Entanglement classification via integer partitions

Abstract

In Walter et al. (Science 340:1205, 2013), they gave a sufficient condition for genuinely entangled pure states and discussed SLOCC classification via polytopes and the eigenvalues of the single-particle states. In this paper, for 4n qubits, we show the invariance of algebraic multiplicities (AMs) and geometric multiplicities (GMs) of eigenvalues and the invariance of sizes of Jordan blocks (JBs) of the coefficient matrices under SLOCC. We explore properties of spectra, eigenvectors, generalized eigenvectors, standard Jordan normal forms (SJNFs), and Jordan chains of the coefficient matrices. The properties and invariance permit a reduction in SLOCC classification of 4n qubits to integer partitions (in number theory) of the number $2^{2n}-k$ and the AMs.

Appendices

Appendix A: A calculation of $Q_{i}Q_{i}^{t}$

We calculate $Q_{i}Q_{i}^{t}$, $i=1,2$, as follows. First we show that

$$\begin{aligned} U^{+}U^{*}=\upsilon ^{\otimes 2n}, \end{aligned}$$

(A1)

where $U^{*}$ is a complex conjugate of U. Equation (A1) holds from $T^{+}T^{*}=\upsilon \otimes \upsilon$ and $U^{+}U^{*}=T^{+}T^{*}\otimes \cdots \otimes T^{+}T^{*}$, where $\upsilon =\left( \begin{array}{cc} 0 &{} 1 \\ -1 &{} 0 \end{array} \right)$. Then, a calculation yields

$$\begin{aligned} Q_{1}Q_{1}^{t}=U\Delta _{1}U^{+}U^{*}\Delta _{1}^{t}U^{t}. \end{aligned}$$

(A2)

Via Eq. (A1),

$$\begin{aligned} Q_{1}Q_{1}^{t}=U\Delta _{1}\upsilon ^{\otimes 2n}\Delta _{1}^{t}U^{t}. \end{aligned}$$

(A3)

Using the definitions for U and $\Delta _{1}$, a straightforward calculation derives

$$\begin{aligned} Q_{1}Q_{1}^{t}= & {} T^{\otimes n}(\otimes _{i=1}^{2n}\mathcal {A} _{q_{i}})\upsilon ^{\otimes 2n}(\otimes _{i=1}^{2n}\mathcal {A} _{q_{i}}^{t})(T^{t})^{\otimes n} \nonumber \\= & {} [T(\mathcal {A}_{q_{1}}\otimes \mathcal {A}_{q_{2}})\upsilon ^{\otimes 2}( \mathcal {A}_{q_{1}}^{t}\otimes \mathcal {A}_{q_{2}}^{t})T^{t}]\otimes \cdots \otimes \nonumber \\&[T(A_{q_{2n-1}}\otimes A_{q_{2n}})\upsilon ^{\otimes 2}(A_{q_{2n-1}}^{t}\otimes A_{q_{2n}}^{t})T^{t}]. \end{aligned}$$

(A4)

Next, we reduce Eq. (A4). It is easy to test

$$\begin{aligned} \mathcal {A}_{i}\upsilon \mathcal {A}_{i}^{t}=(\det \mathcal {A}_{i})\upsilon \end{aligned}$$

(A5)

and

$$\begin{aligned}&(\mathcal {A}_{i}\otimes \mathcal {A}_{j})\upsilon ^{\otimes 2}(\mathcal {A} _{i}^{t}\otimes \mathcal {A}_{j}^{t}) \nonumber \\&\quad =\mathcal {A}_{i}\upsilon \mathcal {A}_{i}^{t}\otimes \mathcal {A} _{j}\upsilon \mathcal {A}_{j}^{t} \end{aligned}$$

(A6)

$$\begin{aligned}&\quad =(\det \mathcal {A}_{i})\upsilon \otimes (\det \mathcal {A}_{j})\upsilon \end{aligned}$$

(A7)

$$\begin{aligned}&\quad =(\det \mathcal {A}_{i})(\det \mathcal {A}_{j})\upsilon ^{\otimes 2}. \end{aligned}$$

(A8)

Thus, via Eqs. (A8), (A4) reduces to

$$\begin{aligned}&Q_{1}Q_{1}^{t} \nonumber \\&\quad =[T(\det \mathcal {A}_{q_{1}}\det \mathcal {A}_{q_{2}})\upsilon ^{\otimes 2}T^{t}]\otimes \cdots \end{aligned}$$

(A9)

$$\begin{aligned}&\qquad \otimes [T(\det A_{q_{2n-1}}\det A_{q_{2n}})\upsilon ^{\otimes 2}T^{t}] \nonumber \\&\quad =[(\det \mathcal {A}_{q_{1}}\det \mathcal {A}_{q_{2}})T\upsilon ^{\otimes 2}T^{t}]\otimes \cdots \end{aligned}$$

(A10)

$$\begin{aligned}&\qquad \otimes [(\det A_{q_{2n-1}}\det A_{q_{2n}})T\upsilon ^{\otimes 2}T^{t}] \nonumber \\&\quad =(\Pi _{i=1}^{2n}\det A_{q_{i}})(T\upsilon ^{\otimes 2}T^{t}\otimes \cdots \otimes T\upsilon ^{\otimes 2}T^{t}). \end{aligned}$$

