SLOCC classification of n qubits invoking the proportional relationships for spectrums and for standard Jordan normal forms

TL;DR

This paper introduces a classification method for n-qubit states based on proportional relationships in spectrums and Jordan normal forms, leading to a detailed partitioning of states under SLOCC, especially for four qubits.

Contribution

It develops a novel classification framework for n-qubit states using spectral and Jordan form relationships, enhancing understanding of SLOCC equivalence classes.

Findings

States of n ≥ 4 qubits are partitioned into 12 groups and 34 families under SLOCC.

The classification method is specifically effective for four qubits.

Proportional relationships in spectrums and Jordan forms are key to the classification.

Abstract

We investigate the proportional relationships for spectrums and for SJNFs (Standard Jordan Normal Forms) of the matrices constructed from coefficient matrices of two SLOCC (stochastic local operations and classical communication) equivalent states of $n$ qubits. Invoking the proportional relationships for spectrums and for SJNFs, pure states of $n$ ($\geq 4$) qubits are partitioned into 12 groups and 34 families under SLOCC, respectively. Specially, it is true for four qubits.

Tables

Table 1. Table 1: SJNFs of SLOCC equivalent states

state	$L_{a_4}\left(a\ne 0\right)$	$L_{a_4}\left(a=1\right)$	$k=a^2$
SJNF	$J_4\left(a^2\right)$	$J_4\left(1\right)$
State	$L_{a_2{0}_{3\oplus 1}}\left(a\ne 0\right)$	$L_{a_2{0}_{3\oplus 1}}\left(a=1\right)$	$k=a^2$
SJNF	$J_2\left(a^2\right)J_2\left(0\right)$	$J_2\left(1\right)J_2\left(0\right)$
State	$L_{a{b}_3}^{\ast }\left(a=0\right)$	$L_{a{b}_3}\left(a=0\right)$	$k=1$
SJNF	$0b^2{J}_2\left(0\right)$	$0b^2{J}_2\left(0\right)$

Table 2. Table 2: 12 types of CPs, 12 types of spectrums, 34 types of the SJNFs for 4 by 4 matrices, and the corresponding states for four qubits.

CPi;spectrum	SJNF	state	SJNF	state
1;${\sigma }_1{\sigma }_1{\sigma }_1{\sigma }_1$	$J_4\left({\sigma }_1\right)$	${\tau }_1$	$J_2\left({\sigma }_1\right)J_2\left({\sigma }_1\right)$	${\eta }_1$
	$J_3\left({\sigma }_1\right){\sigma }_1$	${\theta }_1$	${\sigma }_1{\sigma }_1{J}_2\left({\sigma }_1\right)$	${\zeta }_1$
	${\sigma }_1{\sigma }_1{\sigma }_1{\sigma }_1$	$G_1$
2;${\sigma }_1{\sigma }_2{\sigma }_2{\sigma }_2$	${\sigma }_1{J}_3\left({\sigma }_2\right)$	${\theta }_2$	${\sigma }_1{\sigma }_2{J}_2\left({\sigma }_2\right)$	${\zeta }_2$
	${\sigma }_1{\sigma }_2{\sigma }_2{\sigma }_2$	$G_2$
3;${\sigma }_1{\sigma }_2{\sigma }_3{\sigma }_3$	${\sigma }_1{\sigma }_2{J}_2\left({\sigma }_3\right)$	${\zeta }_3$	${\sigma }_1{\sigma }_2{\sigma }_3{\sigma }_3$	$G_3$
4;${\sigma }_1{\sigma }_1{\sigma }_2{\sigma }_2$	${\sigma }_1{\sigma }_1{\sigma }_2{\sigma }_2$	$G_4$	${\sigma }_1{\sigma }_1{J}_2\left({\sigma }_2\right)$	${\zeta }_4$
	$J_2\left({\sigma }_1\right)J_2\left({\sigma }_2\right)$	${\eta }_2$
5;${\sigma }_1{\sigma }_2{\sigma }_3{\sigma }_4$	${\sigma }_1{\sigma }_2{\sigma }_3{\sigma }_4$	$G_5$
6;0${\sigma }_1{\sigma }_1{\sigma }_1$	$0J_3\left({\sigma }_1\right)$	${\theta }_3$	$0J_2\left({\sigma }_1\right){\sigma }_1$	${\zeta }_6$
	$0{\sigma }_1{\sigma }_1{\sigma }_1$	$G_6$
7;0${\sigma }_1{\sigma }_2{\sigma }_2$	$0{\sigma }_1{J}_2\left({\sigma }_2\right)$	${\zeta }_7$	$0{\sigma }_1{\sigma }_2{\sigma }_2$	$G_7$
8;$0{\sigma }_1{\sigma }_2{\sigma }_3$	$0{\sigma }_1{\sigma }_2{\sigma }_3$	$G_8$
9;$00{\sigma }_1{\sigma }_1$	$J_2\left(0\right)J_2\left({\sigma }_1\right)$	${\kappa }_1$	$J_2\left(0\right){\sigma }_1{\sigma }_1$	${\mu }_2$
	$00J_2\left({\sigma }_1\right)$	${\zeta }_9$	$00{\sigma }_1{\sigma }_1$	${\zeta }_8$
10;$00{\sigma }_1{\sigma }_2$	$J_2\left(0\right){\sigma }_1{\sigma }_2$	${\mu }_1$	$00{\sigma }_1{\sigma }_2$	${\zeta }_{10}$
11;$000{\sigma }_1$	$J_3\left(0\right){\sigma }_1$	${\xi }_1$	$J_2\left(0\right)0{\sigma }_1$	${\theta }_4$
	$000{\sigma }_1$	${\zeta }_{11}$
12;0000	$J_4\left(0\right)$	$L_{0_{7\oplus 1}}$	$J_3\left(0\right)0$	${\xi }_2$
	$J_2\left(0\right)J_2\left(0\right)$	${\tau }_2$	$J_2\left(0\right)00$	${\theta }_5$
	0000	${\zeta }_{12}$

