$4\times 4$ unextendible product basis and genuinely entangled space

Abstract

We show that there are six inequivalent $4\times 4$ unextendible product bases (UPBs) of size eight, when we consider only 4-qubit product vectors. We apply our results to construct positive-partial-transpose entangled states of rank nine. They are at the same time 4-qubit, $2\times 2\times 4$ and $4\times 4$ states, and their ranges have product vectors. One of the six UPBs turns out to be orthogonal to an incompletely genuinely entangled space, in the sense that the latter does not contain $4\times 4$ product vector in any bipartition of 4-qubit systems. We also show that the multipartite UPB orthogonal to a genuinely entangled space exists if and only if the $n\times n\times n$ UPB orthogonal to a genuinely entangled space exists for some n. These results help understand an open problem in Demianowicz and Augusiak (Phys Rev A 98:012313, 2018).

Appendices

Appendix A: The description of six 4-qubit UOMs $F_1,F_2,\ldots ,F_6$

$$\begin{aligned} F_1= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&0&1\\ 1&g_3\ne 0,1&h_3\ne 0,1&i_3\ne 0,1,i_4\\ 1&g_3'&h_3&i_4\ne 0,1\\ f_5\ne 0,1&g_3'&1&i_4'\\ f_5&g_3&1&i_3'\\ f_5'&0&h_3'&1\\ f_5'&1&h_3'&0\\ \end{bmatrix}, \end{aligned}$$

(A1)

$$\begin{aligned} F_1(i_3=i_4')= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&0&1\\ 1&g_3\ne 0,1&h_3\ne 0,1&i_3\ne 0,1\\ 1&g_3'&h_3&i_3'\\ f_5\ne 0,1&g_3'&1&i_3\\ f_5&g_3&1&i_3'\\ f_5'&0&h_3'&1\\ f_5'&1&h_3'&0\\ \end{bmatrix}, \end{aligned}$$

(A2)

$$\begin{aligned} F_2= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&0&i_2\ne 0,1 \\ 1&g_3\ne 0,1&h_3\ne 0,1&i_3\ne i_4,i_4'\\ 1&g_3'&h_3&i_4\\ f_5\ne 0,1&g_3'&1&i_4'\\ f_5&g_3&1&i_3'\\ f_5'&1&h_3'&i_2'\\ f_5'&0&h_3'&1\\ \end{bmatrix}, \nonumber \\\end{aligned}$$

(A3)

$$\begin{aligned} (i) F_2(i_2=i_3, i_4\ne 0,1)= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&0&i_2\ne 0,1,i_4,i_4'\\ 1&g_3\ne 0,1&h_3\ne 0,1&i_2\\ 1&g_3'&h_3&i_4\\ f_5\ne 0,1&g_3'&1&i_4'\\ f_5&g_3&1&i_2'\\ f_5'&1&h_3'&i_2'\\ f_5'&0&h_3'&1\\ \end{bmatrix}, \end{aligned}$$

(A4)

$$\begin{aligned} F_2(i_2=i_3, i_4=0)= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&0&i_2\ne 0,1\\ 1&g_3\ne 0,1&h_3\ne 0,1&i_2\\ 1&g_3'&h_3&0\\ f_5\ne 0,1&g_3'&1&1\\ f_5&g_3&1&i_2'\\ f_5'&1&h_3'&i_2'\\ f_5'&0&h_3'&1\\ \end{bmatrix}, \end{aligned}$$

(A5)

$$\begin{aligned} F_2(i_2=i_3, i_4=1)= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&0&i_2\ne 0,1\\ 1&g_3\ne 0,1&h_3\ne 0,1&i_2\\ 1&g_3'&h_3&1\\ f_5\ne 0,1&g_3'&1&0\\ f_5&g_3&1&i_2'\\ f_5'&1&h_3'&i_2'\\ f_5'&0&h_3'&1\\ \end{bmatrix}, \end{aligned}$$

(A6)

$$\begin{aligned} (ii) F_2(i_2=i_3', i_4\ne 0,1)= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&0&i_2\ne 0,1,i_4,i_4' \\ 1&g_3\ne 0,1&h_3\ne 0,1&i_2'\\ 1&g_3'&h_3&i_4\\ f_5\ne 0,1&g_3'&1&i_4'\\ f_5&g_3&1&i_2\\ f_5'&1&h_3'&i_2'\\ f_5'&0&h_3'&1\\ \end{bmatrix}, \nonumber \\\end{aligned}$$

(A7)

$$\begin{aligned} F_2(i_2=i_3', i_4=0)= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&0&i_2\ne 0,1\\ 1&g_3\ne 0,1&h_3\ne 0,1&i_2'\\ 1&g_3'&h_3&0\\ f_5\ne 0,1&g_3'&1&1\\ f_5&g_3&1&i_2\\ f_5'&1&h_3'&i_2'\\ f_5'&0&h_3'&1\\ \end{bmatrix}, \end{aligned}$$

(A8)

$$\begin{aligned} F_2(i_2=i_3', i_4=1)= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&0&i_2\ne 0,1 \\ 1&g_3\ne 0,1&h_3\ne 0,1&i_2'\\ 1&g_3'&h_3&1\\ f_5\ne 0,1&g_3'&1&0\\ f_5&g_3&1&i_2\\ f_5'&1&h_3'&i_2'\\ f_5'&0&h_3'&1\\ \end{bmatrix}, \end{aligned}$$

(A9)

$$\begin{aligned} (iii) F_2(i_2=i_4, i_3\ne 0,1)= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&0&i_2\ne 0,1,i_3,i_3' \\ 1&g_3\ne 0,1&h_3\ne 0,1&i_3\\ 1&g_3'&h_3&i_2\\ f_5\ne 0,1&g_3'&1&i_2'\\ f_5&g_3&1&i_3'\\ f_5'&1&h_3'&i_2'\\ f_5'&0&h_3'&1\\ \end{bmatrix}, \nonumber \\\end{aligned}$$

(A10)

$$\begin{aligned} F_2(i_2=i_4, i_3=0)= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&0&i_2\ne 0,1\\ 1&g_3\ne 0,1&h_3\ne 0,1&0\\ 1&g_3'&h_3&i_2\\ f_5\ne 0,1&g_3'&1&i_2'\\ f_5&g_3&1&1\\ f_5'&1&h_3'&i_2'\\ f_5'&0&h_3'&1\\ \end{bmatrix}, \end{aligned}$$

(A11)

$$\begin{aligned} F_2(i_2=i_4, i_3=1)= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&0&i_2\ne 0,1\\ 1&g_3\ne 0,1&h_3\ne 0,1&1\\ 1&g_3'&h_3&i_2\\ f_5\ne 0,1&g_3'&1&i_2'\\ f_5&g_3&1&0\\ f_5'&1&h_3'&i_2'\\ f_5'&0&h_3'&1\\ \end{bmatrix}, \end{aligned}$$

(A12)

$$\begin{aligned} (iv) F_2(i_2=i_4', i_3\ne 0,1)= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&0&i_2\ne 0,1,i_3,i_3' \\ 1&g_3\ne 0,1&h_3\ne 0,1&i_3\\ 1&g_3'&h_3&i_2'\\ f_5\ne 0,1&g_3'&1&i_2\\ f_5&g_3&1&i_3'\\ f_5'&1&h_3'&i_2'\\ f_5'&0&h_3'&1\\ \end{bmatrix},\nonumber \\ \end{aligned}$$

(A13)

$$\begin{aligned} F_2(i_2=i_4', i_3=0)= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&0&i_2\ne 0,1 \\ 1&g_3\ne 0,1&h_3\ne 0,1&0\\ 1&g_3'&h_3&i_2'\\ f_5\ne 0,1&g_3'&1&i_2\\ f_5&g_3&1&1\\ f_5'&1&h_3'&i_2'\\ f_5'&0&h_3'&1\\ \end{bmatrix}, \end{aligned}$$

(A14)

$$\begin{aligned} F_2(i_2=i_4', i_3=1)= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&0&i_2\ne 0,1\\ 1&g_3\ne 0,1&h_3\ne 0,1&1\\ 1&g_3'&h_3&i_2'\\ f_5\ne 0,1&g_3'&1&i_2\\ f_5&g_3&1&0\\ f_5'&1&h_3'&i_2'\\ f_5'&0&h_3'&1\\ \end{bmatrix}, \end{aligned}$$

(A15)

$$\begin{aligned} F_3= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&0&i_2\ne 0,1 \\ 1&g_3\ne 0,1&h_3\ne 0,1,h_4&0\\ 1&g_3'&h_4\ne 0,1&i_2\\ f_5\ne 0,1&g_3'&1&i_2'\\ f_5&g_3&1&1\\ f_5'&1&h_3'&i_2'\\ f_5'&0&h_4'&1\\ \end{bmatrix}, \end{aligned}$$

(A16)

$$\begin{aligned} F_4= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&0&i_2\ne 0,1,i_3,i_3' \\ 1&g_3 \ne 0,1,g_4,g_4'&h_3\ne 0,1&i_3\ne 0,1 \\ 1&g_4\ne 0,1&h_3&i_3'\\ f_5\ne 0,1&g_3'&1&i_3\\ f_5&g_4'&1&i_3'\\ f_5'&1&h_3'&i_2'\\ f_5'&0&h_3'&1\\ \end{bmatrix}, \nonumber \\\end{aligned}$$

