# Constructing unextendible product bases from multiqubit ones

Taiyu Zhang<EMAIL_ADDRESS>159th Middle School of Beijing, Xuanwu Men West
Street, Xicheng District, Beijing, 100031, China  Lin Chen
linchen@buaa.edu.cn(corresponding author) LMIB and School of Mathematical
Sciences, Beihang University, Beijing 100191, China  International Research
Institute for Multidisciplinary Science, Beihang University, Beijing 100191,
China

###### Abstract

The construction of multipartite unextendible product bases (UPBs) is a basic
problem in quantum information. We respectively construct two families of
$2\times 2\times 4$ and $2\times 2\times 2\times 4$ UPBs of size eight by
using the existing four-qubit and five-qubit UPBs. As an application, we
construct novel families of multipartite positive-partial-transpose entangled
states, as well as their entanglement properties in terms of the geometric
measure of entanglement.

††preprint: APS/123-QED

##  I Introduction

The unextendible product bases (UPBs) have been applied to various quantum-
information processing in the past decades. First, UPBs help construct
positive-partial-transpose (PPT) entangled states BDF+99 ; dms03 ;
Pittenger2003Unextendible ; Terhal2001A ; SLJM10 , the nonlocality without
entanglement and Bell inequalities ASH+11 ; AL01 ; Fen06 ; Chen2013The ;
Tura2012Four ; Szanto2016Complementary ; Chen2014Unextendible ; Hou2014A ;
2021Strongly ; Shi2021 , genuine entanglement PhysRevA.98.012313 , and Schmidt
number PhysRevA.90.054303 . Multipartite UPBs have been related to the tile
structures and local entanglement-assisted distinguishability 2020Unextendible
. Further, multiqubit UPBs have received extensive attention
Reed2012Realization ; Joh13 due to the role of multiqubit system in quantum
computing and experiments Dicarlo2010Preparation . For example, the three and
four-qubit UPBs have been fully studied in terms of programs
Bravyi2004Unextendible ; Johnston2014The . Some seven-qubit UPBs of size ten
have been constructed 2020The . Further the exclusion of $n$-qubit UPBs of
size $2^{n}-5$ Chen2018Nonexistence has solved an open problem in
Johnston2014The . Nevertheless, the understanding of multipartite UPBs is
still far from a complete picture. In particular, It is an intriguing problem
to find the relation between multiqubit UPBs and multipartite UPBs in high
dimensions, because the latter are usually harder to construct. This is the
first motivation of this paper.

On the other hand, it is known that PPT entangled states are not distillable.
That is, they cannot be asymptotically converted into Bell states under local
operations and classical communications (LOCC), while Bell states are
necessary for most quantum-information tasks such as teleportation and
computing. In contrast, some PPT entangled states can be used to create
distillable key hhh05 . the PPT entangled states are related to the
entanglement distillation problem 2021A ; Sun2020 , the detection of
entanglement Shen2020 ; 2021Detection , as well as multipartite genuinely
entangled states and entanglement-breaking subspaces Sun2020-2 ; Shen2020-2 .
The study of multipartite PPT states has also been devoted to the separability
of completely symmetric states 2019Separability , the separability of
symmetric states and vandermonde decomposition Qian2020 . Hence, it is an
important problem to construct novel PPT entangled states from UPBs, which may
evidently motivated the study of fore-mentioned topics. This is the second
motivation of this paper.

In this paper, we construct a family of $2\times 2\times 4$ UPBs of size eight
by using the merge of some systems of existing four-qubit UPBs of size eight
in Theorem 1, as well as a family of $2\times 2\times 2\times 4$ UPBs of size
eight by using the existing five-qubit UPBs of size eight in Theorem 2. In
particular, we shall show that the resulting set by any one of the four merge
AD, BC, BD, and CD in Eq. (2) is not a $2\times 2\times 4$ UPB, and the
resulting set by any one of the two merge AB and AC in (2) is a $2\times
2\times 4$ UPB of size eight. Further, the resulting set by any one of the
four merge AB, CD, CE and DE in (21) is not a $2\times 2\times 2\times 4$ UPB,
and the resulting set by any one of the six merge AC, AD, AE, BC, BD and BE in
(21) is a $2\times 2\times 2\times 4$ UPB of size eight. As an application, we
construct two families of multipartite PPT entangled states, and evaluate
their entanglement in terms of geometric measurement of entanglement. In
particular, we construct the upper bound of one family of tripartite PPT
entangled states in terms of its parameters. Our work shows the latest
progress on the construction of multipartite UPBs and PPT entangled states by
means of UPBs.

The rest of this paper is organized as follows. In Sec. II we introduce the
preliminary knowledge used in this paper. In Sec. III and IV, we respectively
construct a family of $2\times 2\times 4$ and $2\times 2\times 2\times 4$
UPBs. Using them, we establish two families of PPT entangled states, and
explore the properties of their entanglement in Sec. V. Finally we conclude in
Sec. VI.

##  II Preliminaries

In this section we introduce the preliminary knowledge used in this paper. We
work with an $n$-partite quantum system $A_{1},A_{2},\cdots,A_{n}$ in the
Hilbert space ${\cal H}={\cal H}_{1}\otimes\cdots\otimes{\cal
H}_{n}=\mathbb{C}^{d_{1}}\otimes...\otimes\mathbb{C}^{d_{n}}$. An $n$-partite
product vector in ${\cal H}$ is denoted as
$|{\psi}\rangle=|{a_{1}}\rangle\otimes\cdots\otimes|{a_{n}}\rangle:=|{a_{1},\ldots,a_{n}}\rangle\quad\text{with}\quad|{a_{i}}\rangle\in{\cal
H}_{i}$. Although the normalization factor is necessary for the interpretation
of quantum states, we do not always normalize product vectors for the sake of
mathematical convenience. Next, a set of $n$-partite orthonormal product
vectors $\\{|{a_{i,1}}\rangle,...,|{a_{i,n}}\rangle\\}$ is an unextendible
product basis (UPB) in ${\cal H}$, if there is no $n$-partite product vector
orthogonal to all product vectors in the set at the same time. In particular,
the $n$-qubit UPB exists when $d_{i}=2$ for any $i$. For simplicity when
$n=5$, we refer to $A_{1},A_{2},A_{3},A_{4}$ and $A_{5}$ as $A,B,C,D$ and $E$,
respectively. We have
$\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{2}:={\cal
H}_{A}\otimes{\cal H}_{B}\otimes{\cal H}_{C}\otimes{\cal H}_{D}\otimes{\cal
H}_{E}$, and ${\cal H}_{AB}:={\cal H}_{A}\otimes{\cal H}_{B}$, and so on. We
take the vectors $|{0}\rangle:=\begin{bmatrix}1\\\ 0\end{bmatrix}$ and
$|{1}\rangle:=\begin{bmatrix}0\\\ 1\end{bmatrix}$, so that the set
$\\{|{0}\rangle,|{1}\rangle\\}$ is a qubit orthonormal basis in
$\mathbb{C}^{2}$. We shall denote a general qubit orthonormal basis by
$\\{|{x}\rangle,|{x^{\prime}}\rangle\\}$, with $x=a,b,c$ and so on. It is
straightforward to verify that, a UPB remains a UPB if we switch the systems
or perform any local unitary transformation 2020The . We say that two UPBs are
equivalent if one UPB can be obtained from the other by the switch or local
unitary transformation.

##  III Tripartite UPBs from four-qubit UPBs

In this section, we construct a tripartite UPB of size eight, by using an
existing four-qubit UPB ${\cal S}$ of size eight in the space ${\cal
H}_{A}\otimes{\cal H}_{B}\otimes{\cal H}_{C}\otimes{\cal H}_{D}$. The UPB
${\cal S}$ was constructed in Johnston2014The . For convenience, we describe
the matrix of ${\cal S}$ in (1). Note that the first row of (1) means the
product state $|{0,0,0,0}\rangle$ in ${\cal S}$, and one can similarly figure
out all elements in ${\cal S}$. The matrix representation of a UPB has been
used to characterize the four-qubit orthogonal product bases
1751-8121-50-39-395301 ; Chen2018Multiqubit , and four-qubit UPBs Chen2018The
.

$\displaystyle\begin{bmatrix}0&0&0&0\\\ 1&a&a^{\prime}&a\\\
a&a^{\prime}&1&a^{\prime}\\\ a^{\prime}&1&a&b\\\ 0&a&a^{\prime}&1\\\
1&a&a^{\prime}&a^{\prime}\\\ a&1&a&a\\\ a^{\prime}&a^{\prime}&1&b^{\prime}\\\
\end{bmatrix},$ (1) $\displaystyle\begin{bmatrix}0&0&0&0\\\
1&a&a^{\prime}&a\\\ 0&a&a^{\prime}&1\\\ 1&a&a^{\prime}&a^{\prime}\\\
a&a^{\prime}&1&a^{\prime}\\\ a^{\prime}&1&a&b\\\ a&1&a&a\\\
a^{\prime}&a^{\prime}&1&b^{\prime}\\\ \end{bmatrix}.$ (2)

By switching rows 3 and 5, and rows 4 and 6 in (1), we obtain (2). We merge
two of the systems A, B, C and D, so that (2) corresponds to a set of $2\times
2\times 4$ orthonormal product vectors. For example, the merge of AB in (2)
implies the set
$\\{|{0,0}\rangle|{0}\rangle|{0}\rangle,|{1,a}\rangle|{a^{\prime}}\rangle|{a}\rangle,...,|{a^{\prime},a^{\prime}}\rangle|{1}\rangle|{b}\rangle\\}$.
One can verify that there are six ways for the merge, namely AB, AC, AD, BC,
BD, and CD. We present the following observation.

###### Theorem 1

The resulting set by any one of the four merge AD, BC, BD, and CD in (2) is
not a $2\times 2\times 4$ UPB. The resulting set by any one of the two merge
AB and AC in (2) is a $2\times 2\times 4$ UPB of size eight.

###### Proof.

We prove the claim by six cases (I)-(VI). They respectively work for the
system merge CD, BD, BC, AD, AC and AB in (2). Using the merge, we shall refer
to the set of orthonormal product vectors corresponding to (2) as ${\cal
S}_{1},...,{\cal S}_{6}$, respectively, in the six cases (I)-(VI). Further, if
any two systems of $A,B,C,D$ merge then evidently they are in
$\mathbb{C}^{4}$. For example, the merge of CD makes the set of four-qubit
orthonormal vectors corresponding to (2) become $2\times 2\times 4$
orthonormal vectors in
$\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{4}={\cal
H}_{A}\otimes{\cal H}_{B}\otimes{\cal H}_{CD}$.

(I) When we merge the system CD, one can verify that the set ${\cal S}_{1}$ in
(2) is orthogonal to $|{a,a^{\prime},x_{1}}\rangle_{A:B:CD}\in{\cal
H}_{A}\otimes{\cal H}_{B}\otimes{\cal H}_{CD}$, where $|{x_{1}}\rangle$ is
orthogonal to $|{0,0}\rangle$, $|{1,a^{\prime}}\rangle$ and $|{a,a}\rangle$.
So ${\cal S}_{1}$ is not a UPB in ${\cal H}_{A}\otimes{\cal H}_{B}\otimes{\cal
H}_{CD}$.

(II) When we merge the system BD, one can verify that the set ${\cal S}_{2}$
in (2) is orthogonal to $|{a^{\prime},a,x_{2}}\rangle_{A:C:BD}\in{\cal
H}_{A}\otimes{\cal H}_{C}\otimes{\cal H}_{BD}$, where $|{x_{2}}\rangle$ is
orthogonal to $|{0,0}\rangle$, $|{1,b}\rangle$ and
$|{a^{\prime},b^{\prime}}\rangle$. So ${\cal S}_{2}$ is not a UPB in ${\cal
H}_{A}\otimes{\cal H}_{C}\otimes{\cal H}_{BD}$.

(III) When we merge the system BC, one can verify that the set ${\cal S}_{3}$
in (2) is orthogonal to $|{1,a,x_{3}}\rangle_{A:D:BC}\in{\cal
H}_{A}\otimes{\cal H}_{D}\otimes{\cal H}_{BC}$, where $|{x_{3}}\rangle$ is
orthogonal to $|{a,a^{\prime}}\rangle$, $|{1,a}\rangle$ and
$|{a^{\prime},1}\rangle$. So ${\cal S}_{3}$ is not a UPB in ${\cal
H}_{A}\otimes{\cal H}_{D}\otimes{\cal H}_{BC}$.

(IV) When we merge the system AD, suppose the set ${\cal S}_{4}$ in (2) is
orthogonal to $|{0,a,x_{4}}\rangle_{B:C:AD}\in{\cal H}_{B}\otimes{\cal
H}_{C}\otimes{\cal H}_{AD}$, one can verify that $|{x_{4}}\rangle$ is
orthogonal to $|{0,0}\rangle$, $|{a,a^{\prime}}\rangle$ and
$|{a^{\prime},b^{\prime}}\rangle$. So ${\cal S}_{4}$ is not a UPB in ${\cal
H}_{B}\otimes{\cal H}_{C}\otimes{\cal H}_{AD}$.

(V) Before carrying out the proof, we label $a,a^{\prime}$ in column three and
four of (2) as $a_{3},a_{3}^{\prime}$ and $a_{4},a_{4}^{\prime}$,
respectively. To prove the assertion, it suffices to find some $2\times
2\times 4$ UPBs in (2) by merging system AC. So we construct the following
expressions related to (2).