Table 3. Table 3: SJNFs of S 1 , 2 ( 5 ) superscript subscript 𝑆 1 2 5 S_{1,2}^{(5)} and S 1 , 2 ( 6 ) superscript subscript 𝑆 1 2 6 S_{1,2}^{(6)} .

states	$|{\Psi }_2⟩$	$|{\Psi }_4⟩$	$|{\Psi }_5⟩$	$|{\Psi }_6⟩$
SJNFs	$\pm \frac{1}{2}00$	$0000$	$0000$	$0J_3\left(0\right)$
states	$|{\Xi }_2⟩$	$|{\Xi }_4⟩$	$|{\Xi }_5⟩$	$|{\Xi }_6⟩$	$|{\Xi }_7⟩$
SJNFs	$\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)00$	$0000$	$0000$	$00J_2\left(0\right)$	$0J_3\left(0\right)$

Table 4. Table 4: SLOCC classification of three qubits

states	SJNF of $S_{1,2}^{\left(3\right)}\left(\psi \right)$	SJNF of $S_{1,3}^{\left(3\right)}\left(\psi \right)$
GHZ	$J_1\left(\pm \frac{1}{2}\right)00$	$J_1\left(\pm \frac{1}{2}\right)00$
W	$J_3\left(0\right)0$	$J_3\left(0\right)0$
A-BC	$J_2\left(0\right)J_2\left(0\right)$	$J_2\left(0\right)J_2\left(0\right)$
B-AC	$J_2\left(0\right)J_2\left(0\right)$	$0000$
C-AB	$0000$	$J_2\left(0\right)J_2\left(0\right)$
$|000⟩$	$0000$	$0000$

Table 5. Table 5: Comparison between Theorems 1 and 2

	Theorem 1	Theorem 2
spect. $\psi$	${\lambda }_1$, $\cdots$, ${\lambda }_4$	${\lambda }_1$, $\cdots$, ${\lambda }_4$
spect. ${\psi }^{\prime }$	$k{\lambda }_1$, $\cdots$, $k{\lambda }_4$	${\lambda }_1$, $\cdots$, ${\lambda }_4$
SJNF $\psi$	$J_{{\ell }_1}\left({\lambda }_1\right)\cdots J_{{\ell }_j}\left({\lambda }_j\right)$	$J_{{\ell }_1}\left({\lambda }_1\right)\cdots J_{{\ell }_j}\left({\lambda }_j\right)$
SJNF ${\psi }^{\prime }$	$J_{{\ell }_1}\left(k{\lambda }_1\right)\cdots J_{{\ell }_j}\left(k{\lambda }_j\right)$	$J_{{\ell }_1}\left({\lambda }_1\right)\cdots J_{{\ell }_j}\left({\lambda }_j\right)$

