\rangle+t_{221}\left|221\right\rangle,\,\bar{t}_{211}t_{221}+\bar{t}_{112}t_{122}=0$;
$\sigma_{1}^{(1)2}=\sigma_{1}^{(2)2}=\left|t_{112}\right|^{2}+\left|t_{122}\right|^{2},\,\sigma_{1}^{(3)2}=\sigma_{2}^{(1)2}=\left|t_{211}\right|^{2}+\left|t_{221}\right|^{2}$.
From the ordering property of higher order singular value decomposition, since
$\sigma_{1}^{(3)2}$ is the largest 3-mode singular value, the only way
$\sigma_{1}^{(3)2}=\sigma_{2}^{(1)2}$ can be satisfied is when
$\sigma_{1}^{(3)2}=\frac{1}{2}$, resulting to
$\sigma_{1}^{(1)2}=\sigma_{1}^{(2)2}=\sigma_{1}^{(3)2}=\frac{1}{2}$. This is
not a generic special state that we are looking for.

  2. 2.

$t_{122}=t_{211}=0$
$\left|\text{S}_{2}\right\rangle=t_{111}\left|111\right\rangle+t_{121}\left|121\right\rangle+t_{212}\left|212\right\rangle+t_{222}\left|222\right\rangle,\,\bar{t}_{111}t_{121}+\bar{t}_{212}t_{222}=0$.

  3. 3.

$t_{121}=t_{212}=0$
$\left|\text{S}_{3}\right\rangle=t_{111}\left|111\right\rangle+t_{122}\left|122\right\rangle+t_{211}\left|211\right\rangle+t_{222}\left|222\right\rangle,\,\bar{t}_{111}t_{211}+\bar{t}_{122}t_{222}=0$.