(A17)

$$\begin{aligned} F_5= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&0&i_2 \ne 0,1\\ 1&g_3\ne 0,1&h_3\ne 0,1&1\\ 1&g_3'&h_3&i_2'\\ f_5\ne 0,1,f_6&g_3&1&0\\ f_6 \ne 0,1&g_3'&1&i_2\\ f_5'&1&h_3'&i_2'\\ f_6'&0&h_3'&1\\ \end{bmatrix}, \end{aligned}$$

(A18)

$$\begin{aligned} F_6= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&h_2\ne 0,1&i_2\ne 0,i_3',i_4\\ 1&g_3\ne 0,1&0&i_3\ne 0,i_4\\ 1&g_3'&h_2&i_4\ne 0,1\\ f_5\ne 0,1&1&h_2'&i_3'\\ f_5&0&1&i_4'\\ f_5'&g_3&1&i_2'\\ f_5'&g_3'&h_2'&1\\ \end{bmatrix}, \end{aligned}$$

(A19)

$$\begin{aligned} (i)\ F_6(i_2=1)= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&h_2\ne 0,1&1\\ 1&g_3\ne 0,1&0&i_3\ne 0,1,i_4\\ 1&g_3'&h_2&i_4\ne 0,1\\ f_5\ne 0,1&1&h_2'&i_3'\\ f_5&0&1&i_4'\\ f_5'&g_3&1&0\\ f_5'&g_3'&h_2'&1\\ \end{bmatrix}, \end{aligned}$$

(A20)

$$\begin{aligned} F_6(i_2=1,i_3=i_4')= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&h_2\ne 0,1&1\\ 1&g_3\ne 0,1&0&i_3\\ 1&g_3'&h_2&i_3'\ne 0,1\\ f_5\ne 0,1&1&h_2'&i_3'\\ f_5&0&1&i_3\\ f_5'&g_3&1&0\\ f_5'&g_3'&h_2'&1\\ \end{bmatrix}, \end{aligned}$$

(A21)

$$\begin{aligned} (ii)\ F_6(i_2=i_3)= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&h_2\ne 0,1&i_2\ne 0,1,i_4,i_4'\\ 1&g_3\ne 0,1&0&i_2\\ 1&g_3'&h_2&i_4\ne 0,1\\ f_5\ne 0,1&1&h_2'&i_2'\\ f_5&0&1&i_4'\\ f_5'&g_3&1&i_2'\\ f_5'&g_3'&h_2'&1\\ \end{bmatrix}, \nonumber \\\end{aligned}$$

(A22)

$$\begin{aligned} (iii)\ F_6(i_2=i_4')= & {} \begin{bmatrix} 0&0&0&0\\ 0&1&h_2\ne 0,1&i_2\ne 0,1,i_3,i_3'\\ 1&g_3\ne 0,1&0&i_3\ne 0\\ 1&g_3'&h_2&i_2'\\ f_5\ne 0,1&1&h_2'&i_3'\\ f_5&0&1&i_2\\ f_5'&g_3&1&i_2'\\ f_5'&g_3'&h_2'&1\\ \end{bmatrix}.\nonumber \\ \end{aligned}$$

(A23)

Appendix B:The proof of Lemma 4

In this section, we prove Lemma 4.

(i) Using $F_1$ in “Appendix A” we have

$$\begin{aligned} S_{11}= \begin{bmatrix} 0&1&0&1\\ 1&g_3\ne 0,1&h_3\ne 0,1&i_3\ne 0,1,i_4\\ 1&g_3'&h_3&i_4\ne 0,1\\ f_5\ne 0,1&g_3'&1&i_4'\\ f_5&g_3&1&i_3'\\ f_5'&0&h_3'&1\\ f_5'&1&h_3'&0\\ \end{bmatrix}. \end{aligned}$$

(B1)

Let the $4\times 4$ product state $|x,y\rangle \in {{\mathcal {T}}}_{11}$. Lemma 2 (i) and (ii) imply that we have two cases. First, $|x\rangle$ is orthogonal to four product vectors in $\{|a_{1i},b_{1i}\rangle ,i=2,\ldots ,8\}$, and $|y\rangle$ is orthogonal to three product vectors in $\{|c_{1i},d_{1i}\rangle ,i=2,\ldots ,8\}$. Second, $|x\rangle$ is orthogonal to three product vectors in $\{|a_{1i},b_{1i}\rangle ,i=2,\ldots ,8\}$, and $|y\rangle$ is orthogonal to four product vectors in $\{|c_{1i},d_{1i}\rangle ,i=2,\ldots ,8\}$.

In the first case using Lemma 2 (iii), we obtain

$$\begin{aligned} |0,0,0,0\rangle , |f_5',0,h_3',0\rangle , |f_5,g_3,\beta _3\rangle , |f_5,g_3',\beta _4\rangle \in {{\mathcal {T}}}_{21}, \end{aligned}$$

(B2)

where $\beta _3,\beta _4$ are 2-qubit states in ${\mathbb {H}}_C\otimes {\mathbb {H}}_D$. In the second case, we still use Lemma 2 (iv) and exclude the same states in (B2). If $i_3=i_4'$ then we obtain

$$\begin{aligned} |\beta _5,h_3,i_3\rangle , |\beta _6,h_3,i_3'\rangle \in {{\mathcal {T}}}_{21}. \end{aligned}$$

(B3)

Hence $|{{\mathcal {T}}}_{11}|=4$ or 6. The latter holds if and only if $i_3=i_4'$.

(ii) Using $F_2(i_2=i_3, i_4=0)$ we have

$$\begin{aligned}&S_{21}(i_2=i_3, i_4=0) \\&\quad = \begin{bmatrix} 0&1&0&i_3\ne 0,1\\ 1&g_3\ne 0,1&h_3\ne 0,1&i_3\\ 1&g_3'&h_3&0\\ f_5\ne 0,1&g_3'&1&1\\ f_5&g_3&1&i_3'\\ f_5'&1&h_3'&i_3'\\ f_5'&0&h_3'&1\\ \end{bmatrix}. \end{aligned}$$

Similar to case (i), we obtain $|0,0,0,0\rangle , |f_5',0,h_3',0\rangle , |f_5,g_3,0,i_3'\rangle , |f_5,g_3',\gamma _1\rangle ,$

$|\gamma _2,h_3',i_3\rangle , |f_5',g_3,h_3,i_3'\rangle \in {{\mathcal {T}}}_{21}(i_2=i_3, i_4=0).$ Hence $|{{\mathcal {T}}}_{21}(i_2=i_3, i_4=0)|=6$.

(iii) Using $F_3$, we have $|0,0,0,0\rangle , |f_5',0,\delta _1\rangle , |f_5,g_3,\delta _2\rangle , |f_5,g_3',0,i_2'\rangle \in {{\mathcal {T}}}_{31}.$ Further if $h_3=h_4'$ then $|\delta _3,h_3,0\rangle \in {{\mathcal {T}}}_{31}.$ Hence $|{{\mathcal {T}}}_{31}|=4$ or 5.

(iv) Using $F_4$, we have $|0,0,0,0\rangle , |f_5',0,h_3',0\rangle , |\epsilon _1,h_3,i_3\rangle , |\epsilon _2,h_3,i_3'\rangle \in {{\mathcal {T}}}_{41}.$ Hence $|{{\mathcal {T}}}_{41}|=4$.

(v) Using $F_5$, we have $|0,0,0,0\rangle , |\zeta _1,h_3',i_2'\rangle , |\zeta _2,0,i_2\rangle , |0,g_3',h_3,i_2'\rangle \in {{\mathcal {T}}}_{51}.$ If $f_5=f_6'$ then $|f_5',g_3,\zeta _3\rangle , |f_5,g_3',\zeta _4\rangle \in {{\mathcal {T}}}_{61}.$ Hence $|{{\mathcal {T}}}_{51}|=4$ or 6.

(vi) Using $F_6(i_2=i_3)$, we have $|0,0,0,0\rangle , |f_5',g_3',h_2',0\rangle , |f_5',g_3,0,i_2'\rangle , |f_5,0,\eta _1\rangle ,$

$|\eta _2,h_2',i_2\rangle , |\eta _3,h_2,i_2'\rangle \in {{\mathcal {T}}}_{61}.$ Hence $|{{\mathcal {T}}}_{61}|=6$.

Appendix C: The construction of six 4-qubit UOMs $F_1,F_2,\ldots ,F_6$

We introduce a simple fact from [13, Lemma 2]. It will be used in the proof of Lemma 13.

Lemma 9

• (i) If ${{\mathcal {S}}}\subseteq ({\mathbb {C}}^2)^{\otimes n}$ is a UPB, then for all $|v\rangle \in {{\mathcal {S}}}$ and all integers $1\le j\le n$ there is another product vector $|w\rangle \in {{\mathcal {S}}}$ such that $|v\rangle$ and $|w\rangle$ are orthogonal on the j-th subsystem or j-th qubit.

