# Classifying Density Matrices of 2 and 3 Qubit States Up To LU Equivalence

Isaac Dobes and Naihuan Jing

###### Abstract

In this paper we present a modified version of the proof given Jing-Yang-
Zhao’s paper titled ”Local Unitary Equivalence of Quantum States and
Simultaneous Orthogonal Equivalence,” which established the correspondance
between local unitary equivalence and simultaneous orthogonal equivalence of
$2$-qubits. Our modified proof utilizes a hypermatrix algebra framework, and
through this framework we are able to generalize this correspondence to
$3$-qubits. Finally, we apply a generalization of Specht’s criterion (first
proved in ”Specht’s Criterion for Systems of Linear Mappings” by V. Futorney,
R. A. Horn, and V. V. Sergeichuk) to reduce the problem of local unitary
equivalence of $3$-qubits to checking trace identities and a few other easy-
to-check properties. We also note that all of these results can be extended to
$2$ and $3$ qudits if we relax the notion of LU equivalence to quasi-LU
equivalence, as defined in the aforementioned paper by Jing et. al.

##  1 Introduction

Quantum entanglement is a peculiar phenomenon that has intrigued yet also
eluded researchers for nearly a century [1]. In the last few decades, however,
serious interest and efforts in understanding entanglement has grown rapidly
due to its applications to quantum computing [2, 3]. One important task in the
study of entanglement is the classification of quantum states up to local
unitary (LU) equivalence [3, 4].

Two density matrices $\rho$ and $\widehat{\rho}$ representing two quantum
states in the same quantum system
$\mathbb{C}^{d_{1}}\otimes...\otimes\mathbb{C}^{d_{N}}$ are LU equivalent if
there exist $U_{i}\in SU(d_{i})$, $1\leq i\leq N$, such that

$\widehat{\rho}=(U_{1}\otimes...\otimes U_{N})\rho(U_{1}\otimes...\otimes
U_{N})^{\dagger}.$ (1)

In general, it is difficult to determine LU equivalence between to density
matrices because one does not a priori know anything about what these unitary
operators should be. To this end, in [5], Y. Makhlin provides necessary and
sufficient conditions for the LU equivalence of density matrices of $2$-qubit
states in terms of their Bloch representation. In [4], Jing et. al. consider a
similar representation of density matrices, but more generally for $2$-qudits,
and they establish a correspondence between simultaneous orthogonal (SO)
equivalence and quasi-LU equivalence (a property that follows from and is
closely related to LU equivalence). They then apply a generalization of
Specht’s criterion, first established in [6], to SO equivalent density
matrices, proving that quasi-LU equivalence reduces to checking trace
identities.

One shortcoming with the findings in [4] is that in their definition of quasi-
LU equivalence, they do not require the orthogonal operators to have
determinant $1$. This is not ideal because in the case of $2$-qubits, quasi-LU
equivalence and LU equivalence are in fact equivalent if the orthogonal
operators have determinant $1$. In this paper, we present a modified version
of the proof in [4] which establishes the correspondence between SO and quasi-
LU equivalence, showing that indeed we can require that the orthogonal
operators have determinant $1$. This strengthens their subsequent result,
showing that the problem of determining LU equivalence of density matrices
representing $2$-qubits reduces to checking trace identities.

An additional benefit to our modified proof is that it is written in the
language of hypermatrices and their algebra, and this allows us to generalize
the correspondence between SO and quasi-LU equivalence to $3$-qudits (and LU
equivalence in the case of $3$-qubits). We then apply a more powerful
generalization of Specht’s criterion, first established in [7], to reduce the
problem of determining quasi-LU equivalence of density matrices representing
$3$-qudits (or LU equivalence in the case of $3$-qubits) to trace identities.
Indeed, it is apparent that our proofs can, with relative ease, be further
generalized to $N$-qudits, however, there are computational limits to the
practicality of our results, hence we stop at the $3$-qudit case.

##  2 Preliminaries

###  2.1 Hypermatrix Algebra

Let $A\in F^{n_{1}\times...\times n_{d}}$ and $B\in F^{n_{1}\times...\times
n_{e}}$ be hypermatrices of order $d$ and $e$ respectively
($d,e\in\mathbb{N}$). Recall that the outer product of $A$ and $B$, denoted
$A\circ B$, is the hypermatrix of order $d+e$ whose
$i_{1},...,i_{d},j_{1},...,j_{e}$-coordinate is given by
$a_{i_{1},...,i_{d}}b_{j_{1},...,j_{e}}$. For a hypermatrix $A\in
F^{n_{1}\times n_{2}\times...\times n_{d}}$ and matrices $X_{1}\in
F^{\bullet\times n_{1}},...,X_{d}\in F^{\bullet\times n_{D}}$, the multilinear
matrix multiplication of $(X_{1},...,X_{d})$ with $A$ is defined to be the
hypermatrix $(X_{1},...,X_{d})*A=:A^{\prime}$, where

$A^{\prime}_{i_{1}i_{2}...i_{d}}=\sum\limits_{j_{1},j_{2},...,j_{N}=1}^{n_{1},n_{2},...,n_{N}}(X_{1})_{i_{1}j_{1}}...(X_{d})_{i_{d}j_{d}}A_{j_{1}j_{2}...j_{d}}.$
(2)

Multilinear matrix multiplication is linear in terms of the matrices in both
parts; that is, if $\alpha,\beta\in F$, $X_{1},Y_{1}\in F^{m_{1}\times
n_{1}}$;…; $X_{d},Y_{d}\in F^{m_{d}\times n_{d}}$; and $A,B\in F^{n_{1}\times
n_{2}\times...\times n_{d}}$; then

$(X_{1},...,X_{d})*(\alpha A+\beta
B)=\alpha(X_{1},...,X_{d})*A+\beta(Y_{1},...,Y_{d})*B$ (3)

and

$[\alpha(X_{1},...,X_{d})+\beta(Y_{1}...,Y_{d})]*A=\alpha(X_{1},...,X_{d})*A+\beta(Y_{1},...,Y_{d})*B.$
(4)

Let $k:[d]\rightarrow\mathbb{N}$. The outer product interacts with multilinear
matrix multiplication in the following way

$(X_{1_{1}},...,X_{1_{k(1)}},...,X_{d_{1}},...,X_{d_{k(d)}})*(A_{1}\circ...\circ
A_{d})=(X_{1_{1}},...,X_{1_{k(1)}})*A_{1}\circ...\circ(X_{d_{1}},...,X_{d_{k(d)}})*A_{d}$
(5)

where $X_{i_{j}}\in F^{m^{(i)}_{j}\times n^{(i)}_{j}}$ and $A_{i}\in
F^{n^{(i)}_{1}\times....\times n^{(i)}_{k(i)}}$.

In some instances, it is more convenient to view hypermatrices as matrices.
The $k$-mode unfolding of a hypermatrix $A\in\mathbb{C}^{n_{1}\times
n_{2}\times...\times n_{d}}$ is the
$n_{k}\times(n_{k+1}...n_{d}n_{1}...n_{k-1})$ matrix, denoted $A_{(k)}$, whose
$(i_{k},j)$ entry is given by $(i_{1},...,i_{d})$-entry of $A$, with

$j=1+\sum\limits_{\begin{subarray}{c}l=1\\\ l\neq
k\end{subarray}}^{d}\left[(i_{l}-1)\prod\limits_{\begin{subarray}{c}m=1\\\
m\neq k\end{subarray}}^{l-1}n_{m}\right],$ (6)

or in the case where the index starts at $0$,

$j=\sum\limits_{\begin{subarray}{c}l=1\\\ l\neq
k\end{subarray}}^{d}\left[i_{l}\prod\limits_{\begin{subarray}{c}m=1\\\ m\neq
k\end{subarray}}^{l-1}n_{m}\right].$ (7)

For more information on hypermatrices and their algebra, the reader is
referred to [8, 9].

###  2.2 Quasi-LU Equivalence

Let $\rho$ be the density matrix of a multipartite state on
$\mathbb{C}^{d_{1}}\otimes...\otimes\mathbb{C}^{d_{N}}$ and let
$\\{\lambda_{i}^{k}:0\leq i\leq d_{k}^{2}-1;k=1,...,N\\}$ denote the
generalized Gell-Mann basis for each partite. Then we can express $\rho$ as
the following:

$\displaystyle\quad\frac{1}{d_{1}\cdot...\cdot d_{N}}\bigg{(}I_{2}^{\otimes
N}+\sum\limits_{j_{1}=1}^{N}\sum\limits_{\alpha_{1}=1}^{d_{j_{1}}^{2}-1}T_{j_{1}}^{\alpha_{1}}\lambda_{\alpha_{1}}^{(j_{1})}+\sum\limits_{1\leq
j_{1}<j_{2}\leq
N}\sum\limits_{\alpha_{1}=1}^{d_{j_{1}}^{2}-1}\sum\limits_{\alpha_{2}=1}^{d_{j_{2}}^{2}-1}T_{j_{1}j_{2}}^{\alpha_{1}\alpha_{2}}\lambda_{\alpha_{1}}^{(j_{1})}\lambda_{\alpha_{2}}^{(j_{2})}+....$
$\displaystyle\qquad....+\sum\limits_{1\leq j_{1}<...<j_{M}\leq
N}\sum\limits_{\alpha_{1}=1}^{d_{j_{1}}^{2}-1}\sum\limits_{\alpha_{2}=1}^{d_{j_{2}}^{2}-1}...\sum\limits_{\alpha_{M}=1}^{d_{j_{M}}^{2}-1}T_{j_{1}j_{2}...j_{M}}^{\alpha_{1}\alpha_{2}...\alpha_{M}}\lambda_{\alpha_{1}}^{(1)}\lambda_{\alpha_{2}}^{(2)}...\lambda_{\alpha_{M}}^{j_{M}}+....$
$\displaystyle\qquad....+\sum\limits_{\alpha_{1}=1}^{d_{1}^{2}-1}...\sum\limits_{\alpha_{N}=1}^{d_{N}^{2}-1}T_{12...N}^{\alpha_{1}\alpha_{2}...\alpha_{N}}\lambda_{\alpha_{1}}^{(1)}\lambda_{\alpha_{2}}^{(2)}...\lambda_{\alpha_{N}}^{(N)}\bigg{)}$

where $\lambda_{\alpha_{i}}^{(j_{k})}=I_{2}\otimes...\otimes
I_{2}\otimes\lambda_{\alpha_{i}}^{j_{k}}\otimes I_{2}\otimes...\otimes I_{2}$
(i.e. $\lambda_{\alpha_{i}}$ is the $j_{k}^{th}$ factor in the tensor product
and a basis element of the $j_{k}^{\mathrm{th}}$ partite) and

$T_{j_{1}j_{2}...j_{M}}^{\alpha_{1}\alpha_{2}...\alpha_{M}}=\text{tr}(\rho\sigma_{\alpha_{1}}^{(j_{1})}\lambda_{\alpha_{2}}^{(j_{2})}...\lambda_{\alpha_{M}}^{(j_{M})}),\qquad
M\leq N,$ (8)

are all real coefficients. In particular, the
$T_{j_{1}}=\left[T_{j_{1}}^{1},...,T_{j_{1}}^{d_{j_{1}}^{2}-1}\right]^{t}$ are
vectors, the
$T_{j_{1}j_{2}}=[T_{j_{1}j_{2}}^{\alpha_{1}\alpha_{2}}]_{(d_{j_{1}}^{2}-1)\times(d_{j_{2}}^{2}-1)}$
are matrices, and in general the
$T_{j_{1}j_{2}...j_{M}}=[T_{j_{1}j_{2}...j_{M}}^{\alpha_{1}\alpha_{2}...\alpha_{M}}]_{(d_{j_{1}}^{2}-1)\times...\times(d_{j_{M}}^{2}-1)}$
are hypermatrices of order $M$. These vectors, matrices, and hypermatrices
uniquely define an $N$-qudit state, and so we call the set

$\\{T_{j_{1}...j_{M}}:1\leq j_{1}<...<j_{M}\leq N,M\leq N\\}$

the hypermatrix representation of $\rho$. For convenience, from here on out we
will denote $d_{k}^{2}-1$ as $\delta_{k}$.