$\displaystyle|{0}\rangle=\begin{bmatrix}1\\\
0\end{bmatrix},\quad|{1}\rangle=\begin{bmatrix}0\\\ 1\end{bmatrix},$ (3)
$\displaystyle|{a_{3}}\rangle=\begin{bmatrix}\cos x_{3}\\\ \sin
x_{3}\end{bmatrix},\quad|{a_{3}^{\prime}}\rangle=\begin{bmatrix}\sin x_{3}\\\
-\cos x_{3}\end{bmatrix},$ (4)
$\displaystyle|{a_{4}}\rangle=\begin{bmatrix}\cos x_{4}\\\ \sin
x_{4}\end{bmatrix},\quad|{a_{4}^{\prime}}\rangle=\begin{bmatrix}\sin x_{4}\\\
-\cos x_{4}\end{bmatrix},$ (5)
$\displaystyle|{c_{1}}\rangle=|{c_{2}}\rangle=|{1,a_{4}^{\prime}}\rangle=\begin{bmatrix}0\\\
0\\\ \sin x_{4}\\\ -\cos x_{4}\end{bmatrix},$ (6)
$\displaystyle|{c_{3}}\rangle=|{a_{3},1}\rangle=\begin{bmatrix}0\\\ \cos
x_{3}\\\ 0\\\ \sin x_{3}\end{bmatrix},$ (7)
$\displaystyle|{c_{4}}\rangle=|{a_{3}^{\prime},a_{4}}\rangle=\begin{bmatrix}\sin
x_{3}\cos x_{4}\\\ \sin x_{3}\sin x_{4}\\\ -\cos x_{3}\cos x_{4}\\\ -\cos
x_{3}\sin x_{4}\end{bmatrix},$ (8)
$\displaystyle|{c_{5}}\rangle=|{0,a_{4}^{\prime}}\rangle=\begin{bmatrix}\sin
x_{4}\\\ -\cos x_{4}\\\ 0\\\ 0\end{bmatrix},$ (9)
$\displaystyle|{c_{6}}\rangle=|{0,0}\rangle=\begin{bmatrix}1\\\ 0\\\ 0\\\
0\end{bmatrix},$ (10)
$\displaystyle|{c_{7}}\rangle=|{a_{3},a_{4}}\rangle=\begin{bmatrix}\cos
x_{3}\cos x_{4}\\\ \cos x_{3}\sin x_{4}\\\ \sin x_{3}\cos x_{4}\\\ \sin
x_{3}\sin x_{4}\end{bmatrix},$ (11)
$\displaystyle|{c_{8}}\rangle=|{a_{3}^{\prime},1}\rangle=\begin{bmatrix}0\\\
\sin x_{3}\\\ 0\\\ -\cos x_{3}\end{bmatrix}.$ (12)

Using these equations, we construct the $4\times 7$ matrix
$(|{c_{2}}\rangle,|{c_{3}}\rangle,......,|{c_{8}}\rangle)$ as follows.

$\displaystyle\begin{bmatrix}0&0&\sin x_{3}\cos x_{4}&\sin x_{4}&1&\cos
x_{3}\cos x_{4}&0\\\ 0&\cos x_{3}&\sin x_{3}\sin x_{4}&-\cos x_{4}&0&\cos
x_{3}\sin x_{4}&\sin x_{3}\\\ \sin x_{4}&0&-\cos x_{3}\cos x_{4}&0&0&\sin
x_{3}\cos x_{4}&0\\\ -\cos x_{4}&\sin x_{3}&-\cos x_{3}\sin x_{4}&0&0&\sin
x_{3}\sin x_{4}&-\cos x_{3}\\\ \end{bmatrix}.$ (13)

We merge the system A,C, and suppose that the set ${\cal
S}_{5}=\\{|{a_{j},b_{j},c_{j}}\rangle,j=1,...,8\\}$ in (2) is orthogonal to

$\displaystyle|{u,v,x_{5}}\rangle_{B:D:AC}\in{\cal H}_{B}\otimes{\cal
H}_{D}\otimes{\cal H}_{AC}.$ (14)

Hence $|{u,v}\rangle$ is orthogonal to some of the eight two-qubit product
vectors $|{a_{1},b_{1}}\rangle$,…,$|{a_{8},b_{8}}\rangle$. By checking the
expression of (2), one can show that each column of (2) has at most three
identical elements. In particular, the set $\\{|{a_{j}}\rangle\\}$ has at most
three identical elements, and the set $\\{|{b_{j}}\rangle\\}$ has at most two
identical elements. As a result, $|{u,v}\rangle$ is orthogonal to at most five
of the eight product vectors
$|{a_{1},b_{1}}\rangle$,…,$|{a_{8},b_{8}}\rangle$. We discuss three cases
(V.a), (V.b) and (V.c).

(V.a) Up to the permutation of subscripts in ${\cal S}_{5}$, we may assume
that $|{u,v}\rangle$ is orthogonal to exactly five product vectors
$|{a_{1},b_{1}}\rangle$, $|{a_{2},b_{2}}\rangle$,…, $|{a_{5},b_{5}}\rangle$.
That is, $|{u}\rangle$ is orthogonal to $m$ of the five vectors
$|{a_{1}}\rangle$, $|{a_{2}}\rangle$, …, $|{a_{5}}\rangle$, e.g.
$|{a_{1}}\rangle$, …, $|{a_{m}}\rangle$ and $|{v}\rangle$ is orthogonal to
$|{b_{m+1}}\rangle$, …, $|{b_{5}}\rangle$. Using the property of set ${\cal
S}_{5}$ in (2), one can show that $m=3$, that is, $|{u}\rangle$ is orthogonal
to three identical vectors $|{a_{1}}\rangle,|{a_{2}}\rangle$ and
$|{a_{3}}\rangle\in{\cal H}_{B}$, $|{v}\rangle$ is orthogonal to two identical
vectors $|{b_{3}}\rangle$ and $|{b_{4}}\rangle\in{\cal H}_{D}$. However the
product vectors $|{a_{1},b_{1}}\rangle$, …, $|{a_{4},b_{4}}\rangle$ do not
occupy five rows of the matrix of ${\cal S}$. So $|{u,v}\rangle$ is not
orthogonal to five product vectors in $|{a_{j},b_{j}}\rangle$, which means
this case is not possible.

(V.b) Up to the permutation of subscripts in ${\cal S}_{5}$, we may assume
that $|{u,v}\rangle$ is orthogonal to exactly four product vectors
$|{a_{1},b_{1}}\rangle$, $|{a_{2},b_{2}}\rangle$, $|{a_{3},b_{3}}\rangle$, and
$|{a_{4},b_{4}}\rangle$. So $|{x_{5}}\rangle$ is orthogonal to four of the
eight product vectors $|{c_{1}}\rangle$, … , $|{c_{8}}\rangle$. By observing
them, one can assume that $|{c_{1}}\rangle$=$|{c_{2}}\rangle$. Let the four
product vectors be $|{c_{j_{1}}}\rangle$, $|{c_{j_{2}}}\rangle$,
$|{c_{j_{3}}}\rangle$, $|{c_{j_{4}}}\rangle$, respectively, where
$j_{1},j_{2},j_{3},j_{4}$ are distinct integers in [2,8]. So there are 35
distinct matrices, which are formed by choosing the arrays
$(j_{1},j_{2},j_{3},j_{4})$ as $(2,3,4,5),(2,3,4,6)$,…,$(5,6,7,8)$,
respectively. Because $|{x_{5}}\rangle$ is orthogonal to
$|{c_{j_{1}}}\rangle$, $|{c_{j_{2}}}\rangle$, $|{c_{j_{3}}}\rangle$,
$|{c_{j_{4}}}\rangle$, we obtain that the matrices have determinant zero.
According to our calculation, there are exactly three matrices of determinant
zero, whose arrays are (2,3,5,7), (2,4,5,8), (3,5,6,8), respectively. However
by checking ${\cal S}_{5}$ in (2), one can see that for each array, there is
no $|{u,v,x_{5}}\rangle$ orthogonal to ${\cal S}_{5}$. It is a contradiction
with (14).

(V.c) Up to the permutation of subscripts in ${\cal S}_{5}$, we may assume
that $|{u,v}\rangle$ is orthogonal to at most three product vectors
$|{a_{1},b_{1}}\rangle$, $|{a_{2},b_{2}}\rangle$, $|{a_{3},b_{3}}\rangle$. So
$|{x_{5}}\rangle$ is orthogonal to five of the eight product vectors
$|{c_{1}}\rangle$, … , $|{c_{8}}\rangle$. By observing them, one can assume
that $|{c_{1}}\rangle=|{c_{2}}\rangle$. Let the five product vectors be
$|{c_{j_{1}}}\rangle$, $|{c_{j_{2}}}\rangle$, $|{c_{j_{3}}}\rangle$,
$|{c_{j_{4}}}\rangle$, $|{c_{j_{5}}}\rangle$, respectively, where
$j_{1},j_{2},j_{3},j_{4},j_{5}$ are distinct integers in [2,8]. Because they
are orthogonal to $|{x_{5}}\rangle$, they are linearly dependent. Hence, any
four of $|{c_{j_{1}}}\rangle$, $|{c_{j_{2}}}\rangle$, $|{c_{j_{3}}}\rangle$,
$|{c_{j_{4}}}\rangle$, $|{c_{j_{5}}}\rangle$ form a 4x4 matrix of determinant
zero. So at least five of the 35 matrices in (V.b) have determinant zero. It
is a contradiction with the calculation in (V.b), namely there are only three
matrices of determinant zero. Hence this case is not possible.

To conclude, we have excluded the three cases (V.a), (V.b), and (V.c). So the
set ${\cal S}_{5}=\\{|{a_{j},b_{j},c_{j}}\rangle,j=1,...,8\\}$ in (2) is a
$2\times 2\times 4$ UPB in ${\cal H}_{B}\otimes{\cal H}_{D}\otimes{\cal
H}_{AC}$.

(VI) We convert the set ${\cal S}_{6}$ in (2) in the space ${\cal
H}_{C}\otimes{\cal H}_{D}\otimes{\cal H}_{AB}$ into the UPB ${\cal S}_{5}$ in
case (V), by switching column 2 and 3 and a series of row permutations on the
matrix (2). It implies that the matrix ${\cal S}$ is also a UPB $\in{\cal
H}_{C}\otimes{\cal H}_{D}\otimes{\cal H}_{AB}$. We list the switch and
permutations on (2) as follows.

$\displaystyle\begin{bmatrix}0&0&0&0\\\ 1&a_{2}&a_{3}^{\prime}&a_{4}\\\
0&a_{2}&a_{3}^{\prime}&1\\\ 1&a_{2}&a_{3}^{\prime}&a_{4}^{\prime}\\\
a_{1}&a_{2}^{\prime}&1&a_{4}^{\prime}\\\ a_{1}^{\prime}&1&a_{3}&b_{4}\\\
a_{1}&1&a_{3}&a_{4}\\\ a_{1}^{\prime}&a_{3}^{\prime}&1&b_{4}^{\prime}\\\
\end{bmatrix},$ (15)

$\rightarrow$ switching rows 3 and 5, and rows 4 and 6 $\rightarrow$

$\displaystyle\begin{bmatrix}0&0&0&0\\\ 1&a_{2}&a_{3}^{\prime}&a_{4}\\\
a_{1}&a_{2}^{\prime}&1&a_{4}^{\prime}\\\ a_{1}^{\prime}&1&a_{3}&b\\\
0&a_{2}&a_{3}^{\prime}&1\\\ 1&a_{2}&a_{3}^{\prime}&a_{4}^{\prime}\\\
a_{1}&1&a_{3}&a_{4}\\\ a_{1}^{\prime}&a_{2}^{\prime}&1&b^{\prime}\\\
\end{bmatrix},$ (16)

$\rightarrow$ switch row 1 and 6 $\rightarrow$

$\displaystyle\begin{bmatrix}1&a_{2}&a_{3}^{\prime}&a_{4}^{\prime}\\\
1&a_{2}&a_{3}^{\prime}&a_{4}\\\ a_{1}&a_{2}^{\prime}&1&a_{4}^{\prime}\\\
a_{1}^{\prime}&1&a_{3}&b\\\ 0&a_{2}&a_{3}^{\prime}&1\\\ 0&0&0&0\\\
a_{1}&1&a_{3}&a_{4}\\\ a_{1}^{\prime}&a_{2}^{\prime}&1&b^{\prime}\\\
\end{bmatrix},$ (17)

$\rightarrow$ switch column 2 and 3 $\rightarrow$

$\displaystyle\begin{bmatrix}1&a_{3}^{\prime}&a_{2}&a_{4}^{\prime}\\\
1&a_{3}^{\prime}&a_{2}&a_{4}\\\ a_{1}&1&a_{2}^{\prime}&a_{4}^{\prime}\\\
a_{1}^{\prime}&a_{3}&1&b\\\ 0&a_{3}^{\prime}&a_{2}&1\\\ 0&0&0&0\\\
a_{1}&a_{3}&1&a_{4}\\\ a_{1}^{\prime}&1&a_{2}^{\prime}&b^{\prime}\\\
\end{bmatrix}\in{\cal H}_{ACBD},$ (18)

$\rightarrow$ switch row 1 and 2, 3 and 7, 4 and 8 $\rightarrow$

$\displaystyle\begin{bmatrix}1&a_{3}^{\prime}&a_{2}&a_{4}\\\
1&a_{3}^{\prime}&a_{2}&a_{4}^{\prime}\\\ a_{1}&a_{3}&1&a_{4}\\\
a_{1}^{\prime}&1&a_{2}^{\prime}&b^{\prime}\\\ 0&a_{3}^{\prime}&a_{2}&1\\\
0&0&0&0\\\ a_{1}&1&a_{2}^{\prime}&a_{4}^{\prime}\\\
a_{1}^{\prime}&a_{3}&1&b\\\ \end{bmatrix}.$ (19)

In column 2 and 3 of this matrix, we switch $(a_{3}^{\prime},a_{3})$ and
$(a_{2},a_{2}^{\prime})$, in column 4 we rename
$(a_{4},a_{4}^{\prime},b,b^{\prime})$ as
$(a_{4}^{\prime},a_{4},b^{\prime},b)$. So we obtain

$\displaystyle\begin{bmatrix}1&a_{2}&a_{3}^{\prime}&a_{4}^{\prime}\\\
1&a_{2}&a_{3}^{\prime}&a_{4}\\\ a_{1}&a_{2}^{\prime}&1&a_{4}^{\prime}\\\
a_{1}^{\prime}&1&a_{3}&b\\\ 0&a_{2}&a_{3}^{\prime}&1\\\ 0&0&0&0\\\
a_{1}&1&a_{3}&a_{4}\\\ a_{1}^{\prime}&a_{2}^{\prime}&1&b^{\prime}\\\
\end{bmatrix}\in{\cal H}_{B}\otimes{\cal H}_{D}\otimes{\cal H}_{AC},$ (20)

where the final matrix (20) is exactly the matrix ${\cal S}_{5}$ in (2). So
the set ${\cal S}_{6}$ in (2) is a $2\times 2\times 4$ UPB in ${\cal
H}_{C}\otimes{\cal H}_{D}\otimes{\cal H}_{AB}$.