• (ii) The number of distinct vectors of any qubit in a UPB is an even integer.

The following result from [28, Lemma 5] will be used in the proof of Lemma 14.

Lemma 10

Let X=$[x_{i,j}]$ $\in {{\mathcal {O}}}(m,n)$ be a UOM, and $\mu (x)$ the multiplicity of element x. If $p_j=\sum \mu (x)\mu (x')$, where the summation is over all pairs $\{x,x'\}$ in column j of X, then $\sum p_{j}\ge m(m-1)/2$.

We refer to the positive integer $p_j$ as the o-number of column j of X. It represents the number of all orthogonal pairs in column j of X. We refer readers to [28] for more details on UOMs.

We present a special partition of a positive integer into smaller positive integers. It will be also used in the proof of Lemma 14.

Lemma 11

Suppose p is the sum of 2n positive integers $a_1,a_2,\ldots ,a_{2n}$. Then the maximum of $a_1a_2+a_3a_4+\ldots +a_{2n-1}a_{2n}$ is $\lceil \frac{p-2n+2}{2}\rceil \cdot \lfloor \frac{p-2n+2}{2}\rfloor +n-1$. It is achievable if and only if up to the permutation of subscripts, we have $a_{1}=\lceil \frac{p-2n+2}{2}\rceil$, $a_{2}=\lfloor \frac{p-2n+2}{2}\rfloor$ and $a_i=1$ for $i>2$.

Proof

Let $N=a_1a_2+a_3a_4+\ldots +a_{2n-1}a_{2n}$. We fix the values of $a_1,a_3,\ldots ,a_{2n-1}$ and $a_6, a_8,\ldots ,a_{2n}$, and make $a_1>a_3>\cdots >a_{2n-1}$ by renaming the subscripts. Then we have $a_2+a_4=p-a_1-a_3-a_5-a_6-\ldots -a_{2n}$. Since $a_1>a_3$ is given, so $a_4=1$ and $a_2=p-a_1-a_3-a_5-a_6-\ldots -a_{2n}-1$ come to N greater. When free $a_2$, $a_4$ and $a_6$, using the argument same to freeing $a_2$ and $a_4$, N is greater if $a_4=a_6=1$ and $a_2=p-a_1-a_3-a_5-a_7-a_8-\ldots -a_{2n}-2$ is satisfied . Deducing the rest by this method, one obtain that N reaches the maximum when we fix $a_1$, $a_3$,...,$a_{2n-1}$ and take $a_4=a_6=\ldots =a_{2n}=1$, $a_2=p-a_1-a_3-\ldots -a_{2n-1}-(n-1)$. Using the similar argument to fixing $a_1$, $a_3$,...,$a_{2n-1}$, one can show that N reaches the maximum when we fix $a_2$, $a_4$,...,$a_{2n}$ and take $a_3=a_5=\ldots =a_{2n-1}=1$, $a_1=p-a_2-a_4-\ldots -a_{2n}-(n-1)$. It is to say that, we have $a_3=a_4=\ldots =a_{2n}=1$ and $a_1+a_2=p-(2n-2)$. So N reaches the maximum when $a_1=\lceil \frac{p-2n+2}{2}\rceil$, $a_2=\lfloor \frac{p-2n+2}{2}\rfloor$ and $a_i=1$ for $i=3,4,\ldots ,2n$. $\square$

From now on, we will study 4-qubit UPBs of size 8 and show how to find the UPBs ${{\mathcal {F}}}_1,\ldots ,{{\mathcal {F}}}_6$. First of all, we present the following observation by counting the data in [13].

Lemma 12

Let ${{\mathcal {T}}}_{A:B:C:D}$={$|f_1,g_1,h_1,i_1\rangle$, $|f_2,g_2,h_2,i_2\rangle$,...,$|f_8,g_8,h_8,i_8\rangle$} be a UPB of size 8. If one of the following three conditions holds, then ${{\mathcal {T}}}_{AB:CD}$ is not a UPB in the coarse graining ${\mathbb {C}}^4\otimes {\mathbb {C}}^4$.

• (i) ${{\mathcal {T}}}_{A:B:C:D}$ has a qubit having at least four identical vectors.

• (ii) There are three subscripts $j_1,j_2,j_3$ such that $|f_{j_1}\rangle =|f_{j_2}\rangle =|f_{j_3}\rangle$ and $|g_{j_1}\rangle =|g_{j_2}\rangle =|g_{j_3}\rangle$.

• (iii) There are five distinct subscripts $j_1,j_2,\ldots ,j_5$ such that $|f_{j_1}\rangle =|f_{j_2}\rangle =|f_{j_3}\rangle$ and $|g_{j_4}\rangle$=$|g_{j_5}\rangle$.

Proof

Take ${{\mathcal {T}}}_{AB:CD}$={$|p_1,q_1\rangle$, $|p_2,q_2\rangle$,...,$|p_8,q_8\rangle$}, where $|p_j\rangle =|f_j,g_j\rangle$ and $|q_j\rangle =|h_j,i_j\rangle$ for $j=1,2,\ldots ,8$.

• (i) By renaming the subscripts and permuting the qubits, we may assume that $|f_1\rangle =|f_2\rangle =|f_3\rangle =|f_4\rangle$. Also $|g_1\rangle$, $|g_2\rangle$, $|g_3\rangle$, $|g_4\rangle$ is linearly dependent. So $|p_1\rangle$, $|p_2\rangle$, $|p_3\rangle$, $|p_4\rangle$ are linearly dependent. Then the space spanned by $|p_1\rangle$, $|p_2\rangle$, $|p_3\rangle$, $|p_4\rangle$ has dimension at most two. Therefore, there exists a $|p\rangle \in {\mathbb {C}}^4$ orthogonal to $|p_1\rangle$, $|p_2\rangle$, $|p_3\rangle$, $|p_4\rangle$ and $|p_5\rangle$. Moreover, there is a $|q\rangle \in {\mathbb {C}}^4$ orthogonal to $|q_6\rangle$, $|q_7\rangle$ and $|q_8\rangle$. So $|p,q\rangle$ is orthogonal to ${{\mathcal {T}}}_{AB:CD}$. By definition ${{\mathcal {T}}}_{AB:CD}$ is not a UPB in ${\mathbb {C}}^4\otimes {\mathbb {C}}^4$.

• (ii) We have $|p_{j_1}\rangle$=$|p_{j_2}\rangle$=$|p_{j_3}\rangle$. Therefore, the space spanned by $|p_{j_1}\rangle$, $|p_{j_2}\rangle$, $|p_{j_3}\rangle$, $|p_{j_4}\rangle$ and $|p_{j_5}\rangle$ has dimension at most three. Besides the space spanned by $|q_{j_6}\rangle$, $|q_{j_7}\rangle$ and $|q_{j_8}\rangle$ also has dimension at most three. Then, there exist $|p\rangle$, $|q\rangle$ $\in {\mathbb {C}}^4$ such that $|p,q\rangle$ is orthogonal to ${{\mathcal {T}}}_{AB:CD}$.

• (iii) Since the space spanned by $|q_7\rangle$, $|q_8\rangle$ and $|q_9\rangle$ has dimension at most three, there is a $|q\rangle \in {\mathbb {C}}^4$ orthogonal to $|q_7\rangle$, $|q_8\rangle$ and $|q_9\rangle$. Then $|f_{i_1}',g_{i_2}',q\rangle$ is orthogonal to ${{\mathcal {T}}}_{AB:CD}$. $\square$

Lemma 13

Let ${{\mathcal {T}}}_{A:B:C:D}$={$|f_1,g_1,h_1,i_1\rangle$, $|f_2,g_2,h_2,i_2\rangle$,...,$|f_8,g_8,h_8,i_8\rangle$} be a UPB of size 8. If there are $|f_1\rangle =|f_2\rangle =|f_3\rangle$ and $|g_2\rangle =|g_3\rangle =|g_4\rangle$, then the remaining vectors $|f_4\rangle$, $|f_5\rangle$,...,$|f_8\rangle$ are pairwise linearly independent or ${{\mathcal {T}}}_{AB:CD}$ is no longer a UPB in ${{\mathcal {H}}}_{AB}:{{\mathcal {H}}}_{CD}$. Similarly, the remaining vectors $|g_1\rangle$, $|g_5\rangle$, $|g_6\rangle$,...,$|g_8\rangle$ are pairwise linearly independent or ${{\mathcal {T}}}_{AB:CD}$ is no longer a UPB in ${{\mathcal {H}}}_{AB}:{{\mathcal {H}}}_{CD}$.

Proof

First of all, we prove that either the vectors $|f_4\rangle$, $|f_5\rangle$,...,$|f_8\rangle$ are pairwise linearly independent or ${{\mathcal {T}}}_{AB:CD}$ is no longer a UPB in ${{\mathcal {H}}}_{AB}:{{\mathcal {H}}}_{CD}$. If the set {$f_4$, $f_5$,..., $f_8$} has two identical elements, then the same element has multipicity two, three, four or five. When the element has multiplicity three, four or five, ${{\mathcal {T}}}_{AB:CD}$ is no longer a UPB from Lemma 12 (iii). Also a qubit has an even number of distinct elements from Lemma 9 (ii). So when the set {$f_4$, $f_5$,..., $f_8$} has an element of multiplicity two, it must contain two different elements of both multiplicity two. It is a contradiction with Lemma 12 (iii). Now we have proved it.