Let $\rho$ and $\widehat{\rho}$ be two density matrices of $N$-qudit states in
$\mathbb{C}^{d_{1}}\otimes...\otimes\mathbb{C}^{d_{N}}$. $\rho$ and
$\widehat{\rho}$ are locally unitary (LU) equivalent if there exist $U_{i}\in
SU(d_{i})$, $1\leq i\leq N$ such that

$\widehat{\rho}=(U_{1}\otimes...\otimes U_{N})\rho(U_{1}\otimes...\otimes
U_{N})^{\dagger}.$ (9)

Given basis element $\lambda_{i}^{k}$, we have that

$U_{k}\lambda_{i}^{k}U_{k}^{\dagger}=\sum\limits_{j=1}^{\delta_{k}}X_{ij}\lambda_{j}^{k}$
(10)

for some matrix $X=[X_{ij}]_{\delta_{k}\times\delta_{k}}$. Since the
generalized Gell-Mann matrices are Hermitian,

$U_{k}\lambda_{i}^{k}U_{k}^{\dagger}=U\lambda_{i}^{k}U_{k}^{\dagger}$ (11)

from which it follows that the coefficients $x_{ij}$ are real numbers.
Moreover,

$\Tr((U_{k}\lambda_{i}^{k}U_{k}^{\dagger})(U_{k}\lambda_{j}^{k}U_{k}^{\dagger}))=\Tr(\lambda_{i}^{k}\lambda_{j}^{k}U^{\dagger}U)=\Tr(\lambda_{i}^{k}\lambda_{j}^{k})=\delta_{ij},$
(12)

where $\delta_{ij}$ denotes the Kronecker delta and the last equality follows
from the fact that the Gell-Mann matrices are orthogonal with respect to the
Hilbert-Schmidt inner product. Thus, it follows that the matrix $X$ is
orthogonal; denote it as $\widetilde{O}_{k}$. Then from the linearity of the
Kronecker product and the identity below:

$(A_{1}\otimes A_{2}\otimes...\otimes A_{n})(B_{1}\otimes
B_{2}\otimes...\otimes
B_{n})=(A_{1}B_{1})\otimes(A_{2}B_{2})\otimes...\otimes(A_{n}B_{n}),$ (13)

for each $M$ we have that

$\displaystyle(U_{1}\otimes...\otimes U_{N})\left(\sum\limits_{1\leq
j_{1}<...<j_{M}\leq
N}\sum\limits_{\alpha_{1}=1,...,\alpha_{M}=1}^{\delta_{j_{1}},...,\delta_{j_{M}}}T_{j_{1}...j_{M}}^{\alpha_{1}...\alpha_{M}}\lambda_{\alpha_{1}}^{(1)}...\lambda_{\alpha_{M}}^{j_{M}}\right)(U_{1}\otimes...\otimes
U_{N})^{\dagger}$ $\displaystyle=\sum\limits_{1\leq j_{1}<...<j_{M}\leq
N}\sum\limits_{\alpha_{1}=1,...,\alpha_{M}=1}^{\delta_{j_{1}},...,\delta_{j_{M}}}T_{j_{1}...j_{M}}^{\alpha_{1}...\alpha_{M}}(U_{1}\otimes...\otimes
U_{N})\lambda_{\alpha_{1}}^{(1)}...\lambda_{\alpha_{M}}^{j_{M}}(U_{1}\otimes...\otimes
U_{N})^{\dagger}$ $\displaystyle=\sum\limits_{1\leq j_{1}<...<j_{M}\leq
N}\sum\limits_{\alpha_{1}=1,...,\alpha_{M}=1}^{\delta_{j_{1}},...,\delta_{j_{M}}}T_{j_{1}...j_{M}}^{\alpha_{1}...\alpha_{M}}(U_{j_{1}}\lambda_{\alpha_{1}}U_{j_{1}}^{\dagger})^{(j_{1})}...(U_{j_{M}}\lambda_{\alpha_{M}}U_{j_{M}}^{\dagger})^{(j_{M})}\qquad\text{by
equation }\eqref{transformation}$ $\displaystyle=\sum\limits_{1\leq
j_{1}<...<j_{M}\leq
N}\sum\limits_{\alpha_{1}=1,...,\alpha_{M}=1}^{\delta_{j_{1}},...,\delta_{j_{M}}}T_{j_{1}...j_{M}}^{\alpha_{1}...\alpha_{M}}\left(\sum\limits_{k_{1}}^{\delta_{j_{1}}}(\widetilde{O}_{j_{1}})_{\alpha_{1}k_{1}}\lambda_{k_{1}}\right)^{(j_{1})}...\left(\sum\limits_{k_{M}=1}^{\delta_{j_{M}}}(\widetilde{O}_{j_{M}})_{\alpha_{M}k_{M}}\lambda_{k_{M}}\right)^{(j_{M})}$
$\displaystyle=\sum\limits_{1\leq j_{1}<...<j_{M}\leq
N}\left(\sum\limits_{k_{1}=1,...,k_{M}=1}^{\delta_{j_{1}},...,\delta_{j_{M}}}\right)\left(\sum\limits_{\alpha_{1}=1,...,\alpha_{M}=1}^{\delta_{j_{1}},...,\delta_{j_{M}}}(\widetilde{O}_{j_{1}}^{t})_{k_{1}\alpha_{1}}...(\widetilde{O}_{j_{M}}^{t})_{k_{M}\alpha_{M}}T_{j_{1}...j_{M}}^{\alpha_{1}...\alpha_{M}}\right)\lambda_{k_{1}}^{(1)}...\lambda_{k_{M}}^{(j_{M})}$
$\displaystyle=\sum\limits_{1\leq j_{1}<...<j_{M}\leq
N}\left(\sum\limits_{k_{1}=1,...,k_{M}=1}^{\delta_{j_{1}},...,\delta_{j_{M}}}\right)\left((O_{j_{1}},...,O_{j_{M}})*T_{j_{1}...j_{M}}\right)^{k_{1}...k_{M}}\lambda_{k_{1}}^{(j_{1})}...\lambda_{k_{M}}^{(j_{M})},\quad\text{setting
}\widetilde{O}_{j_{i}}^{t}=:O_{j_{i}}.$

Thus, we say that $\rho$ and $\widehat{\rho}$ are quasi-LU equivalent if there
exists $O_{j_{1}}\in O(\delta_{j_{1}}),...,O_{j_{M}}\in O(\delta_{j_{M}})$
such that

$\widehat{T}_{j_{1}...j_{M}}=(O_{j_{1}},...,O_{j_{M}})*T_{j_{1}...j_{M}}$ (14)

for $1\leq M\leq N$. As shown above, LU equivalence implies quasi-LU
equivalence. In the case of $N$-qubits, (1) defines a surjective map
$SU(2)\rightarrow SO(3)$, and so in this instance we may assume that the
matrices are special orthogonal and that the converse holds as well (i.e.
quasi-LU equivalence implies LU equivalence).

##  3 2-Qudits

The density matrix $\rho$ of a $2$-qudit on
$\mathbb{C}^{d_{1}}\otimes\mathbb{C}^{d_{2}}$ is given by

$\frac{1}{d_{1}d_{2}}\left(I_{2}\otimes
I_{2}+\sum\limits_{i=1}^{\delta_{1}}T_{1}^{i}\lambda_{i}^{1}\otimes
I_{2}+\sum\limits_{i=1}^{\delta_{2}}T_{2}^{i}I_{2}\otimes\lambda_{i}^{2}+\sum\limits_{i,j=1}^{\delta_{1},\delta_{2}}T_{12}^{ij}\lambda_{i}^{1}\otimes\lambda_{j}^{2}\right),$

and so its matrix representation is given by $\\{T_{1},T_{2},T_{12}\\}$. We
say that the matrix representations $\\{T_{1},T_{2},T_{12}\\}$ and
$\\{\widehat{T}_{1},\widehat{T}_{2},\widehat{T}_{12}\\}$ of $\rho$ and
$\widehat{\rho}$ (respectively) are simultaneously orthogonal (SO) equivalent
if there exists orthogonal matrices $O_{1}\in O(\delta_{1}),O_{2}\in
O(\delta_{2})$ such that

$\displaystyle\widehat{T}_{12}=(O_{1},O_{2})*T_{12}\quad\text{and}\quad\widehat{T}_{1}\circ\widehat{T}_{2}=(O_{1},O_{2})*(T_{1}\circ
T_{2}).$ (15)

In the case of $2$-qubits, we additionally assume that $O_{1},O_{2}$ have
determinant $1$ (i.e. $O_{1},O_{2}\in SO(3)$).

In [4], it is shown that for $2$-qudit states, quasi-LU equivalence is
equivalent to SO equivalence. However, in the special case of $2$-qubits, it
is not assumed that $O_{1},O_{2}\in SO(3)$, only that they are in $O(3)$.
Below, we present a modified proof of their result, showing that indeed in the
case of $2$-qubits quasi-LU equivalence is equivalent to SO equivalence with
$O_{1},O_{2}\in SO(3)$, and so consequently both are equivalent to LU
equivalence. Another benefit of our modified proof -as we will see later- is
that it extends to $3$-qudits (and indeed possibly $n$-qudits). Before we
state the result and give our proof, we first note that if
$v\in\mathbb{R}^{l}$ and $M\in\mathbb{R}^{m\times n}$, then direct computation
yields

$(v\circ M)_{(1)}=v\circ\mathrm{vec}(M),$ (16)

where $\mathrm{vec}$ is the vectorization operator which transforms the matrix
$M=[m_{ij}]_{m\times n}$ into the vector

$\left[\begin{array}[]{cccccccc}m_{11}&m_{21}&...m_{m1}&m_{12}&...&...&m_{mn}\end{array}\right]^{t}.$

Furthermore, by direct computation, we also have that

$\mathrm{vec}((X_{1},X_{2})*M)=(X_{2}\otimes X_{1})\mathrm{vec}(M)$ (17)

for any matrices $X_{1}\in\mathbb{R}^{\bullet\times m}$ and
$X_{2}\in\mathbb{R}^{\bullet\times n}$.

######  Theorem 1 (quasi-LU Equivalence and SO Equivalence: 2-Qudits).

Let $\rho$ and $\widehat{\rho}$ be density matrices of $2$-qudit states on
$\mathbb{C}^{d_{1}}\otimes\mathbb{C}^{d_{2}}$ with matrix representations
$\\{T_{1},T_{2},T_{12}\\}$ and
$\\{\widehat{T}_{1},\widehat{T}_{2},\widehat{T}_{12}\\}$ (respectively), and
assume that $T_{1},T_{2},T_{12}\neq\mathbf{0}$. $\rho$ and $\widehat{\rho}$
are quasi-LU equivalent if and only if they are SO equivalent and
$\|\widehat{T}_{1}\|=\|T_{1}\|$ or $\|\widehat{T}_{2}\|=\|T_{2}\|$.
Furthermore, in the case of $2$-qubits, they are LU equivalent.

###### Proof.

$\rho$ and $\widehat{\rho}$ are quasi-LU equivalent if and only if there
exists orthogonal matrices $O_{1}\in O(\delta_{1}),O_{2}\in O(\delta_{2})$
such that

$\widehat{T}_{1}=O_{1}*T_{1}=O_{1}T_{1},\qquad\widehat{T}_{2}=O_{2}*T_{2}=O_{2}T_{2},\qquad\text{and}\qquad\widehat{T}_{12}=(O_{1},O_{2})*T_{12}=O_{1}T_{12}O_{2}^{t}.$
(18)

Assuming quasi-LU equivalence, we immediately have
$\widehat{T}_{12}=(O_{1},O_{2})*T_{12}$; moreover,

$\widehat{T}_{1}\circ\widehat{T}_{2}=O_{1}T_{1}\circ
O_{2}T_{2}=O_{1}T_{1}(O_{2}T_{2})^{t}=O_{1}(T_{1}T_{2}^{t})O_{2}^{t}=(O_{1},O_{2})*T_{1}\circ
T_{2},$ (19)

and

$\|\widehat{T}_{i}\|=\|O_{i}*T_{i}\|=\|O_{i}T_{i}\|=\|T_{i}\|$ (20)

for $i=1,2$. Thus, forward implication is proven.

Conversely, suppose $\rho$ and $\widehat{\rho}$ are SO equivalent and without
loss of generality assume that $\|\widehat{T}_{1}\|=\|T_{1}\|$. First, note
that $T_{i},T_{12}\neq\mathbf{0}$ implies that
$\widehat{T}_{i},\widehat{T}_{12}\neq\mathbf{0}$; consequently, for $i=1,2$ we
have:

$\displaystyle 0<\|\widehat{T}_{i}\|^{2}\|\widehat{T}_{12}\|^{2}$
$\displaystyle=(\widehat{T}_{i}^{t}\widehat{T}_{i})(\text{vec}(\widehat{T}_{12})^{t}\text{vec}(\widehat{T}_{12}))$
(21)
$\displaystyle=\widehat{T}_{i}^{t}(\widehat{T}_{i}\circ\text{vec}(\widehat{T}_{12}))\text{vec}(\widehat{T}_{12})$
(22)
$\displaystyle=\widehat{T}_{i}^{t}(\widehat{T}_{i}\circ\widehat{T}_{12})_{(1)}\mathrm{vec}(\widehat{T}_{12})$
(23)
$\displaystyle=\widehat{T}_{i}^{t}\big{(}(O_{i}*T_{i})\circ((O_{1},O_{2})*T_{12})\big{)}_{(1)}\text{vec}(\widehat{T}_{12})$
(24)
$\displaystyle=\widehat{T}_{i}^{t}\big{(}O_{i}T_{i}\circ\text{vec}((O_{1},O_{2})*T_{12})\big{)}\mathrm{vec}(\widehat{T}_{12})$
(25) $\displaystyle=\widehat{T}_{i}^{t}\big{(}O_{i}T_{i}\circ(O_{2}\otimes
O_{1})\text{vec}(T_{12})\big{)}\text{vec}(\widehat{T}_{12})$ (26)
$\displaystyle=\widehat{T}_{i}^{t}O_{i}T_{i}\mathrm{vec}(T_{12})^{t}(O_{2}\otimes
O_{1})^{t}\text{vec}(\widehat{T}_{12})$ (27)