Equations

$$|{\psi }^{\prime }⟩={\mathcal{A}}_1\otimes {\mathcal{A}}_2\otimes \cdots \otimes {\mathcal{A}}_n|\psi ⟩.$$

$${\Omega }_{q_1,...,q_i}^{(n)}(|\psi ⟩)$$

$${\Omega }_{q_1{q}_2}^{(n)}(|{\psi }^{\prime }⟩)$$

$$T=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cccc}1&0&0&1\\ 0&i&i&0\\ 0&-1&1&0\\ i&0&0&-i\end{array}\right).$$

$$S_{q_1{q}_2}^{(n)}({\psi }^{\prime })$$

$$S_{q_1{q}_2}^{(n)}(\psi )=[T{C}_{q_1{q}_2}^{(n)}(|\psi ⟩)]{\upsilon }^{\otimes (n-2)}[T{C}_{q_1{q}_2}^{(n)}(|\psi ⟩){]}^t.$$

$$S_{q_1{q}_2}^{(4)}(\psi )=(T{C}_{q_1{q}_2}^{(4)}{T}^{+})(T{C}_{q_1{q}_2}^{(4)}{T}^{+}{)}^t.$$

$$S_{q_1{q}_2}^{(3)}(\psi )=[T{C}_{q_1{q}_2}^{(3)}(|\psi ⟩)]\upsilon [T{C}_{q_1{q}_2}^{(3)}(|\psi ⟩){]}^t.$$

$$S_{1,2}^{(2)}(\psi )=[T{C}_{1,2}^{(2)}(|\psi ⟩)][T{C}_{1,2}^{(2)}(|\psi ⟩){]}^t.$$

$$S_{q_1{q}_2}^{(n)}({\psi }^{\prime })=G_1{S}_{q_1{q}_2}^{(n)}(\psi )G_1^{-1}.$$

Full text

I Introduction

Quantum entanglement is an essential resource in quantum teleportation, quantum cryptography, and quantum information and computation Nielsen . A key task of the entanglement theory is to classify different types of entanglement. SLOCC classification is very significant because the states in the same SLOCC class are able to perform the same QIT-tasks Dur Verstraete . It is well known that two-qubit states were partitioned into two SLOCC classes, three-qubit states were partitioned into six SLOCC classes, and there are infinitely many SLOCC classes for $n$ ($\geq 4$) qubits Dur . It is highly desirable to partition these infinite classes into a finite number of families according to a SLOCC invariant criterion. In the pioneering work of Verstraete et al. Verstraete , by using their general singular value decomposition Verstraete *et al. *partitioned pure four-qubit states into nine SLOCC inequivalent families: $G_{abcd}$, $L_{abc_{2}}$, $L_{a_{2}b_{2}}$, $L_{ab_{3}}$, $L_{a_{4}}$,$\ L_{a_{2}0_{3\oplus 1}}$, $L_{0_{5\oplus 3}}$, $L_{0_{7\oplus 1}}$, and $L_{0_{3\oplus 1}0_{3\oplus 1}}$ Verstraete . Since then, the extensive efforts have contributed to studying entanglement classification of four qubits Verstraete ; Miyake ; Cao ; LDF07b ; Chterental ; Lamata ; LDFQIC09 ; Borsten ; Viehmann ; Buniy ; Sharma12 .

Recently, considerable efforts have been devoted to find SLOCC invariant polynomials in the coefficients of states for classifications and measures of entanglement of$\ n$ qubits Wong ; Luque ; Leifer ; Levay ; LDF07a ; Osterloh09 ; Viehmann ; Eltschka ; Gour ; LDFPRA13 . It is well known that the concurrence and the 3-tangle are invariant polynomials of degrees 2 and 4 for two and three qubits, respectively Coffman . Explicit and simple expresses of invariant polynomials of degrees 2 for even $n$ qubits LDF07a , 4 for odd $n$ ($\geq 4$) qubits LDF07a , 4 for even $n$ ($\geq 4$) qubits LDFPRA13 , were presented.

Very recently, SLOCC invariant ranks of the coefficient matrices were proposed for SLOCC classification LDFPRL12 ; LDFPRA12 ; Wang ; Fan ; LDFPRA15 .

In this paper, for two SLOCC equivalent states of $n$ qubits, we show that the matrices constructed from coefficient matrices of the two states have proportional spectrums and proportional SJNFs. Invoking the proportional relationships for spectrums pure states of $n$ ($\geq 4$) qubits are partitioned into 12 groups under SLOCC, and invoking the proportional relationships for SJNFs pure states of $n$ ($\geq 4$) qubits are partitioned into 34 families under SLOCC. Specially, for four qubits, we obtain new SLOCC classifications.