Using the similar argument, one may show that the vectors $|g_1\rangle$, $|g_5\rangle$, $|g_6\rangle$,...,$|g_8\rangle$ are pairwise linearly independent or ${{\mathcal {T}}}_{AB:CD}$ is no longer a UPB in ${{\mathcal {H}}}_{AB}:{{\mathcal {H}}}_{CD}$. $\square$

Lemma 14

Let ${{\mathcal {T}}}_{A:B:C:D}$={$|f_1,g_1,h_1,i_1\rangle$, $|f_2,g_2,h_2,i_2\rangle$,...,$|f_8,g_8,h_8,i_8\rangle$} be a 4-qubit UPB of size 8. If the first and second qubit respectively have three identical vectors, then ${{\mathcal {T}}}_{A:B:C:D}$ is not a UPB of size 8 in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$.

Proof

Suppose $|a_1\rangle$, $|a_2\rangle$, $|a_3\rangle$ are three identical vectors among $|f_1\rangle$, $|f_2\rangle$,...,$|f_8\rangle$ and $|b_1\rangle$, $|b_2\rangle$, $|b_3\rangle$ are three identical vectors among $|g_1\rangle$, $|g_2\rangle$,...,$|g_8\rangle$. If $|a_1\rangle$, $|a_2\rangle$, $|a_3\rangle$, $|b_1\rangle$, $|b_2\rangle$, $|b_3\rangle$ are in three, five or six distinct product vectors of ${{\mathcal {T}}}_{A:B:C:D}$, then ${{\mathcal {T}}}_{AB:CD}$ is not a UPB from Lemma 12 (ii) and (iii).

We only need to investigate the case that $|a_1\rangle$, $|a_2\rangle$, $|a_3\rangle$, $|b_1\rangle$, $|b_2\rangle$, $|b_3\rangle$ are in four distinct product vectors. Denote by U the UOM over the UPB ${{\mathcal {T}}}_{A:B:C:D}$. Moreover, $p_j$ are o-number of column j of the U for $j=1,2,3,4$. Then U has $p_1=5$ and $p_2=5$ in condition of Lemma 13. Also $p_1+p_2+p_3+p_4\ge 8(8-1)/2$ from Lemma 10 (vi). So $p_3+p_4\ge 18$. We have $p_3$, $p_4$ $\le (\frac{8-2n+2}{2})^2=n^2-9n+24$ from Lemma 11 and $p=8$ in condition of Lemma 11. Then the possible value of n is 1, 2, 3, 4. If $n=1$, then $a_1=a_2=\frac{8-2\times 1+2}{2}=4$ in Lemma 11. That is, ${{\mathcal {T}}}_{A:B:C:D}$ has a qubit having four identical vectors. Then ${{\mathcal {T}}}_{A:B:C:D}$ is not a UPB from Lemma 12 (i). If $n=3$ or 4, then $p_3, p_4\le 6$ or 4. It makes a contradiction with $p_3+p_4\ge 18$. If $n=2$, then $p_3, p_4\le 10$. So the case $n=2$ is the only case that satisfies the condition $p_3+p_4\ge 18$. To satisfy $p_3+p_4\ge 18$, one of them must be $3\times 3+1\times 1$ and the other one could be either of $3\times 3+1\times 1$, $3\times 2+2\times 1$, $2\times 2+ 2\times 2$. There is no harm in supposing $p_3=3\times 3+1\times 1$. Then $p_4=3\times 3+1\times 1$, $3\times 2+2\times 1$ or $2\times 2+ 2\times 2$. We can obtain $h_{i_1}=h_{i_2}=h_{i_3}$ and $i_{i_4}=i_{i_5}$ for five distinct descripts $i_1$, $i_2$, $i_3$, $i_4$, $i_5$ $\in \{1, 2, 3, 4, 5, 6, 7, 8\}$. So ${{\mathcal {T}}}_{A:B:C:D}$ is not a UPB from Lemma 12 (iii). $\square$

Lemma 15

Let ${{\mathcal {T}}}_{A:B:C:D}$={$|f_1,g_1,h_1,i_1\rangle$, $|f_2,g_2,h_2,i_2\rangle$,...,$|f_8,g_8,h_8,i_8\rangle$} be a 4-qubit UPB of size 8. Then ${{\mathcal {T}}}_{AB:CD}$ is not a UPB of size 8 in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$ when one of the following three conditions is satisfied.

• (i) There are three subscripts $j_1,j_2,j_3$ such that $|f_{j_1}\rangle =|f_{j_2}\rangle =|f_{j_3}\rangle$ and $|h_{j_1}\rangle =|h_{j_2}\rangle =|h_{j_3}\rangle$.

• (ii) There are three subscripts $j_1,j_2,j_3$ such that $|f_{j_1}\rangle =|f_{j_2}\rangle =|f_{j_3}\rangle$, $|g_{j_1}\rangle =|g_{j_2}\rangle$ and $|h_{j_1}\rangle =|h_{j_2}\rangle$.

• (iii) There are three subscripts $j_1,j_2,j_3$ such that $|f_{j_1}\rangle =|f_{j_2}\rangle$, $|g_{j_1}\rangle =|g_{j_2}\rangle$ and $|h_{j_1}\rangle =|h_{j_2}\rangle =|h_{j_3}\rangle$.

Proof

(i) We prove the assertion by contradiction. Suppose ${{\mathcal {T}}}_{AB:CD}$ is a UPB of size 8 in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$. Up to the equivalence, we can assume $j_1=1$, $j_2=2$ and $j_3=3$, and $f_1=f_2=f_3=h_1=h_2=h_3=0$. We express the UOM of ${{\mathcal {T}}}_{A:B:C:D}$ as

$$\begin{aligned} U_1= \begin{bmatrix} 0&\quad g_1&\quad 0&\quad i_1\\ 0&\quad g_2&\quad 0&\quad i_2\\ 0&\quad g_3&\quad 0&\quad i_3\\ f_4&\quad g_4&\quad h_4&\quad i_4\\ f_5&\quad g_5&\quad h_5&\quad i_5\\ f_6&\quad g_6&\quad h_6&\quad i_6\\ f_7&\quad g_7&\quad h_7&\quad i_7\\ f_8&\quad g_8&\quad h_8&\quad i_8\\ \end{bmatrix}. \end{aligned}$$

(C1)

Since the first three rows of $U_1$ correspond to three orthogonal product vectors, we obtain that $|g_1,i_1\rangle ,|g_2,i_2\rangle ,|g_3,i_3\rangle$ are orthogonal. Up to equivalence we may assume that $g_1=g_2=0$ and $g_3=1$. Since ${{\mathcal {T}}}_{AB:CD}$ is a UPB of size 8 in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$, Lemma 12 (iii) shows that $g_4,g_5,\ldots ,g_8$ are distinct. In addition, we have $g_4,g_5,\ldots ,g_8\ne 0$ from Lemma 14. So one of them must be 1 by Lemma 9 (ii). So $U_1$ is equivalent to

$$\begin{aligned} \begin{bmatrix} 0&\quad 0&\quad 0&\quad i_1\\ 0&\quad 0&\quad 0&\quad i_1'\\ 0&\quad 1&\quad 0&\quad i_3\\ f_4&\quad 1&\quad h_4&\quad i_4\\ f_5&\quad g_5&\quad h_5&\quad i_5\\ f_6&\quad g_5'&\quad h_6&\quad i_6\\ f_7&\quad g_7&\quad h_7&\quad i_7\\ f_8&\quad g_7'&\quad h_8&\quad i_8\\ \end{bmatrix}. \end{aligned}$$

(C2)

We have $g_5,g_7\ne 0,1$ from the discussion in the paragraph above (C2). Since the first and second row vectors are orthogonal to the last four row vectors of $U_2$, we obtain that $|0,0\rangle \in {{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_C$ is orthogonal to $|f_j,h_j\rangle$ for $j=5,6,7,8$. Since ${{\mathcal {T}}}_{AB:CD}$ is a UPB of size 8, Lemma 12 (iii) shows that $f_5,f_6,f_7,f_8$ contain exactly two 1’s, and so do $h_5,h_6,h_7,h_8$. So the matrix in (C2) is equivalent to

$$\begin{aligned} U_{11}= \begin{bmatrix} 0&\quad 0&\quad 0&\quad i_1\\ 0&\quad 0&\quad 0&\quad i_1'\\ 0&\quad 1&\quad 0&\quad i_3\\ f_4&\quad 1&\quad h_4&\quad i_4\\ 1&\quad g_5&\quad h_5&\quad i_5\\ 1&\quad g_5'&\quad h_6&\quad i_6\\ f_7&\quad g_7&\quad 1&\quad i_7\\ f_8&\quad g_7'&\quad 1&\quad i_8\\ \end{bmatrix} \quad \text { or}\quad U_{12}= \begin{bmatrix} 0&\quad 0&\quad 0&\quad i_1\\ 0&\quad 0&\quad 0&\quad i_1'\\ 0&\quad 1&\quad 0&\quad i_3\\ f_4&\quad 1&\quad h_4&\quad i_4\\ 1&\quad g_5&\quad h_5&\quad i_5\\ f_6&\quad g_5'&\quad 1&\quad i_6\\ 1&\quad g_7&\quad h_7&\quad i_7\\ f_8&\quad g_7'&\quad 1&\quad i_8\\ \end{bmatrix}, \end{aligned}$$

(C3)

where $g_5\ne g_7,g_7'$. For $U_{11}$, Lemma 12 shows that $h_5,f_7,f_8\ne 0$. Since row 5 is orthogonal to row 7 and 8, we have $i_5'=i_7=i_8$. So column 3 and 4 of $U_{11}$ shows a contradiction with Lemma 12 (iii) and the fact that ${{\mathcal {T}}}_{AB:CD}$ is a UPB of size 8.