Now, $\widehat{T}_{i}^{t}O_{i}T_{i}$ and $\text{vec}(T_{jk})^{t}(O_{2}\otimes
O_{1})^{t}\text{vec}(\widehat{T}_{12})$ are just numbers, so equations (21)
and (27) imply that

$\alpha_{i}:=\frac{\widehat{T}_{i}^{t}O_{i}T_{i}}{\widehat{T}_{i}^{t}\widehat{T}_{i}}=\frac{\text{vec}(\widehat{T}_{jk})^{t}\text{vec}(\widehat{T}_{12})}{\text{vec}(T_{12})^{t}(O_{2}\otimes
O_{1})^{t}\text{vec}(\widehat{T}_{12})}\neq 0.$ (28)

By assumption we also have that

$(\widehat{T}_{i}\circ\widehat{T}_{12})_{(1)}=((O_{i},O_{1},O_{2})*T_{i}\circ
T_{12})_{(1)}.$ (29)

The left-hand side of the above equation can be written as
$\widehat{T}_{i}\mathrm{vec}(\widehat{T}_{12})^{t}$, and the right-hand side
can be expressed as

$\displaystyle\begin{split}((O_{i}\boldsymbol{\cdot}T_{i}\circ(O_{1},O_{2})*T_{12}))_{(1)}&=(O_{i}*T_{i})\circ\mathrm{vec}((O_{1},O_{2})*T_{12})\\\
&=O_{i}T_{i}\circ(O_{2}\otimes O_{1})\mathrm{vec}(T_{12})\\\
&=O_{i}T_{i}\mathrm{vec}(T_{12})(O_{2}\otimes O_{1})^{t}\end{split}$ (30)

which implies that

$\widehat{T}_{i}\underbrace{\text{vec}(\widehat{T}_{12})^{t}\text{vec}(\widehat{T}_{12})}_{>0}=O_{i}T_{i}\text{vec}(T_{12})^{t}(O_{2}\otimes
O_{1})^{t}\text{vec}(\widehat{T}_{12})$ (31)

Hence,

$\widehat{T}_{i}=\frac{O_{i}T_{i}\text{vec}(T_{12})^{t}(O_{2}\otimes
O_{1})^{t}\text{vec}(\widehat{T}_{12})}{\mathrm{vec}(\widehat{T}_{12})^{t}\mathrm{vec}(\widehat{T}_{12})}$
(32)

and so consequently by (28)

$\alpha_{i}\widehat{T}_{i}=\frac{\text{vec}(\widehat{T}_{12})^{t}\text{vec}(\widehat{T}_{12})}{\text{vec}(T_{12})^{t}(O_{2}\otimes
O_{1})^{t}\text{vec}(\widehat{T}_{12})}\cdot\frac{O_{i}T_{i}\text{vec}(T_{12})^{t}(O_{2}\otimes
O_{1})^{t}\text{vec}(\widehat{T}_{12})}{\text{vec}(\widehat{T}_{12})^{t}\text{vec}(\widehat{T}_{12})}=O_{i}T_{i}.$

Similarly,

$\underbrace{\widehat{T}_{i}^{t}\widehat{T}_{i}}_{>0}\mathrm{vec}(\widehat{T}_{12})^{t}=\widehat{T}_{i}^{t}O_{i}T_{i}\text{vec}(T_{12})^{t}(O_{2}\otimes
O_{1})^{t}$ (33)

and so

$\text{vec}(\widehat{T}_{12})^{t}=\frac{\widehat{T}_{i}^{t}O_{i}T_{i}\text{vec}(T_{12})^{t}(O_{2}\otimes
O_{1})^{t}}{\widehat{T}_{i}^{t}\widehat{T}_{i}}\Rightarrow\text{vec}(\widehat{T}_{12})=\frac{(O_{2}\otimes
O_{1})\text{vec}(T_{12})T_{i}^{t}O_{i}^{t}\widehat{T}_{i}}{\widehat{T}_{i}^{t}\widehat{T}_{i}};$
(34)

therefore by (28)

$\alpha_{i}^{-1}\text{vec}(\widehat{T}_{12})=\frac{\widehat{T}_{i}^{t}\widehat{T}_{i}}{\widehat{T}_{i}^{t}O_{i}T_{i}}\cdot\frac{(O_{2}\otimes
O_{1})\text{vec}(T_{12})T_{i}^{t}O_{i}^{t}\widehat{T}_{i}}{\widehat{T}_{i}^{t}\widehat{T}_{i}}=(O_{2}\otimes
O_{1})\text{vec}(T_{12}),$ (35)

or equivalently,

$\alpha_{i}^{-1}\widehat{T}_{12}=(O_{1},O_{2})*T_{12}.$ (36)

By assumption, either $\|\widehat{T}_{i}\|=\|T_{i}\|$ or
$\|\widehat{T}_{12}\|=\|T_{12}\|$, and in either case it follows that
$|\alpha_{i}|=1$, hence $\alpha_{i}=\pm 1$ since the $T_{i}$ and $T_{12}$ are
real, proving SO equivalence for $2$-qudits ($d>2$).

Now in the case of qubits, since by assumption
$\widehat{T}_{12}=(O_{1},O_{2})*T_{12}$, it must be the case that
$\alpha_{i}=1$, for otherwise we would have
$\widehat{T}_{12}=-\widehat{T}_{12}$ implying $\widehat{T}_{12}=\mathbf{0}$,
contradicting the assumption that $\widehat{T}_{12}\neq\mathbf{0}$. Thus,
$\alpha_{i}=1$ and this proves that SO equivalence for $2$-qubits is
equivalent to quasi-LU equivalence, and thus LU equivalence. ∎

###  3.1 SO Equivalence and Trace Identities for 2-Qudits

In [6, Theorem 3.4], Jing generalizes Specht’s criterion, showing that for any
two sets $\\{A_{i}\\}$ and $\\{B_{i}\\}$ of (real) $m\times n$ matrices, the
following are equivalent:

  1. 1.

$\\{A_{i}\\}$ is simultaneously orthogonal equivalent to $\\{B_{i}\\}$ (i.e.
for each $i$, $B_{i}=OA_{i}P^{t}$ for some orthogonal matrices $O$ and $P$);

  2. 2.

$\\{A_{i}A_{i}^{t}:i\leq j\\}$ is simultaneously orthogonal similar to
$\\{B_{i}B_{j}^{t}:i\leq j\\}$ (i.e. for each $i$ and $j$,
$B_{i}B_{j}=OA_{i}A_{j}O^{t}$ for some orthogonal matrix $O$);

  3. 3.

$\Tr(w\\{A_{i}A_{j}^{t}\\})=\Tr(w\\{B_{i}B_{j}^{t}\\})$ for any word $w$ in
respective alphabets.

In general, it is difficult to show that two states are quasi-LU equivalent
because we do not a priori know anything about the orthogonal matrices that
relate the two states. However, the result above and the previous theorem
imply the following:

######  Theorem 2 (Characterization of quasi-LU Equivalence: 2-Qudits).

Suppose $\rho$ and $\widehat{\rho}$ be two bipartite density matrices over the
same Hilbert space with respective matrix representations
$\\{T_{1},T_{2},T_{12}\\}$ and
$\\{\widehat{T}_{1},\widehat{T}_{2},\widehat{T}_{12}\\}$. Let
$\\{A_{1},A_{2}\\}=\\{T_{1}\circ T_{2},T_{12}\\}$ and
$\\{B_{1},B_{2}\\}=\\{\widehat{T}_{1}\circ\widehat{T}_{2},\widehat{T}_{12}\\}$.
Then $\rho$ and $\widehat{\rho}$ are quasi-LU equivalent if and only if

$\Tr(A_{i_{1}}A_{j_{1}}^{t}...A_{i_{k}}A_{j_{k}}^{t})=\Tr(B_{i_{1}}B_{j_{1}}^{t}...B_{i_{k}}B_{j_{k}}^{t})$

for any compositions $i_{1},...,i_{k}$ and $j_{1},...,j_{k}$ such that $1\leq
i_{1}\leq j_{1}\leq 2$,…, $1\leq i_{k}\leq j_{k}\leq 2$, and
$\|T_{1}\|=\|\widehat{T}_{1}\|$ or $\|T_{2}\|=\|\widehat{T}_{2}\|$.
Furthermore, in the case of $2$-qubits, they are LU equivalent.

Thus, the problem of quasi-LU equivalence, or LU equivalence in the case of
$2$-qubits, reduces to checking trace identities, which are easy to compute
and terminate in polynomial time (in particular we only need to check the
trace identities for words of length at most $16(\delta_{1}+\delta_{2})^{2}$;
see appendix or [7]).

##  4 3-Qudits

The density matrix $\rho$ of a $3$-qudit on
$\mathbb{C}^{d_{1}}\otimes\mathbb{C}^{d_{2}}\otimes\mathbb{C}^{d_{3}}$ is
given by

$\displaystyle\quad\frac{1}{d_{1}d_{2}d_{3}}\bigg{(}I_{2}\otimes I_{2}\otimes
I_{2}+\sum\limits_{i=1}^{\delta_{1}}T_{1}^{i}\lambda_{i}\otimes I_{2}\otimes
I_{2}+\sum\limits_{i=1}^{\delta_{2}}T_{2}^{i}I_{2}\otimes\lambda_{i}\otimes
I_{2}+\sum\limits_{i=1}^{\delta_{3}}T_{3}^{i}I_{2}\otimes
I_{2}\otimes\lambda_{i}$
$\displaystyle\qquad+\sum\limits_{i,j=1}^{\delta_{1},\delta_{2}}T_{12}^{ij}\lambda_{i}\otimes\lambda_{j}\otimes
I_{2}+\sum\limits_{i,j=1}^{\delta_{1},\delta_{3}}T_{13}^{ij}\lambda_{i}\otimes
I_{2}\otimes\lambda_{j}+\sum\limits_{i,j=1}^{\delta_{2},\delta_{3}}T_{23}^{ij}I_{2}\otimes\lambda_{i}\otimes\lambda_{j}+\sum\limits_{i,j,k=1}^{\delta_{1},\delta_{2},\delta_{3}}T_{123}^{ijk}\lambda_{i}\otimes\lambda_{j}\otimes\lambda_{k}\bigg{)}$

and so its hypermatrix representation is given by
$\\{T_{1},T_{2},T_{3},T_{12},T_{13},T_{23},T_{123}\\}$. For convenience set
$T:=T_{123}$ and denote the $(i,j,k)$ entry of $T$ as $t_{ijk}$. Then by
unfolding $T$, we may view it as any one of the following three matrices

$T_{(1)}:=\left[\begin{array}[]{ccccccccccccc}t_{111}&t_{121}&...&t_{1\delta_{2}1}&t_{112}&t_{122}&...&t_{1\delta_{2}2}&t_{113}&t_{123}&...&...&t_{1\delta_{2}\delta_{3}}\\\
t_{211}&t_{221}&...&t_{1\delta_{2}1}&t_{212}&t_{222}&...&t_{2\delta_{2}2}&t_{213}&t_{223}&...&...&t_{2\delta_{2}\delta_{3}}\\\
\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\ddots&\vdots\\\
t_{\delta_{1}11}&t_{\delta_{1}21}&...&t_{\delta_{1}\delta_{1}1}&t_{\delta_{1}12}&t_{\delta_{1}22}&...&t_{\delta_{1}\delta_{2}2}&t_{\delta_{1}13}&t_{\delta_{1}23}&...&...&t_{\delta_{1}\delta_{2}\delta_{3}}\end{array}\right]_{\delta_{1}\times\delta_{2}\delta_{3}},$
$T_{(2)}:=\left[\begin{array}[]{ccccccccccccc}t_{111}&t_{211}&...&t_{\delta_{1}11}&t_{112}&t_{212}&...&t_{\delta_{1}12}&t_{113}&t_{213}&...&...&t_{\delta_{1}1\delta_{3}}\\\
t_{121}&t_{221}&...&t_{\delta_{1}21}&t_{122}&t_{222}&...&t_{\delta_{1}22}&t_{123}&t_{223}&...&...&t_{\delta_{1}2\delta_{3}}\\\
\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\ddots&\vdots\\\
t_{1\delta_{2}1}&t_{2\delta_{2}1}&...&t_{\delta_{1}\delta_{2}1}&t_{1\delta_{2}2}&t_{2\delta_{2}2}&...&t_{\delta_{1}\delta_{2}2}&t_{1\delta_{2}3}&t_{2\delta_{2}3}&...&...&t_{\delta_{1}\delta_{2}\delta_{3}}\\\
\end{array}\right]_{\delta_{2}\times\delta_{1}\delta_{3}},$
$T_{(3)}:=\left[\begin{array}[]{ccccccccccccc}t_{111}&t_{211}&...&t_{\delta_{1}11}&t_{121}&t_{221}&...&t_{\delta_{1}21}&t_{131}&t_{231}&...&...&t_{\delta_{1}\delta_{2}1}\\\
t_{112}&t_{212}&...&t_{\delta_{1}12}&t_{122}&t_{222}&...&t_{\delta_{1}22}&t_{132}&t_{232}&...&...&t_{\delta_{1}\delta_{2}2}\\\
\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\ddots&\vdots\\\
t_{11\delta_{3}}&t_{21\delta_{3}}&...&t_{\delta_{1}1\delta_{3}}&t_{12\delta_{3}}&t_{22\delta_{3}}&...&t_{\delta_{1}2\delta_{3}}&t_{13\delta_{3}}&t_{23\delta_{3}}&...&...&t_{\delta_{1}\delta_{2}\delta_{3}}\end{array}\right]_{\delta_{3}\times\delta_{1}\delta_{2}}.$