II **SLOCC classification of $n$ qubits **

II.1 The proportional relationships for spectrums and for SJNFs

Let $|\psi\rangle=\sum_{i=0}^{2^{n}-1}a_{i}|i\rangle$ be an $n$-qubit pure state. It is well known that two $n$-qubit pure states $|\psi\rangle$ and $|\psi^{\prime}\rangle$ are SLOCC equivalent if and only if there are invertible local operators $\mathcal{A}_{i}\in GL(2,C)$, $i=1,\cdots,n$, such that Dur

[TABLE]

To any state $|\psi\rangle$ of $n$ qubits, we associate a $2^{\ell}$ by $2^{n-\ell}$ matrix $C_{q_{1}\cdots q_{\ell}}^{(n)}(|\psi\rangle)$ whose entries are the coefficients $a_{0},a_{1},\cdots,a_{2^{n}-1}$ of the state $|\psi\rangle$, where $q_{1},\cdots,q_{\ell}$ are chosen as the row bits LDFPRL12 ; LDFPRA12 . In LDFPRA15 , in terms of the coefficient matrix $C_{q_{1},...,q_{i}}^{(n)}$ we constructed a $2^{i}$ by $2^{i}$ matrix $\Omega_{q_{1},...,q_{i}}^{(n)}$

[TABLE]

where $\upsilon=\sqrt{-1}\sigma_{y}$ and \sigma_{y}\is the Pauli operator, and $C^{t}$ is the transpose of $C$.

From LDFPRA15 , when $q_{1}$ and $q_{2}$ are chosen as the row bits, we can show that if $n$-qubit states $|\psi^{\prime}\rangle$ and $|\psi\rangle$ are SLOCC equivalent, then

[TABLE]

Let the unitary matrix

[TABLE]

Let $G_{1}=\ \ T(\mathcal{A}_{q_{1}}\otimes\mathcal{A}_{q_{2}})T^{+}$, where $T^{+}$ is the Hermitian transpose of $T$. It is easy to check that $G_{1}G_{1}^{t}=(\det\mathcal{A}_{q_{1}}\ast\det\mathcal{A}_{q_{2}})I$. Let $S_{q_{1}q_{2}}^{(n)}(\psi^{\prime})=T\Omega_{q_{1}q_{2}}^{(n)}(|\psi^{\prime}\rangle)T^{t}$. Then, from Eq. (LABEL:g-4) we obtain

[TABLE]

where $T^{\ast}$ is a conjugate matrix, $T^{t}T^{\ast}=I$, and $k=\Pi_{\ell=1}^{n}\det\mathcal{A}_{\ell}$. Note that $S_{q_{1}q_{2}}^{(n)}(\psi^{\prime})$ and $S_{q_{1}q_{2}}^{(n)}(\psi)$ are $4$ by $4$ matrices.

In this paper, we write the direct sum of standard Jordan blocks $J_{n_{1}}(\lambda_{1})$,$\cdots$, and $J_{n_{j}}(\lambda_{j})$ as $J_{n_{1}}(\lambda_{1})\cdots J_{n_{j}}(\lambda_{j})$. The Jordan block $J_{1}(a)$ is simply written as $a$. We define that the two SJNFs $J_{n_{1}}(\lambda_{1})\cdots J_{n_{j}}(\lambda_{j})$ and $J_{n_{1}}(k\lambda_{1})\cdots J_{n_{j}}(k\lambda_{j})$, where $k\neq 0$, are proportional.

Eq. (5) leads to the following theorem 1.

Theorem 1. If the states $|\psi^{\prime}\rangle$ and $|\psi\rangle$ of $n$ qubits satisfy Eq. (1), i.e. the state $|\psi^{\prime}\rangle$ is SLOCC equivalent to $|\psi\rangle$, then

(1) if $S_{q_{1}q_{2}}^{(n)}(\psi)$ has the spectrum $\lambda_{1}$, $\cdots$, $\lambda_{4}$, then $S_{q_{1}q_{2}}^{(n)}(\psi^{\prime})$ has the spectrum $k\lambda_{1}$, $\cdots$, $k\lambda_{4}$, where $k=\Pi_{\ell=1}^{n}\det\mathcal{A}_{\ell}$.

(2) if $S_{q_{1}q_{2}}^{(n)}(\psi)$ has the SJNF $J_{n_{1}}(\lambda_{1})\cdots J_{n_{j}}(\lambda_{j})$, then $S_{q_{1}q_{2}}^{(n)}(\psi^{\prime})$ has the SJNF $J_{n_{1}}(k\lambda_{1})\cdots J_{n_{j}}(k\lambda_{j})$, where $k=\Pi_{\ell=1}^{n}\det\mathcal{A}_{\ell}$.