(A11)

One can check that

$$\begin{aligned} T\upsilon ^{\otimes 2}T^{t}=I_{4}. \end{aligned}$$

(A12)

Thus, from Eqs. (A11, A12), we obtain

$$\begin{aligned} Q_{1}Q_{1}^{t}=(\Pi _{i=1}^{2n}\det A_{q_{i}})I_{2^{2n}}. \end{aligned}$$

(A13)

A calculation also yields

$$\begin{aligned} Q_{1}Q_{1}^{t}=Q_{1}^{t}Q_{1}. \end{aligned}$$

(A14)

Similarly,

$$\begin{aligned} Q_{2}Q_{2}^{t}=Q_{2}^{t}Q_{2}=(\Pi _{i=1}^{2n}\det A_{q_{2n+i}})I_{2^{2n}}. \end{aligned}$$

(A15)

Appendix B: Proportional relations

Let

$$\begin{aligned} M_{2n}= & {} \left( \begin{array}{cc} 0 &{} m \\ m^{t} &{} 0 \end{array} \right) , \end{aligned}$$

(B1)

$$\begin{aligned} D_{2n}= & {} \left( \begin{array}{cc} 0 &{} hm \\ gm^{t} &{} 0 \end{array} \right) , \end{aligned}$$

(B2)

where m is an n by n matrix and g and h are nonzero complex numbers.

Property (1). $\det (\lambda I-D_{2n})=\det (\lambda ^{2}I-ghmm^{t})$. Let $\lambda ^{2}=\sigma$. Then, we obtain $\det (\sigma I-ghmm^{t})$. Let $a^{2}$ be an eigenvalue of $mm^{t}$. Then, $gha^{2}$ is an eigenvalue of $ghmm^{t}$, $\pm a$ are eigenvalues of $M_{2n}$, and $\pm \sqrt{gh}a$ are eigenvalues of $D_{2n}$. Therefore, spectra of $D_{2n}$ and $M_{2n}$ are proportional.

Property (2.1). Let

$$\begin{aligned} V=\left( \begin{array}{c} v^{\prime } \\ v^{\prime \prime } \end{array} \right) , \end{aligned}$$

(B3)

where $v^{\prime }$ and $v^{\prime \prime }$ are $n\times 1$ vectors, be an eigenvector of $M_{2n}$ corresponding to the eigenvalue $\lambda \ne 0$. Then,

$$\begin{aligned} M_{2n}V=\lambda V. \end{aligned}$$

(B4)

Let

$$\begin{aligned} W=\left( \begin{array}{c} \sqrt{h/g}v^{\prime } \\ v^{\prime \prime } \end{array} \right) . \end{aligned}$$

(B5)

Then, via Eq. (B4) one can check that

$$\begin{aligned} D_{2n}W=\sqrt{gh}\lambda W. \end{aligned}$$

(B6)

It means that W is an eigenvector of $D_{2n}$ corresponding to the eigenvalue $\sqrt{gh}\lambda$.

One can see that if there are s linearly independent eigenvectors $\left( \begin{array}{c} v_{i}^{\prime } \\ v_{i}^{\prime \prime } \end{array} \right)$, $i=1,2,\ldots ,s$, corresponding to the eigenvalue $\lambda$ of $M_{2n}$, then there are s linearly independent eigenvectors $\left( \begin{array}{c} \sqrt{h/g}v_{i}^{\prime } \\ v_{i}^{\prime \prime } \end{array} \right)$, $i=1,2,\ldots ,s$, corresponding to the eigenvalue $\sqrt{gh} \lambda$ of $D_{2n}$. Therefore, the eigenvalue $\lambda$ of $M_{2n}$ and the eigenvalue$\sqrt{gh}\lambda$ of $D_{2n}$ possess the same geometry multiplicity. It implies that $M_{2n}$ and $D_{2n}$ have the same number of JBs corresponding to the eigenvalues $\lambda$ and $\sqrt{gh}\lambda$.

Property (2.2). Let V be an eigenvector of $M_{2n}$ corresponding to the zero eigenvalue. Then, one can check that V is also an eigenvector of $D_{2n}$ corresponding to the zero eigenvalue. It means that the zero eigenvalue of $M_{2n}$ and the zero eigenvalue of $D_{2n}$ possess the same eigenspace and of course the same geometry multiplicity. Thus, $M_{2n}$ and $D_{2n}$ have the same number of JBs corresponding to the zero eigenvalue.

Property (3). $M_{2n}$ has a JB with the size of r corresponding to the eigenvalue $\lambda$ if and only if $D_{2n}$ has a JB with the size of r corresponding to the eigenvalue $\sqrt{gh}\lambda$. The property is also true when $\lambda =0$.