In conclusion, we have proven the assertion in terms of the six cases
(I)-(VI). This completes the proof.  $\square$

##  IV Four-partite UPBs from five-qubit UPBs

In this section, we construct a four-partite UPB of size eight, by using an
existing five-qubit UPB ${\cal T}$ of size eight in the space ${\cal
H}_{A}\otimes{\cal H}_{B}\otimes{\cal H}_{C}\otimes{\cal H}_{D}\otimes{\cal
H}_{E}$. The UPB ${\cal T}$ was constructed in Johnston2014The . For
convenience, we describe the matrix of ${\cal T}$ in (21). Note that the first
row of (21) means the product state $|{0,0,0,0,0}\rangle\in{\cal T}$, and one
can similarly figure out all elements in ${\cal T}$.

$\displaystyle\begin{bmatrix}0&0&0&0&0\\\ 0&0&1&a_{4}&a_{5}\\\
a_{1}&a_{2}&a_{3}&1&a_{5}^{\prime}\\\
a_{1}&a_{2}&a_{3}^{\prime}&a_{4}^{\prime}&1\\\
1&a_{2}^{\prime}&b_{3}&b_{4}&b_{5}\\\
1&a_{2}^{\prime}&b_{3}^{\prime}&c_{4}&c_{5}\\\
a_{1}^{\prime}&1&c_{3}&b_{4}^{\prime}&c_{5}^{\prime}\\\
a_{1}^{\prime}&1&c_{3}^{\prime}&c_{4}^{\prime}&b_{5}^{\prime}\\\
\end{bmatrix}.$ (21)

We merge two of the systems A, B, C, D and E, so that (21) corresponds to a
set of $2\times 2\times 2\times 4$ orthogonal product vectors. For example,
the merge of AB in (21) implies the set
$\\{|{0,0}\rangle|{0}\rangle|{0}\rangle|{0}\rangle,...,|{a_{1}^{\prime},1}\rangle|{c_{3}^{\prime}}\rangle|{c_{4}^{\prime}}\rangle|{b_{5}^{\prime}}\rangle\\}$.
One can verify that there are ten ways for the merge, namely AB, AC, AD, AE,
BC, BD, BE, CD, CE and DE. We present the following observation.

###### Theorem 2

Suppose (21) is real. The resulting set by any one of the four merge AB, CD,
CE and DE in (21) is not a $2\times 2\times 2\times 4$ UPB. The resulting set
by any one of the six merge AC, AD, AE, BC, BD and BE in (21) is a $2\times
2\times 2\times 4$ UPB of size eight.

###### Proof.

We prove the claim by ten cases (I)-(X). They respectively work for the system
merge AB, CD, CE, DE, AC, AD, AE, BC, BD and BE in (21). Using the merge, we
shall refer to the set of orthonormal product vectors corresponding to (21) as
${\cal T}_{1},...,{\cal T}_{10}$, respectively, in the ten cases (I)-(X).
Further, if any two systems of A, B, C, D, E merge then evidently they are in
$\mathbb{C}^{4}$ . For example, the merge of DE makes the set of four-qubit
orthonormal vectors corresponding to (21) become $2\times 2\times 4$
orthonormal vectors in
$\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{4}={\cal
H}_{A}\otimes{\cal H}_{B}\otimes{\cal H}_{C}\otimes{\cal H}_{DE}$.

(I) When we merge the system AB, one can verify that the set ${\cal T}_{1}$ in
(21) is orthogonal to
$|{1,a_{4}^{\prime},a_{5},y_{4}}\rangle_{C:D:E:AB}\in{\cal H}_{C}\otimes{\cal
H}_{D}\otimes{\cal H}_{E}\otimes{\cal H}_{AB},$ where $|{y_{4}}\rangle$ is
orthogonal to $|{a_{1},a_{2}}\rangle$, $|{1,a_{2}^{\prime}}\rangle$ and
$|{a_{1}^{\prime},1}\rangle$. So ${\cal T}_{1}$ is not a UPB in ${\cal
H}_{C}\otimes{\cal H}_{D}\otimes{\cal H}_{E}\otimes{\cal H}_{AB}$.

(II) When we merge the system CD, one can verify that the set ${\cal T}_{2}$
in (21) is orthogonal to
$|{a_{1}^{\prime},a_{2},c_{5},y_{1}}\rangle_{A:B:E:CD}\in{\cal
H}_{A}\otimes{\cal H}_{B}\otimes{\cal H}_{E}\otimes{\cal H}_{CD}$, where
$|{y_{1}}\rangle$ is orthogonal to $|{0,0}\rangle$, $|{1,a_{4}}\rangle$ and
$|{c_{3}^{\prime},c_{4}^{\prime}}\rangle$. So ${\cal T}_{2}$ is not a UPB in
${\cal H}_{A}\otimes{\cal H}_{B}\otimes{\cal H}_{E}\otimes{\cal H}_{CD}$.

(III) When we merge the system CE, one can verify that the set ${\cal T}_{3}$
in (21) is orthogonal to
$|{a_{1}^{\prime},a_{2},b_{4},y_{2}}\rangle_{A:B:D:CE}\in{\cal
H}_{A}\otimes{\cal H}_{B}\otimes{\cal H}_{D}\otimes{\cal H}_{CE}$, where
$|{y_{2}}\rangle$ is orthogonal to $|{0,0}\rangle$, $|{1,a_{5}}\rangle$ and
$|{c_{3}^{\prime},b_{5}^{\prime}}\rangle$. So ${\cal T}_{3}$ is not a UPB in
${\cal H}_{A}\otimes{\cal H}_{B}\otimes{\cal H}_{D}\otimes{\cal H}_{CE}$.

(IV) When we merge the system DE, one can verify that the set ${\cal T}_{4}$
in (21) is orthogonal to
$|{a_{1}^{\prime},a_{2},c_{3}^{\prime},y_{3}}\rangle_{A:B:C:DE}\in{\cal
H}_{A}\otimes{\cal H}_{B}\otimes{\cal H}_{C}\otimes{\cal H}_{DE}$, where
$|{y_{3}}\rangle$ is orthogonal to $|{0,0}\rangle$, $|{a_{4},a_{5}}\rangle$
and $|{c_{4}^{\prime},b_{5}^{\prime}}\rangle$. So ${\cal T}_{4}$ is not a UPB
in ${\cal H}_{A}\otimes{\cal H}_{B}\otimes{\cal H}_{C}\otimes{\cal H}_{DE}$.

In the following, we investigate the merge of system AC, AD, AE, BC, BD and BE
in (21), respectively. For this purpose, we consider the positive numbers
$x_{j},y_{j},w_{j}\in(0,\pi/2)$ and

$\displaystyle|{0}\rangle=\begin{bmatrix}1\\\
0\end{bmatrix},\quad|{1}\rangle=\begin{bmatrix}0\\\ 1\end{bmatrix},$ (22)
$\displaystyle|{a_{1}}\rangle=\begin{bmatrix}\cos x_{1}\\\ \sin
x_{1}\end{bmatrix},\quad|{a_{1}^{\prime}}\rangle=\begin{bmatrix}\sin x_{1}\\\
-\cos x_{1}\end{bmatrix},$ $\displaystyle|{a_{2}}\rangle=\begin{bmatrix}\cos
x_{2}\\\ \sin
x_{2}\end{bmatrix},\quad|{a_{2}^{\prime}}\rangle=\begin{bmatrix}\sin x_{2}\\\
-\cos x_{2}\end{bmatrix},$ $\displaystyle|{a_{3}}\rangle=\begin{bmatrix}\cos
x_{3}\\\ \sin
x_{3}\end{bmatrix},\quad|{a_{3}^{\prime}}\rangle=\begin{bmatrix}\sin x_{3}\\\
-\cos x_{3}\end{bmatrix},$ $\displaystyle|{a_{4}}\rangle=\begin{bmatrix}\cos
x_{4}\\\ \sin
x_{4}\end{bmatrix},\quad|{a_{4}^{\prime}}\rangle=\begin{bmatrix}\sin x_{4}\\\
-\cos x_{4}\end{bmatrix},$ $\displaystyle|{a_{5}}\rangle=\begin{bmatrix}\cos
x_{5}\\\ \sin
x_{5}\end{bmatrix},\quad|{a_{5}^{\prime}}\rangle=\begin{bmatrix}\sin x_{5}\\\
-\cos x_{5}\end{bmatrix},$ $\displaystyle|{b_{3}}\rangle=\begin{bmatrix}\cos
y_{3}\\\ \sin
y_{3}\end{bmatrix},\quad|{b_{3}^{\prime}}\rangle=\begin{bmatrix}\sin y_{3}\\\
-\cos y_{3}\end{bmatrix},$ $\displaystyle|{b_{4}}\rangle=\begin{bmatrix}\cos
y_{4}\\\ \sin
y_{4}\end{bmatrix},\quad|{b_{4}^{\prime}}\rangle=\begin{bmatrix}\sin y_{4}\\\
-\cos y_{4}\end{bmatrix},$ $\displaystyle|{b_{5}}\rangle=\begin{bmatrix}\cos
y_{5}\\\ \sin
y_{5}\end{bmatrix},\quad|{b_{5}^{\prime}}\rangle=\begin{bmatrix}\sin y_{5}\\\
-\cos y_{5}\end{bmatrix},$ $\displaystyle|{c_{3}}\rangle=\begin{bmatrix}\cos
w_{3}\\\ \sin
w_{3}\end{bmatrix},\quad|{c_{3}^{\prime}}\rangle=\begin{bmatrix}\sin w_{3}\\\
-\cos w_{3}\end{bmatrix},$ $\displaystyle|{c_{4}}\rangle=\begin{bmatrix}\cos
w_{4}\\\ \sin
w_{4}\end{bmatrix},\quad|{c_{4}^{\prime}}\rangle=\begin{bmatrix}\sin w_{4}\\\
-\cos w_{4}\end{bmatrix},$ $\displaystyle|{c_{5}}\rangle=\begin{bmatrix}\cos
w_{5}\\\ \sin
w_{5}\end{bmatrix},\quad|{c_{5}^{\prime}}\rangle=\begin{bmatrix}\sin w_{5}\\\
-\cos w_{5}\end{bmatrix}.$ $\displaystyle\begin{bmatrix}0&0&0&0&0\\\
0&1&0&a_{4}&a_{5}\\\ a_{1}&a_{3}&a_{2}&1&a_{5}^{\prime}\\\
a_{1}&a_{3}^{\prime}&a_{2}&a_{4}^{\prime}&1\\\
1&b_{3}&a_{2}^{\prime}&b_{4}&b_{5}\\\
1&b_{3}^{\prime}&a_{2}^{\prime}&c_{4}&c_{5}\\\
a_{1}^{\prime}&c_{3}&1&b_{4}^{\prime}&c_{5}^{\prime}\\\
a_{1}^{\prime}&c_{3}^{\prime}&1&c_{4}^{\prime}&b_{5}^{\prime}\\\
\end{bmatrix}.$ (23)

(V) When we merge the system AC of the set ${\cal T}_{5}$ in (21), we switch
column B and C, and obtain the matrix in (23). The first two columns of (23)
consists of the following eight rows.

$\displaystyle|{e_{1}}\rangle=|{0,0}\rangle=\begin{bmatrix}1\\\ 0\\\ 0\\\
0\end{bmatrix},$ (24)
$\displaystyle|{e_{2}}\rangle=|{0,1}\rangle=\begin{bmatrix}0\\\ 1\\\ 0\\\
0\end{bmatrix},$
$\displaystyle|{e_{3}}\rangle=|{a_{1},a_{3}}\rangle=\begin{bmatrix}\cos
x_{1}\cos x_{3}\\\ \cos x_{1}\sin x_{3}\\\ \sin x_{1}\cos x_{3}\\\ \sin
x_{1}\sin x_{3}\end{bmatrix},$
$\displaystyle|{e_{4}}\rangle=|{a_{1},a_{3}^{\prime}}\rangle=\begin{bmatrix}\cos
x_{1}\sin x_{3}\\\ -\cos x_{1}\cos x_{3}\\\ \sin x_{1}\sin x_{3}\\\ -\sin
x_{1}\cos x_{3}\end{bmatrix},$
$\displaystyle|{e_{5}}\rangle=|{1,b_{3}}\rangle=\begin{bmatrix}0\\\ 0\\\ \cos
y_{3}\\\ \sin y_{3}\end{bmatrix},$
$\displaystyle|{e_{6}}\rangle=|{1,b_{3}^{\prime}}\rangle=\begin{bmatrix}0\\\
0\\\ \sin y_{3}\\\ -\cos y_{3}\end{bmatrix},$
$\displaystyle|{e_{7}}\rangle=|{a_{1}^{\prime},c_{3}}\rangle=\begin{bmatrix}\sin
x_{1}\cos w_{3}\\\ \sin x_{1}\sin w_{3}\\\ -\cos x_{1}\cos w_{3}\\\ -\cos
x_{1}\sin w_{3}\end{bmatrix},$
$\displaystyle|{e_{8}}\rangle=|{a_{1}^{\prime},c_{3}^{\prime}}\rangle=\begin{bmatrix}\sin
x_{1}\sin w_{3}\\\ -\sin x_{1}\cos w_{3}\\\ -\cos x_{1}\sin w_{3}\\\ \cos
x_{1}\cos w_{3}\end{bmatrix}.$