We give our argument as follows. Let $\Gamma=G_{1}S_{q_{1}q_{2}}^{(n)}(\psi)G_{1}^{-1}$. Then, $S_{q_{1}q_{2}}^{(n)}(\psi^{\prime})=k\Gamma$. Clearly, $\Gamma$ is similar to $S_{q_{1}q_{2}}^{(n)}(\psi)$. Therefore, $\Gamma$ and $S_{q_{1}q_{2}}^{(n)}(\psi)$ have the same spectrum and SJNF.

(1). It is clear that if $\Gamma$ has the spectrum $\lambda_{1}$,$\cdots$, $\lambda_{4}$, then $S_{q_{1}q_{2}}^{(n)}(\psi^{\prime})$ has the spectrum $k\lambda_{1}$,$\cdots$, $k\lambda_{4}$.

(2). There is an invertible matrix $H$ such that $\Gamma=HJH^{-1}$, where the SJNF $J=J_{n_{1}}(\lambda_{1})\cdots J_{n_{j}}(\lambda_{j})$. Then, $k\Gamma=HkJH^{-1}$. It is not hard to see that the SJNF of $kJ$ is $J_{n_{1}}(k\lambda_{1})\cdots J_{n_{j}}(k\lambda_{j})$.

Example 1. We have the following SLOCC equivalent states of four qubits: $L_{a_{4}}(a\neq 0)$ and $L_{a_{4}}(a=1)$ LDFQIC09 ; $L_{a_{2}0_{3\oplus 1}}(a\neq 0)$ and $L_{a_{2}0_{3\oplus 1}}(a=1)$ LDFQIC09 ; and $L_{ab_{3}}^{\ast}(a=0)$ and $L_{ab_{3}}(a=0)$ LDFPRA12 . We list the SJNFs of $S_{1,2}^{(4)}$ of the states in Table 1.

Example 2. For four qubits, let $\zeta_{4}=a(|0\rangle+|15\rangle)+b(|5\rangle+|10\rangle)+|6\rangle$, and $\zeta_{5}=b(|0\rangle+|15\rangle)+a(|5\rangle+|10\rangle)+|6\rangle$, where $a\neq b$. The SJNF of $S_{1,2}^{(4)}(\zeta_{4})$ is $J_{2}(b)aa$ while the SJNF of $S_{1,2}^{(4)}(\zeta_{5})$ is $J_{2}(a)bb$. So, by (2) of Theorem 1 the two states $\zeta_{4}$ and $\zeta_{5}$ are SLOCC inequivalent.

We can rewrite $S_{q_{1}q_{2}}^{(n)}(\psi)$ as

[TABLE]

II.2 Partition pure states of $n$ ($\geq 4$) qubits into 12 groups

and 34 families

Theorem 1 permits a reduction of SLOCC classification of $n$ ($\geq 4$) qubits to a classification of $4$ by $4$ complex matrices. For $4$ by $4$ matrices, a calculation yields 12 types of CPs (characteristic polynomials), 12 types of spectrums, and 34 types of SJNFs in Table 2. It is easy to see that CPs and spectrums have the same effect for SLOCC classification.

Note that in Table 2, $\sigma_{i}\neq 0$, $\sigma_{i}\neq\sigma_{j}$ when $i\neq j$. Next, we give 12 types of CPs of $4$ by $4$ matrices as follows.

CP${}_{1}:(\sigma-\sigma_{1})^{4}$; CP${}_{2}:(\sigma-\sigma_{1})(\sigma-\sigma_{2})^{3}$; CP${}_{3}:(\sigma-\sigma_{1})(\sigma-\sigma_{2})(\sigma-\sigma_{3})^{2}$; CP${}_{4}:(\sigma-\sigma_{1})^{2}(\sigma-\sigma_{2})^{2}$; CP${}_{5}:\Pi_{i=1}^{4}(\sigma-\sigma_{i})$; CP${}_{6}:\sigma(\sigma-\sigma_{1})^{3}$; CP${}_{7}:\sigma(\sigma-\sigma_{1})(\sigma-\sigma_{2})^{2}$; CP${}_{8}:\sigma\Pi_{i=1}^{3}(\sigma-\sigma_{i})$; CP${}_{9}:\sigma^{2}(\sigma-\sigma_{1})^{2}$; CP${}_{10}:\sigma^{2}(\sigma-\sigma_{1})(\sigma-\sigma_{2})$; CP${}_{11}:\sigma^{3}(\sigma-\sigma_{1})$; CP${}_{12}:\sigma^{4}$.