Suppose that $M_{2n}$ has a JB with the size of r corresponding to the eigenvalue $\lambda$ ($\lambda$ may be zero). Then, there exists a Jordan chain with the size of r corresponding to the eigenvalue $\lambda$ [37]. Let the Jordan chain be

$$\begin{aligned} v_{i}=\left( \begin{array}{c} v_{i}^{\prime } \\ v_{i}^{\prime \prime } \end{array} \right) ,i=1,2,\ldots ,r, \end{aligned}$$

(B7)

where $v_{i}^{\prime }$ and $v_{i}^{\prime \prime }$ are $n\times 1$ vectors, $v_{1}$ is the eigenvector, and $v_{i}$ satisfy

$$\begin{aligned} \left( M_{2n}-\lambda I_{2n}\right) v_{i}=v_{i-1},i=2,\ldots ,r. \end{aligned}$$

(B8)

From the Jordan chain, we construct the following chain:

$$\begin{aligned} z_{1}= & {} \left( \begin{array}{c} \sqrt{h/g}v_{1}^{\prime } \\ v_{1}^{\prime \prime } \end{array} \right) , \end{aligned}$$

(B9)

$$\begin{aligned} z_{i}= & {} \left( \begin{array}{c} \frac{1}{g(\sqrt{gh})^{i-2}}v_{i}^{\prime } \\ \frac{1}{(\sqrt{gh})^{i-1}}v_{i}^{\prime \prime } \end{array} \right) ,i=2,\ldots ,r. \end{aligned}$$

(B10)

One can test that $z_{1}$ is an eigenvector of $D_{2n}$ corresponding to the eigenvalue $\sqrt{gh}\lambda$. A calculation yields that

$$\begin{aligned} D_{2n}z_{i}= & {} \left( \begin{array}{cc} 0 &{} hm \\ gm^{t} &{} 0 \end{array} \right) \left( \begin{array}{c} \frac{1}{g(\sqrt{gh})^{i-2}}v_{i}^{\prime } \\ \frac{1}{(\sqrt{gh})^{i-1}}v_{i}^{\prime \prime } \end{array} \right) \nonumber \\= & {} \left( \begin{array}{c} \frac{h}{(\sqrt{gh})^{i-1}}mv_{i}^{\prime \prime } \\ \frac{1}{(\sqrt{gh})^{i-2}}m^{t}v_{i}^{\prime } \end{array} \right) \end{aligned}$$

(B11)

$$\begin{aligned}= & {} \left( \begin{array}{c} \frac{gh}{g(\sqrt{gh})^{i-1}}mv_{i}^{\prime \prime } \\ \frac{1}{(\sqrt{gh})^{i-2}}m^{t}v_{i}^{\prime } \end{array} \right) \end{aligned}$$

(B12)

$$\begin{aligned}= & {} \left( \begin{array}{c} \frac{1}{g(\sqrt{gh})^{i-3}}mv_{i}^{\prime \prime } \\ \frac{1}{(\sqrt{gh})^{i-2}}m^{t}v_{i}^{\prime } \end{array} \right) , \end{aligned}$$

(B13)

$$\begin{aligned} \sqrt{gh}\lambda I_{2n}z_{i}= & {} \sqrt{gh}\lambda \left( \begin{array}{c} \frac{1}{g(\sqrt{gh})^{i-2}}v_{i}^{\prime } \\ \frac{1}{(\sqrt{gh})^{i-1}}v_{i}^{\prime \prime } \end{array} \right) \nonumber \\= & {} \left( \begin{array}{c} \frac{1}{g(\sqrt{gh})^{i-3}}\lambda v_{i}^{\prime } \\ \frac{1}{(\sqrt{gh})^{i-2}}\lambda v_{i}^{\prime \prime } \end{array} \right) , \end{aligned}$$

(B14)

and

$$\begin{aligned} z_{i-1}=\left( \begin{array}{c} \frac{1}{g(\sqrt{gh})^{i-3}}v_{i-1}^{\prime } \\ \frac{1}{(\sqrt{gh})^{i-2}}v_{i-1}^{\prime \prime } \end{array} \right) . \end{aligned}$$

(B15)

From Eqs. (B8, B13, B14, B15), we can show that

$$\begin{aligned} \left( D_{2n}-\sqrt{gh}\lambda I_{2n}\right) z_{i}=z_{i-1},i=2,\ldots ,r. \end{aligned}$$

(B16)

Thus, we obtain a Jordan chain $z_{1}$, $\ldots$, $z_{r}$ corresponding to the eigenvalue $\sqrt{gh}\lambda$ of $D_{2n}$. It means that the two Jordan chains have the same size. Note that the Jordan chain $z_{1}$, $\ldots$, $z_{r}$ corresponds to the JB of size r corresponding to the eigenvalue $\sqrt{gh}\lambda$ of $D_{2n}$. Conversely, it is also true.

Appendix C: Properties of the matrix $M_{2n}$

Let

$$\begin{aligned} M_{2n}=\left( \begin{array}{cc} 0 &{} m \\ m^{t} &{} 0 \end{array} \right) , \end{aligned}$$

(C1)

where m is an n by n matrix. We calculate the characteristic polynomial of $M_{2n}$ below.

$$\begin{aligned}&\det (\lambda I_{2n}-M_{2n}) \nonumber \\&\quad =\det \left( \lambda ^{2}I_{n}-mm^{t}\right) =\det \left( \lambda ^{2}I_{n}-m^{t}m\right) . \end{aligned}$$

(C2)

Equation (C2) leads to Property 1.