Let $[|{e_{1}}\rangle,...,|{e_{8}}\rangle]$=

$\displaystyle\begin{bmatrix}1&0&\cos{x_{1}}\cos{x_{3}}&\cos{x_{1}}\sin{x_{3}}&0&0&\sin{x_{1}}\cos{w_{3}}&\sin{x_{1}}\sin{w_{3}}\\\
0&1&\cos{x_{1}}\sin{x_{3}}&-\cos{x_{1}}\cos{x_{3}}&0&0&\sin{x_{1}}\sin{w_{3}}&-\sin{x_{1}}\cos{w_{3}}\\\
0&0&\sin{x_{1}}\cos{x_{3}}&\sin{x_{1}}\sin{x_{3}}&\cos{y_{3}}&\sin{y_{3}}&-\cos{x_{1}}\cos{w_{3}}&-\cos{x_{1}}\sin{w_{3}}\\\
0&0&\sin{x_{1}}\sin{x_{3}}&-\sin{x_{1}}\cos{x_{3}}&\sin{y_{3}}&-\cos{y_{3}}&-\cos{x_{1}}\sin{w_{3}}&\cos{x_{1}}\cos{w_{3}}\\\
\end{bmatrix}.$

Suppose that the five-qubit set

${\cal T}_{5}=\\{|{a_{j},b_{j},c_{j},e_{j}}\rangle,j=1,...,8\\}$

in (23) is orthogonal to the product vector
$|{u,v,w,y_{5}}\rangle_{B:D:E:AC}\in{\cal H}_{B}\otimes{\cal
H}_{D}\otimes{\cal H}_{E}\otimes{\cal
H}_{AC}=\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{4}$.
Hence $|{u,v,w}\rangle$ is orthogonal to some of the eight three-qubit product
vectors $|{a_{1},b_{1},c_{1}}\rangle,...,|{a_{8},b_{8},c_{8}}\rangle$. Up to
the permutation of subscripts, we may assume that $|{u,v,w}\rangle$ is
orthogonal to k product vectors
$|{a_{1},b_{1},c_{1}}\rangle,...,|{a_{k},b_{k},c_{k}}\rangle$, and
$|{y_{5}}\rangle$ is orthogonal to (8-k) product vectors
$|{f_{k+1}}\rangle,...,|{f_{8}}\rangle$. By checking the expression of ${\cal
T}_{5}$, one can show the property that each column of the matrix of ${\cal
T}_{5}$ has at most two identical elements. In particular, the set
$\\{|{a_{j}}\rangle\\}$ has at most two identical elements, and the set
$\\{|{b_{j}}\rangle\\}$, $\\{|{c_{j}}\rangle\\}$ has no identical elements. So
we obtain $k<5$, and it suffices to study the case k=4. Hence,
$|{u,v,w}\rangle$ is orthogonal to exactly four of the eight product vectors
$|{a_{1},b_{1},c_{1}}\rangle,...,|{a_{8},b_{8},c_{8}}\rangle$. At the same
time, $|{y_{5}}\rangle$ is orthogonal to four of the eight product vectors
$|{e_{1}}\rangle,...,|{e_{8}}\rangle$. So the four vectors form a $4\times 4$
singular matrix, which has rank at most three. However, by applying Matlab to
matrix (IV), we have shown that such a matrix does not exist. Hence, the
product vector $|{u,v,w,y_{5}}\rangle_{B:D:E:AC}$ does not exist. We have
proven that ${\cal T}_{5}=\\{|{a_{j},b_{j},c_{j},e_{j}}\rangle,j=1,...,8\\}$
is a four-partite UPB in ${\cal H}_{B}\otimes{\cal H}_{D}\otimes{\cal
H}_{E}\otimes{\cal H}_{AC}$.

(VI) When we merge the system AD of the set ${\cal T}_{6}$ in (21), we switch
column D $\rightarrow$ B $\rightarrow$ C $\rightarrow$ D, and obtain the
matrix in (26).

$\displaystyle\begin{bmatrix}0&0&0&0&0\\\ 0&a_{4}&0&1&a_{5}\\\
a_{1}&1&a_{2}&a_{3}&a_{5}^{\prime}\\\
a_{1}&a_{4}^{\prime}&a_{2}&a_{3}^{\prime}&1\\\
1&b_{4}&a_{2}^{\prime}&b_{3}&b_{5}\\\
1&c_{4}&a_{2}^{\prime}&b_{3}^{\prime}&c_{5}\\\
a_{1}^{\prime}&b_{4}^{\prime}&1&c_{3}&c_{5}^{\prime}\\\
a_{1}^{\prime}&c_{4}^{\prime}&1&c_{3}^{\prime}&b_{5}^{\prime}\\\
\end{bmatrix}.$ (26)

The first two columns of (26) consists of the following eight rows.

$\displaystyle|{f_{1}}\rangle=|{0,0}\rangle=\begin{bmatrix}1\\\ 0\\\ 0\\\
0\end{bmatrix},$ (27)
$\displaystyle|{f_{2}}\rangle=|{0,a_{4}}\rangle=\begin{bmatrix}\cos x_{4}\\\
\sin x_{4}\\\ 0\\\ 0\end{bmatrix},$
$\displaystyle|{f_{3}}\rangle=|{a_{1},1}\rangle=\begin{bmatrix}0\\\ \cos
x_{1}\\\ 0\\\ \sin x_{1}\end{bmatrix},$
$\displaystyle|{f_{4}}\rangle=|{a_{1},a_{4}^{\prime}}\rangle=\begin{bmatrix}\cos
x_{1}\sin x_{4}\\\ -\cos x_{1}\cos x_{4}\\\ \sin x_{1}\sin x_{4}\\\ -\sin
x_{1}\cos x_{4}\end{bmatrix},$
$\displaystyle|{f_{5}}\rangle=|{1,b_{4}}\rangle=\begin{bmatrix}0\\\ 0\\\ \cos
y_{4}\\\ \sin y_{4}\end{bmatrix},$
$\displaystyle|{f_{6}}\rangle=|{1,c_{4}}\rangle=\begin{bmatrix}0\\\ 0\\\ \cos
w_{4}\\\ \sin w_{4}\end{bmatrix},$
$\displaystyle|{f_{7}}\rangle=|{a_{1}^{\prime},b_{4}^{\prime}}\rangle=\begin{bmatrix}\sin
x_{1}\sin y_{4}\\\ -\sin x_{1}\cos y_{4}\\\ -\cos x_{1}\sin y_{4}\\\ \cos
x_{1}\cos y_{4}\end{bmatrix},$
$\displaystyle|{f_{8}}\rangle=|{a_{1}^{\prime},c_{4}^{\prime}}\rangle=\begin{bmatrix}\sin
x_{1}\sin w_{4}\\\ -\sin x_{1}\cos w_{4}\\\ -\cos x_{1}\sin w_{4}\\\ \cos
x_{1}\cos w_{4}\end{bmatrix}.$

Let $[|{f_{1}}\rangle,...,|{f_{8}}\rangle]$ =

$\displaystyle\begin{bmatrix}1&\cos x_{4}&0&\cos x_{1}\sin x_{4}&0&0&\sin
x_{1}\sin y_{4}&\sin x_{1}\sin w_{4}\\\ 0&\sin x_{4}&\cos x_{1}&-\cos
x_{1}\cos x_{4}&0&0&-\sin x_{1}\cos y_{4}&-\sin x_{1}\cos w_{4}\\\ 0&0&0&\sin
x_{1}\sin x_{4}&\cos y_{4}&\cos w_{4}&-\cos x_{1}\sin y_{4}&-\cos x_{1}\sin
w_{4}\\\ 0&0&\sin x_{1}&-\sin x_{1}\cos x_{4}&\sin y_{4}&\sin w_{4}&\cos
x_{1}\cos y_{4}&\cos x_{1}\cos w_{4}\\\ \end{bmatrix}$ (28)

Suppose that the five-qubit set

${\cal T}_{6}=\\{|{a_{j},b_{j},c_{j},f_{j}}\rangle,j=1,...,8\\}$

in (26) is orthogonal to the product vector
$|{u,v,w,y_{6}}\rangle_{B:C:E:AD}\in\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{4}$.
Hence $|{u,v,w}\rangle$ is orthogonal to some of the eight three-qubit product
vectors $|{a_{1},b_{1},c_{1}}\rangle,...,|{a_{8},b_{8},c_{8}}\rangle$. Up to
the permutation of subscripts, we may assume that $|{u,v,w}\rangle$ is
orthogonal to k product vectors
$|{a_{1},b_{1},c_{1}}\rangle,...,|{a_{k},b_{k},c_{k}}\rangle$, and
$|{y_{6}}\rangle$ is orthogonal to (8-k) product vectors
$|{f_{k+1}}\rangle,...,|{f_{8}}\rangle$. By checking the expression of ${\cal
T}_{6}$, one can show the property that each column of the matrix of ${\cal
T}_{6}$ has at most two identical elements. In particular, the set
$\\{|{a_{j}}\rangle\\}$ has at most two identical elements, and the set
$\\{|{b_{j}}\rangle\\}$, $\\{|{c_{j}}\rangle\\}$ has no identical elements. So
we obtain $k<5$, and it suffices to study the case $k=4$. Hence,
$|{u,v,w}\rangle$ is orthogonal to exactly four of the eight product vectors
$|{a_{1},b_{1},c_{1}}\rangle,...,|{a_{8},b_{8},c_{8}}\rangle$. At the same
time, $|{y_{6}}\rangle$ is orthogonal to four of the eight product vectors
$|{f_{1}}\rangle,...,|{f_{8}}\rangle$. So the four vectors form a $4\times 4$
singular matrix, which has rank at most three. However, by applying Matlab to
matrix (28), we have shown that such a matrix does not exist. Hence, the
product vector $|{u,v,w,y_{6}}\rangle_{B:C:E:AD}$ does not exist. We have
proven that ${\cal T}_{6}=\\{|{a_{j},b_{j},c_{j},f_{j}}\rangle,j=1,...,8\\}$
is a four-partite UPB in ${\cal
H}_{B:C:E:AD}=\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{4}$.

(VII) When we merge the system AD of the set ${\cal T}_{7}$ in (21), we switch
column E $\rightarrow$ B $\rightarrow$ C $\rightarrow$ D $\rightarrow$ E, and
obtain the matrix in (29).

$\displaystyle\begin{bmatrix}0&0&0&0&0\\\ 0&a_{5}&0&1&a_{4}\\\
a_{1}&a_{5}^{\prime}&a_{2}&a_{3}&1\\\
a_{1}&1&a_{2}&a_{3}^{\prime}&a_{4}^{\prime}\\\
1&b_{5}&a_{2}^{\prime}&b_{3}&b_{4}\\\
1&c_{5}&a_{2}^{\prime}&b_{3}^{\prime}&c_{4}\\\
a_{1}^{\prime}&c_{5}^{\prime}&1&c_{3}&b_{4}^{\prime}\\\
a_{1}^{\prime}&b_{5}^{\prime}&1&c_{3}^{\prime}&c_{4}^{\prime}\\\
\end{bmatrix}.$ (29)

The first two columns of (29) consists of the following eight rows.

$\displaystyle|{g_{1}}\rangle=|{0,0}\rangle=\begin{bmatrix}1\\\ 0\\\ 0\\\
0\end{bmatrix},$ (30)
$\displaystyle|{g_{2}}\rangle=|{0,a_{5}}\rangle=\begin{bmatrix}\cos x_{5}\\\
\sin x_{5}\\\ 0\\\ 0\end{bmatrix},$
$\displaystyle|{g_{3}}\rangle=|{a_{1},a_{5}^{\prime}}\rangle=\begin{bmatrix}\cos
x_{1}\sin x_{5}\\\ -\cos x_{1}\cos x_{5}\\\ \sin x_{1}\sin x_{5}\\\ -\sin
x_{1}\cos x_{5}\end{bmatrix},$
$\displaystyle|{g_{4}}\rangle=|{a_{1},1}\rangle=\begin{bmatrix}0\\\ \cos
x_{1}\\\ 0\\\ \sin x_{1}\end{bmatrix},$
$\displaystyle|{g_{5}}\rangle=|{1,b_{5}}\rangle=\begin{bmatrix}0\\\ 0\\\ \cos
y_{5}\\\ \sin y_{5}\end{bmatrix},$
$\displaystyle|{g_{6}}\rangle=|{1,c_{5}}\rangle=\begin{bmatrix}0\\\ 0\\\ \cos
w_{5}\\\ \sin w_{5}\end{bmatrix},$
$\displaystyle|{g_{7}}\rangle=|{a_{1}^{\prime},c_{5}^{\prime}}\rangle=\begin{bmatrix}\sin
x_{1}\sin w_{5}\\\ -\sin x_{1}\cos w_{5}\\\ -\cos x_{1}\sin w_{5}\\\ \cos
x_{1}\cos w_{5}\end{bmatrix},$
$\displaystyle|{g_{8}}\rangle=|{a_{1}^{\prime},b_{5}^{\prime}}\rangle=\begin{bmatrix}\sin
x_{1}\sin y_{5}\\\ -\sin x_{1}\cos y_{5}\\\ -\cos x_{1}\sin y_{5}\\\ \cos
x_{1}\cos y_{5}\end{bmatrix}.$

Let $|{g_{1}}\rangle$, …, $|{g_{8}}\rangle$ =

$\displaystyle\begin{bmatrix}1&\cos x_{5}&\cos x_{1}\sin x_{5}&0&0&0&\sin
x_{1}\sin w_{5}&\sin x_{1}\sin y_{5}\\\ 0&\sin x_{5}&-\cos x_{1}\cos
x_{5}&\cos x_{1}&0&0&-\sin x_{1}\cos w_{5}&-\sin x_{1}\cos y_{5}\\\ 0&0&\sin
x_{1}\sin x_{5}&0&\cos y_{5}&\cos w_{5}&-\cos x_{1}\sin w_{5}&-\cos x_{1}\sin
y_{5}\\\ 0&0&-\sin x_{1}\cos x_{5}&\sin x_{1}&\sin y_{5}&\sin w_{5}&\cos
x_{1}\cos w_{5}&\cos x_{1}\cos y_{5}\\\ \end{bmatrix}.$ (31)