Given $\delta_{i}\times\delta_{i}$ matrices $A_{i}$ ($1\leq i\leq 3$), direct
computation yields

$((A_{1},A_{2},A_{3})*T)_{(1)}=A_{1}T_{(1)}(A_{3}\otimes A_{2})^{t}$ (37)
$((A_{1},A_{2},A_{3})*T)_{(2)}=A_{2}T_{(2)}(A_{3}\otimes A_{1})^{t}$ (38)
$((A_{1},A_{2},A_{3})*T)_{(3)}=A_{3}T_{(3)}(A_{2}\otimes A_{1})^{t}.$ (39)

As before, if $v\in\mathbb{R}^{\delta_{1}}$ and
$M\in\mathbb{R}^{\delta_{2}\delta_{3}}$, then direct computation yields

$(v\circ M)_{(1)}=v\circ\mathrm{vec}(M)$ (40)

and additionally, we have

$(M\circ v)_{(3)}^{t}=\mathrm{vec}(M)\circ v$ (41)

as well as

$\mathrm{vec}((A_{1},A_{2})*M)=(A_{2}\otimes A_{1})\mathrm{vec}(M).$ (42)

Now, if $\rho$ and $\widehat{\rho}$ are density matrices for $3$-qubit states
with hypermatrix representations
$\\{T_{1},T_{2},T_{3},T_{12},T_{13},T_{23},T_{123}\\}$ and
$\\{\widehat{T}_{1},\widehat{T}_{2},\widehat{T}_{3},\widehat{T}_{12},\widehat{T}_{13},\widehat{T}_{23},\widehat{T}_{123}\\}$,
then we say they are simultaneously orthogonal (SO) equivalent if there exists
$O_{i}\in O(\delta_{i})$ such that

$\displaystyle\begin{split}\widehat{T}_{123}=(O_{1},O_{2},O_{3})*T_{123}&&\widehat{T}_{1}\circ\widehat{T}_{23}=(O_{1},O_{2},O_{3})*(T_{1}\circ
T_{23})\\\
\widehat{T}_{2}\circ\widehat{T}_{13}=(O_{2},O_{1},O_{3})*(T_{2}\circ
T_{13})&&\widehat{T}_{12}\circ\widehat{T}_{3}=(O_{1},O_{2},O_{3})*(T_{12}\circ
T_{3}).\end{split}$ (43)

As in the case of Theorem 1, we will establish that quasi-LU equivalence and
SO equivalence are nearly equivalent notions. This will then allow us to apply
a broader generalization of Specht’s criterion to reduce quasi-LU equivalence
to checking trace identities.

######  Theorem 3 (quasi-LU Equivalence and SO Equivalence: 3-Qudits).

Let $\rho$ and $\widehat{\rho}$ be density matrices of $3$-qudit states on
$\mathbb{C}^{d_{1}}\otimes\mathbb{C}^{d_{2}}\otimes\mathbb{C}^{d_{3}}$ with
hypermatrix representations
$\\{T_{1},T_{2},T_{3},T_{12},T_{13},T_{23},T_{123}\\}$ and
$\\{\widehat{T}_{1},\widehat{T}_{2},\widehat{T}_{3},\widehat{T}_{12},\widehat{T}_{13},\widehat{T}_{23},\widehat{T}_{123}\\}$
respectively, and assume that $T_{1},T_{2},T_{3}$ and $T_{12},T_{13},T_{23}$
are nonzero. Then $\rho$ and $\widehat{\rho}$ are quasi-LU equivalent if and
only if they are SO equivalent and $\|\widehat{T}_{1}\|=\|T_{1}\|$ or
$\|\widehat{T}_{23}\|=\|T_{23}\|$, $\|\widehat{T}_{2}\|=\|T_{2}\|$ or
$\|\widehat{T}_{13}\|=\|T_{13}\|$, and $\|\widehat{T}_{3}\|=\|T_{3}\|$ or
$\|\widehat{T}_{12}\|=\|T_{12}\|$.

Additionally, in the case of $3$-qubits, if we assume that one of
$\widehat{T}_{1}^{t}\widehat{T}_{12}\widehat{T}_{2}$,
$\widehat{T}_{2}^{t}\widehat{T}_{23}\widehat{T}_{3}$, or
$\widehat{T}_{3}^{t}\widehat{T}_{13}^{t}\widehat{T}_{1}$ has the same sign as
$T_{1}^{t}T_{12}T_{2}$, $\widehat{T}_{2}^{t}T_{23}T_{3}$, or
$T_{3}^{t}T_{13}^{t}T_{1}$ (respectively), and
$\det(\widehat{T}_{jk})=\det(T_{jk})\neq 0$ for $(j,k)=(1,2),(1,3),(23)$, then
$\rho$ and $\widehat{\rho}$ are LU equivalent.

###### Proof.

If $\rho$ and $\widehat{\rho}$ are quasi-LU equivalent, then there exists
$O_{i}\in O(\delta_{i})$ such that

$\displaystyle\widehat{T}_{i}=O_{i}*T_{i},\quad\widehat{T}_{jk}=(O_{j},O_{k})*T_{jk},\quad\text{and}\quad\widehat{T}_{123}=(O_{1},O_{2},O_{3})*T_{123}$
(44)

for $1\leq i\leq 3$ and $1\leq j<k\leq 3$. It immediately follows that
$\|\widehat{T}_{i}\|=\|T_{i}\|$ and $\|\widehat{T}_{jk}\|=\|T_{jk}\|$.
Moreover,

$\widehat{T}_{i}\circ\widehat{T}_{jk}=(O_{i}*T_{i})\circ((O_{j},O_{k})*T_{jk})=(O_{i},O_{j},O_{k})*(T_{i}\circ
T_{jk}),$ (45)

showing that $\rho$ and $\widehat{\rho}$ are SO-equivalent.

Conversely, similar to as before, for $(i,j,k)=(1,2,3)$ and $(i,j,k)=(2,1,3)$
we have

$\displaystyle 0<\|\widehat{T}_{i}\|^{2}\|\widehat{T}_{jk}\|^{2}$
$\displaystyle=(\widehat{T}_{i}^{t}\widehat{T}_{i})(\text{vec}(\widehat{T}_{jk})^{t}\text{vec}(\widehat{T}_{jk}))$
(46)
$\displaystyle=\widehat{T}_{i}^{t}(\widehat{T}_{i}\circ\text{vec}(\widehat{T}_{jk}))\text{vec}(\widehat{T}_{jk})$
(47)
$\displaystyle=\widehat{T}_{i}^{t}((\widehat{T}_{i}\circ\widehat{T}_{jk})_{(1)})\text{vec}(\widehat{T}_{jk})$
(48)
$\displaystyle=\widehat{T}_{i}^{t}\Big{(}\big{(}(O_{i},O_{j},O_{k})*(T_{i}\circ
T_{jk})\big{)}_{(1)}\Big{)}\text{vec}(\widehat{T}_{jk})$ (49)
$\displaystyle=\widehat{T}_{i}^{t}\Big{(}\big{(}(O_{i}*T_{i})\circ((O_{j},O_{k})*T_{jk})\big{)}_{(1)}\Big{)}\text{vec}(\widehat{T}_{jk})$
(50)
$\displaystyle=\widehat{T}_{i}^{t}\big{(}(O_{i}*T_{i})\circ\text{vec}((O_{j},O_{k})*T_{jk})\big{)}\text{vec}(\widehat{T}_{jk})$
(51)
$\displaystyle=\widehat{T}_{i}^{t}\Big{(}(O_{i}T_{i})\circ\big{(}(O_{k}\otimes
O_{j})\text{vec}(T_{jk})\big{)}\Big{)}\text{vec}(\widehat{T}_{jk})$ (52)
$\displaystyle=\widehat{T}_{i}^{t}O_{i}T_{i}\text{vec}(T_{jk})^{t}(O_{k}\otimes
O_{j})^{t}\text{vec}(\widehat{T}_{jk})$ (53)

Again, note that $\widehat{T}_{i}^{t}O_{i}T_{i}$ and
$\text{vec}(T_{jk})^{t}(O_{k}\otimes O_{j})^{t}\text{vec}(\widehat{T}_{jk})$
are just numbers, so equations (46) and (53) imply that

$\alpha_{i}:=\frac{\widehat{T}_{i}^{t}O_{i}T_{i}}{\widehat{T}_{i}^{t}\widehat{T}_{i}}=\frac{\text{vec}(\widehat{T}_{jk})^{t}\text{vec}(\widehat{T}_{jk})}{\text{vec}(T_{jk})^{t}(O_{k}\otimes
O_{j})^{t}\text{vec}(\widehat{T}_{jk})}\neq 0.$ (54)

By assumption, we also have that

$(\widehat{T}_{i}\circ\widehat{T}_{jk})_{(1)}=\big{(}(O_{i},O_{j},O_{k})*(T_{i}\circ
T_{jk})\big{)}_{(1)}$ (55)

The left-hand side of the above equation can be expressed as
$\widehat{T}_{i}\mathrm{vec}(\widehat{T}_{jk})^{t}$ and the right-hand side
may be expressed as

$\begin{split}\big{(}(O_{i},O_{j},O_{k})*(T_{i}\circ
T_{jk})\big{)}_{(1)}&=\big{(}(O_{i}*T_{i})\circ((O_{j},O_{k})*T_{jk})\big{)}_{(1)}\\\
&=(O_{i}*T_{i})\circ\mathrm{vec}((O_{j},O_{k})*T_{jk})\\\
&=O_{i}T_{i}\circ(O_{k}\otimes O_{j})\mathrm{vec}(T_{jk})\\\
&=O_{i}T_{i}\mathrm{vec}(T_{jk})^{t}(O_{k}\otimes O_{j})^{t}\end{split}$ (56)

which implies that

$\widehat{T}_{i}\underbrace{\text{vec}(\widehat{T}_{jk})^{t}\text{vec}(\widehat{T}_{jk})}_{>0}=O_{i}T_{i}\text{vec}(T_{jk})^{t}(O_{k}\otimes
O_{j})^{t}\text{vec}(\widehat{T}_{jk})$ (57)

hence

$\widehat{T}_{i}=\frac{O_{i}T_{i}\text{vec}(T_{jk})^{t}(O_{k}\otimes
O_{j})^{t}\text{vec}(\widehat{T}_{jk})}{\text{vec}(\widehat{T}_{jk})^{t}\text{vec}(\widehat{T}_{jk})}.$
(58)

Therefore by (54)

$\alpha_{i}\widehat{T}_{i}=\frac{\mathrm{vec}(\widehat{T}_{jk})^{t}\mathrm{vec}(\widehat{T}_{jk})}{\text{vec}(T_{jk})^{t}(O_{k}\otimes
O_{j})^{t}\mathrm{vec}(\widehat{T}_{jk})}\cdot\frac{O_{i}T_{i}\mathrm{vec}(T_{jk})^{t}(O_{k}\otimes
O_{j})^{t}\mathrm{vec}(\widehat{T}_{jk})}{\mathrm{vec}(\widehat{T}_{jk})^{t}\mathrm{vec}(\widehat{T}_{jk})}=O_{i}T_{i}.$
(59)

Similarly,

$\underbrace{\widehat{T}_{i}^{t}\widehat{T}_{i}}_{>0}\mathrm{vec}(\widehat{T}_{jk})^{t}=\widehat{T}_{i}^{t}O_{i}T_{i}\text{vec}(T_{jk})^{t}(O_{k}\otimes
O_{j})^{t}$ (60)

and so

$\mathrm{vec}(\widehat{T}_{jk})^{t}=\frac{\widehat{T}_{i}^{t}O_{i}T_{i}\mathrm{vec}(T_{jk})^{t}(O_{k}\otimes
O_{j})^{t}}{\widehat{T}_{i}^{t}\widehat{T}_{i}}\Rightarrow\mathrm{vec}(\widehat{T}_{jk})=\frac{(O_{k}\otimes
O_{j})\mathrm{vec}(T_{jk})T_{i}^{t}O_{i}^{t}\widehat{T}_{i}}{\widehat{T}_{i}^{t}\widehat{T}_{i}}.$
(61)

Therefore by (54)

$\alpha_{i}^{-1}\mathrm{vec}(\widehat{T}_{jk})=\frac{\widehat{T}_{i}^{t}\widehat{T}_{i}}{\widehat{T}_{i}^{t}O_{i}T_{i}}\cdot\frac{(O_{k}\otimes
O_{j})\mathrm{vec}(T_{jk})T_{i}^{t}O_{i}^{t}\widehat{T}_{i}}{\widehat{T}_{i}^{t}\widehat{T}_{i}}=(O_{k}\otimes
O_{j})\mathrm{vec}(T_{jk})$ (62)

which implies that

$\alpha_{i}^{-1}\widehat{T}_{jk}=(O_{j},O_{k})*T_{jk}$ (63)

By assumption, either $\|\widehat{T}_{i}\|=\|T_{i}\|$ or
$\|\widehat{T}_{jk}\|=\|T_{jk}\|$, and in either case it follows that
$|\alpha_{i}|=1$, hence $\alpha_{i}=\pm 1$ since the $T_{i}$ and $T_{jk}$ are
real.