Property 1

• 1.1. $\lambda$ is an eigenvalue of $M_{2n}$ if and only if $\lambda ^{2}$ is an eigenvalue of $m^{t}m$ and $mm^{t}$, respectively. Thus, the nonzero eigenvalues of $M_{2n}$ are $\pm \lambda _{i}$, $i=1,2,$ ....

• 1.2. The AM of the zero eigenvalue of $M_{2n}$ is even.

Property 2

If V in Eq. (B3) is an eigenvector of $M_{2n}$ corresponding to the zero eigenvalue, then $V_{1}=\left( \begin{array}{c} v^{\prime } \\ 0 \end{array} \right)$ (if $v^{\prime }\ne 0$) and $V_{2}=\left( \begin{array}{c} 0 \\ v^{\prime \prime } \end{array} \right)$ (if $v^{\prime \prime }\ne 0$) are also eigenvectors of $M_{2n}$ corresponding to the zero eigenvalue. Clearly, V is a linear combination of $V_{1}$ and $V_{2}$, i.e. $V=V_{1}+V_{2}$.

Proof

From that $M_{2n}V=0$, we obtain

$$\begin{aligned} mv^{\prime \prime }= & {} 0, \end{aligned}$$

(C3)

$$\begin{aligned} m^{t}v^{\prime }= & {} 0. \end{aligned}$$

(C4)

It is easy to verify that $V_{1}=\left( \begin{array}{c} v^{\prime } \\ 0 \end{array} \right)$ (if $v^{\prime }\ne 0$) and $V_{2}=\left( \begin{array}{c} 0 \\ v^{\prime \prime } \end{array} \right)$ (if $v^{\prime \prime }\ne 0$) are also eigenvectors of $M_{2n}$ corresponding to the zero eigenvalue. $\square$

Property 3

The GM of the zero eigenvalue of $M_{2n}$ is $2(n-rk(m))$, where rk stands for “rank”. Thus, there are $2(n-rk(m))$ JBs corresponding to the zero eigenvalue of $M_{2n}$.

Proof

From the linear algebra, it is easy to see that Property 3 holds. We want to prove it differently next. From [37], we know that the generalized eigenvector of rank 1 is just an eigenvector. For $M_{2n}$, let $\chi _{1}$ be the number of linear independent generalized eigenvectors of rank 1 corresponding to the zero eigenvalue. Then, from [37]

$$\begin{aligned} \chi _{1}=2n-rk(M_{2n}). \end{aligned}$$

(C5)

It is easy to see that $rk(M_{2n})=2*rk(m)$. $\square$

Property 4

A basis of the zero eigenspace of $M_{2n}$ can be obtained via the bases of the zero eigenspaces of m and $m^{t}$ as follows. Let $v_{1}^{\prime }$, $v_{2}^{\prime }$, $\ldots$, $v_{n-rk(m)}^{\prime }$ be all the linearly independent eigenvectors of $m^{t}$ corresponding to the zero eigenvalue and $v_{1}^{\prime \prime }$, $v_{2}^{\prime \prime }$, $\ldots$, $v_{n-rk(R)}^{\prime \prime }$ be all the linearly independent eigenvectors of m corresponding to the zero eigenvalue. Then,

$$\begin{aligned}&\left\{ \left( \begin{array}{c} v_{1}^{\prime } \\ 0 \end{array} \right) ,\cdots ,\left( \begin{array}{c} v_{n-rk(m)}^{\prime } \\ 0 \end{array} \right) , \right. \nonumber \\&\quad \left. \left( \begin{array}{c} 0 \\ v_{1}^{\prime \prime } \end{array} \right) ,\cdots ,\left( \begin{array}{c} 0 \\ v_{n-rk(m)}^{\prime \prime } \end{array} \right) \right\} \end{aligned}$$

(C6)

is a basis of the zero eigenspace of $M_{2n}$.

Proof

Let V in Eq. (B3) be an eigenvector of $M_{2n}$ corresponding to the zero eigenvalue. Then, by Eqs. (C3, C4), $v^{\prime \prime }$ is an eigenvector of m corresponding to the zero eigenvalue if $v^{\prime \prime }\ne 0$ and $v^{\prime }$ is an eigenvector of $m^{t}$ corresponding to the zero eigenvalue if $v^{\prime }\ne 0$. Conversely, if $v^{\prime }$ (resp. $v^{\prime \prime }$) is an eigenvector of $m^{t}$ (resp. m) corresponding to the zero eigenvalue, then $\left( \begin{array}{c} v^{\prime } \\ 0 \end{array} \right)$ (resp. $\left( \begin{array}{c} 0 \\ v^{\prime \prime } \end{array} \right)$) is an eigenvector of $M_{2n}$ corresponding to the zero eigenvalue. From Eqs. (C3, C4), we know that m and $m^{t}$ have $n-rk(m)$ linearly independent eigenvectors corresponding to the zero eigenvalue, respectively. Thus, Property 4 holds and we have Property 3 again. $\square$

Property 5.1

For $M_{2n}$, let $\chi _{\ell }$ be the number of linear independent generalized eigenvectors of rank $\ell$ corresponding to the zero eigenvalue [37]. Then, $\chi _{2k}+\chi _{2k+1}$, where $k\ge 1$, must be even.