For each state of $n$ ($\geq 4$) qubits, the spectrum of S_{q_{1}q_{2}}^{(n)}\must belong to one of the 12 types of the spectrums in Table 2. Let the states of $n$ ($\geq 4$) qubits, for which spectrums of $S_{q_{1}q_{2}}^{(n)}$ possess the same type in Table 2, belong to the same group. Thus, the states of $n$ ($\geq 4$) qubits are partitioned into 12 groups. In light of Theorem 1, the states belonging to different groups must be SLOCC inequivalent.

For each state of $n$ ($\geq 4$) qubits, the SJNF of $S_{q_{1}q_{2}}^{(n)}$ up to the order of the standard Jordan blocks must belong to one of the 34 types of the SJNFs in Table 2. Let the states of $n$ ($\geq 4$) qubits with the same type of SJNFs of $S_{q_{1}q_{2}}^{(n)}$ in Table 2 up to the order of the Jordan blocks belong to the same family. Thus, we partition the states of $n$ ($\geq 4$) qubits into 34 families. In light of Theorem 1, the states belonging to different families must be SLOCC inequivalent.

Example 3. For the maximally entangled states $|\Psi_{2}\rangle$, $|\Psi_{4}\rangle-|\Psi_{6}\rangle$ of five qubits and $|\Xi_{2}\rangle$, $|\Xi_{4}\rangle-|\Xi_{7}\rangle$ of six qubits Osterloh06 , SJNFs of $S_{1,2}^{(5)}$ partition $|\Psi_{2}\rangle$, $|\Psi_{4}\rangle-|\Psi_{6}\rangle$ into three families, and SJNFs of $S_{1,2}^{(6)}$ partition $|\Xi_{2}\rangle$, $|\Xi_{4}\rangle-|\Xi_{7}\rangle$ into four families. See Table 3.

III **SLOCC classification of two, three, and four qubits **

III.1 SLOCC classification of four qubits

For four qubits, invoking the fact that $T^{+}T^{\ast}=\upsilon^{\otimes 2}$ Eq. (6) reduces to

[TABLE]

From the above discussion, in light of Theorem 1 pure states of four qubits are partitioned into 12 groups and 34 families in Table 2. Furthermore, for each type of spectrums, CPs, and SJNFs in Table 2, we give a state in Table 2 and the appendix for which $S_{1,2}^{(4)}$ has the corresponding type. For example, S_{1,2}^{(4)}\of the state $\theta_{1}$ has the spectrum $\sigma_{1}$, $\sigma_{1}$, $\sigma_{1}$, $\sigma_{1}$, the CP $(\sigma-\sigma_{1})^{4}$, and the SJNF $J_{3}(\sigma_{1})\sigma_{1}$. It is plain to see that 12 groups and 34 families in Table 2 are both complete for four qubits.

Here, we make a comparison to Verstraete et al.’s nine families. They showed that for a complex $n$ by $n$ matrix, there are complex orthogonal matrices $O_{1}$ and $O_{2}$ such that $R=O_{1}R^{\prime}O_{2}$, where $R^{\prime}$ is a direct sum of blocks defined in Verstraete . Note that the blocks are not standard Jordan blocks. The decomposition was called a generalization of the singular value decomposition and used to partition pure states of four qubits into nine families Verstraete .

Recently, Chterental and Djoković pointed out an error in Verstraete et al.’s nine families by indicating that the family$\ L_{ab_{3}}$ is SLOCC equivalent to the subfamily $L_{abc_{2}}(a=c)$ of the family $L_{abc_{2}}$ Chterental LDFPRA15 . Thus, the classification for the nine families is incomplete. The need to redo this classification of four qubits was proposed Chterental .

III.2 SLOCC classification of three qubits

For three qubits, Eq. (6) reduces to

[TABLE]

Let $\lambda^{2}=[(c_{0}c_{7}-c_{1}c_{6})-(c_{2}c_{5}-c_{3}c_{4})]^{2}-4(c_{0}c_{3}-c_{1}c_{2})(c_{4}c_{7}-c_{5}c_{6})$. Note that $\lambda^{2}$ is just the $3$-tangle. The spectrum of $S_{1,2}^{(3)}(\psi)$ is $\pm\lambda,0,0$. We list the SJNFs of $S_{1,2}^{(3)}(\psi)$ and $S_{1,3}^{(3)}(\psi)$ in the Table 4. In light of Theorem 1, we can distinguish the six SLOCC classes of three qubits.