Now, we still need to check the case when $(i,j,k)=(3,1,2)$. This case is very
similar to the above but with some subtle differences. Observe,

$\displaystyle 0<\|\widehat{T}_{12}\|^{2}\|\widehat{T}_{3}\|^{2}$
$\displaystyle=\mathrm{vec}(\widehat{T}_{12})^{t}\mathrm{vec}(\widehat{T}_{12})\widehat{T}_{3}^{t}\widehat{T}_{3}$
(64)
$\displaystyle=\mathrm{vec}(\widehat{T}_{12})^{t}(\mathrm{vec}(\widehat{T}_{12})\circ\widehat{T}_{3})\widehat{T}_{3}$
(65)
$\displaystyle=\mathrm{vec}(\widehat{T}_{12})^{t}((\widehat{T}_{12}\circ\widehat{T}_{3})^{t}_{(3)})\widehat{T}_{3}$
(66)
$\displaystyle=\mathrm{vec}(\widehat{T}_{12})^{t}\Big{(}\big{(}(O_{1},O_{2},O_{3})*(T_{12}\circ
T_{3})\big{)}_{(3)}^{t}\Big{)}\widehat{T}_{3}$ (67)
$\displaystyle=\mathrm{vec}(\widehat{T}_{12})^{t}\Big{(}\big{(}((O_{1},O_{2})*T_{12})\circ(O_{3}*T_{3})\big{)}_{(3)}^{t}\Big{)}\widehat{T}_{3}$
(68)
$\displaystyle=\mathrm{vec}(\widehat{T}_{12})^{t}\big{(}\mathrm{vec}((O_{1},O_{2})*T_{12})\circ(O_{3}*T_{3})\big{)}\widehat{T}_{3}$
(69)
$\displaystyle=\mathrm{vec}(\widehat{T}_{12})^{t}\Big{(}\big{(}(O_{2}\otimes
O_{1})\mathrm{vec}(T_{12})\big{)}\circ(O_{3}T_{3})\Big{)}\widehat{T}_{3}$ (70)
$\displaystyle=\mathrm{vec}(\widehat{T}_{12})^{t}(O_{2}\otimes
O_{1})\mathrm{vec}(T_{12})T_{3}^{t}O_{3}^{t}\widehat{T}_{3}.$ (71)

As before, we note that $\mathrm{vec}(\widehat{T}_{12})^{t}(O_{2}\otimes
O_{1})\mathrm{vec}(T_{12})$ and $T_{3}^{t}O_{3}\widehat{T}_{3}$ are just
numbers, so equations (64) and (71) imply that

$\alpha_{3}:=\frac{T_{3}^{t}O_{3}^{t}\widehat{T}_{3}}{\widehat{T}_{3}^{t}\widehat{T}_{3}}=\frac{\mathrm{vec}(\widehat{T}_{12})^{t}\mathrm{vec}(\widehat{T}_{12})}{\mathrm{vec}(\widehat{T}_{12})^{t}(O_{2}\otimes
O_{1})\mathrm{vec}(T_{12})}\neq 0.$ (72)

By assumption, we also have that

$(\widehat{T}_{12}\circ\widehat{T}_{3})_{(3)}^{t}=\big{(}(O_{1},O_{2},O_{3})*(T_{12}\circ
T_{3})\big{)}_{(3)}^{t}.$ (73)

The left-hand side of the above equation may be expressed as
$\mathrm{vec}(\widehat{T}_{12})\widehat{T}_{3}^{t}$, and the right-hand side
of the above equation may be expressed as

$\begin{split}\big{(}(O_{1},O_{2},O_{3})*(T_{12}\circ
T_{3})\big{)}_{(3)}^{t}&=\big{(}((O_{1},O_{2})*T_{12})\circ(O_{3}*T_{3})\big{)}_{(3)}^{t}\\\
&=\mathrm{vec}((O_{1},O_{2})*T_{12})\circ(O_{3}*T_{3})\\\ &=(O_{2}\otimes
O_{1})\mathrm{vec}(T_{12})\circ(O_{3}T_{3})\\\ &=(O_{2}\otimes
O_{1})\mathrm{vec}(T_{12})T_{3}^{t}O_{3}^{t}\end{split}$ (74)

which implies that

$\underbrace{\mathrm{vec}(\widehat{T}_{12})^{t}\mathrm{vec}(\widehat{T}_{12})}_{>0}\widehat{T}_{3}^{t}=\mathrm{vec}(\widehat{T}_{12})^{t}(O_{2}\otimes
O_{1})\mathrm{vec}(T_{12})T_{3}^{t}O_{3}^{t}$ (75)

hence

$\widehat{T}_{3}=\frac{O_{3}T_{3}\mathrm{vec}(T_{12})^{t}(O_{2}\otimes
O_{1})^{t}\mathrm{vec}(\widehat{T}_{12})}{\mathrm{vec}(\widehat{T}_{12})^{t}\mathrm{vec}(\widehat{T}_{12})}.$
(76)

Therefore by (72)

$\begin{split}\alpha_{3}\widehat{T}_{3}&=\frac{\mathrm{vec}(\widehat{T}_{12})^{t}\mathrm{vec}(\widehat{T}_{12})}{\mathrm{vec}(\widehat{T}_{12})^{t}(O_{2}\otimes
O_{1})\mathrm{vec}(T_{12})}\cdot\frac{O_{3}T_{3}\mathrm{vec}(T_{12})^{t}(O_{2}\otimes
O_{1})^{t}\mathrm{vec}(\widehat{T}_{12})}{\mathrm{vec}(\widehat{T}_{12})^{t}\mathrm{vec}(\widehat{T}_{12})}\\\
&=\frac{O_{3}T_{3}\mathrm{vec}(T_{12})^{t}(O_{2}\otimes
O_{1})^{t}\mathrm{vec}(\widehat{T}_{12})}{\mathrm{vec}(T_{12})^{t}(O_{2}\otimes
O_{1})^{t}\mathrm{vec}(\widehat{T}_{12})}\qquad\text{since
}\alpha_{3}=\alpha_{3}^{t}\text{ because }\alpha_{3}\text{ is just a
number}\\\ &=O_{3}T_{3}.\end{split}$ (77)

Similarly,

$\mathrm{vec}(\widehat{T}_{12})\underbrace{\widehat{T}_{3}^{t}\widehat{T}_{3}}_{>0}=(O_{2}\otimes
O_{1})\mathrm{vec}(T_{12})T_{3}^{t}O_{3}^{t}\widehat{T}_{3}$ (78)

and so

$\mathrm{vec}(\widehat{T}_{12})=\frac{(O_{2}\otimes
O_{1})\mathrm{vec}(T_{12})T_{3}^{t}O_{3}^{t}\widehat{T}_{3}}{\widehat{T}_{3}^{t}\widehat{T}_{3}}.$
(79)

Therefore by (72)

$\alpha_{3}^{-1}\mathrm{vec}(\widehat{T}_{12})=\frac{\widehat{T}_{3}^{t}\widehat{T}_{3}}{T_{3}^{t}O_{3}^{t}\widehat{T}_{3}}\cdot\frac{(O_{2}\otimes
O_{1})\mathrm{vec}(T_{12})T_{3}^{t}O_{3}^{t}\widehat{T}_{3}}{\widehat{T}_{3}^{t}\widehat{T}_{3}}=\frac{(O_{2}\otimes
O_{1})\mathrm{vec}(T_{12})T_{3}^{t}O_{3}^{t}\widehat{T}_{3}}{T_{3}^{t}O_{3}^{t}\widehat{T}_{3}}=(O_{2}\otimes
O_{1})\mathrm{vec}(T_{12})$ (80)

which implies that

$\alpha_{3}^{-1}\widehat{T}_{12}=(O_{1},O_{2})*T_{12}.$ (81)

Furthermore, by assumption, either $\|\widehat{T}_{3}\|=\|T_{3}\|$ or
$\|\widehat{T}_{12}\|=\|T_{12}\|$ and in either case it follows that
$|\alpha_{3}|=1$, hence $\alpha_{3}=\pm 1$ since $T_{3}$ and $T_{12}$ are
real.

Thus, for $(i,j,k)=(1,2,3),(2,1,3),(3,1,2)$ we have that

$\widehat{T}_{i}=\alpha_{i}^{-1}O_{i}*T_{i},\quad\widehat{T}_{jk}=\alpha_{i}(O_{j},O_{k})*T_{jk},\quad\text{and}\quad\widehat{T}_{123}=(O_{1},O_{2},O_{3})*T_{123},$
(82)

with $\alpha_{i}=\pm 1$ and $O_{i}\in O(\delta_{i})$, proving $SO$ equivalence
for $3$-qudits ($d>2$). Now lastly, in the case of $3$-qubits, if in addition,
we assume (without loss of generality) that
$\widehat{T}_{1}^{t}\widehat{T}_{12}\widehat{T}_{2}$ has the same sign as
$T_{1}^{t}T_{12}T_{2}$, and that $\det(\widehat{T}_{jk})=\det(T_{jk})\neq 0$
for $(j,k)=(1,2),(1,3),(2,3)$, then since

$\widehat{T}_{1}^{t}\widehat{T}_{12}\widehat{T}_{2}=(\alpha_{1}^{-1}O_{1}T_{1})^{t}(\alpha_{3}O_{1}T_{12}O_{2}^{t})(\alpha_{2}^{-1}O_{2}T_{2})=\alpha_{1}\alpha_{2}\alpha_{3}T_{1}^{t}T_{12}T_{2}$
(83)

it follows that $\alpha_{1}\alpha_{2}\alpha_{3}$ is positive. Thus, we have
two possible cases:

  1. 1.

all of the $\alpha_{i}$ are positive and hence $1$, proving LU equivalence, or

  2. 2.

exactly two of the $\alpha_{i}$ are negative and hence $-1$.

In the latter case, if $\alpha_{i},\alpha_{j}=-1$ and $\alpha_{k}=1$, then

$\det(\widehat{T}_{jk})=\det(\alpha_{i}O_{j}T_{jk}O_{k}^{t})=\alpha_{i}^{3}\det(O_{j}T_{jk}O_{k}^{t})=\alpha_{i}\det(T_{jk})=-\det(T_{jk})$
(84)

which contradicts the assumption that $\det(\widehat{T}_{jk})=\det(T_{jk})\neq
0$. Consequently, we must have the first case, in which $LU$ equivalence is
established. This completes the proof. ∎

There was a different approach to local unitary invariants of 3-qubits in [13,
14], and our current studies are suitable for generalizations.

###  4.1 SO-Equivalence and Trace Identities for 3-Qudits

We would now like to apply Theorem 3.4 from [6] like how we did in the
$2$-qubit case, reducing the quasi-LU equivalence to checking trace
identities; however, there are some issues. Suppose $\rho$ and
$\widehat{\rho}$ are $3$-qubit density matrices with hypermatrix
representations $\\{T_{1},T_{2},T_{3},T_{12},T_{13},T_{23},T_{123}\\}$ and
$\\{\widehat{T}_{1},\widehat{T}_{2},\widehat{T}_{3},\widehat{T}_{12},\widehat{T}_{13},\widehat{T}_{23},\widehat{T}_{123}\\}$
respectively. Define

$\\{A_{1},A_{2},A_{3},A_{4}\\}:=\\{(T_{123})_{(1)},(T_{1}\circ
T_{23})_{(1)},(T_{2}\circ T_{13})_{(2)},(T_{12}\circ T_{3})_{(3)}\\}$

and similarly

$\\{B_{1},B_{2},B_{3},B_{4}\\}:=\\{(\widehat{T}_{123})_{(1)},(\widehat{T}_{1}\circ\widehat{T}_{23})_{(1)},(\widehat{T}_{2}\circ\widehat{T}_{13})_{(2)},(\widehat{T}_{12}\circ\widehat{T}_{3})_{(3)}^{t}\\}$

(note that these are all $\delta_{1}\times\delta_{2}\delta_{3}$ matrices).
Then by Theorem 3.4 in [6], there exists $O_{1}\in O(\delta_{1})$ and
$\widetilde{O}\in O(\delta_{2}\delta_{3})$ such that

$\displaystyle\begin{split}(\widehat{T}_{123})_{(1)}&=O_{1}(T_{123})_{(1)}\widetilde{O}^{t}&&&(\widehat{T}_{1}\circ\widehat{T}_{23})_{(1)}&=O_{1}(T_{1}\circ
T_{23})_{(1)}\widetilde{O}^{t}\\\
(\widehat{T}_{2}\circ\widehat{T}_{13})_{(2)}&=O_{1}(T_{2}\circ
T_{13})_{(2)}\widetilde{O}^{t}&&&(\widehat{T}_{12}\circ\widehat{T}_{3})_{(1)}&=O_{1}(T_{12}\circ
T_{3})_{(1)}\widetilde{O}^{t}\end{split}$ (85)

if and only if

$\Tr(A_{i_{1}}A_{j_{1}}^{t}...A_{i_{k}}A_{j_{k}}^{t})=\Tr(B_{i_{1}}B_{j_{1}}^{t}...B_{i_{k}}B_{j_{k}}^{t})$
(86)

for any compositions $i,...,i_{k}$ and $j_{1},...,j_{k}$ such that $1\leq
i_{1}\leq j_{1}\leq 4$,…, $\leq i_{k}\leq j_{k}\leq 4$ and
$\|T_{1}\|=\|\widehat{T}_{1}\|$ or $\|T_{23}\|=\|\widehat{T}_{23}\|$,
$\|T_{2}\|=\|\widehat{T}_{2}\|$ or $\|T_{13}\|=\|\widehat{T}_{13}\|$, and
$\|T_{3}\|=\|\widehat{T}_{3}\|$ or $\|T_{12}\|=\|\widehat{T}_{12}\|$. But this
is not good enough!