Proof

From [37],

$$\begin{aligned} \chi _{2k}=rk(M_{2n}^{2k-1})-rk(M_{2n}^{2k}) \end{aligned}$$

(C7)

and

$$\begin{aligned} \chi _{2k+1}=rk(M_{2n}^{2k})-rk(M_{2n}^{2k+1}), \end{aligned}$$

(C8)

where $k\ge 1$. Then,

$$\begin{aligned} \chi _{2k}+\chi _{2k+1}=rk(M_{2n}^{2k-1})-rk(M_{2n}^{2k+1}). \end{aligned}$$

(C9)

Let us compute $M_{2n}^{2k+1}$ next.

$$\begin{aligned} M_{2n}^{3}= & {} \left( \begin{array}{cc} &{} mm^{t}m \\ m^{t}mm^{t} &{} \end{array} \right) , \end{aligned}$$

(C10)

$$\begin{aligned} M_{2n}^{2k+1}= & {} \left( \begin{array}{cc} &{} M^{\prime } \\ (M^{\prime })^{t} &{} \end{array} \right) , \end{aligned}$$

(C11)

where $M^{\prime }=mm^{t}mm^{t}\cdots mm^{t}m$.

It is easy to check that $rk(M_{2n}^{2k+1})=2*rk(M^{\prime })$. Similarly, the number $rk(M_{2n}^{2k-1})$ is even. Therefore, $\chi _{2k}+\chi _{2k+1}$ is even. Specially, $\chi _{2}+\chi _{3}$ is even. $\square$

Property 5.2

The number of the occurrences of JBs with the same odd size corresponding to the zero eigenvalue of $M_{2n}$ may be even or odd.

Proof

For the JB $J_{2k+1}(0)$ corresponding to the eigenvector $x_{1}$, there is a Jordan chain $x_{1}$, $x_{2}$, $\ldots$, $x_{2k+1}$ corresponding to the zero eigenvalue, where $x_{i}$ is the generalized eigenvector of rank i of $M_{2n}$. Clearly, $x_{2j}$ is the generalized eigenvector of rank 2j and $x_{2j+1}$ is the one with rank $2j+1$, where $j=1,\ldots ,k$. Thus, the chain adds 1 to $\chi _{2j}$ and 1 to $\chi _{2j+1}$, respectively, $j=1,\ldots ,k$. That is, the chain adds 2 to the number $\chi _{2j}+\chi _{2j+1}$, $j=1,\ldots ,k$. One can know that any number of occurrences of JBs with the same odd size corresponding to the zero eigenvalue will not change the parity of $\chi _{2k}+\chi _{2k+1}$. Therefore, Property 5.2 holds. $\square$

Property 5.3

The number of the occurrences of the JBs with the same even size corresponding to the zero eigenvalue must be even.

Proof

For the JB $J_{2k}(0)$ with $k\ge 1$ corresponding to the eigenvector $y_{1}$, there is a Jordan chain $y_{1}$, $y_{2}$, $\ldots$, $y_{2k}$ corresponding to the zero eigenvalue, where $y_{i}$ is the generalized eigenvector of rank i of $M_{2n}$. Thus, $y_{2j}$ is the generalized eigenvector of rank 2j, while $y_{2j+1}$ is the one with rank $2j+1$, where $j=1,\ldots ,k-1$. Thus, the chain adds 1 to $\chi _{2j}$ and 1 to $\chi _{2j+1}$, respectively, $j=1,\ldots ,k-1$. $\square$

Clearly, $y_{2k}$ is the generalized eigenvector of rank 2k. Thus, it adds 1 to $\chi _{2k}$. But the chain does not include the generalized eigenvector of rank $2k+1$. Thus, it adds 0 to $\chi _{2k+1}$. It means that the chain will change the parity of $\chi _{2k}+\chi _{2k+1}$.

Accordingly, for the $2l+1$ occurrences of the JB $J_{2k}(0)$ with $k\ge 1$ , the corresponding $2l+1$ Jordan chains include $2l+1$ generalized eigenvectors of the same rank 2k, but the chains do not have any generalized eigenvector of rank $2k+1$. Thus, in light of Property 5.2, the number $\chi _{2k}+\chi _{2k+1}$ will be an odd number. It does not satisfy Property 5.1.

For the 2l occurrences of the JB $J_{2k}(0)$ with $k\ge 1$, the corresponding 2l Jordan chains include 2l generalized eigenvectors of the same rank 2k, but the chains do not have any generalized eigenvector of rank $2k+1$. Thus, in light of Property 5.2, the number $\chi _{2k}+\chi _{2k+1}$ will be an even number.

One can see that $\chi _{2k}+\chi _{2k+1}$ is even permits that the size of a JB with the zero eigenvalue is odd or even.

For example, a calculation shows that for four qubits, $\Phi _{8}$ has the SJNFs $J_{4}(0)^{\oplus 2}$, $J_{2}(0)^{\oplus 2}\oplus J_{1}(0)^{\oplus 4}$ , $J_{2}(0)^{\oplus 2}\oplus J_{3}(0)\oplus J_{1}(0)$ and $J_{2}(0)^{\oplus 4}$ for the states $L_{a_{4}}(a=0)$, $L_{abc_{2}}(a=b=c=0)$, $L_{a_{2}0_{3\oplus 1}}(a=0)$ and $L_{a_{2}b_{2}}(a=b=0)$, respectively. In detail, $J_{4}(0)$ occurs twice, and $J_{2}(0)$ occurs twice, twice, and for four times in the above SJNFs. See Table 2. For these SJNFs, $\chi _{2k}+\chi _{2k+1}$ is even.