Suppose that the five-qubit set

${\cal T}_{7}=\\{|{a_{j},b_{j},c_{j},g_{j}}\rangle,j=1,...,8\\}$

in (29) is orthogonal to the product vector
$|{u,v,w,y_{7}}\rangle_{B:C:D:AE}\in\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{4}$.
Hence $|{u,v,w}\rangle$ is orthogonal to some of the eight three-qubit product
vectors $|{a_{1},b_{1},c_{1}}\rangle,...,|{a_{8},b_{8},c_{8}}\rangle$. Up to
the permutation of subscripts, we may assume that $|{u,v,w}\rangle$ is
orthogonal to k product vectors
$|{a_{1},b_{1},c_{1}}\rangle,...,|{a_{k},b_{k},c_{k}}\rangle$, and
$|{y_{7}}\rangle$ is orthogonal to (8-k) product vectors
$|{g_{k+1}}\rangle,...,|{g_{8}}\rangle$. By checking the expression of ${\cal
T}_{7}$, one can show the property that each column of the matrix of ${\cal
T}_{7}$ has at most two identical elements. In particular, the set
$\\{|{a_{j}}\rangle\\}$ has at most two identical elements, and the set
$\\{|{b_{j}}\rangle\\}$, $\\{|{c_{j}}\rangle\\}$ has no identical elements. So
we obtain $k<5$, and it suffices to study the case $k=4$. Hence,
$|{u,v,w}\rangle$ is orthogonal to exactly four of the eight product vectors
$|{a_{1},b_{1},c_{1}}\rangle,...,|{a_{8},b_{8},c_{8}}\rangle$. At the same
time, $|{y_{7}}\rangle$ is orthogonal to four of the eight product vectors
$|{g_{1}}\rangle,...,|{g_{8}}\rangle$. So the four vectors form a $4\times 4$
singular matrix, which has rank at most three. However, by applying Matlab to
matrix (31), we have shown that such a matrix does not exist. Hence, the
product vector $|{u,v,w,y_{7}}\rangle_{B:D:E:AC}$ does not exist. We have
proven that ${\cal T}_{7}=\\{|{a_{j},b_{j},c_{j},g_{j}}\rangle,j=1,...,8\\}$
is a four-partite UPB in ${\cal
H}_{B:C:D:AE}=\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{4}$.

(VIII) When we merge the system BC of the set ${\cal T}_{8}$ in (21), we
switch column $C\rightarrow B\rightarrow A\rightarrow C$, and obtain the
matrix in (32).

$\displaystyle\begin{bmatrix}0&0&0&0&0\\\ 0&1&0&a_{4}&a_{5}\\\
a_{2}&a_{3}&a_{1}&1&a_{5}^{\prime}\\\
a_{2}&a_{3}^{\prime}&a_{1}&a_{4}^{\prime}&1\\\
a_{2}^{\prime}&b_{3}&1&b_{4}&b_{5}\\\
a_{2}^{\prime}&b_{3}^{\prime}&1&c_{4}&c_{5}\\\
1&c_{3}&a_{1}^{\prime}&b_{4}^{\prime}&c_{5}^{\prime}\\\
1&c_{3}^{\prime}&a_{1}^{\prime}&c_{4}^{\prime}&b_{5}^{\prime}\\\
\end{bmatrix}.$ (32)

The first two columns of (32) consists of the following eight rows.

$\displaystyle|{h_{1}}\rangle=|{0,0}\rangle=\begin{bmatrix}1\\\ 0\\\ 0\\\
0\end{bmatrix},$ (33)
$\displaystyle|{h_{2}}\rangle=|{0,1}\rangle=\begin{bmatrix}0\\\ 1\\\ 0\\\
0\end{bmatrix},$
$\displaystyle|{h_{3}}\rangle=|{a_{2},a_{3}}\rangle=\begin{bmatrix}\cos
x_{2}\cos x_{3}\\\ \cos x_{2}\sin x_{3}\\\ \sin x_{2}\cos x_{3}\\\ \sin
x_{2}\sin x_{3}\end{bmatrix},$
$\displaystyle|{h_{4}}\rangle=|{a_{2},a_{3}^{\prime}}\rangle=\begin{bmatrix}\cos
x_{2}\sin x_{3}\\\ -\cos x_{2}\cos x_{3}\\\ \sin x_{2}\sin x_{3}\\\ -\sin
x_{2}\cos x_{3}\end{bmatrix},$
$\displaystyle|{h_{5}}\rangle=|{a_{2}^{\prime},b_{3}}\rangle=\begin{bmatrix}\sin
x_{2}\cos y_{3}\\\ \sin x_{2}\sin y_{3}\\\ -\cos x_{2}\cos y_{3}\\\ -\cos
x_{2}\sin y_{3}\end{bmatrix},$
$\displaystyle|{h_{6}}\rangle=|{a_{2}^{\prime},b_{3}^{\prime}}\rangle=\begin{bmatrix}\sin
x_{2}\sin y_{3}\\\ -\sin x_{2}\cos y_{3}\\\ -\cos x_{2}\sin y_{3}\\\ \cos
x_{2}\cos y_{3}\end{bmatrix},$
$\displaystyle|{h_{7}}\rangle=|{1,c_{3}}\rangle=\begin{bmatrix}0\\\ 0\\\ \cos
w_{3}\\\ \sin w_{3}\end{bmatrix},$
$\displaystyle|{h_{8}}\rangle=|{1,c_{3}^{\prime}}\rangle=\begin{bmatrix}0\\\
0\\\ \sin w_{3}\\\ -\cos w_{3}\end{bmatrix}.$

Let $[|{h_{1}}\rangle,...,|{h_{8}}\rangle]$=

$\displaystyle\begin{bmatrix}1&0&\cos{x_{2}}\cos{x_{3}}&\cos{x_{2}}\sin{x_{3}}&\sin{x_{2}}\cos{y_{3}}&\sin{x_{2}}\sin{y_{3}}&0&0\\\
0&1&\cos{x_{2}}\sin{x_{3}}&-\cos{x_{2}}\cos{x_{3}}&\sin{x_{2}}\sin{y_{3}}&-\sin{x_{2}}\cos{y_{3}}&0&0\\\
0&0&\sin{x_{2}}\cos{x_{3}}&\sin{x_{2}}\sin{x_{3}}&-\cos{x_{2}}\cos{y_{3}}&-\cos{x_{2}}\sin{y_{3}}&\cos{w_{3}}&\sin{w_{3}}\\\
0&0&\sin{x_{2}}\sin{x_{3}}&-\sin{x_{2}}\cos{x_{3}}&-\cos{x_{2}}\sin{y_{3}}&\cos{x_{2}}\cos{y_{3}}&\sin{w_{3}}&-\cos{w_{3}}\\\
\end{bmatrix}.$

Suppose that the five-qubit set

${\cal T}_{8}=\\{|{a_{j},b_{j},c_{j},h_{j}}\rangle,j=1,...,8\\}$

in (32) is orthogonal to the product vector
$|{u,v,w,y_{8}}\rangle_{A:D:E:BC}\in\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{4}$.
Hence $|{u,v,w}\rangle$ is orthogonal to some of the eight three-qubit product
vectors $|{a_{1},b_{1},c_{1}}\rangle,...,|{a_{8},b_{8},c_{8}}\rangle$. Up to
the permutation of subscripts, we may assume that $|{u,v,w}\rangle$ is
orthogonal to k product vectors
$|{a_{1},b_{1},c_{1}}\rangle,...,|{a_{k},b_{k},c_{k}}\rangle$, and
$|{y_{8}}\rangle$ is orthogonal to (8-k) product vectors
$|{h_{k+1}}\rangle,...,|{h_{8}}\rangle$. By checking the expression of ${\cal
T}_{8}$, one can show the property that each column of the matrix of ${\cal
T}_{8}$ has at most two identical elements. In particular, the set
$\\{|{a_{j}}\rangle\\}$ has at most two identical elements, and the set
$\\{|{b_{j}}\rangle\\}$, $\\{|{c_{j}}\rangle\\}$ has no identical elements. So
we obtain $k<5$, and it suffices to study the case $k=4$. Hence,
$|{u,v,w}\rangle$ is orthogonal to exactly four of the eight product vectors
$|{a_{1},b_{1},c_{1}}\rangle,...,|{a_{8},b_{8},c_{8}}\rangle$. At the same
time, $|{y_{8}}\rangle$ is orthogonal to four of the eight product vectors
$|{h_{1}}\rangle,...,|{h_{8}}\rangle$. So the four vectors form a $4\times 4$
singular matrix, which has rank at most three. However, by applying Matlab to
matrix (IV), we have shown that such a matrix does not exist. Hence, the
product vector $|{u,v,w,y_{8}}\rangle_{A:D:E:BC}$ does not exist. We have
proven that ${\cal T}_{8}=\\{|{a_{j},b_{j},c_{j},h_{j}}\rangle,j=1,...,8\\}$
is a four-partite UPB in ${\cal
H}_{A:D:E:BC}=\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{4}$.

(IX) When we merge the system BD of the set ${\cal T}_{9}$ in (21), we switch
column $D\rightarrow B\rightarrow A\rightarrow C\rightarrow D$, and obtain the
matrix in (35).

$\displaystyle\begin{bmatrix}0&0&0&0&0\\\ 0&a_{4}&0&1&a_{5}\\\
a_{2}&1&a_{1}&a_{3}&a_{5}^{\prime}\\\
a_{2}&a_{4}^{\prime}&a_{1}&a_{3}^{\prime}&1\\\
a_{2}^{\prime}&b_{4}&1&b_{3}&b_{5}\\\
a_{2}^{\prime}&c_{4}&1&b_{3}^{\prime}&c_{5}\\\
1&b_{4}^{\prime}&a_{1}^{\prime}&c_{3}&c_{5}^{\prime}\\\
1&c_{4}^{\prime}&a_{1}^{\prime}&c_{3}^{\prime}&b_{5}^{\prime}\\\
\end{bmatrix}.$ (35)

The first two columns of (35) consists of the following eight rows.

$\displaystyle|{j_{1}}\rangle=|{0,0}\rangle=\begin{bmatrix}1\\\ 0\\\ 0\\\
0\end{bmatrix},$ (36)
$\displaystyle|{j_{2}}\rangle=|{0,a_{4}}\rangle=\begin{bmatrix}\cos x_{4}\\\
\sin x_{4}\\\ 0\\\ 0\end{bmatrix},$
$\displaystyle|{j_{3}}\rangle=|{a_{2},1}\rangle=\begin{bmatrix}0\\\ \cos
x_{2}\\\ 0\\\ \sin x_{2}\end{bmatrix},$
$\displaystyle|{j_{4}}\rangle=|{a_{2},a_{4}^{\prime}}\rangle=\begin{bmatrix}\cos
x_{2}\sin x_{4}\\\ -\cos x_{2}\cos x_{4}\\\ \sin x_{2}\sin x_{4}\\\ -\sin
x_{2}\cos x_{4}\end{bmatrix},$
$\displaystyle|{j_{5}}\rangle=|{a_{2}^{\prime},b_{4}}\rangle=\begin{bmatrix}\sin
x_{2}\cos y_{4}\\\ \sin x_{2}\sin y_{4}\\\ -\cos x_{2}\cos y_{4}\\\ -\cos
x_{2}\sin y_{4}\end{bmatrix},$
$\displaystyle|{j_{6}}\rangle=|{a_{2}^{\prime},c_{4}}\rangle=\begin{bmatrix}\sin
x_{2}\cos w_{4}\\\ \sin x_{2}\sin w_{4}\\\ -\cos x_{2}\cos w_{4}\\\ -\cos
x_{2}\sin w_{4}\end{bmatrix},$
$\displaystyle|{j_{7}}\rangle=|{1,b_{4}^{\prime}}\rangle=\begin{bmatrix}0\\\
0\\\ \sin y_{4}\\\ -\cos y_{4}\end{bmatrix},$
$\displaystyle|{j_{8}}\rangle=|{1,c_{4}^{\prime}}\rangle=\begin{bmatrix}0\\\
0\\\ \sin w_{4}\\\ -\cos w_{4}\end{bmatrix}.$

Let $[|{j_{1}}\rangle,...,|{j_{8}}\rangle]$=

$\displaystyle\begin{bmatrix}1&\cos
x_{4}&0&\cos{x_{2}}\sin{x_{4}}&\sin{x_{2}}\cos{y_{4}}&\sin{x_{2}}\cos{w_{4}}&0&0\\\
0&\sin
x_{4}&\cos{x_{2}}&-\cos{x_{2}}\cos{x_{4}}&\sin{x_{2}}\sin{y_{4}}&\sin{x_{2}}\sin{w_{4}}&0&0\\\
0&0&0&\sin{x_{2}}\sin{x_{4}}&-\cos{x_{2}}\cos{y_{4}}&-\cos{x_{2}}\cos{w_{4}}&\sin{y_{4}}&\sin{w_{4}}\\\
0&0&\sin{x_{2}}&-\sin{x_{2}}\cos{x_{4}}&-\cos{x_{2}}\sin{y_{4}}&-\cos{x_{2}}\sin{w_{4}}&-\cos{y_{4}}&-\cos{w_{4}}\\\
\end{bmatrix}.$

Suppose that the five-qubit set

${\cal T}_{9}=\\{|{a_{j},b_{j},c_{j},m_{j}}\rangle,j=1,...,8\\}$

in (35) is orthogonal to the product vector
$|{u,v,w,y_{9}}\rangle_{A:C:E:BD}\in\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{4}$.
Hence $|{u,v,w}\rangle$ is orthogonal to some of the eight three-qubit product
vectors $|{a_{1},b_{1},c_{1}}\rangle,...,|{a_{8},b_{8},c_{8}}\rangle$. Up to
the permutation of subscripts, we may assume that $|{u,v,w}\rangle$ is
orthogonal to k product vectors
$|{a_{1},b_{1},c_{1}}\rangle,...,|{a_{k},b_{k},c_{k}}\rangle$, and
$|{y_{9}}\rangle$ is orthogonal to (8-k) product vectors
$|{m_{k+1}}\rangle,...,|{m_{8}}\rangle$. By checking the expression of ${\cal
T}_{9}$, one can show the property that each column of the matrix of ${\cal
T}_{9}$ has at most two identical elements. In particular, the set
$\\{|{a_{j}}\rangle\\}$ has at most two identical elements, and the set
$\\{|{b_{j}}\rangle\\}$, $\\{|{c_{j}}\rangle\\}$ has no identical elements. So
we obtain $k<5$, and it suffices to study the case $k=4$. Hence,
$|{u,v,w}\rangle$ is orthogonal to exactly four of the eight product vectors
$|{a_{1},b_{1},c_{1}}\rangle,...,|{a_{8},b_{8},c_{8}}\rangle$. At the same
time, $|{y_{9}}\rangle$ is orthogonal to four of the eight product vectors
$|{m_{1}}\rangle,...,|{m_{8}}\rangle$. So the four vectors form a $4\times 4$
singular matrix, which has rank at most three. However, by applying Matlab to
matrix (IV), we have shown that such a matrix does not exist. Hence, the
product vector $|{u,v,w,y_{9}}\rangle_{A:C:E:BD}$ does not exist. We have
proven that ${\cal T}_{9}=\\{|{a_{j},b_{j},c_{j},m_{j}}\rangle,j=1,...,8\\}$
is a four-partite UPB in ${\cal
H}_{A:C:E:BD}=\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{4}$.