On the other hand for $U_{12}$, similar to the above argument for $U_{11}$ one can show that $i_8=i_5'$, $i_7=i_6'$, $f_8=f_6'$ and $h_7=h_5'$. Then row 4 of $U_{12}$ is not orthogonal to all four bottom row vectors of $U_{12}$. It is a contradiction with the fact that $U_{12}$ is a UOM. We have proven that ${{\mathcal {T}}}_{AB:CD}$ is not a UPB of size 8 in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$.

(ii) We prove the assertion by contradiction. Suppose ${{\mathcal {T}}}_{AB:CD}$ is a UPB of size 8 in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$. Up to the equivalence, we can assume $j_1=1$, $j_2=2$ and $j_3=3$ and $f_1=f_2=f_3=g_1=g_2=h_1=h_2=0$. We express the UOM of ${{\mathcal {T}}}_{A:B:C:D}$ as

$$\begin{aligned} U_1= \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ 0&\quad g_3&\quad h_3&\quad i_3\\ f_4&\quad g_4&\quad h_4&\quad i_4\\ f_5&\quad g_5&\quad h_5&\quad i_5\\ f_6&\quad g_6&\quad h_6&\quad i_6\\ f_7&\quad g_7&\quad h_7&\quad i_7\\ f_8&\quad g_8&\quad h_8&\quad i_8\\ \end{bmatrix}. \end{aligned}$$

(C4)

We claim that there are only two cases $U_{11}$ and $U_{12}$ in (C5), where $f_5$ may be $f_7$ or $f_7'$ and $f_5, f_6, f_7\ne 0,1$ is satisfied.

$$\begin{aligned} U_{11}= \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ 0&\quad g_3&\quad h_3&\quad i_3\\ 1&\quad g_4&\quad h_4&\quad i_4\\ f_5&\quad g_5&\quad h_5&\quad i_5\\ f_5'&\quad g_6&\quad h_6&\quad i_6\\ f_7&\quad g_7&\quad h_7&\quad i_7\\ f_7'&\quad g_8&\quad h_8&\quad i_8\\ \end{bmatrix}, {\quad } U_{12}= \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ 0&\quad g_3&\quad h_3&\quad i_3\\ 1&\quad g_4&\quad h_4&\quad i_4\\ 1&\quad g_5&\quad h_5&\quad i_5\\ f_6&\quad g_6&\quad h_6&\quad i_6\\ f_6&\quad g_7&\quad h_7&\quad i_7\\ f_6'&\quad g_8&\quad h_8&\quad i_8\\ \end{bmatrix}. \end{aligned}$$

(C5)

We can obtain that k of $f_4, f_5, f_6, f_7, f_8$ of $U_1$ equal to 1 from Lemma9(i), where $1\le k\le 5$ and k is a positive integer. Moreover, we have $f_4, f_5, f_6, f_7, f_8 \ne 0$ and $1\le k\le 3$ from Lemma 12(i). However, if $k=3$, then column 1, 2 of $U_1$ make a contradiction with the fact that $U_1$ is a UOM and Lemma 12(iii). Then we have $k=1,2$. Namely, up to equivalence we obtain two cases, $f_4=1,f_5,f_6,f_7,f_8\ne 1$ and $f_4=f_5=1,f_6,f_7,f_8\ne 1$. For $f_4=1,f_5,f_6,f_7,f_8\ne 1$, at most two of $f_5,f_6,f_7,f_8$ are the same from 12(iii). Moreover, from Lemma 9(i) we can obtain $f_6=f_5',f_8=f_7'$ up to equivalent, where $f_5$ may be $f_7$ or $f_7'$. So we have proved the claim in the line above (C5).

For $U_{11}$ in (C5), we claim that there are two cases, $U_{111}$ and $U_{112}$. We can obtain $f_5,f_5',f_7,f_7'\ne 1$ from $f_5,f_7\ne 0,1$ in the line above (C5). Since row 1, 2 are orthogonal to row 3, 5, 6, 7 and 8 of $U_{11}$, we obtain that $|0,0\rangle \in {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C$ is orthogonal to $g_j,h_j$ for $j=3,5,6,7,8$. First, one can show at most two of $g_3,g_5,g_6,g_7,g_8$ are 1’s. Otherwise, row 1 and 2 of $U_{11}$ is a contradiction with Lemma 12(iii) and the fact that $U_{11}$ is a UOM by the assumption $U_1$ in (C4) is a UOM. Second, one can show at most three of $h_3,h_5,h_6,h_7,h_8$ are 1’s from Lemma 12(i). Then we have shown that two of $g_3,g_5,g_6,g_7,g_8$ are 1’s and three of $h_3,h_5,h_6,h_7,h_8$ are 1’s in $U_{11}$. If $g_3=1$, up to equivalence we can assume $g_5=1$. We directly obtain $h_6=h_7=h_8=1$. Then $U_{11}$ becomes $U_{111}$. On the other hand for $g_3\ne 1$, that is $h_3=1$, up to equivalence we can assume $g_5=g_6=1$. We directly obtain $h_7=h_8=1$. Then $U_{11}$ becomes $U_{112}$. Now we have proved the claim at the beginning of this paragraph.

There is $|1,0,0,i_4'\rangle \in {{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C\otimes {{\mathcal {H}}}_D$ orthogonal to all row vectors of $U_{111}$ and $U_{112}$. It is a contradiction with the fact $U_{111}$ and $U_{112}$ are UOMs of size 8 in ${{\mathcal {H}}}_{CD}\otimes {{\mathcal {H}}}_{CD}$ by the assumption $U_1$ in (C4) is a UOM.

$$\begin{aligned} U_{111}= \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ 0&\quad 1&\quad h_3&\quad i_3\\ 1&\quad g_4&\quad h_4&\quad i_4\\ f_5&\quad 1&\quad h_5&\quad i_5\\ f_5'&\quad g_6&\quad 1&\quad i_6\\ f_7&\quad g_7&\quad 1&\quad i_7\\ f_7'&\quad g_8&\quad 1&\quad i_8\\ \end{bmatrix}, {\quad } U_{112}= \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ 0&\quad g_3&\quad 1&\quad i_3\\ 1&\quad g_4&\quad h_4&\quad i_4\\ f_5&\quad 1&\quad h_5&\quad i_5\\ f_5'&\quad 1&\quad h_6&\quad i_6\\ f_7&\quad g_7&\quad 1&\quad i_7\\ f_7'&\quad g_8&\quad 1&\quad i_8\\ \end{bmatrix}. \end{aligned}$$

(C6)

For $U_{12}$ in (C5), we claim that there are four cases $U_{121}$, $U_{122}$, $U_{123}$ and $U_{124}$ in (C7). Since row 1, 2 are orthogonal to row 3, 6, 7 and 8 of $U_{11}$, we obtain that $|0,0\rangle \in {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C$ is orthogonal to $|g_j,h_j\rangle$ for $j=3,6,7,8$. First, one can show at most two of $g_3,g_6,g_7,g_8$ are 1’s. Otherwise, rows 1 and 2 of $U_{11}$ are a contradiction with Lemma 12(iii) and the fact that $U_{12}$ is a UOM by the assumption that $U_1$ in (C4) is a UOM. Second, one can show at most three of $h_3,h_6,h_7,h_8$ are 1’s from Lemma 12(i). If three of $h_3,h_6,h_7,h_8$ are 1’s, then there exists $|0,1,0,i_3'\rangle$ orthogonal to $U_{12}$ for $g_3=1$ and there exists $|0,1,0,i_j\rangle$ orthogonal to $U_{12}$ for $h_3=1$ and $g_j=1$ for $j=6$, or 7, or 8. It is a contradiction with the definition of UPB and the fact $U_{12}$ is a UOM by the assumption $U_1$ in (C4) is a UOM. Then we have shown that two of $g_3,g_6,g_7,g_8$ are 1’s and two of $h_3,h_6,h_7,h_8$ are 1’s in $U_{12}$. For $g_3=1$, we have two cases, $g_6=1$ and $g_8=1$. In case one, one can directly obtain $h_7=h_8=1$. Then $U_{12}$ becomes $U_{121}$. In case two, one can directly obtain $h_6=h_7=1$. Then $U_{12}$ becomes $U_{122}$. On the other hand for $g_3\ne 1$, that is $h_3=1$, we also have two cases, $g_6=g_7=1$ and $g_6=g_8=1$. In case one, one can directly obtain $h_8=1$. Then $U_{12}$ becomes $U_{123}$. In case two, one can directly obtain $h_7=1$. Then $U_{12}$ becomes $U_{124}$. Now we have proved the claim at the beginning of this paragraph.