III.3 **SLOCC classification of two qubits **

For two qubits, Eq. (6) reduces to

[TABLE]

The spectrum of $S_{1,2}^{(2)}(\psi)$ is 0, 0, 0, $\lambda^{\prime}$, where $\lambda^{\prime}=2(a_{0}a_{3}-a_{1}a_{2})$. There are two cases for SJNFs. Case 1. For which $a_{0}a_{3}=a_{1}a_{2}$ (it is a separate state), the SJNF of $S_{1,2}^{(2)}(\psi)$ is $J_{2}(0)00$. Case 2. For which $a_{0}a_{3}\neq a_{1}a_{2}$ (it is an entangled state), the SJNF of $S_{1,2}^{(2)}(\psi)$ is $\lambda^{\prime}000$. Thus, in light of Theorem 1, we can distinguish two-qubit states into two SLOCC classes.

IV SLOCC classification of $n$ qubits under $\mathcal{A}_{i}\in SL(2,C)$

SLOCC classification under $A_{i}\in SL(2,C)$ or the classification under determinant one SLOCC operations was discussed in previous articles Verstraete Luque . Note that under $\mathcal{A}_{i}\in SL(2,C)$, $G_{1}\in SO(4,C)$ and Eq. (5) reduces to

[TABLE]

Thus, Eq. (10) leads to the following theorem.

Theorem 2. If the states $|\psi^{\prime}\rangle$ and $|\psi\rangle$ of $n$ qubits are SLOCC equivalent under $\mathcal{A}_{i}\in SL(2,C)$, then $S_{q_{1}q_{2}}^{(n)}(\psi^{\prime})$ is orthogonally similar to $S_{q_{1}q_{2}}^{(n)}(\psi)$. The similarity implies that $S_{q_{1}q_{2}}^{(n)}(\psi^{\prime})$ and $S_{q_{1}q_{2}}^{(n)}(\psi)$ have the same CP, spectrum, and SJNF up to the order of the standard Jordan blocks.

Example 4. $L_{ab_{3}}^{\ast}(a=0)$ is SLOCC equivalent to $L_{ab_{3}}(a=0)$ under $A_{i}\in SL(2,C)$ LDFPRA12 . The SJNFs of $S_{1,2}^{(4)}$ are both $0b^{2}J_{2}(0)$.

Restated in the contrapositive the theorem reads: If two matrices $S_{q_{1}q_{2}}^{(n)}$ associated with two n-qubit pure states differ in their CPs, spectrums, or SJNFs, then the two states are SLOCC inequivalent under $\mathcal{A}_{i}\in SL(2,C)$. From Example 2, by Theorem 2 the two states $\zeta_{4}$ and $\zeta_{5}$ are SLOCC inequivalent under $A_{i}\in SL(2;C)$ because SJNFs of $S_{1,2}^{(4)}(\zeta_{4})$ and $S_{1,2}^{(4)}(\zeta_{5})$ are different.

Note that a SLOCC equivalent class may include infinite SLOCC equivalent classes under $\mathcal{A}_{i}\in SL(2,C)$.

V Conclusion

In Theorem 1, we demonstrate that for two SLOCC equivalent states, the spectrums and SJNFs of the matrices $S_{q_{1}q_{2}}^{(n)}$ have proportional relationships. Invoking the proportional relationships, we partition pure states of $n$ ($\geq 4$) qubits into 12 groups and 34 families under SLOCC, respectively.

In Theorem 2, we deduce that for two equivalent states under determinant one SLOCC operations, the spectrums, CPs, SJNFs of $S_{q_{1}q_{2}}^{(n)}$ are invariant. The invariance can be used for SLOCC classification of $n$ qubits under determinant one SLOCC operations.

To make a comparison, we list the differences between Theorems 1 and 2 in Table 5.

It is known that SJNF is used to solve a system of linear differential equations. The  classification of SJNFs under SLOCC in this paper seems to be useful for classifying linear differential systems.

Acknowledgement—This work was supported by NSFC (Grant No. 10875061) and Tsinghua National Laboratory for Information Science and Technology.

VI Appendix Corresponding states of four qubits

Using $G_{abcd}$ we obtain the following 8 states.