One can know that $\Phi _{8}$ does not have SJNFs $\pm \lambda J_{2}(0)J_{4}(0)$, $J_{2}(0)J_{6}(0)$ or $00J_{2}(0)J_{4}(0)$ because for these SJNFs $\chi _{2}+\chi _{3}$ is odd. Note that $J_{2}(0)$, $J_{4}(0)$, and $J_{6}(0)$ occur once in the above different SJNFs.

Property 6

Let V in Eq. (B3) be an eigenvector of $M_{2n}$ corresponding to the nonzero eigenvalue $\lambda$. Then, $v^{\prime }\ne 0$ and $v^{\prime \prime }\ne 0$.

Proof

From the equation $\left( M_{2n}-\lambda I_{2n}\right) V=0$, we obtain

$$\begin{aligned} mv^{\prime \prime }= & {} \lambda v^{\prime }\text {, } \end{aligned}$$

(C12)

$$\begin{aligned} m^{t}v^{\prime }= & {} \lambda v^{\prime \prime }\text {.} \end{aligned}$$

(C13)

Then from Eqs. (C12, C13), it is easy to show that $v^{\prime }\ne 0$ and $v^{\prime \prime }\ne 0$. In other words, the vectors of the forms $\left( \begin{array}{c} v^{\prime } \\ 0 \end{array} \right)$ or $\left( \begin{array}{c} 0 \\ v^{\prime \prime } \end{array} \right)$ are not eigenvectors of $M_{2n}$ corresponding to nonzero eigenvalues.

Property 7

The GMs of the nonzero eigenvalues $\pm \lambda$ of $M_{2n}$ both are $n-rk(m^{t}m-\lambda ^{2}I_{n})$. Thus, there are $n-rk(m^{t}m-\lambda ^{2}I_{n})$ JBs corresponding to the nonzero eigenvalues $\pm \lambda$ of $M_{2n}$, respectively.

Proof

Let $\chi _{1}(\lambda )$ (resp. $\chi _{1}(-\lambda )$) be the number of linear independent generalized eigenvectors of rank 1 corresponding to the nonzero eigenvalue $\lambda$ (resp. $-\lambda$). One can know that $\chi _{1}(\lambda )$ (resp. $\chi _{1}(-\lambda )$) is just the GMs of the nonzero eigenvalues $\lambda$ (resp. $-\lambda$) of $M_{2n}$. Then, from [37]

$$\begin{aligned} \chi _{1}(\lambda )= & {} 2n-rk(M_{2n}-\lambda I_{2n}) \nonumber \\= & {} 2n-rk\left( \begin{array}{cc} -\lambda I_{n} &{} m \\ m^{t} &{} -\lambda I_{n} \end{array} \right) . \end{aligned}$$

(C14)

$$\begin{aligned} \chi _{1}(-\lambda )= & {} 2n-rk(M_{2n}+\lambda I_{2n}) \nonumber \\= & {} 2n-rk\left( \begin{array}{cc} \lambda I_{n} &{} m \\ m^{t} &{} \lambda I_{n} \end{array} \right) . \end{aligned}$$

(C15)

To calculate the ranks of the matrices $\left( \begin{array}{cc} -\lambda I_{n} &{} m \\ m^{t} &{} -\lambda I_{n} \end{array} \right)$ and $\left( \begin{array}{cc} \lambda I_{n} &{} m \\ m^{t} &{} \lambda I_{n} \end{array} \right)$, we do the following operations:

$$\begin{aligned}&\left( \begin{array}{cc} I_{n} &{} 0 \\ \frac{1}{\lambda }m^{t} &{} I_{n} \end{array} \right) \left( \begin{array}{cc} -\lambda I_{n} &{} m \\ m^{t} &{} -\lambda I_{n} \end{array} \right) \nonumber \\&\quad =\left( \begin{array}{cc} -\lambda I_{n} &{} m \\ 0 &{} \frac{1}{\lambda }m^{t}m-\lambda I_{n} \end{array} \right) \end{aligned}$$

(C16)

and

$$\begin{aligned}&\left( \begin{array}{cc} -I_{n} &{} 0 \\ \frac{1}{\lambda }m^{t} &{} -I_{n} \end{array} \right) \left( \begin{array}{cc} \lambda I_{n} &{} m \\ m^{t} &{} \lambda I_{n} \end{array} \right) \nonumber \\&\quad =\left( \begin{array}{cc} -\lambda I_{n} &{} -m \\ 0 &{} \frac{1}{\lambda }m^{t}m-\lambda I_{n} \end{array} \right) . \end{aligned}$$

(C17)