(X) When we merge the system BE of the set ${\cal T}_{10}$ in (21), we switch
column $E\rightarrow B\rightarrow A\rightarrow C\rightarrow D\rightarrow E$ of
the set ${\cal T}$, and obtain the matrix in (38).

$\displaystyle\begin{bmatrix}0&0&0&0&0\\\ 0&a_{5}&0&1&a_{4}\\\
a_{2}&a_{5}^{\prime}&a_{1}&a_{3}&1\\\
a_{2}&1&a_{1}&a_{3}^{\prime}&a_{4}^{\prime}\\\
a_{2}^{\prime}&b_{5}&1&b_{3}&b_{4}\\\
a_{2}^{\prime}&c_{5}&1&b_{3}^{\prime}&c_{4}\\\
1&c_{5}^{\prime}&a_{1}^{\prime}&c_{3}&b_{4}^{\prime}\\\
1&b_{5}^{\prime}&a_{1}^{\prime}&c_{3}^{\prime}&c_{4}^{\prime}\\\
\end{bmatrix}.$ (38)

The first two columns of (38) consists of the following eight rows.

$\displaystyle|{l_{1}}\rangle=|{0,0}\rangle=\begin{bmatrix}1\\\ 0\\\ 0\\\
0\end{bmatrix},$ (39)
$\displaystyle|{l_{2}}\rangle=|{0,a_{5}}\rangle=\begin{bmatrix}\cos x_{5}\\\
\sin x_{5}\\\ 0\\\ 0\end{bmatrix},$
$\displaystyle|{l_{3}}\rangle=|{a_{2},a_{5}^{\prime}}\rangle=\begin{bmatrix}\cos
x_{2}\sin x_{5}\\\ -\cos x_{2}\cos x_{5}\\\ \sin x_{2}\sin x_{5}\\\ -\sin
x_{2}\cos x_{5}\end{bmatrix},$
$\displaystyle|{l_{4}}\rangle=|{a_{2},1}\rangle=\begin{bmatrix}0\\\ \cos
x_{2}\\\ 0\\\ \sin x_{2}\end{bmatrix},$
$\displaystyle|{l_{5}}\rangle=|{a_{2}^{\prime},b_{5}}\rangle=\begin{bmatrix}\sin
x_{2}\cos y_{5}\\\ \sin x_{2}\sin y_{5}\\\ -\cos x_{2}\cos y_{5}\\\ -\cos
x_{2}\sin y_{5}\end{bmatrix},$
$\displaystyle|{l_{6}}\rangle=|{a_{2}^{\prime},c_{5}}\rangle=\begin{bmatrix}\sin
x_{2}\cos w_{5}\\\ \sin x_{2}\sin w_{5}\\\ -\cos x_{2}\cos w_{5}\\\ -\cos
x_{2}\sin w_{5}\end{bmatrix},$
$\displaystyle|{l_{7}}\rangle=|{1,c_{5}^{\prime}}\rangle=\begin{bmatrix}0\\\
0\\\ \sin w_{5}\\\ -\cos w_{5}\end{bmatrix},$
$\displaystyle|{l_{8}}\rangle=|{1,b_{5}^{\prime}}\rangle=\begin{bmatrix}0\\\
0\\\ \sin y_{5}\\\ -\cos y_{5}\end{bmatrix}.$

Let $[|{l_{1}}\rangle,...,|{l_{8}}\rangle]$=

$\displaystyle\begin{bmatrix}1&\cos
x_{5}&\cos{x_{2}}\sin{x_{5}}&0&\sin{x_{2}}\cos{y_{5}}&\sin{x_{2}}\cos{w_{5}}&0&0\\\
0&\sin x_{5}&-\cos{x_{2}}\cos{x_{5}}&\cos
x_{2}&\sin{x_{2}}\sin{y_{5}}&\sin{x_{2}}\sin{w_{5}}&0&0\\\
0&0&\sin{x_{2}}\sin{x_{5}}&0&-\cos{x_{2}}\cos{y_{5}}&-\cos{x_{2}}\cos{w_{5}}&\sin{w_{5}}&\sin{y_{5}}\\\
0&0&-\sin{x_{2}}\cos{x_{5}}&\sin
x_{2}&-\cos{x_{2}}\sin{y_{5}}&-\cos{x_{2}}\sin{w_{5}}&-\cos{w_{5}}&-\cos{y_{5}}\\\
\end{bmatrix}.$

Suppose that the five-qubit set

${\cal T}_{10}=\\{|{a_{j},b_{j},c_{j},n_{j}}\rangle,j=1,...,8\\}$

in (38) is orthogonal to the product vector
$|{u,v,w,y_{10}}\rangle_{A:C:D:BE}\in\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{4}$.
Hence $|{u,v,w}\rangle$ is orthogonal to some of the eight three-qubit product
vectors $|{a_{1},b_{1},c_{1}}\rangle,...,|{a_{8},b_{8},c_{8}}\rangle$. Up to
the permutation of subscripts, we may assume that $|{u,v,w}\rangle$ is
orthogonal to k product vectors
$|{a_{1},b_{1},c_{1}}\rangle,...,|{a_{k},b_{k},c_{k}}\rangle$, and
$|{y_{10}}\rangle$ is orthogonal to (8-k) product vectors
$|{n_{k+1}}\rangle,...,|{n_{8}}\rangle$. By checking the expression of ${\cal
T}_{10}$, one can show the property that each column of the matrix of ${\cal
T}_{10}$ has at most two identical elements. In particular, the set
$\\{|{a_{j}}\rangle\\}$ has at most two identical elements, and the set
$\\{|{b_{j}}\rangle\\}$, $\\{|{c_{j}}\rangle$} has no identical elements. So
we obtain $k<5$, and it suffices to study the case $k=4$. Hence,
$|{u,v,w}\rangle$ is orthogonal to exactly four of the eight product vectors
$|{a_{1},b_{1},c_{1}}\rangle,...,|{a_{8},b_{8},c_{8}}\rangle$. At the same
time, $|{y_{10}}\rangle$ is orthogonal to four of the eight product vectors
$|{n_{1}}\rangle,...,|{n_{8}}\rangle$. So the four vectors form a $4\times 4$
singular matrix, which has rank at most three. However, by applying Matlab to
matrix (IV), we have shown that such a matrix does not exist. Hence, the
product vector $|{u,v,w,y_{10}}\rangle_{A:C:D:BE}$ does not exist. We have
proven that ${\cal T}_{10}=\\{|{a_{j},b_{j},c_{j},n_{j}}\rangle,j=1,...,8\\}$
is a four-partite UPB in ${\cal
H}_{A:C:D:BE}=\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{4}$.
$\square$

So we have managed to construct a family of real four-partite UPBs of size
eight, by using the existing five-qubit UPBs. It remains to show whether the
real UPB in Theorem 2 can be extended to complex field. Our primary
investigation shows that it might be true by using the similar technique in
the above proof.

##  V Application

In this section, we apply the results in the last two sections. We shall
construct two families of multipartite entangled states, which have positive
partial transpose (PPT). A bipartite state $\rho\in{\cal
B}(\mathbb{C}^{m}\otimes\mathbb{C}^{n}):={\cal B}({\cal H}_{A}\otimes{\cal
H}_{B})$ is a positive semidefinite matrix of trace one. The partial transpose
of $\rho$ is defined as
$\rho^{\Gamma}=\sum_{j,k}(|{j}\rangle\\!\langle{k}|\otimes
I)\rho(|{j}\rangle\\!\langle{k}|\otimes I)$ with the computational basis
$\\{|{j}\rangle\\}\in\mathbb{C}^{m}$. We say that $\rho$ is a PPT state when
$\rho^{\Gamma}$ is also positive semidefinite. For example, the separable
state is PPT, because it is the convex sum of product states. On the other
hand, the PPT state may be not separable when $\mathop{\rm min}\\{m,n\\}>2$
peres1996 ; hhh96 . The examples of two-qutrit PPT entangled states were
constructed in 1980s, and then introduced in quantum information horodecki1997
; MD1975Completely ; E1982Decomposable . Next, two-qutrit PPT entangled states
of rank four were fully characterized in cd11JMP and skowronek11 ,
respectively. In contrast, the construction of multipartite PPT entangled
states and their investigation in terms of entanglement theory is more
involved. A systematic method for such a construction is to employ an
$n$-partite UPB
$\\{|{u_{j}}\rangle\\}_{j=1,2,...,m}\in\mathbb{C}^{d_{1}}\otimes...\otimes\mathbb{C}^{d_{n}}$.
That is, one can show that

$\displaystyle\rho=\frac{1}{n}\bigg{(}I-\sum^{m}_{j=1}|{u_{j}}\rangle\langle{u_{j}}|\bigg{)},$
(41)

is an $n$-partite PPT entangled state of rank $d_{1}...d_{n}-m$. In Theorems 1
and 2, we have constructed respectively two and six UPBs. Every one of them
can be applied to (41) and generate a PPT entangled state $\rho$. For example,
we demonstrate a $4\times 2\times 2$ PPT entangled state of rank eight from
Theorem 1 as follows.

$\displaystyle\rho=$
$\displaystyle\frac{1}{8}\bigg{(}I-|{0,0,0,0}\rangle\langle{0,0,0,0}|-|{1,a_{2},a_{3}^{\prime},a_{4}}\rangle\langle{1,a_{2},a_{3}^{\prime},a_{4}}|-|{0,a_{2},a_{3}^{\prime},1}\rangle\langle{0,a_{2},a_{3}^{\prime},1}|$
$\displaystyle-$
$\displaystyle|{1,a_{2},a_{3}^{\prime},a_{4}^{\prime}}\rangle\langle{1,a_{2},a_{3}^{\prime},a_{4}^{\prime}}|-|{a_{1},a_{2}^{\prime},1,a_{4}^{\prime}}\rangle\langle{a_{1},a_{2}^{\prime},1,a_{4}^{\prime}}|-|{a_{1}^{\prime},1,a_{3},b_{4}}\rangle\langle{a_{1}^{\prime},1,a_{3},b_{4}}|$
$\displaystyle-$
$\displaystyle|{a_{1},1,a_{3},a_{4}}\rangle\langle{a_{1},1,a_{3},a_{4}}|-|{a_{1}^{\prime},a_{2}^{\prime},1,b_{4}^{\prime}}\rangle\langle{a_{1}^{\prime},a_{2}^{\prime},1,b_{4}^{\prime}}|\bigg{)}$
$\displaystyle:=$ $\displaystyle\frac{1}{8}\bigg{(}I-P\bigg{)},$ (42)

where system $A,B$ are merged. Further, we assume that $\alpha$ is the four-
qubit PPT entangled state having the same expression as that of $\rho$.