$$\begin{aligned} U_{121}= & {} \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ 0&\quad 1&\quad h_3&\quad i_3\\ 1&\quad g_4&\quad h_4&\quad i_4\\ 1&\quad g_5&\quad h_5&\quad i_5\\ f_6&\quad 1&\quad h_6&\quad i_6\\ f_6&\quad g_7&\quad 1&\quad i_7\\ f_6'&\quad g_8&\quad 1&\quad i_8\\ \end{bmatrix}, {\quad } U_{122}= \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ 0&\quad 1&\quad h_3&\quad i_3\\ 1&\quad g_4&\quad h_4&\quad i_4\\ 1&\quad g_5&\quad h_5&\quad i_5\\ f_6&\quad g_6&\quad 1&\quad i_6\\ f_6&\quad g_7&\quad 1&\quad i_7\\ f_6'&\quad 1&\quad h_8&\quad i_8\\ \end{bmatrix}, \nonumber \\ {\quad } U_{123}= & {} \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ 0&\quad g_3&\quad 1&\quad i_3\\ 1&\quad g_4&\quad h_4&\quad i_4\\ 1&\quad g_5&\quad h_5&\quad i_5\\ f_6&\quad 1&\quad h_6&\quad i_6\\ f_6&\quad 1&\quad h_7&\quad i_7\\ f_6'&\quad g_8&\quad 1&\quad i_8\\ \end{bmatrix}, {\quad } U_{124}= \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ 0&\quad g_3&\quad 1&\quad i_3\\ 1&\quad g_4&\quad h_4&\quad i_4\\ 1&\quad g_5&\quad h_5&\quad i_5\\ f_6&\quad 1&\quad h_6&\quad i_6\\ f_6&\quad g_7&\quad 1&\quad i_7\\ f_6'&\quad 1&\quad h_8&\quad i_8\\ \end{bmatrix}. \end{aligned}$$

(C7)

In the following, we show that neither of $U_{121}$, $U_{122}$, $U_{123}$, $U_{124}$ is a UOM in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$. This will prove the claim of (ii).

For $U_{121}$ in (C7), we have $f_6, f_6'\ne 1$ from the line above (C5). We obtain $g_7, g_8\ne 0$ from Lemma 14 and $h_3\ne 0$ from Lemma 15 (i). Since row 3 is orthogonal to row 7, 8, we can obtain $i_7=i_8=i_3'$. Then $U_{121}$ becomes $U_{1211}$ in (C8). So there exists $|0,0,1,i_3\rangle \in {{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C\otimes {{\mathcal {H}}}_D$ orthogonal to all row vectors of $U_{1211}$. It shows a contradiction with the definition of UOM and the fact that $U_{1211}$ is a UOM by the assumption $U_1$ in (C4) is a UOM.

For $U_{122}$ in (C7), we have $f_6\ne 1$ from the line above (C5). We obtain $g_6, g_7\ne 0$ from Lemma 14 and $h_3\ne 0$ from Lemma 15 (i). Since row 3 is orthogonal to row 6, 7, we can obtain $i_6=i_7=i_3'$. Then $U_{122}$ becomes $U_{1221}$ in (C8). So there exists $|0,0,1,i_3\rangle \in {{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C\otimes {{\mathcal {H}}}_D$ orthogonal to all row vectors of $U_{1221}$. It shows a contradiction with the definition of UPB and the fact that $U_{1221}$ is a UOM.

$$\begin{aligned} U_{1211}= \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ 0&\quad 1&\quad h_3&\quad i_3\\ 1&\quad g_4&\quad h_4&\quad i_4\\ 1&\quad g_5&\quad h_5&\quad i_5\\ f_6&\quad 1&\quad h_6&\quad i_6\\ f_6&\quad g_7&\quad 1&\quad i_3'\\ f_6'&\quad g_8&\quad 1&\quad i_3'\\ \end{bmatrix}, {\quad } U_{1221}= \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ 0&\quad 1&\quad h_3&\quad i_3\\ 1&\quad g_4&\quad h_4&\quad i_4\\ 1&\quad g_5&\quad h_5&\quad i_5\\ f_6&\quad g_6&\quad 1&\quad i_3'\\ f_6&\quad g_7&\quad 1&\quad i_3'\\ f_6'&\quad 1&\quad h_8&\quad i_8\\ \end{bmatrix}. \end{aligned}$$

(C8)

For $U_{123}$ in (C7), we have $f_6\ne 0,1$ from the line above (C5). We obtain $g_3,g_4,g_5\ne 0$ from Lemma 14. In the following, we show $h_4,h_5,h_7\ne 1$. First, one can obtain $h_4\ne 1$. Otherwise, we have $|1,0,0,i_5'\rangle \in {{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C\otimes {{\mathcal {H}}}_D$ is orthogonal to all row vectors of $U_{123}$. Second, one can obtain $h_5\ne 1$. Otherwise, we have $|1,0,0,i_4'\rangle \in {{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C\otimes {{\mathcal {H}}}_D$ is orthogonal to all row vectors of $U_{123}$. Lastly, one can obtain $h_7\ne 1$. Otherwise, we have $|0,1,0,i_6'\rangle \in {{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C\otimes {{\mathcal {H}}}_D$ is orthogonal to all row vectors of $U_{123}$. Then we have proved

$$\begin{aligned} f_6,f_6'\ne 0,1,\ g_3,g_4,g_5\ne 0,1\ h_4,h_5,h_7\ne 1. \end{aligned}$$

(C9)

We claim that $U_{123}$ in (C7) has two cases $U_{1231}$ and $U_{1232}$ in (C10). Since row 3 is orthogonal to row 6, 7, we can obtain $|1,i_3\rangle$ is orthogonal to $|h_6,i_6\rangle$ and $|h_7,i_7\rangle$ from $f_6\ne 1,g_3\ne 0$ by (C9). One can show that $h_6,h_7$ are not equal to 0 at the same time. Otherwise, column 3 of $U_{123}$ shows a contradiction with Lemma 12 (i) and the fact $U_{123}$ is a UOM by the assumption $U_1$ in (C4) is a UOM. Then we have $h_6=0,h_7\ne 0$ or $h_6\ne 0,h_7=0$ or $h_6,h_7\ne 0$. Since $f_6=f_7,g_6=g_7$ and $i_6,i_7$ is undetermined, one can obtain the first two cases $h_6=0,h_7\ne 0$ and $h_6\ne 0,h_7=0$ are equivalent. Up to equivalence, we have two cases $h_6=0,h_7\ne 0$ or $h_6,h_7\ne 0$ for $U_{123}$. For $h_6=0,h_7\ne 0$ in $U_{123}$, since $|1,i_3\rangle$ is orthogonal to $|h_7,i_7\rangle$ from the second line in this paragraph, we obtain $i_7=i_3'$. Since $h_7\ne 1$ by (C9) and row 5 is orthogonal to row 6, we have $i_6=i_7'=i_3$. Since $f_6\ne 0,g_4,g_5\ne 0, h_4,h_5\ne 1$ by (C9) and row 6 is orthogonal to row 4, 5, we have $i_4=i_5=i_6'=i_3'$. Then $U_{123}$ becomes $U_{1231}$ in (C10). For $h_6,h_7\ne 0$ in $U_{123}$, since $|1,i_3\rangle$ is orthogonal to $|h_6,i_6\rangle$ and $|h_7,i_7\rangle$ from the second line in this paragraph, we can obtain $i_6=i_7=i_3'$. Then $U_{123}$ becomes $U_{1232}$ in (C10). We have proved the claim at the beginning of this paragraph.

$$\begin{aligned} U_{1231}= \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ 0&\quad g_3&\quad 1&\quad i_3\\ 1&\quad g_4&\quad h_4&\quad i_3'\\ 1&\quad g_5&\quad h_5&\quad i_3'\\ f_6&\quad 1&\quad 0&\quad i_3\\ f_6&\quad 1&\quad h_7\ne 0&\quad i_3'\\ f_6'&\quad g_8&\quad 1&\quad i_8\\ \end{bmatrix}, {\quad } U_{1232}= \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ 0&\quad g_3&\quad 1&\quad i_3\\ 1&\quad g_4&\quad h_4&\quad i_4\\ 1&\quad g_5&\quad h_5&\quad i_5\\ f_6&\quad 1&\quad h_6\ne 0&\quad i_3'\\ f_6&\quad 1&\quad h_7\ne 0&\quad i_3'\\ f_6'&\quad g_8&\quad 1&\quad i_8\\ \end{bmatrix}.\nonumber \\ \end{aligned}$$

(C10)

For $U_{1231}$ in (C10), there exists $|1,0,0,i_3\rangle \in {{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C\otimes {{\mathcal {H}}}_D$ is orthogonal to all row vectors of $U_{1231}$. It shows a contradiction with the definition of UOM and the fact $U_{1231}$ is a UOM in ${{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C\otimes {{\mathcal {H}}}_D$ by the assumption $U_1$ in (C4) is a UOM. That is, $U_{1231}$ is not a UOM in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$.