Indeed, we need $\widetilde{O}^{t}$ to be a tensor product of a
$\delta_{2}\times\delta_{2}$ orthogonal matrix with a
$\delta_{3}\times\delta_{3}$ orthogonal matrix; otherwise, we do not have SO
equivalence and hence neither quasi-LU equivalence. To guarantee that
$\widetilde{O}^{t}\in O(\delta_{2})\otimes O(\delta_{3})$, we need some
additional assumptions and a more powerful generalization of Specht’s
criterion.

####  4.1.1 Applying the Partial Trace and Futorny et al’s Theorem

Let $\rho$ be the density matrix of a $3$-qudit state over
$\mathbb{C}^{d_{1}}\otimes\mathbb{C}^{d_{2}}\otimes\mathbb{C}^{d_{3}}$ with
hypermatrix representation
$\\{T_{1},T_{2},T_{3},T_{12},T_{13},T_{23},T_{123}\\}$. The partial trace over
the first system denoted $\Tr_{1}$, is defined as

$\Tr_{1}(\rho):=\sum\limits_{i=0}^{d_{3}-1}(\bra{i}\otimes I_{d_{2}}\otimes
I_{d_{3}}\rho(\ket{i}\otimes I_{d_{2}}\otimes I_{d_{3}})$

where $\\{\ket{i}\\}$ forms a basis for $\mathbb{C}^{d_{1}}$. This reduces to
a density matrix of the following $2$-qubit state:

$\frac{1}{d_{2}d_{3}}\left(I_{d_{2}}\otimes
I_{d_{3}}+\sum\limits_{i=1}^{d_{2}^{2}-1}T_{1}^{i}\lambda_{i}^{2}\otimes
I_{d_{3}}+\sum\limits_{i=1}^{d_{3}^{2}-1}T_{2}^{i}I_{d_{2}}\otimes\lambda_{i}^{3}+\sum\limits_{i,j=1}^{d_{2}^{2}-1,d_{3}^{2}-1}T_{12}^{ij}\lambda_{i}^{2}\otimes\lambda_{j}^{3}\right)=:\rho_{23}.$

The partial traces over the second and third subsystems, $\Tr_{2}$ and
$\Tr_{3}$ are analogously defined. Note that it is immediately clear that if
two states $\rho$ and $\widehat{\rho}$ are quasi-LU equivalent, then their
partial traces must also be quasi-LU equivalent.

In [7, Corollary 1], Futorney et. al. apply the theory of quiver
representations to generalize Theorem 3.4 in [6]. We present their result and
the necessary notions from quiver representation theory to understand it in
the appendix. In particular, if we consider the following quiver:

${1}$${3}$${2}$$\scriptstyle{\vdots}$$\scriptstyle{\vdots}$

which has $k>0$ arrows from $1$ to $3$ and $l-k>0$ arrows from $2$ to $3$, and
we assign to it a matrix representation $A$ with dimension
$\dim(n_{1},n_{2},m)$, then by Corollary 1 from [7] we obtain the following
result:

######  Lemma 1.

If $A_{1},...,A_{k},A_{k+1},...,A_{l}$ and $B_{1},...,B_{k},B_{k+1},...,B_{l}$
are real matrices each with $m$ rows, and with
$A_{1},...,A_{k},B_{1},...,B_{k}$ having $n_{1}$ columns and with
$A_{k+1},...,A_{l},B_{k+1},...,B_{l}$ having $n_{2}$ columns, then the
following two statements are equivalent

  * •

there exists orthogonal matrices $O\in O(m)$, $\widetilde{O}_{1}\in O(n_{1})$,
and $\widetilde{O}_{2}\in O(n_{2})$ such that

$(B_{1},...,B_{k},B_{k+1},...,B_{l})=(OA_{1}\widetilde{O}_{1},...,OA_{k}\widetilde{O}_{1},OA_{k+1}\widetilde{O}_{2},...,OA_{l}\widetilde{O}_{2})$
(87)

  * •

$\begin{split}&\Tr(w\\{A_{i_{1}}A_{j_{1}}^{t},A_{i_{2}}A_{j_{2}}^{t}\big{|}1\leq
i_{1}\leq j_{1}\leq k,k+1\leq i_{2}\leq j_{2}\leq l\\})\\\
&=\Tr(w\\{B_{i_{1}}B_{j_{1}}^{t},B_{i_{2}}B_{j_{2}}^{t}\big{|}1\leq i_{1}\leq
j_{1}\leq k,k+1\leq i_{2}\leq j_{2}\leq l\\})\end{split}$ (88)

for every word $w$ in noncommuting variables.

Furthermore, it suffices to check (88) for all words of length at most
$[(r+2)(n_{1}+n_{2}+m)]^{2}$ where $r$ is the smallest positive integer such
that $\frac{r(r+1)}{2}\geq\max\\{k,l-k\\}$.

Therefore, is we set

$(A_{1},A_{2},A_{3},A_{4},A_{5},A_{6}):=((T_{123})_{(1)},(T_{1}\circ
T_{23})_{(1)},(T_{2}\circ T_{13})_{(2)},(T_{12}\circ T_{3})_{(1)},(T_{1}\circ
T_{2}\circ T_{3})_{(1)},T_{1})$

and

$(B_{1},B_{2},B_{3},B_{4},B_{5},B_{6}):=((\widehat{T}_{123})_{(1)},(\widehat{T}_{1}\circ\widehat{T}_{23})_{(1)},(\widehat{T}_{2}\circ\widehat{T}_{13})_{(2)},(\widehat{T}_{12}\circ\widehat{T}_{3})_{(1)},(\widehat{T}_{1}\circ\widehat{T}_{2}\circ\widehat{T}_{3})_{(1)},\widehat{T}_{1}),$

then by lemma 1 there exists $O_{1}\in O(\delta_{1})$, $\widetilde{O}\in
O(\delta_{2}\delta_{3})$, and $O\in O(1)$ (hence $O=[\pm 1]$ and without loss
of generality we may assume that $O=[1]$) such that

$\displaystyle\begin{split}(\widehat{T}_{123})_{(1)}&=O_{1}(T_{123})_{(1)}\widetilde{O}^{t}&&&\widehat{T}_{1}\circ\widehat{T}_{23})_{(1)}&=O_{1}(T_{1}\circ
T_{23})_{(1)}\widetilde{O}^{t}\\\
(\widehat{T}_{2}\circ\widehat{T}_{13})_{(2)}&=O_{1}(T_{2}\circ
T_{13})_{(2)}\widetilde{O}^{t}&&&\widehat{T}_{12}\circ\widehat{T}_{3})_{(1)}&=O_{1}(T_{12}\circ
T_{3})_{(1)}\widetilde{O}^{t}\\\
(\widehat{T}_{1}\circ\widehat{T}_{2}\circ\widehat{T}_{3})_{(1)}&=O_{1}(T_{1}\circ
T_{2}\circ
T_{3})_{(1)}\widetilde{O}^{t}&&&\widehat{T}_{1}&=O_{1}T_{1}\end{split}$ (89)

if and only if

$\Tr(w\\{A_{i}A_{j}^{t},A_{6}A_{6}^{t}\\})=\Tr(w\\{B_{i}B_{j}^{t},B_{6}B_{6}^{t}\\})$
(90)

with $1\leq i\leq j\leq 5$, for every word $w$ in noncommuting variables of
length at most $25(1+\delta_{1}+\delta_{2}\delta_{3})^{2}$.

Suppose (99) holds for all such words. If in addition we assume that
$\Tr_{1}(\rho)$ is quasi-LU equivalent to $\Tr_{1}(\widehat{\rho})$, then
there exists $O_{2}\in O(\delta_{2})$ and $O_{3}\in O(\delta_{3})$ such that

$\widehat{T}_{2}=O_{2}*T_{2},\qquad\widehat{T}_{3}=O_{3}*T_{3},\qquad\text{and}\qquad\widehat{T}_{23}=(O_{2},O_{3})*T_{12}.$
(91)

Therefore,

$\begin{split}(\widehat{T}_{1}\circ\widehat{T}_{23})_{(1)}&=\Big{(}\big{(}O_{1}*T_{1}\big{)}\circ\big{(}(O_{2},O_{3})*T_{23}\big{)}\Big{)}_{(1)}\\\
&=\big{(}(O_{1},O_{2},O_{3})*(T_{1}\circ T_{23})\big{)}_{(1)}\\\
&=O_{1}(T_{1}\circ T_{23})_{(1)}(O_{3}\otimes O_{2})^{t},\end{split}$ (92)

and similarly

$\begin{split}(\widehat{T}_{1}\circ\widehat{T}_{2}\circ\widehat{T}_{3})_{(1)}&=\big{(}(O_{1}*T_{1})\circ(O_{2}*T_{2})\circ(O_{3}*T_{3})\big{)}_{(1)}\\\
&=\big{(}(O_{1},O_{2},O_{3})*T_{1}\circ T_{2}\circ T_{3}\big{)}_{(1)}\\\
&=O_{1}(T_{1}\circ T_{2}\circ T_{3})_{(1)}(O_{3}\otimes
O_{2})^{t}.\end{split}$ (93)

Hence,

$(T_{1}\circ T_{23})_{(1)}^{t}(T_{1}\circ
T_{23})_{(1)}\widetilde{O}^{t}=(T_{1}\circ T_{23})_{(1)}^{t}(T_{1}\circ
T_{23})_{(1)}(O_{3}\otimes O_{2})^{t},$ (94)

and

$(T_{1}\circ T_{2}\circ T_{3})_{(1)}^{t}(T_{1}\circ T_{2}\circ
T_{3})_{(1)}\widetilde{O}^{t}=(T_{1}\circ T_{2}\circ
T_{3})_{(1)}^{t}(T_{1}\circ T_{2}\circ T_{3})_{(1)}(O_{3}\otimes O_{2})^{t}.$
(95)

If either $(T_{1}\circ T_{23})_{(1)}^{t}(T_{1}\circ T_{23})_{(1)}$ or
$(T_{1}\circ T_{2}\circ T_{3})_{(1)}^{t}(T_{1}\circ T_{2}\circ T_{3})_{(1)}$
is invertible, then it follows that

$\widetilde{O}=O_{3}\otimes O_{2}.$ (96)

Thus,

$\displaystyle\begin{split}(\widehat{T}_{123})_{(1)}&=O_{1}(T_{123})_{(1)}(O_{3}\otimes
O_{2})^{t}&&&(\widehat{T}_{1}\circ\widehat{T}_{23})_{(1)}&=O_{1}(T_{1}\circ
T_{23})_{(1)}(O_{3}\otimes O_{2})^{t}\\\
(\widehat{T}_{2}\circ\widehat{T}_{13})_{(2)}&=O_{1}(T_{2}\circ
T_{13})_{(2)}(O_{3}\otimes
O_{2})^{t}&&&(\widehat{T}_{12}\circ\widehat{T}_{3})_{(1)}&=O_{1}(T_{12}\circ
T_{3})_{(1)}(O_{3}\otimes O_{2})^{t}\end{split}$ (97)

Proving that $\rho$ and $\widehat{\rho}$ are SO equivalent (and hence quasi-LU
equivalent). Thus, we have the following theorem which characterizes quasi-LU
equivalence:

######  Theorem 4 (Characterizing quasi-LU Equivalence: Version 1).

Suppose $\rho$ and $\widehat{\rho}$ are tripartite density matrices over the
same Hilbert space
$\mathbb{C}^{d_{1}}\otimes\mathbb{C}^{d_{2}}\otimes\mathbb{C}^{d_{3}}$ with
respective matrix representations
$\\{T_{1},T_{2},T_{3},T_{12},T_{13},T_{23},T_{123}\\}$ and
$\\{\widehat{T}_{1},\widehat{T}_{2},\widehat{T}_{2},\widehat{T}_{12},\widehat{T}_{13},\widehat{T}_{23},\widehat{T}_{123}\\}$.
Let

$(A_{1},A_{2},A_{3},A_{4},A_{5},A_{6}):=((T_{123})_{(1)},(T_{1}\circ
T_{23})_{(1)},(T_{2}\circ T_{13})_{(2)},(T_{12}\circ T_{3})_{(1)},(T_{1}\circ
T_{2}\circ T_{3})_{(1)},T_{1})$

and

$(B_{1},B_{2},B_{3},B_{4},B_{5},B_{6}):=((\widehat{T}_{123})_{(1)},(\widehat{T}_{1}\circ\widehat{T}_{23})_{(1)},(\widehat{T}_{2}\circ\widehat{T}_{13})_{(2)},(\widehat{T}_{12}\circ\widehat{T}_{3})_{(1)},(\widehat{T}_{1}\circ\widehat{T}_{2}\circ\widehat{T}_{3})_{(1)},\widehat{T}_{1}).$

Then $\rho$ and $\widehat{\rho}$ are quasi-LU equivalent if the following
conditions are met:

  1. 1.