We investigate the geometric measure of entanglement of states $\rho$ and
$\alpha$ wg2003 ; cxz2010 . For any $n$-partite quantum state $\sigma$, the
measure is defined as

$\displaystyle G(\sigma):=-\log_{2}\mathop{\rm
max}_{\delta_{1},...,\delta_{n}}\langle{\delta_{1},...,\delta_{n}}|\sigma|{\delta_{1},...,\delta_{n}}\rangle,$
(43)

where $|{\delta_{1},...,\delta_{n}}\rangle$ is a normalized product state in
$\mathbb{C}^{d_{1}}\otimes...\otimes\mathbb{C}^{d_{n}}$. In particular,
$\rho\in{\cal B}(\mathbb{C}^{4}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{2})$
means that $d_{1}=4$ and $d_{2}=d_{3}=2$ with $n=3$, and $\alpha\in{\cal
B}(\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{2})$
means that $d_{1}=d_{2}=d_{3}=d_{4}=2$ with $n=4$. Hence we obtain that

$\displaystyle G(\rho)\leq G(\alpha).$ (44)

So $G(\rho)$ is upper bounded by $G(\alpha)$. However it is not easy to
compute, and also might not be a tight upper bound. Eqs. (V) and (43) imply
that, the evaluation of $G(\rho)$ is equivalent to

$\displaystyle\mathop{\rm
min}_{|{\delta_{1},\delta_{2},\delta_{3}}\rangle\in\mathbb{C}^{4}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{2}}\langle{\delta_{1},\delta_{2},\delta_{3}}|P|{\delta_{1},\delta_{2},\delta_{3}}\rangle,$
(45)

for $P\in{\cal B}(\mathbb{C}^{4}\otimes\mathbb{C}^{2}\otimes\mathbb{C}^{2})$
in (V). To find an upper bound, we only consider
$|{\delta_{1},\delta_{2},\delta_{3}}\rangle\in\mathbb{R}^{4}\otimes\mathbb{R}^{2}\otimes\mathbb{R}^{2}$
in (45). We assume that

$\displaystyle|{\delta_{1}}\rangle=\begin{bmatrix}\cos\nu_{1}\cos\mu_{1}\\\
\cos\nu_{1}\sin\mu_{1}\\\ \sin\nu_{1}\cos\mu_{2}\\\
\sin\nu_{1}\sin\mu_{2}\end{bmatrix},\quad|{\delta_{2}}\rangle=\begin{bmatrix}\cos\nu_{2}\\\
\sin\nu_{2}\end{bmatrix},\quad|{\delta_{3}}\rangle=\begin{bmatrix}\cos\nu_{3}\\\
\sin\nu_{3}\end{bmatrix},$ (46)

where $\nu_{j},\mu_{j}\in[0,2\pi)$. So (45) is upper bounded by

$\displaystyle\mathop{\rm
min}_{|{\delta_{1},\delta_{2},\delta_{3}}\rangle\in\mathbb{R}^{4}\otimes\mathbb{R}^{2}\otimes\mathbb{R}^{2}}\langle{\delta_{1},\delta_{2},\delta_{3}}|P|{\delta_{1},\delta_{2},\delta_{3}}\rangle.$
(47)

Using Eqs. (3) -(12), (V) and (46), we can write up the target function in
(47) as

$\displaystyle\langle{\delta_{1},\delta_{2},\delta_{3}}|\bigg{(}|{0,0,0,0}\rangle\langle{0,0,0,0}|+|{1,a_{2},a_{3}^{\prime},a_{4}}\rangle\langle{1,a_{2},a_{3}^{\prime},a_{4}}|+|{0,a_{2},a_{3}^{\prime},1}\rangle\langle{0,a_{2},a_{3}^{\prime},1}|$
$\displaystyle+$
$\displaystyle|{1,a_{2},a_{3}^{\prime},a_{4}^{\prime}}\rangle\langle{1,a_{2},a_{3}^{\prime},a_{4}^{\prime}}|+|{a_{1},a_{2}^{\prime},1,a_{4}^{\prime}}\rangle\langle{a_{1},a_{2}^{\prime},1,a_{4}^{\prime}}|+|{a_{1}^{\prime},1,a_{3},b_{4}}\rangle\langle{a_{1}^{\prime},1,a_{3},b_{4}}|$
$\displaystyle+$
$\displaystyle|{a_{1},1,a_{3},a_{4}}\rangle\langle{a_{1},1,a_{3},a_{4}}|+|{a_{1}^{\prime},a_{2}^{\prime},1,b_{4}^{\prime}}\rangle\langle{a_{1}^{\prime},a_{2}^{\prime},1,b_{4}^{\prime}}|\bigg{)}|{\delta_{1},\delta_{2},\delta_{3}}\rangle$
$\displaystyle=$
$\displaystyle(\cos\nu_{1}\cos\mu_{1}\cos\nu_{2}\cos\nu_{3})^{2}$
$\displaystyle+$ $\displaystyle\bigg{(}(\cos x_{2}\sin\nu_{1}\cos\mu_{2}+\sin
x_{2}\sin\nu_{1}\sin\mu_{2})$ $\displaystyle(\sin x_{3}\cos\nu_{2}-\cos
x_{3}\sin\nu_{2})(\cos x_{4}\cos\nu_{3}+\sin x_{4}\sin\nu_{3})\bigg{)}^{2}$
$\displaystyle+$ $\displaystyle\bigg{(}(\cos x_{2}\cos\nu_{1}\cos\mu_{1}+\sin
x_{2}\cos\nu_{1}\sin\mu_{1})$ $\displaystyle(\sin x_{3}\cos\nu_{2}-\cos
x_{3}\sin\nu_{2})\sin\nu_{3}\bigg{)}^{2}$ $\displaystyle+$
$\displaystyle\bigg{(}(\cos x_{2}\sin\nu_{1}\cos\mu_{2}+\sin
x_{2}\sin\nu_{1}\sin\mu_{2})$ $\displaystyle(\sin x_{3}\cos\nu_{2}-\cos
x_{3}\sin\nu_{2})(\sin x_{4}\cos\nu_{3}-\cos x_{4}\sin\nu_{3})\bigg{)}^{2}$
$\displaystyle+$ $\displaystyle\bigg{(}(\cos x_{1}\sin
x_{2}\cos\nu_{1}\cos\mu_{1}-\cos x_{1}\cos x_{2}\cos\nu_{1}\sin\mu_{1}+$
$\displaystyle\sin x_{1}\sin x_{2}\sin\nu_{1}\cos\mu_{2}-\sin x_{1}\cos
x_{2}\sin\nu_{1}\sin\mu_{2})(\sin x_{4}\cos\nu_{3}-\cos
x_{4}\sin\nu_{3})\sin\nu_{2}\bigg{)}^{2}$ $\displaystyle+$
$\displaystyle\bigg{(}(\sin x_{1}\cos\nu_{1}\sin\mu_{1}-\cos
x_{1}\sin\nu_{1}\sin\mu_{2})$ $\displaystyle+$ $\displaystyle(\cos
x_{3}\cos\nu_{2}+\sin x_{3}\sin\nu_{2})(\cos y_{4}\cos\nu_{3}+\sin
y_{4}\sin\nu_{3})\bigg{)}^{2}$ $\displaystyle+$ $\displaystyle\bigg{(}(\cos
x_{1}\cos\nu_{1}\sin\mu_{1}+\sin x_{1}\sin\nu_{1}\sin\mu_{2})$
$\displaystyle+$ $\displaystyle(\cos x_{3}\cos\nu_{2}+\sin
x_{3}\sin\nu_{2})(\cos x_{4}\cos\nu_{3}+\sin x_{4}\sin\nu_{3})\bigg{)}^{2}$
$\displaystyle+$ $\displaystyle\bigg{(}(\sin x_{1}\sin
x_{2}\cos\nu_{1}\cos\mu_{1}-\sin x_{1}\cos x_{2}\cos\nu_{1}\sin\mu_{1}-$
$\displaystyle\cos x_{1}\sin x_{2}\sin\nu_{1}\cos\mu_{2}+\cos x_{1}\cos
x_{2}\sin\nu_{1}\sin\mu_{2})(\sin y_{4}\cos\nu_{3}-\cos
y_{4}\sin\nu_{3})\sin\nu_{2}\bigg{)}^{2}.$

Due to the difficulty of simplifying the function
$f(\nu_{1},\nu_{2},\nu_{3},\mu_{1},\mu_{2})$ in Eq. (V), we aim to find an
upper bound of the function, over the parameters $\nu_{j},\mu_{j}\in[0,2\pi)$
and real constant $x_{j},y_{j}$. In particular, one can obtain that

$\displaystyle f(0,0,\pi/2,0,\mu_{2})=(\cos x_{2})^{2}(\sin x_{3})^{2},$ (49)
$\displaystyle f(0,0,0,\pi/2,\mu_{2})=(\cos x_{1})^{2}(\cos x_{3})^{2}(\cos
x_{4})^{2}+(\cos x_{3})^{2}(\cos y_{4})^{2}(\sin x_{1})^{2},$ (50)
$\displaystyle f(\pi/2,0,\pi/2,\pi/2,0)=(\cos x_{2})^{2}(\sin x_{3})^{2}(\cos
x_{4})^{2}+(\cos x_{2})^{2}(\sin x_{3})^{2}(\sin x_{4})^{2},$ (51)
$\displaystyle
f(\nu_{1},0,\nu_{3},0,0)=(\cos\nu_{1})^{2}(\cos\nu_{3})^{2}+(\sin\nu_{1})^{2}(\cos
x_{2})^{2}(\sin x_{3})^{2}+$ $\displaystyle(\cos\nu_{1})^{2}(\cos
x_{2})^{2}(\sin\nu_{3})^{2}(\sin x_{3})^{2}.$ (52)

One can show that $f(\nu_{1},0,\nu_{3},0,0)\geq f(\nu_{1},0,\pi/2,0,0)=(\cos
x_{2})^{2}(\sin x_{3})^{2}$. Hence Eq. (45) is upper bounded by the minimum of
(49)-(51), denoted as ${\cal M}$. As a result, the geometric measure of
entanglement of $\rho$, namely $G(\rho)$ in (43) is upper bounded by
$-\log_{2}(1-{\cal M})$. So we have managed to evaluate the geometric measure
of the PPT entangled state $\rho$ from a family of $2\times 2\times 4$ UPBs we
have constructed in this paper. The more explicit evaluation requires a better
understanding of (V).

##  VI Conclusions

We have constructed two families of multipartite UPBs. They are respectively a
$2\times 2\times 4$ UPB of size eight by using the four-qubit UPBs of size
eight, as well as a $2\times 2\times 2\times 4$ UPB of size eight by using the
five-qubit UPBs of size eight. As a byproduct, we have constructed two novel
multipartite PPT entangled states and evaluate their entanglement. A question
arising from this paper is to construct more multipartite high-dimensional
UPBs using the existing multiqubit UPBs, as it is always a technical challenge
with extensive applications in various quantum information tasks.

## Acknowledgments

LC was supported by the NNSF of China (Grant No. 11871089), and the
Fundamental Research Funds for the Central Universities (Grant Nos.
KG12040501, ZG216S1810 and ZG226S18C1).

## Conflict of interest statement

On behalf of all authors, the corresponding author states that there is no
conflict of interest.

## Data availability statement

All data, models, and code generated or used during the study appear in the
submitted article.

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# Speech Representation Learning through
Self-supervised Pretraining and Multi-task Finetuning

###### Abstract

Speech representation learning plays a vital role in speech processing. Among
them, self-supervised learning (SSL) has become an important research
direction. It has been shown that an SSL pretraining model can achieve
excellent performance in various downstream tasks of speech processing. On the
other hand, supervised multi-task learning (MTL) is another representation
learning paradigm, which has been proven effective in computer vision (CV) and
natural language processing (NLP). However, there is no systematic research on
the general representation learning model trained by supervised MTL in speech
processing. In this paper, we show that MTL finetuning can further improve SSL
pretraining. We analyze the generalizability of supervised MTL finetuning to
examine if the speech representation learned by MTL finetuning can generalize
to unseen new tasks.

Index Terms—  Speech representation learning, self-supervised learning, multi-
task learning, transfer learning

##  1 Introduction

Recently many SSL approaches have been proposed for pretraining models for
speech processing tasks, including pretraining with generative losses [1, 2,
3, 4] or discriminative losses [5, 6, 7, 8, 9]. Some approaches use multi-task
learning with multiple SSL objectives [10, 11]. After a shared model is
pretrained with SSL to extract general representations, it can then be
specialized on downstream tasks with task-specific head models and simple
finetuning. This method achieves state-of-the-art performance in many
applications.

To fairly evaluate the generalizability of SSL approaches without further
heavy downstream task-specific finetuning, the SUPERB benchmark [12] is
proposed. The SUPERB benchmark measures the performance of a shared model
across a wide range of speech processing tasks without heavy finetuning. Ten
tasks are included to investigate four aspects of speech: content, speaker,
semantics, and paralinguistics. To evaluate a general model trained with SSL,
the pretrained model parameters are frozen, and the fixed representations are
extracted and fed into each task-specific prediction head (small downstream
model) for training. During the evaluation, the pretrained shared model and
trained prediction heads are used on all tasks. In the above scenario, some
self-supervised models show outstanding performances on all the ten tasks in
SUPERB.

Supervised multi-task learning (MTL) is to train a shared model on various
downstream tasks [13, 14, 15, 16, 17]. This paper wants to investigate if MTL
on various downstream tasks can further improve the representations from SSL.
In CV [18, 19] and NLP [20, 21], general models trained by MTL approaches can
be evaluated on benchmarks that include various tasks. However, in speech,
there has not been a systematic study of general representation learning
models trained by MTL of various speech processing tasks. SpeechNet [22]
proposes a general modularized model to perform a variety of speech processing
tasks, but the purpose of SpeechNet is to allow different tasks to utilize
different subsets of the general modules in SpeechNet, rather than proposing
one shared model to extract general representations for training task-specific
heads of downstream tasks.

In this paper, we investigate two MTL training scenarios and also one task
transfer learning scenario. For the two MTL scenarios, we select a state-of-
the-art SSL pretrained shared model in the SUPERB benchmark as the starting
point for MTL. Then we train the shared model with MTL in two different
scenarios:

  * •

All-task MTL Finetuning: Finetune the SSL pretrained shared model with all
tasks in SUPERB. It serves as a strong baseline for SSL approaches and the
following scenarios.

  * •

Leave-one-out MTL Finetuning: Finetune the SSL pretrained shared model with
all but one tasks in SUPERB. We can observe the influence of removing one task
on the learned representations and their performance on the other tasks.

To further examine if the representations learned with supervised MTL can
generalize to an unseen new task, we have an additional Task Transfer Learning
scenario.

  * •

We take a shared model from the Leave-one-out MTL scenario and freeze its
parameters. We extract the representations with this shared model for training
the prediction head of the remaining task that is not involved in MTL
finetuning.

Through the different training scenarios above, we perform a preliminary study
of the generalizability of representation learning by MTL of various speech
processing tasks on a standard benchmark. The code is released for
reproduction and future extension111https://github.com/s3prl/s3prl/tree/multi-
task-distributed.

Fig. 1: Four different training scenarios. Scenarios (b-1) and (b-2) require
the pretrained shared model from (a). Scenario (c) requires the finetuned
shared model from (b-2). The parameters of the shared model in scenarios (a)
and (c) are frozen.

##  2 Training Scenarios

In this section, we describe four related training scenarios in the following
subsections: SSL Pretraining, All-task MTL Finetuning, Leave-one-out MTL
Finetuning, and Task Transfer Learning. The two MTL finetuning scenarios
require the SSL pretrained shared model, and the Task Transfer Learning
scenario requires the shared model from the Leave-one-out MTL Finetuning
scenario.