From the linear algebra, since $\left( \begin{array}{cc} I_{n} &{} 0 \\ \frac{1}{\lambda }m^{t} &{} I_{n} \end{array} \right)$ is full rank, via Eq. (C16) we obtain

$$\begin{aligned} rk\left( \begin{array}{cc} -\lambda I_{n} &{} m \\ m^{t} &{} -\lambda I_{n} \end{array} \right)= & {} rk\text { }\left( \begin{array}{cc} -\lambda I_{n} &{} m \\ 0 &{} \frac{1}{\lambda }m^{t}m-\lambda I_{n} \end{array} \right) \nonumber \\= & {} n+rk\left( \frac{1}{\lambda }m^{t}m-\lambda I_{n}\right) \nonumber \\= & {} n+rk\left( \frac{1}{\lambda }(m^{t}m-\lambda ^{2}I_{n})\right) \nonumber \\= & {} n+rk(m^{t}m-\lambda ^{2}I_{n}). \end{aligned}$$

(C18)

From the linear algebra, since $\left( \begin{array}{cc} -I_{n} &{} 0 \\ \frac{1}{\lambda }m^{t} &{} -I_{n} \end{array} \right)$ is full rank, via Eq. (C17) we obtain

$$\begin{aligned} rk\left( \begin{array}{cc} \lambda I_{n} &{} m \\ m^{t} &{} \lambda I_{n} \end{array} \right)= & {} rk\left( \begin{array}{cc} -\lambda I_{n} &{} -m \\ 0 &{} \frac{1}{\lambda }m^{t}m-\lambda I_{n} \end{array} \right) \nonumber \\= & {} n+rk\left( \frac{1}{\lambda }m^{t}m-\lambda I_{n}\right) \nonumber \\= & {} n+rk(m^{t}m-\lambda ^{2}I_{n}). \end{aligned}$$

(C19)

From Eqs. (C14, C18), $\chi _{1}(\lambda )=2n-[n+rk(m^{t}m-\lambda ^{2}I_{n})]=n-rk(m^{t}m-\lambda ^{2}I_{n})$. Clearly, $m^{t}m-\lambda ^{2}I_{n}$ is a characteristic matrix of $m^{t}m$ in $\lambda ^{2}$. From Eqs. (C15, C19), $\chi _{1}(-\lambda )=n-rk(m^{t}m-\lambda ^{2}I_{n})$. Therefore, $\chi _{1}(\lambda )=\chi _{1}(-\lambda )$ and then Property 7 holds.

By Property 1.1, when $\lambda$ is an eigenvalue of $M_{2n}$, then $\lambda ^{2}$ is an eigenvalue of $m^{t}m$. It is well known that roots of the equation $\det (m^{t}m-\lambda ^{2}I_{n})=0$ are eigenvalues of $m^{t}m$. Thus, $0\le rk(m^{t}m-\lambda ^{2}I_{n})<n$, and then, $0<\chi _{1}(\lambda )\le n$. When $\lambda ^{2}$ is not an eigenvalue of $m^{t}m$, i.e. $\lambda$ is not an eigenvalue of $M_{2n}$, then $\det (m^{t}m-\lambda ^{2}I_{n})\ne 0$, i.e. $rk(m^{t}m-\lambda ^{2}I_{n})=n$. Thus, $\chi _{1}(\lambda )=0$. $\square$

Property 8

The Jordan chain with the nonzero eigenvalue $\lambda$ corresponding to the eigenvector $\left( \begin{array}{c} v_{1}^{\prime } \\ v_{1}^{\prime \prime } \end{array} \right)$ and the Jordan chain with the nonzero eigenvalue $-\lambda$ corresponding to the eigenvector $\left( \begin{array}{c} -v_{1}^{\prime } \\ v_{1}^{\prime \prime } \end{array} \right)$ have the same size. Thus, their corresponding JBs have the same size.

Proof

Let

$$\begin{aligned} v_{i}=\left( \begin{array}{c} v_{i}^{\prime } \\ v_{i}^{\prime \prime } \end{array} \right) ,i=1,2,\ldots ,r, \end{aligned}$$

(C20)

where $v_{i}^{\prime }$ and $v_{i}^{\prime \prime }$ are $n\times 1$ vectors, be a Jordan chain with the nonzero eigenvalue $\lambda$ corresponding to the eigenvector $v_{1}=\left( \begin{array}{c} v_{1}^{\prime } \\ v_{1}^{\prime \prime } \end{array} \right)$. By Property 6, $v_{1}^{\prime }\ne 0$ and $v_{1}^{\prime \prime }\ne 0$. Then, by the definition of Jordan chain [37],

$$\begin{aligned} (M_{2n}-\lambda I_{2n})v_{1}=0 \end{aligned}$$

(C21)

and

$$\begin{aligned} (M_{2n}-\lambda I_{2n})v_{k}=v_{k-1},k\ge 2. \end{aligned}$$

(C22)

Let

$$\begin{aligned} \omega _{1}= & {} \left( \begin{array}{c} -v_{1}^{\prime } \\ v_{1}^{\prime \prime } \end{array} \right) , \end{aligned}$$

(C23)

$$\begin{aligned} \omega _{2}= & {} \left( \begin{array}{c} v_{2}^{\prime } \\ -v_{2}^{\prime \prime } \end{array} \right) , \end{aligned}$$

(C24)

$$\begin{aligned}&\vdots \nonumber \\ \omega _{l}= & {} (-1)^{l+1}\left( \begin{array}{c} -v_{l}^{\prime } \\ v_{l}^{\prime \prime } \end{array} \right) ,l\ge 2. \end{aligned}$$