$\|\widehat{T}_{1}\|=\|T_{1}\|$ or $\|\widehat{T}_{23}\|=\|T_{23}\|$,
$\|\widehat{T}_{2}\|=\|T_{2}\|$ or $\|\widehat{T}_{13}\|=\|T_{13}\|$, and
$\|\widehat{T}_{3}\|=\|T_{3}\|$ or $\|\widehat{T}_{12}\|=\|T_{12}\|$;

  2. 2.

$(T_{1}\circ T_{23})_{(1)}^{t}(T_{1}\circ T_{23})_{(1)}$ or $(T_{1}\circ
T_{2}\circ T_{3})_{(1)}^{t}(T_{1}\circ T_{2}\circ T_{3})_{(1)}$ is invertible;

  3. 3.

The partial traces $\Tr_{1}(\rho)$ and $\Tr_{1}(\widehat{\rho})$ are quasi-LU
equivalent; and lastly

  4. 4.

$\Tr(w\\{A_{i}A_{j}^{t},A_{6}A_{6}^{t}\\})=\Tr(w\\{B_{i}B_{j}^{t},B_{6}B_{6}^{t}\\})$
with $1\leq i\leq j\leq 5$, for all words $w$ in noncommuting variables of
length at most $25(1+\delta_{1}+\delta_{23})^{2}$

Moreover in the case of $3$-qubits, if in addition to the above conditions we
also assume that one of $\widehat{T}_{1}^{t}\widehat{T}_{12}\widehat{T}_{2}$,
$\widehat{T}_{2}^{t}\widehat{T}_{23}\widehat{T}_{3}$, or
$\widehat{T}_{3}^{t}\widehat{T}_{13}^{t}\widehat{T}_{1}$ has the same sign as
$T_{1}^{t}T_{12}T_{2}$, $\widehat{T}_{2}^{t}T_{23}T_{3}$, or
$T_{3}^{t}T_{13}^{t}T_{1}$ (respectively), and
$\det(\widehat{T}_{jk})=\det(T_{jk})\neq 0$ for $(j,k)=(1,2),(1,3),(23)$, then
$\rho$ and $\widehat{\rho}$ are LU equivalent.

With this theorem, we have reduced the difficult problem of determining quasi-
LU equivalence between $3$-qudit states (or LU equivalence in the case of
$3$-qubit states) to checking norms, traces, the invertibility of a matrix,
and quasi-LU equivalence of partial traces, which from theorem 2 just reduces
to checking norms and traces. That is, with our result, we have shown that
determining LU/quasi-LU equivalence nearly reduces to checking norms and
traces, which are easy to compute and are LU invariants. One potential issue
is the invertibility check of $(T_{1}\circ T_{23})_{(1)}^{t}(T_{1}\circ
T_{23})_{(1)}$ or $(T_{1}\circ T_{2}\circ T_{3})_{(1)}^{t}(T_{1}\circ
T_{2}\circ T_{3})_{(1)}$. Not all density matrices of $3$-qudits states
possess the property that $(T_{1}\circ T_{23})_{(1)}^{t}(T_{1}\circ
T_{23})_{(1)}$ or $(T_{1}\circ T_{2}\circ T_{3})_{(1)}^{t}(T_{1}\circ
T_{2}\circ T_{3})_{(1)}$ is invertible, and this is not an LU/quasi-LU
invariant.

One, albeit incomplete, remedy to this potential issue is to instead define

$(A_{1},A_{2},A_{3},A_{4},A_{5},A_{6}):=((T_{123})_{(2)},(T_{1}\circ
T_{23})_{(2)},(T_{2}\circ T_{13})_{(1)},(T_{12}\circ T_{3})_{(2)},(T_{1}\circ
T_{2}\circ T_{3})_{(2)},T_{2})$

and

$(B_{1},B_{2},B_{3},B_{4},B_{5},B_{6}):=((\widehat{T}_{123})_{(2)},(\widehat{T}_{1}\circ\widehat{T}_{23})_{(2)},(\widehat{T}_{2}\circ\widehat{T}_{13})_{(1)},(\widehat{T}_{12}\circ\widehat{T}_{3})_{(2)},(\widehat{T}_{1}\circ\widehat{T}_{2}\circ\widehat{T}_{3})_{(2)},\widehat{T}_{2}).$

Then by applying Lemma 1. to these matrices, there exists $O_{2}\in
O(\delta_{2})$, $\widetilde{O}\in O(\delta_{1}\delta_{3})$, and $O\in O(1)$
(hence $O=[\pm 1]$ and so without loss of generality we may assume that
$O=[1]$) such that

$\displaystyle\begin{split}(\widehat{T}_{123})_{(2)}&=O_{2}(T_{123})_{(2)}\widetilde{O}^{t}&&&\widehat{T}_{1}\circ\widehat{T}_{23})_{(2)}&=O_{2}(T_{1}\circ
T_{23})_{(2)}\widetilde{O}^{t}\\\
(\widehat{T}_{2}\circ\widehat{T}_{13})_{(1)}&=O_{1}(T_{2}\circ
T_{13})_{(1)}\widetilde{O}^{t}&&&\widehat{T}_{12}\circ\widehat{T}_{3})_{(2)}&=O_{1}(T_{12}\circ
T_{3})_{(2)}\widetilde{O}^{t}\\\
(\widehat{T}_{1}\circ\widehat{T}_{2}\circ\widehat{T}_{3})_{(2)}&=O_{1}(T_{1}\circ
T_{2}\circ
T_{3})_{(2)}\widetilde{O}^{t}&&&\widehat{T}_{1}&=O_{2}T_{2}\end{split}$ (98)

if and only if

$\Tr(w\\{A_{i}A_{j}^{t},A_{6}A_{6}^{t}\\})=\Tr(w\\{B_{i}B_{j}^{t},B_{6}B_{6}^{t}\\})$
(99)

with $1\leq i\leq j\leq 5$, for every word $w$ in noncommuting variables of
length at most $25(1+\delta_{2}+\delta_{1}\delta_{3})^{2}$. Additionally if we
assume that $\Tr_{2}(\rho)$ is quasi-LU equivalent to
$\Tr_{2}(\widehat{\rho})$, then there exists $O_{1}\in O(\delta_{1})$ and
$O_{3}\in O(\delta_{3})$ such that

$\widehat{T}_{1}=O_{1}*T_{1},\quad\widehat{T}_{3}=O_{3}*T_{3},\quad\text{and}\quad\widehat{T}_{13}=(O_{1},O_{3})*T_{13}.$
(100)

It then follows from calculations similar to those in equations (91) to (95),
that if $(T_{2}\circ T_{13})_{(1)}^{t}(T_{2}\circ T_{13})_{(1)}$ or
$(T_{1}\circ T_{2}\circ T_{3})_{(2)}^{t}(T_{1}\circ T_{2}\circ T_{3})_{(2)}$
is invertible, then

$\displaystyle\begin{split}(\widehat{T}_{123})_{(2)}&=O_{2}(T_{123})_{(2)}(O_{3}\otimes
O_{1})^{t}&&&(\widehat{T}_{1}\circ\widehat{T}_{23})_{(2)}&=O_{2}(T_{1}\circ
T_{23})_{(2)}(O_{3}\otimes O_{1})^{t}\\\
(\widehat{T}_{2}\circ\widehat{T}_{13})_{(1)}&=O_{2}(T_{2}\circ
T_{13})_{(1)}(O_{3}\otimes
O_{1})^{t}&&&(\widehat{T}_{12}\circ\widehat{T}_{3})_{(2)}&=O_{2}(T_{12}\circ
T_{3})_{(1)}(O_{3}\otimes O_{1})^{t},\end{split}$ (101)

proving that $\rho$ and $\widehat{\rho}$ are SO equivalent (and hence quasi-LU
equivalent). Thus, we have the following theorem, which gives an alternative
characterization of quasi-LU equivalence:

######  Theorem 5 (Characterizing quasi-LU Equivalence: Version 2).

Suppose $\rho$ and $\widehat{\rho}$ are tripartite density matrices over the
same Hilbert space
$\mathbb{C}^{d_{1}}\otimes\mathbb{C}^{d_{2}}\otimes\mathbb{C}^{d_{3}}$ with
respective matrix representations
$\\{T_{1},T_{2},T_{3},T_{12},T_{13},T_{23},T_{123}\\}$ and
$\\{\widehat{T}_{1},\widehat{T}_{2},\widehat{T}_{2},\widehat{T}_{12},\widehat{T}_{13},\widehat{T}_{23},\widehat{T}_{123}\\}$.
Let

$(A_{1},A_{2},A_{3},A_{4},A_{5},A_{6}):=((T_{123})_{(2)},(T_{1}\circ
T_{23})_{(2)},(T_{2}\circ T_{13})_{(1)},(T_{12}\circ T_{3})_{(2)},(T_{1}\circ
T_{2}\circ T_{3})_{(2)},T_{1})$

and

$(B_{1},B_{2},B_{3},B_{4},B_{5},B_{6}):=((\widehat{T}_{123})_{(2)},(\widehat{T}_{1}\circ\widehat{T}_{23})_{(2)},(\widehat{T}_{2}\circ\widehat{T}_{13})_{(1)},(\widehat{T}_{12}\circ\widehat{T}_{3})_{(2)},(\widehat{T}_{1}\circ\widehat{T}_{2}\circ\widehat{T}_{3})_{(2)},\widehat{T}_{1}).$

Then $\rho$ and $\widehat{\rho}$ are quasi-LU equivalent if the following
conditions are met:

  1. 1.

$\|\widehat{T}_{1}\|=\|T_{1}\|$ or $\|\widehat{T}_{23}\|=\|T_{23}\|$,
$\|\widehat{T}_{2}\|=\|T_{2}\|$ or $\|\widehat{T}_{13}\|=\|T_{13}\|$, and
$\|\widehat{T}_{3}\|=\|T_{3}\|$ or $\|\widehat{T}_{12}\|=\|T_{12}\|$;

  2. 2.

$(T_{2}\circ T_{13})_{(1)}^{t}(T_{2}\circ T_{13})_{(1)}$ or $(T_{1}\circ
T_{2}\circ T_{3})_{(2)}^{t}(T_{1}\circ T_{2}\circ T_{3})_{(2)}$ is invertible;

  3. 3.

The partial traces $\Tr_{2}(\rho)$ and $\Tr_{2}(\widehat{\rho})$ are quasi-LU
equivalent; and lastly

  4. 4.

$\Tr(w\\{A_{i}A_{j}^{t},A_{6}A_{6}^{t}\\})=\Tr(w\\{B_{i}B_{j}^{t},B_{6}B_{6}^{t}\\})$
with $1\leq i\leq j\leq 5$, for all words $w$ in noncommuting variables of
length at most $25(1+\delta_{1}+\delta_{23})^{2}$

Moreover in the case of $3$-qubits, if in addition to the above conditions we
also assume that one of $\widehat{T}_{1}^{t}\widehat{T}_{12}\widehat{T}_{2}$,
$\widehat{T}_{2}^{t}\widehat{T}_{23}\widehat{T}_{3}$, or
$\widehat{T}_{3}^{t}\widehat{T}_{13}^{t}\widehat{T}_{1}$ has the same sign as
$T_{1}^{t}T_{12}T_{2}$, $\widehat{T}_{2}^{t}T_{23}T_{3}$, or
$T_{3}^{t}T_{13}^{t}T_{1}$ (respectively), and
$\det(\widehat{T}_{jk})=\det(T_{jk})\neq 0$ for $(j,k)=(1,2),(1,3),(23)$, then
$\rho$ and $\widehat{\rho}$ are LU equivalent.

This does not solve the issue of needing invertibility of a matrix, but it at
least gives us more options since now we only need one of $(T_{1}\circ
T_{23})_{(1)}^{t}(T_{1}\circ T_{23})_{(1)}$, $(T_{2}\circ
T_{13})_{(1)}^{t}(T_{2}\circ T_{13})_{(1)}$, or $(T_{1}\circ T_{2}\circ
T_{3})_{(k)}^{t}(T_{1}\circ T_{2}\circ T_{3})_{(k)}$ ($k\in\\{1,2\\}$) to be
invertible. Indeed, by choosing a different subset and/or matrix unfolding of
the tensors in the hypermatrix representation, it may be possible to derive
other characterizations of quasi-LU equivalence similar to that of the
previous two theorems.

##  5 Conclusion

Our proofs of Theorem 1 and Theorem 3 can be further generalized to the
$4$-qudits and beyond, thus establishing a near equivalence of quasi-LU and SO
equivalence for $N$-qudits. Furthermore, due to the broad scope of Futorney’s
generalization of Specht’s criterion, similar reasoning to that given in the
establishment of Theorems 4 and 5 can theoretically be applied to reduce the
problem of determining quasi-LU equivalence of $N$-qudits (or LU equivalence
in the case of $N$-qubits) to checking trace identities and possibly a few
other easy-to-check properties. Nonetheless, the number of trace identities
grows rapidly each time we go up in several qudits, and so as a matter of
practicality, we stop at $3$-qudits since this seems to be the case where our
results would be most useful.