###  2.1 SSL Pretraining

Many SSL approaches are evaluated and compared on the ten tasks in SUPERB
[12]. For each SSL approach, we first pretrain a model with SSL objectives.
Then we use this pretrained model as the shared model to extract
representations for all downstream tasks. The parameters of the pretrained
model are frozen. Then we train each task-specific prediction head (small
downstream model) with the fixed representations, as illustrated in Figure 1
(a).

###  2.2 All-task MTL Finetuning

We take the shared model pretrained in Subsection 2.1 as the starting point.
Then we further finetune the shared model with MTL by jointly training it with
downstream task-specific heads of all tasks in SUPERB, as illustrated in
Figure 1 (b-1). In this way, the shared model can be updated by the gradients
of all tasks to fit the respective objectives of each task. Therefore, the
representations extracted from the shared model can perform well on the tasks
involved in MTL. It serves as a strong baseline for SSL approaches and the
following scenarios. To further examine the generalizability of representation
learning by supervised MTL, we have two additional training scenarios below.

###  2.3 Leave-one-out MTL Finetuning

Similarly, we take the shared model pretrained in Subsection 2.1 as the
starting point. Then we further finetune the shared model with MTL by jointly
training it with downstream task-specific heads of all but one tasks in
SUPERB, as illustrated in Figure 1 (b-2). Compared to the finetuned shared
model in Subsection 2.2, we can observe the influence of removing one task on
the learned representations and their performance on the other tasks.

###  2.4 Task Transfer Learning

We take the finetuned shared model in Subsection 2.3 as the pretrained shared
model in this scenario. Then we freeze the shared model, and only train the
downstream head model of the remaining task that is not used in MTL
finetuning, as illustrated in Figure 1 (c).

Comparing the representations in this scenario with those learned with only
SSL approaches in Subsection 2.1, we can observe the generalizability of the
representations learned with MTL on a new task compared to SSL only. On the
other hand, in comparison with the representations learned with All-task MTL
in Subsection 2.2, we can observe how the performance of a task is influenced
if this task is not involved in the MTL finetuning.

##  3 Experimental Setup

Table 1: Experimental Results of training scenarios described in Section 2. We have the numbers in Scenario (b-2) be bold or underlined if they are better or worse than (b-1) respectively. Scenario | Tasks for MTL | ASR | PR | SF | SD | ER | IC | KS | ASV | SID
---|---|---|---|---|---|---|---|---|---|---
Finetuning | WER$\downarrow$ | PER$\downarrow$ | F1$\uparrow$ | CER$\downarrow$ | DER$\downarrow$ | ACC$\uparrow$ | ACC$\uparrow$ | ACC$\uparrow$ | EER$\downarrow$ | ACC$\uparrow$
(a) SSL | N/A | 6.42 | 5.41 | 88.53 | 25.20 | 5.88 | 64.24 | 98.34 | 96.30 | 5.11 | 81.42
(b-1) SSL+MTL | all | 6.22 | 3.61 | 87.56 | 26.76 | 4.93 | 67.28 | 99.60 | 97.34 | 6.76 | 90.86
(b-2): SSL+MTL | all but ASR | X | 3.63 | 87.28 | 27.11 | 4.89 | 65.07 | 99.63 | 97.57 | 7.78 | 90.69
all but PR | 6.79 | X | 86.94 | 27.66 | 4.81 | 66.73 | 99.66 | 97.44 | 7.94 | 91.16
all but SF | 6.10 | 3.39 | X | X | 4.73 | 65.71 | 99.58 | 97.18 | 7.61 | 90.70
all but SD | 6.28 | 3.54 | 87.94 | 26.31 | X | 66.73 | 99.63 | 97.11 | 7.49 | 90.79
all but ER | 6.17 | 3.40 | 87.45 | 26.90 | 4.77 | X | 99.55 | 97.27 | 7.19 | 90.51
all but IC | 6.13 | 3.34 | 87.65 | 26.94 | 4.78 | 66.08 | X | 97.27 | 6.74 | 90.55
all but KS | 6.17 | 3.55 | 87.83 | 26.88 | 4.91 | 66.27 | 99.71 | X | 7.86 | 90.67
all but ASV | 5.90 | 2.79 | 87.88 | 26.52 | 3.61 | 64.88 | 99.58 | 97.44 | X | 85.06
all but SID | 5.95 | 3.25 | 87.33 | 27.39 | 4.50 | 68.66 | 99.55 | 97.27 | 9.00 | X
(c) Task Transfer | N/A | 6.27 | 5.79 | 88.14 | 26.24 | 5.80 | 64.24 | 97.42 | 96.33 | 7.55 | 62.05

The improvement relations of tasks.

The hurt relations of tasks.

Fig. 2: Two relation graphs of tasks. If a task A performs worse/better after
removing a task B in MTL, we connect an edge from B to A in the
improvement/hurt graphs, indicating B can improves/hurts the performance of A.
The width of an edge is thin/medium/thick if the relative score change is in
the range [0.5%, 2%), [2%, 8%), or larger than 8%. If the the relative score
change is less than 0.5%, we consider it negligible and no edge is connected.

###  3.1 Tasks In SUPERB

Ten tasks in SUPERB can be used to investigate four aspects of speech: content
(Phoneme Recognition (PR), Automatic Speech Recognition (ASR), Keyword
Spotting (KS), and Query by Example Spoken Term Detection (QbE)), speaker
(Speaker Identification (SID), Automatic Speaker Verification (ASV), and
Speaker Diarization (SD)), semantics (Intent Classification (IC) and Slot
Filling (SF)), and paralinguistics (Emotion Recognition (ER)). Since no
downstream model training is required in QbE, we only perform MTL experiments
and compare the results on the other nine tasks.

  * •

PR converts an utterance into a sequence of phonemes. Alignment modeling is
included in the PR task to avoid the potential inaccurate forced alignment.
The evaluation metric is the phone error rate (PER).

  * •

ASR transcribes an utterance into a sequence of words. While PR analyzes the
performance of modeling phonetics, ASR reflects the performance of recognizing
more common text units in a real-world scenario. The evaluation metric is the
word error rate (WER).

  * •

KS identifies preregistered keywords in an utterance by classifying the
utterance into a predefined set of words. The task is important for on-device
speech processing and requires low response time. The evaluation metric is the
accuracy (ACC).

  * •

SID classifies the speaker identity of an utterance in a multi-class
classification setting, where the set of speakers are the same for both
training and testing. The evaluation metric is the accuracy (ACC).

  * •

ASV verifies whether the speakers of a pair of utterances match in a binary
classification setting. Different from SID, the speakers in the testing set
may not appear in the training set. Therefore, ASV is more challenging than
SID. The evaluation metric is the equal error rate (EER).

  * •

SD segments an utterance and classifies the segments into speaker identities,
i.e., who is speaking when. Multiple speakers can speak simultaneously. Rich
and various speaker characteristics should be encoded in the extracted
representations for each frame to represent mixtures of signals. The
evaluation metric is the diarization error rate (DER).

  * •

IC classifies an utterance into predefined classes of speaker intents. The
evaluation metric is the accuracy (ACC).

  * •

SF converts an utterance into a sequence of semantic slot-type classes. For
example, FromLocation can be a slot-type for a spoken word Taipei, which is
known as a slot-value. Both slot-types and slot-values are essential for an
SLU system. Therefore, we use two evaluation metrics for slot-types and slot-
values respectively: the slot-type F1 score (F1) and the slot-value character
error rate (CER).

  * •

ER predicts an emotion class for each utterance. The evaluation metric is
accuracy (ACC).

As for the datasets and splits used for each task, we follow the original
settings in SUPERB (PR [23], ASR [23], KS [24], SID [25], ASV [25], SD [26],
IC [27], SF [28], and ER [29]).

###  3.2 The SSL Pretraining Approach In Experiments

Many SSL approaches are evaluated and compared on the tasks in SUPERB. Among
them, HuBERT [9] achieves the overall best performance. Therefore, we select
HuBERT as the SSL pretraining approach across all of our experiments.

HuBERT utilizes an offline clustering algorithm on hidden representations to
provide aligned target labels for a BERT-like [30] prediction. The clustered
labels at the masked locations serve as the prediction targets. We use a
weighted sum of hidden representations of all layers in the HuBERT model as
the representations for downstream heads, as in SUPERB.

###  3.3 Model Architecture and Implementation Details

Since MTL requires more computational resources than single-task training, we
adopt HuBERT Base rather than Large in SUPERB as our shared model
architecture. For task-specific head architectures, we simply follow the
settings in SUPERB. We use a batch size of 2 for MTL finetuning in training
scenarios described in Subsections 2.2 and 2.3, and a batch size of 8 for
downstream head training in the training scenario described in Subsection 2.4.
Each model is trained with an Adam optimizer with a linearly warmup learning
rate from 0 to 1e-5 for the first 5000 steps and then a linearly decaying
learning rate to 0 for 195000 steps. For more implementation details, please
refer to the released code.

##  4 Experimental Results

All experimental results are presented in Table 1. An up-/down-arrow beside an
evaluation metric means that better performance results in a higher/lower
number of that metric. The results are grouped according to four different
training scenarios corresponding to the subsections in Section 2 respectively.

###  4.1 The Performance Of All-task MTL Finetuning

From the comparison between Scenario (a) and Scenario (b-1), we observe that
the performance of the shared model finetuned with All-task MTL is better in
the tasks ASR, PR, SD, ER, IC, KS, SID, and worse only in the tasks SF and ASV
than SSL pretraining. It indicates that MTL is a strong baseline for SSL
pretraining or other representation learning approaches.

One thing worth noting is that all tasks except for ASV do not suffer from
overfitting in terms of the validation scores during training. However, the
model is prone to overfit on ASV. In Scenario (a), the model checkpoint of a
downstream head can be determined by the best validation score during training
for each task respectively. Yet in Scenario (b-1), since the shared model is
jointly trained with the downstream heads of all tasks, it is hard to select
the model checkpoint based on the validation scores of all tasks. In this
paper, we simply select the last model checkpoint after 200,000 training steps
for all tasks. It may be a reason for the worse performance of MTL finetuning
on ASV. We leave exploring a better method to select the model checkpoint with
MTL in future work.

###  4.2 The Influence Of Removing One Task In MTL Finetuning

The rows in Scenario (b-2) show the results of finetuning the shared model
with Leave-one-out MTL. To compare these results with Scenario (b-1) more
clearly, we have the numbers in Scenario (b-2) be bold or underlined if they
are better or worse than (b-1) respectively. Furthermore, we calculate the
relative score increases/decreases of Scenario (b-2) compared to Scenario
(b-1), and plot two relation graphs in Figure 2 accordingly. For SF, we use
CER to plot the edges because the relative changes are small in terms of F1.
If a task A improves/regresses after removing another task B in MTL, it means
that B can hurt/help A in MTL.

  * •

For ASR, PR is an important auxiliary task for ASR, and SD helps a little. ASR
performs better after removing any of the other tasks.

  * •

For PR, ASR helps in MTL while the other tasks hurt the performance of PR. We
can observe that content recognition tasks such as ASR and PR are hurt the
most by speaker recognition tasks such as ASV and SID.

  * •

For SF, ASR, PR, ER, and SID help in MTL. IC and KS help a little in terms of
F1 but hurt in terms of CER. SD and ASV hurt SF.

  * •

For SD, all of the other tasks hurt SD. SD is also hurt the most by speaker
recognition tasks such as ASV and SID. Although SD, ASV and SID are all
related to speaker characteristics, SID and ASV focus on the utterance-level
embedding, while SD aims to distinguish frame-level speaker characteristics.
Therefore, the fine-grained information needed by SD may be lost when jointly
trained with ASV or SID.

  * •

For ER, all of the other tasks except for SID help ER.

  * •

For IC and KS, the influences of the other tasks are negligible.

  * •

For ASV, all of the other tasks except for IC help ASV. As discussed in
Subsection 4.1, ASV suffers from severe overfitting. Therefore, jointly
learning ASV with other tasks can mitigate this issue, especially with SID.

  * •

For SID, all of the other tasks except for PR help SID. ASV especially helps a
lot while the others help a little.

From another perspective, the results may provide a different insight in
addition to the evaluation of supervised MTL as a representation learning. If
we focus on a certain primary task, we may select proper auxiliary tasks to
assist the primary task based on these MTL experimental results. For example,
if we want to have a better performance on ER, we can jointly train ER with
all of the other tasks except for SID.

The transitive relations of tasks in MTL need to be further verified. For
example, suppose we know jointly training a task A with another task B can
improve the performance of A; and we also know jointly training A with another
task C can improve the performance of A. Yet, we cannot conclude that jointly
training A with B and C simultaneously can improve A’s performance. We leave
more in-depth research of MTL and its optimization on speech processing tasks
in future work.

###  4.3 The Performance Of Task Transfer Learning

To further examine the generalizability of representation learning by MTL, we
present the results of the Task Transfer Learning scenario in Table 1 (c).
Compared to Table 1 (a), all of the tasks except for ASR and KS in Scenario
(c) perform worse. It indicates SSL pretraining is still a more generalizable
representation learning approach for a new downstream task. On the other hand,
compared to Table 1 (b-1), all of the tasks except for SF in Scenario (c)
perform worse. It indicates whether a task is involved in MTL is crucial to
the performance of this task.

To obtain better generalizability, it is worth trying to train the shared
model with both SSL and MTL simultaneously as semi-supervised MTL
representation learning. We leave this exploration in future work.

##  5 Conclusion And Discussion

In this work, we investigate different training scenarios of supervised MTL as
a speech representation learning approach along with SSL pretraining on a
benchmark with various speech processing tasks. We analyze the
generalizability of representations learned with supervised MTL empirically.

This paper is only a preliminary study of MTL with various speech processing
tasks. The performance of MTL is dependent on many factors, such as the amount
of data, task relationships, noise and so on. These factors should be isolated
and investigated with more theoretical analyses and empirical experiments in
the future.

## References

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