For $U_{1232}$ in (C10), there exists $|0,1,0,i_3\rangle \in {{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C\otimes {{\mathcal {H}}}_D$ is orthogonal to all row vectors of $U_{1232}$. It shows a contradiction with the definition of UOM and the fact $U_{1232}$ is a UOM in ${{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C\otimes {{\mathcal {H}}}_D$ by the assumption $U_1$ in (C4) is a UOM. That is, $U_{1232}$ is not a UOM in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$.

Therefore, $U_{123}$ in (C10) is not a UOM in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$.

For $U_{124}$ in (C7), we have $f_6,f_6'\ne 0,1$ from the line above (C5). We obtain $g_3,g_4,g_5\ne 0$ from Lemma 14. In the following, we show that $h_4,h_5\ne 1$. First, one can obtain $h_4\ne 1$. Otherwise, we have $|1,0,0,i_5'\rangle \in {{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C\otimes {{\mathcal {H}}}_D$ is orthogonal to all row vectors of $U_{124}$. Second, one can obtain $h_5\ne 1$. Otherwise, we have $|1,0,0,i_4'\rangle \in {{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C\otimes {{\mathcal {H}}}_D$ is orthogonal to all row vectors of $U_{124}$. Then we have proved

$$\begin{aligned} f_6,f_6'\ne 0,1,\ g_3,g_4,g_5\ne 0,\ h_4,h_5\ne 1. \end{aligned}$$

(C11)

We claim that $U_{124}$ in (C7) has three cases $U_{1241}$, $U_{1242}$ and $U_{1243}$ in (C12). Since row 3 is orthogonal to row 6, 8, we can obtain $|1,i_3\rangle$ is orthogonal to $|h_6,i_6\rangle$ and $|h_8,i_8\rangle$ from $f_6,f_6'\ne 1,g_3\ne 0$ by (C11). One can show that $h_6,h_8$ are not equal to 0 at the same time. Otherwise, column 3 of $U_{124}$ shows a contradiction with Lemma 12 (i) and the fact $U_{124}$ is a UOM by the assumption $U_1$ in (C4) is a UOM. Then we have three cases $h_6=0,h_8\ne 0$ or $h_6\ne 0,h_8=0$ or $h_6,h_8\ne 0$ for $U_{124}$. For $h_6=0,h_8\ne 0$ in $U_{124}$, since $f_6\ne 0$, $g_4,g_5\ne 0$, $h_4,h_5\ne 1$ by (C11) and row 6 is orthogonal to row 4, 5, we have $i_4=i_5=i_6'$. Then $U_{124}$ becomes $U_{1241}$ in (C12). For $h_6\ne 0,h_8=0$ in $U_{124}$, since $f_6'\ne 0$, $g_4,g_5\ne 0$, $h_4,h_5\ne 1$ by (C11) and row 8 is orthogonal to row 4, 5, we have $i_4=i_5=i_8'$. Then $U_{124}$ becomes $U_{1242}$ in (C12). For $h_6,h_8\ne 0$ in $U_{124}$, since $|1,i_3\rangle$ is orthogonal to $|h_6,i_6\rangle$ and $|h_7,i_7\rangle$ from the second line in this paragraph, we can obtain $i_6=i_8=i_3'$. Then $U_{124}$ becomes $U_{1243}$ in (C12). We have proved the claim at the beginning of this paragraph.

$$\begin{aligned} U_{1241}= & {} \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ 0&\quad g_3&\quad 1&\quad i_3\\ 1&\quad g_4&\quad h_4&\quad i_6'\\ 1&\quad g_5&\quad h_5&\quad i_6'\\ f_6&\quad 1&\quad 0&\quad i_6\\ f_6&\quad g_7&\quad 1&\quad i_7\\ f_6'&\quad 1&\quad h_8\ne 0&\quad i_8\\ \end{bmatrix}, {\quad } U_{1242}= \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ 0&\quad g_3&\quad 1&\quad i_3\\ 1&\quad g_4&\quad h_4&\quad i_8'\\ 1&\quad g_5&\quad h_5&\quad i_8'\\ f_6&\quad 1&\quad h_6\ne 0&\quad i_6\\ f_6&\quad g_7&\quad 1&\quad i_7\\ f_6'&\quad 1&\quad 0&\quad i_8\\ \end{bmatrix}, \nonumber \\ {\quad } U_{1243}= & {} \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ 0&\quad g_3&\quad 1&\quad i_3\\ 1&\quad g_4&\quad h_4&\quad i_4\\ 1&\quad g_5&\quad h_5&\quad i_5\\ f_6&\quad 1&\quad h_6\ne 0&\quad i_3'\\ f_6&\quad g_7&\quad 1&\quad i_7\\ f_6'&\quad 1&\quad h_8\ne 0&\quad i_3'\\ \end{bmatrix}. \end{aligned}$$

(C12)

For $U_{1241}$ in (C12), there exists $|1,0,0,i_6\rangle \in {{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C\otimes {{\mathcal {H}}}_D$ is orthogonal to all row vectors of $U_{1241}$. It shows a contradiction with the definition of UOM and the fact $U_{1241}$ is a UOM in ${{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C\otimes {{\mathcal {H}}}_D$ by the assumption $U_1$ in (C4) is a UOM. That is, $U_{1241}$ is not a UOM in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$.

For $U_{1242}$ in (C12), there exists $|1,0,0,i_8\rangle \in {{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C\otimes {{\mathcal {H}}}_D$ is orthogonal to all row vectors of $U_{1242}$. It shows a contradiction with the definition of UOM and the fact $U_{1242}$ is a UOM in ${{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C\otimes {{\mathcal {H}}}_D$ by the assumption $U_1$ in (C4) is a UOM. That is, $U_{1242}$ is not a UOM in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$.

For $U_{1243}$ in (C12), there exists $|0,1,0,i_3\rangle \in {{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C\otimes {{\mathcal {H}}}_D$ is orthogonal to all row vectors of $U_{1243}$. It shows a contradiction with the definition of UOM and the fact $U_{1243}$ is a UOM in ${{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C\otimes {{\mathcal {H}}}_D$ by the assumption $U_1$ in (C4) is a UOM. That is, $U_{1243}$ is not a UOM in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$.

So $U_{124}$ in (C10) is not a UOM in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$.

We have proved the claim below (C7).

Therefore, we have proved that ${{\mathcal {T}}}_{AB:CD}$ is not a UPB of size 8 in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$.

(iii) We prove the assertion by contradiction. Suppose ${{\mathcal {T}}}_{AB:CD}$ is a UPB of size 8 in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$. Up to the equivalence, we can assume $j_1=1$, $j_2=2$ and $j_3=3$ and $f_1=f_2=g_1=g_2=h_1=h_2=h_3=0$. We express the UOM of ${{\mathcal {T}}}_{A:B:C:D}$ as

$$\begin{aligned} U_{1}= \begin{bmatrix} 0&0&0&0\\ 0&0&0&1\\ f_3&g_3&0&i_3\\ f_4&g_4&h_4&i_4\\ f_5&g_5&h_5&i_5\\ f_6&g_6&h_6&i_6\\ f_7&g_7&h_7&i_7\\ f_8&g_8&h_8&i_8\\ \end{bmatrix}. \end{aligned}$$

(C13)

We claim that there are only two cases $U_{11}$ and $U_{12}$, where $c_1$, $c_2$, $c_3$, $c_4$, $c_5$ are not 0’s or 1’s.

$$\begin{aligned} U_{11}= \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ f_3&\quad g_3&\quad 0&\quad i_3\\ f_4&\quad g_4&\quad 1&\quad i_4\\ f_5&\quad g_5&\quad c_1&\quad i_5\\ f_6&\quad g_6&\quad c_2&\quad i_6\\ f_7&\quad g_7&\quad c_3&\quad i_7\\ f_8&\quad g_8&\quad c_4&\quad i_8\\ \end{bmatrix}, {\quad } U_{12}= \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ f_3&\quad g_3&\quad 0&\quad i_3\\ f_4&\quad g_4&\quad 1&\quad i_4\\ f_5&\quad g_5&\quad 1&\quad i_5\\ f_6&\quad g_6&\quad c_5&\quad i_6\\ f_7&\quad g_7&\quad c_5&\quad i_7\\ f_8&\quad g_8&\quad c_5'&\quad i_8\\ \end{bmatrix}. \end{aligned}$$

(C14)

In fact, if one of $c_1$, $c_2$, $c_3$, $c_4$, $c_5$ is 0 or 1 then ${{\mathcal {T}}}_{AB:CD}$ is not a UPB from Lemma 12 (i) or $U_{11}=U_{12}$. If two of $c_1$, $c_2$, $c_3$, $c_4$ are 1’s, we can assume $c_1=c_2=1$. Then the space spanned by $|h_4,i_4\rangle$, $|h_5,i_5\rangle$, $|h_6,i_6\rangle$, $|h_7,i_7\rangle$ has dimension at most three. Also the space spanned by $|f_1,g_1\rangle$, $|f_2,g_2\rangle$, $|f_3,g_3\rangle$, $|f_8,g_8\rangle$ has dimension at most three. So there exist $\phi$, $\psi \in {\mathbb {C}}^4$ such that $\phi$ and $\psi$ is respectively orthogonal to $|f_1,g_1\rangle$, $|f_2,g_2\rangle$, $|f_3,g_3\rangle$, $|f_8,g_8\rangle$ and $|h_4,i_4\rangle$, $|h_5,i_5\rangle$, $|h_6,i_6\rangle$, $|h_7,i_7\rangle$. It makes a contradiction with the assumption that ${{\mathcal {T}}}_{AB:CD}$ is a UPB. Therefore, the form $U_{11}$ and $U_{12}$ are all possible cases. We have proved the claim above (C14).