##  Appendix A Specht’s Criterion and Generalizations

The following results in this appendix are valid in complex inner product
spaces and complex Euclidean spaces, however, for our purposes, we will only
consider real Euclidean spaces. We note, however, that the following results
can be extended to the aforementioned spaces by replacing ”real” with
”complex”, ”orthogonal” with ”unitary”, and in the case of complex inner
product spaces, ”$t$” with ”$\dagger$”.

The original version of Specht’s criterion can be found in [10]. For our
purposes, we only need the real version of Specht’s criterion, which was
proven by Carl Pearcy in [11]. It is stated below:

######  Proposition 1 (Specht’s criterion (real version)).

If $A$ and $B$ are real $n\times n$ matrices, then they are orthogonally
similar, i.e. $B=O^{t}AO$ for some $n\times n$ orthogonal matrix $O$, if and
only if

$\Tr(w\\{A,A^{t}\\})=\Tr(w\\{B,B^{t}\\})$ (102)

for every word $w(x,y)$ in two noncommuting variables.

One issue with Specht’s Criterion is that as it is stated, it requires
infinitely many trace identities to check. However, in 1962 C. Pearcy proved
in [11] that it suffices to verify conditions (102) for all words of length at
most $2n^{2}$, and in 1986 T. Laffey showed in [12] that it suffices to verify
conditions (102) for all words of length at most $\frac{2}{3}(n^{2}+2)$. Other
tighter upper bounds have been derived, but note that
$\frac{2}{3}(n^{2}+2)<n^{2}$ for all positive integers $n>2$, so for our
purposes we conclude that (by Laffey’s upper bound) it is enough to check
(102) for all words of length at most $n^{2}$.

In this paper we utilize two generalizations of Specht’s Criterion, due to
Jing [6] and Futorney et. al. [7]. Jing’s generalization is in fact a special
case of Futorney’s, so the rest of this appendix will be dedicated to
reviewing Futorney’s result. To better understand what Futorney et. al.
proved, we first review the basics of quiver representation theory.

###  A.1 Quiver Representations and Futorney’s Theorem

A quiver is a directed graph (loops and multiple arrows are allowed) used to
represent vector spaces and algebras. A representation
$\mathcal{A}=(\mathcal{A}_{\alpha},\mathcal{U}_{v})$ of a quiver $Q$ over a
field $F$ is given by assigning to each vertex $v$ a vector space
$\mathcal{U}_{v}$ over $F$ to each arrow $\alpha:u\rightarrow v$ a linear
transformation $\mathcal{A}_{\alpha}:u\rightarrow v$. The vector
$\dim(\mathcal{A}):=(\dim(\mathcal{U}_{1}),...,\dim(\mathcal{U}_{t}))$ is
called the dimension of the representation $\mathcal{A}$.

An oriented cycle $\pi$ of length $l\geq 1$ in a quiver $Q$ is a sequence of
arrows of the form

${\pi:}$${v_{1}}$${v_{2}}$${...}$${v_{l}}$$\scriptstyle{\alpha_{l}}$$\scriptstyle{\alpha_{1}}$$\scriptstyle{\alpha_{2}}$$\scriptstyle{\alpha_{l-1}}$

That is, it is a closed-directed walk (note that some vertices and arrows may
repeat due to possible loops). For each representation $\mathcal{A}$ of a
quiver $Q$ and any cycle $\pi$, we define $\mathcal{A}(\pi)$ to be the cycle
of linear transformations

${\mathcal{A}(\pi):}$${\mathcal{U}_{v_{1}}}$${\mathcal{U}_{v_{2}}}$${...}$${\mathcal{U}_{v_{l}}}$$\scriptstyle{\mathcal{A}_{\alpha_{l}}}$$\scriptstyle{\mathcal{A}_{\alpha_{1}}}$$\scriptstyle{\mathcal{A}_{\alpha_{2}}}$$\scriptstyle{\mathcal{A}_{\alpha_{l-1}}}$

The trace of $\mathcal{A}(\pi)$ is defined as
$\mathrm{trace}(\mathcal{A}(\pi)):=\mathrm{trace}(\mathcal{A}_{\alpha_{1}}\mathcal{A}_{\alpha_{2}}...\mathcal{A}_{\alpha_{l}})$.
Note that $\mathrm{trace}(\mathcal{A}(\pi))$ does not depend on the choice of
the initial vertex $v_{1}$ in the cycle since the trace is invariant under
cyclic permutations.

For each linear transformation $\mathcal{A}:\mathcal{U}\rightarrow\mathcal{V}$
between (real) Euclidean spaces $\mathcal{U}$ and $\mathcal{V}$, the adjoint
map $\mathcal{A}^{*}:\mathcal{V}\rightarrow\mathcal{U}$ is given by
$\langle\mathcal{A}x,y\rangle=\langle x,\mathcal{A}^{*}y\rangle$ for all
$x\in\mathcal{U}$ and $y\in\mathcal{V}$. For a quiver $Q$ with vertices
$v_{1},...,v_{t}$, a (real) Euclidean representation
$\mathcal{A}=(\mathcal{A}_{\alpha},\mathcal{U}_{v})$ is given by assigning to
each vertex $v$ a real Euclidean space $\mathcal{U}_{v}$, and to each arrow
$\alpha:u\rightarrow v$ a linear transformation
$\mathcal{A}_{\alpha}:\mathcal{U}_{u}\rightarrow\mathcal{U}_{v}$. Two (real)
Euclidean representations $\mathcal{A}=(\mathcal{U}_{\alpha},\mathcal{U}_{v})$
and $\mathcal{B}=(\mathcal{B}_{\alpha},\mathcal{V}_{v})$ of $Q$ are isometric
if there exists a family isometries (i.e. linear isomorphisms that preserve
inner products) $\varphi_{1}:\mathcal{U}_{1}\rightarrow\mathcal{V}_{1}$,…,
$\varphi_{t}:\mathcal{U}_{t}\rightarrow\mathcal{V}_{t}$ such that the diagram

${\mathcal{U}_{u}}$${\mathcal{U}_{v}}$${\mathcal{V}_{u}}$${\mathcal{V}_{V}}$$\scriptstyle{\mathcal{A}_{\alpha}}$$\scriptstyle{\varphi_{u}}$$\scriptstyle{\varphi_{v}}$$\scriptstyle{\mathcal{B}_{\alpha}}$

commutes (i.e
$\varphi_{v}\mathcal{A}_{\alpha}=\mathcal{B}_{\alpha}\varphi_{u})$ for each
arrow $\alpha:u\rightarrow v$.

Now, for each quiver $Q$, we denote $\widetilde{Q}$ to be the quiver with
double the number of arrows in $Q$, obtained from $Q$ by attaching the arrow
$\alpha^{*}:v\rightarrow u$ for each arrow $\alpha:u\rightarrow v$ in $Q$. For
each (real) Euclidean representation $\mathcal{A}$ of $Q$, we define the
(real) Euclidean representation $\widetilde{A}$ of $\widetilde{Q}$ that
coincides with $\mathcal{A}$ on $Q\subset\widetilde{Q}$ and that assigns to
each new arrow $\alpha^{*}:v\rightarrow u$ the linear transformation
$\widetilde{\mathcal{A}}_{\alpha^{*}}:=\mathcal{A}_{\alpha}^{*}:\mathcal{U}_{v}\rightarrow\mathcal{U}_{u}$
(i.e. the adjoint of $\mathcal{A}_{\alpha}$). For example, if $Q$ is given by

${Q:}$${u}$${v}$$\scriptstyle{\beta}$$\scriptstyle{\alpha}$

and $\mathcal{A}$ is a representation on $Q$ given by

${\mathcal{A}:}$${\mathcal{U}_{u}}$${\mathcal{U}_{v}}$$\scriptstyle{\mathcal{A}_{\beta}}$$\scriptstyle{\mathcal{A}_{\alpha}}$

then $\widetilde{Q}$ is given by

${\widetilde{Q}:}$${u}$${v}$$\scriptstyle{\beta}$$\scriptstyle{\beta^{*}}$$\scriptstyle{\alpha}$$\scriptstyle{\alpha^{*}}$

then $\widetilde{\mathcal{A}}$ is given by

${\widetilde{\mathcal{A}}:}$${\mathcal{U}_{u}}$${\mathcal{U}_{v}}$$\scriptstyle{\mathcal{A}_{\beta}}$$\scriptstyle{\widetilde{\mathcal{A}}_{\beta^{*}}}$$\scriptstyle{\mathcal{A}_{\alpha}}$$\scriptstyle{\widetilde{\mathcal{A}}_{\alpha^{*}}}$

Representations of quivers can be expressed in terms of matrices. If $[x]$ is
the coordinate vector of $x\in\mathcal{U}$ in some orthonormal basis, then
$\langle x,y\rangle=[x]^{T}[y]$ for all $x,y\in\mathcal{U}$. Furthermore, if
$A$ is the matrix of the linear transformation
$\mathcal{A}:\mathcal{U}\rightarrow\mathcal{V}$ in some orthonormal bases for
$\mathcal{U}$ and $\mathcal{V}$, then $A^{t}$ is the matrix of the adjoint
transformation $\mathcal{A}^{*}:\mathcal{V}\rightarrow\mathcal{U}$. A matrix
representation $A$ of dimension $(d_{1},...,d_{t})$ of a quiver $Q$ is given
by assigning to each arrow $\alpha:u\rightarrow v$ a matrix $A_{\alpha}$ of
size $d_{v}\times d_{u}$ (note that we take $d_{i}:=0$ if the vertex $i$ does
not have arrows). Two (real) Euclidean matrix representations $A$ and $B$ of
$Q$ are isometric if there exists orthogonal matrices $O_{1},...,O_{t}$ such
that

$B_{\alpha}=O_{v}^{-1}A_{\alpha}O_{u}=O_{v}^{t}A_{\alpha}O_{u}$ (103)

for every arrow $\alpha:u\rightarrow v$. Note that $\mathcal{A}$ and
$\mathcal{B}$ are isometric if and only if $A$ and $B$ are isometric.
Furthermore, if $\widetilde{\mathcal{A}}$ is the corresponding representation
obtained from a representation $\mathcal{A}$ as described in the previous
paragraph, then $\widetilde{A}$ is the matrix form of
$\widetilde{\mathcal{A}}$. Lastly, for each oriented cycle $\pi$ in a quiver
$Q$ and each matrix representation $A$ of $Q$, we denote
$A(\pi):=A_{\alpha_{1}}A_{\alpha_{2}}...A_{\alpha_{l}}$.

With this quiver representation framework, Futorney et. al. then prove [7] the
following important theorem.

######  Theorem 6.

Two (real) Euclidean matrix representations $A$ and $B$ of a quiver $Q$ are
isometric if and only if

$\mathrm{trace}(\widetilde{A}(\pi))=\mathrm{trace}(\widetilde{B}(\pi))$ (104)

for each oriented cycle $\pi$ in the quiver $\widetilde{Q}$. Moreover, it
suffices to verify (104) for all cycles $\pi$ of length at most

$\varphi((r+2)(d_{1}+...+d_{t})),$

where $\varphi(n)$ is any bound for the sufficient word length in Specht’s
criterion (e.g. $\varphi(n)=n^{2}$) and $r$ is the minimal natural number such
that

$\frac{r(r+1)}{2}\geq\max\\{m_{ij}|i\text{ and }j\text{ are vertices of }Q\\}$

in which $m_{ij}$ is the number of arows from $j$ to $i$ in $Q$.

Note that if we assign a matrix representation $A$ of dimension $n$ to the
following quiver

${Q:}$${1}$

and then apply Theorem 6, then we obtain Specht’s Criterion. In [6], the
author proves the following generalization of Specht’s Criterion:

######  Proposition 2.

Let $(A_{1},...,A_{k})$ and $(B_{1},..,B_{k})$ be two $k$-tuples of $m\times
n$ matrices. Then there exists orthogonal matrices $O$ and $P$ such that

$(B_{1},...,B_{k})=(OA_{1}P,...,OA_{k}P)$ (105)

if and only if

$\Tr(w\\{A_{1}^{t}A_{1},...,A_{i}^{t}A_{j},...,A_{k}^{t}A_{k}\\})=\Tr(w\\{B_{1}^{t}B_{1},...,B_{i}^{t}B_{j},...,B_{k}^{t}B_{k}\\})$
(106)

for every word $w(x_{11},...,x_{ij},...,x_{kk})$ in $k^{2}$ noncommuting
variables.

Indeed, if we assign two matrix representations $A$ and $B$ of dimension
$(n,m)$ on the following quiver

${1}$${2}$$\scriptstyle{\vdots}$

and then apply Theorem 6, then we precisely get Jing’s generalization of
Specht’s Criterion. Lastly, we note that in our proof of Theorem’s 5 and 6, we
apply Futorny et al’s Theorem to the following quiver

${1}$${3}$${2}$

attaching two different pairs of matrix representations $A$ and $B$ of
dimensions $(\delta_{2}\delta_{3},1,\delta_{1})$ and
$(\delta_{1}\delta_{3},1,\delta_{2})$ (respectively) to it.

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