$G_{1}=G_{abcd}(a=b=c=d\neq 0)$; (we will omit $G_{abcd}$ next);

$G_{2}:abcd\neq$ $0;b=c=d$ but $a\neq b$;

$G_{3}:abcd\neq$ [math], two of $a,b,c$, and $d$ are equal while the other two are not equal;

$G_{4}:abcd\neq 0$, $a,b$, $c$, and $d$ consists of two pairs of equal numbers;

$G_{5}:abcd\neq$ $0,$ $a,b$, $c$, and $d$ are distinct;

$G_{6}:$ only one of $a,b$, $c$, and $d$ is zero and other three are equal;

$G_{7}:$ only one of $a,b$, $c$, and $d$ is zero and only two of them are equal;

$G_{8}:$ only one of $a,b$, $c$, and $d$ is zero and the other three are distinct.

Using $L_{abc_{2}}$ we obtain the following 11 states.

$\zeta_{1}=L_{abc_{2}}($ $a=b=c\neq$ $0);$

$\zeta_{2}=L_{abc_{2}}(abc\neq 0$ and one of $a$ and $b$ equals c$)$;

$\zeta_{3}=L_{abc_{2}}(abc\neq 0$ and $a,b,c$ are distinct.$)$;

$\zeta_{4}=a(|0\rangle+|15\rangle)+b(|5\rangle+|10\rangle)+|6\rangle$, where $a\neq b$;

$\zeta_{6}=L_{abc_{2}}($ only one of $a$ and $b$ is zero while the other is equal to $c$.$)$;

$\zeta_{7}=L_{abc_{2}}(c\neq 0$ and only one of $a$ and $b$ is zero while the other is not equal to $c$.$)$;

$\zeta_{8}=L_{abc_{2}}(c=0$ and $a=b\neq 0)$; $\zeta_{9}=L_{abc_{2}}(c\neq 0$ and $a=b=0)$; $\zeta_{10}=L_{abc_{2}}(c=0$ while $ab\neq 0$ and $a\neq b)$; $\zeta_{11}=L_{abc_{2}}(c=0$ while only one of $a$ and $b$ is zero$)$; $\zeta_{12}=L_{abc_{2}}(a=b=c=0)$.

Using $L_{a_{2}b_{2}}$ we obtain the following two states.

$\eta_{1}=L_{a_{2}b_{2}}(a=b\neq 0)$; $\eta_{2}=L_{a_{2}b_{2}}(ab\neq 0$ and $a\neq b)$.

Let $L_{ab_{3}}^{\prime}=b(|0\rangle+|15\rangle)+\frac{b+a}{2}(|5\rangle+|10\rangle)+\frac{b-a}{2}(|6\rangle+|9\rangle)+\frac{i}{\sqrt{2}}(|1\rangle+|2\rangle-|7\rangle-|11\rangle)$. Using $L_{ab_{3}}^{\prime}$ we obtain the following five states.

$\theta_{1}=L_{ab_{3}}^{\prime}(a=b\neq 0)$;

$\theta_{2}=L_{ab_{3}}^{\prime}(ab\neq 0$ and $a\neq b)$;

$\theta_{3}=a(|0\rangle+|15\rangle)+\frac{a}{2}(|5\rangle+|10\rangle+|6\rangle+|9\rangle)+\frac{i}{\sqrt{2}}(|1\rangle+|2\rangle-|7\rangle-|11\rangle)$ (obtained from $L_{ab_{3}}^{\prime}(a=0$ but $b\neq 0)$);

$\theta_{4}=L_{ab_{3}}^{\prime}(b=0$ but $a\neq 0)$; $\theta_{5}=L_{ab_{3}}^{\prime}(a=b=0)$.

Using $L_{a_{4}}$ we obtain the following two states.

$\tau_{1}=L_{a_{4}}(a\neq 0)$; $\tau_{2}=L_{a_{4}}(a=0)$.

Using $L_{a_{2}0_{3\oplus 1}}$ we obtain the following one state.

$\kappa_{1}=L_{a_{2}0_{3\oplus 1}}(a\neq 0)$.

Let $L_{ab0_{3\oplus 1}}=\frac{a+b}{2}(|0\rangle+|15\rangle)+\frac{a-b}{2}(|3\rangle+|12\rangle)+|5\rangle+|6\rangle$. Using $L_{ab0_{3\oplus 1}}$ we obtain the following two states.

$\mu_{1}=L_{ab0_{3\oplus 1}}(ab\neq 0$ and $a\neq b)$; $\mu_{2}=L_{ab0_{3\oplus 1}}(a=b\neq 0)$ ;

Let $\xi=\frac{a}{2}(|0\rangle+|3\rangle+|12\rangle+|15\rangle)+i|1\rangle-i|13\rangle+|10\rangle$. Using $\xi$ we obtain the following two states.

$\xi_{1}=\xi(a\neq 0)$; $\xi_{2}=\xi(a=0)$.