For $U_{11}$, since the first two rows are orthogonal to row 3, 5, 6, 7, 8, we obtain that $|0,0\rangle \in {{\mathcal {H}}}_{AB}$ is orthogonal to $|f_j,g_j\rangle$ for $j=3,5,6,7,8$. Namely, three of $f_3$, $f_5$, $f_6$, $f_7$, $f_8$ or $g_3$, $g_5$, $g_6$, $g_7$, $g_8$ are 1’s. It’s a contradiction with Lemma 12 (iii) and the fact that ${{\mathcal {T}}}_{AB:CD}$ is a UPB of size 8.

For $U_{12}$, we have $f_3, g_3, f_4, g_4, f_5, g_5, f_6, g_6, f_7, g_7, f_8, g_8\ne 0$ from Lemma 15(ii). And the first rows are orthogonal to each of the last six rows of $U_{12}$. So $|0,0\rangle \in {{\mathcal {H}}}_{AB}$ is orthogonal to each of $|f_3,g_3\rangle$, $|f_6,g_6\rangle$, $|f_7,g_7\rangle$, $|f_8,g_8\rangle$. Also the first two rows both have at most two identical product vectors from Lemma 12 (iii). So we have the two cases $U_{121}$ and $U_{122}$.

$$\begin{aligned} U_{121}= \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ 1&\quad g_3&\quad 0&\quad i_3\\ f_4&\quad g_4&\quad 1&\quad i_4\\ f_5&\quad g_5&\quad 1&\quad i_5\\ 1&\quad g_6&\quad a_5&\quad i_6\\ f_7&\quad 1&\quad a_5&\quad i_7\\ f_8&\quad 1&\quad a_5'&\quad i_8\\ \end{bmatrix}, {\quad } U_{122}= \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ 1&\quad g_3&\quad 0&\quad i_3\\ f_4&\quad g_4&\quad 1&\quad i_4\\ f_5&\quad g_5&\quad 1&\quad i_5\\ f_6&\quad 1&\quad a_5&\quad i_6\\ f_7&\quad 1&\quad a_5&\quad i_7\\ 1&\quad g_8&\quad a_5'&\quad i_8\\ \end{bmatrix}. \end{aligned}$$

(C15)

For $U_{121}$, row 3 is orthogonal to row 7 and 8, so $|i_3\rangle$ is orthogonal to $|i_7\rangle$ and $|i_8\rangle$. That is, $i_7=i_8=i_3'$. Using the similar argument, we can obtain $i_6=i_7=i_3'$ in $U_{122}$ because row 3 is orthogonal to row 6 and 7. So $U_{121}$ and $U_{122}$ are, respectively, equivalent to $U_{1211}$ and $U_{1221}$.

$$\begin{aligned} U_{1211}= \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ 1&\quad g_3&\quad 0&\quad i_3\\ f_4&\quad g_4&\quad 1&\quad i_4\\ f_5&\quad g_5&\quad 1&\quad i_5\\ 1&\quad g_6&\quad a_5&\quad i_6\\ f_7&\quad 1&\quad a_5&\quad i_3'\\ f_8&\quad 1&\quad a_5'&\quad i_3'\\ \end{bmatrix}, {\quad } U_{1221}= \begin{bmatrix} 0&\quad 0&\quad 0&\quad 0\\ 0&\quad 0&\quad 0&\quad 1\\ 1&\quad g_3&\quad 0&\quad i_3\\ f_4&\quad g_4&\quad 1&\quad i_4\\ f_5&\quad g_5&\quad 1&\quad i_5\\ f_6&\quad 1&\quad a_5&\quad i_3'\\ f_7&\quad 1&\quad a_5&\quad i_3'\\ 1&\quad g_8&\quad a_5'&\quad i_8\\ \end{bmatrix}. \end{aligned}$$

(C16)

So column 3 and 4 of $U_{1211}$ and $U_{1221}$ shows a contradiction with Lemma 12 (iii) and the fact that ${{\mathcal {T}}}_{AB:CD}$ is a UPB of size 8. $\square$

The following proposition can be proven using similar argument to Lemma 12, 13, 14 and 15 , so we omit its proof. The conclusions in Proposition  are proven step by step. So the conclusion of Proposition  (viii) is the strongest, and it manages to find out all six UOMs $F_1,\ldots ,F_6$ in “Appendix A.” In fact, one can show that any UOM of size 8 has a column having two identical elements, by using the definition of UPB. So $F_1, F_2, F_3, F_4, F_5, F_6$ are exactly all UOMs of size 8 in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$ generated by 4-qubit UPBs.

Proposition 16

Let ${{\mathcal {T}}}_{A:B:C:D}$={$|f_1,g_1,h_1,i_1\rangle$, $|f_2,g_2,h_2,i_2\rangle$,...,$|f_8,g_8,h_8,i_8\rangle$} be a 4-qubit UPB of size 8 in ${{\mathcal {H}}}_A\otimes {{\mathcal {H}}}_B\otimes {{\mathcal {H}}}_C\otimes {{\mathcal {H}}}_D$.

• (i) If there are two subscripts $j_1,j_2$ such that $|f_{j_1}\rangle =|f_{j_2}\rangle$, $|g_{j_1}\rangle =|g_{j_2}\rangle$ and $|h_{j_1}\rangle =|h_{j_2}\rangle$, then ${{\mathcal {T}}}_{A:B:C:D}$ is no longer a UPB in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$.

• (ii) If there are two subscripts $j_1,j_2$ such that $|f_{j_1}\rangle =|f_{j_2}\rangle$, $|g_{j_1}\rangle =|g_{j_2}\rangle$, $|h_{j_1}\rangle =|h_{j_2}'\rangle$ and $|i_{j_1}\rangle =|i_{j_2}'\rangle$, then ${{\mathcal {T}}}_{A:B:C:D}$ is no longer a UPB in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$.

• (iii) If there are two subscripts $j_1,j_2$ such that $|f_{j_1}\rangle =|f_{j_2}\rangle$, $|g_{j_1}\rangle =|g_{j_2}'\rangle$, $|h_{j_1}\rangle =|h_{j_2}\rangle$ and $|i_{j_1}\rangle =|i_{j_2}'\rangle$, then $F_1$ in “Appendix A” happens to be only UOM in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$ satisfying this condition.

• (iv) If there are two subscripts $j_1,j_2$ such that $|f_{j_1}\rangle =|f_{j_2}\rangle$ and $|g_{j_1}\rangle =|g_{j_2}\rangle$, then ${{\mathcal {T}}}_{A:B:C:D}$ is no longer UPB in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$.

• (v) If there are two subscripts $j_1,j_2$ such that $|f_{j_1}\rangle =|f_{j_2}\rangle$ and $|h_{j_1}\rangle =|h_{j_2}\rangle$, then $F_1,F_2,F_3,F_4,F_5$ in “Appendix A” happen to be all UOMs in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$ satisfying this condition.

• (vi) If there are five distinct subscripts $j_1,j_2,j_3,j_4,j_5$ such that $|f_{j_1}\rangle =|f_{j_2}\rangle =|f_{j_3}\rangle$ and $|h_{j_3}\rangle =|h_{j_4}\rangle =|h_{j_5}\rangle$, then $F_1,F_2,F_3,F_4,F_5$ in “Appendix A” happen to be all UOMs in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$ satisfying this condition.

• (vii) If there are three distinct subscripts $j_1,j_2,j_3$ such that $|f_{j_1}\rangle =|f_{j_2}\rangle =|f_{j_3}\rangle$, then $F_1,F_2,F_3,F_4,F_5$ in “Appendix A” happen to be all UOMs in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$ satisfying this condition.

• (viii) If there are two subscripts $j_1,j_2$ such that $|f_{j_1}\rangle =|f_{j_2}\rangle$, then $F_1,F_2,F_3,F_4,F_5, F_6$ in “Appendix A” happen to be all UOMs in ${{\mathcal {H}}}_{AB}\otimes {{\mathcal {H}}}_{CD}$ satisfying this condition.