(C25)

It is easy to check that

$$\begin{aligned} (M_{2n}+\lambda I_{2n})\omega _{1}=0 \end{aligned}$$

(C26)

and

$$\begin{aligned} (M_{2n}+\lambda I_{2n})\omega _{k}=\omega _{k-1},k\ge 2. \end{aligned}$$

(C27)

Here, $\omega _{1}$ is an eigenvector of $M_{2n}$ corresponding to $-\lambda$. Let s be the size of the Jordan chain with the nonzero eigenvalue $-\lambda$ corresponding to the eigenvector $\omega _{1}$. Clearly, $s\ge r$ . Conversely, similarly, we can show that $r\ge s$. Thus, $s=r$. $\square$

Appendix D: The number of different lists of partitions of AMs

We define a product of sets $\overline{L}$ and $\overline{M}$ as $\overline{L }\times \overline{M}=[\{l,m\}|l\in$ $\overline{l}$ and $m\in \overline{M}]$, and we define that $\{l,m\}$ is an unordered list of partitions. Thus, $\{l,m\}=\{m,l\}$. By the definition, $\overline{2}\times \overline{2} =[\{(2),(2)\}$, $\{(2),(1,1)\}$, $\{(1,1),(1,1)\}]$. Note that $\{(1,1),(2)\}=\{(2),(1,1)\}$.

From Eq. (30), let

$$\begin{aligned} \Gamma =\widetilde{2k}\times \overline{\ell _{1}}\times \overline{\ell _{2}} \cdots \times \overline{\ell _{s}}. \end{aligned}$$

(D1)

From the above discussion, we consider that $\overline{\ell _{1}}\times \overline{\ell _{2}}\cdots \times \overline{\ell _{s}}$ is an unordered list of partitions. Note that some $\ell _{i}$ in a set of AMs $\{2k;\ell _{1},\ell _{2},\ldots ,\ell _{s}\}$ from Eq. (26) may occur twice or more. For example, $\Phi _{8}$ has the spectrum $\{(\pm \lambda _{1}{})^{\circledcirc 2},(\pm \lambda _{2}{})^{\circledcirc 2}\}$ and the set of the AM is $\{0;2,2\}$.

First, let us compute how many different lists of partitions there are from the product set $\underbrace{\overline{l}\times \cdots \times \overline{l}} _{j}$. We consider distributing j indistinguishable balls into P(l) distinguishable boxes. Let $\rho (l,j)=\left( \begin{array}{c} j+P(l)-1 \\ j \end{array} \right)$. Thus, there are $\rho (l,j)$ distributing ways without exclusion [36]. Via the probability model, $\underbrace{\overline{l}\times \cdots \times \overline{l}}_{j}$ has $\rho (l,j)$ different lists of partitions. Specially, $\overline{2}\times \overline{2}$ has $\rho (2,2)$ ($=3$) different lists of partitions.

It is easy to check that $\overline{2}\times \overline{3}$ has $P(2)P(3)=6$ different lists of partitions. When l, k, and m are distinct from each other, $\overline{l}\times \overline{k}\times \overline{m}$ has P(l)P(k)P(m) different lists of partitions.

Let us compute how many different lists of partitions there are from the product set $\Gamma$ in Eq. (D1) for all k. For the sake of clarity, we rewrite $\Gamma$ in Eq. (D1) as follows:

$$\begin{aligned} \Gamma =\widetilde{2k}\times \overline{\varkappa _{1}}\times \cdots \times \overline{\varkappa _{i}}\times \underbrace{\overline{\theta }\times \cdots \times \overline{\theta }}_{j}\times \cdots \times \underbrace{\overline{ \varsigma }\times \cdots \times \overline{\varsigma }}_{m}, \end{aligned}$$

(D2)

where $\varkappa _{1}$, $\ldots$, $\varkappa _{i}$, $\theta$, $\ldots$, and $\varsigma$ are different from each other. From Eq. (D2), for all k we obtain

$$\begin{aligned} \eta =\sum _{k=0}^{2^{2n}}P^{*}(2k)\sum _{\varpi }P(\varkappa _{1})\ldots P(\varkappa _{i})\rho (\theta ,j)\ldots \rho (\varsigma ,m) \end{aligned}$$

(D3)

different lists of partitions, where $\varpi =\{\varkappa _{1},\ldots ,\varkappa _{i},\underbrace{\theta ,\ldots ,\theta }_{j},\ldots ,\underbrace{ \varsigma ,\ldots ,\varsigma }_{m}\}$, which is a partition of $2^{2n}-k$, and the second sum is evaluated over all the partitions of $2^{2n}-k$.

To compute $P^{*}(2^{2n+1})$, from $\widetilde{2^{2n+1}}$ we should remove the partition $(\underbrace{1,\ldots ,1}_{2^{2n+1}})$, which means that the SJNF of the $\Phi _{2^{2n+1}}(|\psi \rangle )$ in Eq. (10 ) is the zero matrix, and then, $\Phi _{2^{2n+1}}(|\psi \rangle )$ in Eq. ( 10) is the zero matrix. Therefore, in total we obtain $\eta -1$ different lists of partitions of AMs in $(2k;\ell _{1},\ell _{2},\ldots ,\ell _{s})$.