N-qudit SLOCC equivalent W states are determined by their bipartite reduced density matrices with tree form

Abstract

It has been proved that N-qudit (i.e., d-level subsystems) generalized W states are determined by their bipartite reduced density matrices. In this paper, we prove that only $(N-1)$ of the bipartite reduced density matrices are sufficient. Furthermore, we find that N-qudit W states preserve their determinability under stochastic local operation and classical communication (SLOCC). That is, all multipartite pure states that are SLOCC equivalent to N-qudit W states can be uniquely determined (among pure, mixed states) by their $(N-1)$ of the bipartite reduced density matrices, if the $(N-1)$ pairs of qudits constitute a tree graph on N vertices, where each pair of qudits represents an edge.

Appendix

In this section, we prove for any $(s,t)\in E$ and $i_t,j_s \ne 0$, if

$$\begin{aligned} {{r}_{\left( {{i'}_{1}}\cdots {{i'}_{s-1}}{{0}_{s}}{{i'}_{s+1}}\cdots {{i'}_{t-1}}{{i}_{t}}{{i'}_{t+1}}\cdots {{i'}_{N}} \right) \left( {{i'}_{1}}\cdots {{i'}_{s-1}}{{0}_{s}}{{i'}_{s+1}}\cdots {{i'}_{t-1}}{{i}_{t}}{{i'}_{t+1}}\cdots {{i'}_{N}} \right) }}=0, \end{aligned}$$

then

$$\begin{aligned} {{r}_{\left( {{i'}_{1}}\cdots {{i'}_{s-1}}{{j}_{s}}{{i'}_{s+1}}\cdots {{i'}_{t-1}}{{0}_{t}}{{i'}_{t+1}}\cdots {{i'}_{N}} \right) \left( {{i'}_{1}}\cdots {{i'}_{s-1}}{{j}_{s}}{{i'}_{s+1}}\cdots {{i'}_{t-1}}{{0}_{t}}{{i'}_{t+1}}\cdots {{i'}_{N}} \right) }}=0. \end{aligned}$$

Proof

For any $(s,t)\in E$, comparing the coefficients of $|0_si_t\rangle \langle 0_si_t|$, $|j_s0_t\rangle \langle j_s0_t|$ and $|0_si_t\rangle \langle j_s0_t|$ from $\rho _M^{st}$ and $\rho _{SW}^{st}$, since $i_t,j_s \ne 0$, we have

$$\begin{aligned} \sum \limits _{{{i}_{V\backslash \{s,t\}}}=0}^{d-1}{{{r}_{\left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) }}}\text {=}{{\left| {{c}_{ti}} \right| }^{2}}, \end{aligned}$$

(14)

$$\begin{aligned} \sum \limits _{{{i}_{V\backslash \{s,t\}}}=0}^{d-1}{{{r}_{\left( {{i}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i}_{N}} \right) }}}\text {=}{{\left| {{c}_{sj}} \right| }^{2}}, \end{aligned}$$

(15)

and

$$\begin{aligned} \sum \limits _{{{i}_{V\backslash \{s,t\}}}=0}^{d-1}{{{r}_{\left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i}_{N}} \right) }}}\text {=}{{c}_{ti}}{{\bar{c}}_{sj}}. \end{aligned}$$

(16)

As ${{r}_{\left( {{i'}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i'}_{N}} \right) \left( {{i'}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i'}_{N}} \right) }}=0$, by property (ii) of positive semidefinite matrices, we can get ${{r}_{\left( {{i'}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i'}_{N}} \right) \left( {{i'}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i'}_{N}} \right) }}=0$ too. Then we can rewrite Eqs. (14), (15) and (16) as follows

$$\begin{aligned}&\sum \limits _{{{i}_{V\backslash \{s,t\}}}\in I}{{{r}_{\left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) }}}\text {=}{{\left| {{c}_{ti}} \right| }^{2}}, \end{aligned}$$

(17)

$$\begin{aligned}&\sum \limits _{{{i}_{V\backslash \{s,t\}}}\in I}{{{r}_{\left( {{i}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i}_{N}} \right) }}}\text {+}{{r}_{\left( {{{{i}'}}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{{{i}'}}_{N}} \right) \left( {{{{i}'}}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{{{i}'}}_{N}} \right) }}\text {=}{{\left| {{c}_{sj}} \right| }^{2}}, \end{aligned}$$

(18)

and

$$\begin{aligned} \sum \limits _{{{i}_{V\backslash \{s,t\}}}\in I}{{{r}_{\left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i}_{N}} \right) }}}\text {=}{{c}_{ti}}{{\bar{c}}_{sj}}, \end{aligned}$$

(19)

where the sum $\sum \limits _{{{i}_{V\backslash \{s,t\}}}\in I}$ varies over $0,1,\ldots ,d-1$ at the $1,\ldots ,s-1,s+1,\ldots ,t-1,t+1,\ldots ,N$ parties, excepts for ${{i'}_{1}}\cdots {{i'}_{s-1}}{{i'}_{s+1}}\cdots {{i'}_{t-1}}{{i'}_{t+1}}\cdots {{i'}_{N}}$. Therefore, from the above three equations, we can get

$$\begin{aligned} \begin{aligned}&{{ \left| \sum \limits _{{{i}_{V\backslash \{s,t\}}}\in I}{{{r}_{\left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i}_{N}} \right) }}} \right| }^{2}}\\ =&\Bigg ( \sum \limits _{{{i}_{V\backslash \{s,t\}}}\in I}{{{r}_{\left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) }}} \Bigg ) \Bigg ( \sum \limits _{{{i}_{V\backslash \{s,t\}}}\in I}{{{r}_{\left( {{i}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i}_{N}} \right) }}}ht.\\&+ {{r}_{\left( {{{{i}'}}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{{{i}'}}_{N}} \right) \left( {{{{i}'}}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{{{i}'}}_{N}} \right) }} \Bigg ) \\ =&\bigg (\left. \sum \limits _{{{i}_{V\backslash \{s,t\}}}\in I}{{{r}_{\left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) }}} \Bigg )\Bigg ( \sum \limits _{{{i}_{V\backslash \{s,t\}}}\in I}{{{r}_{\left( {{i}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i}_{N}} \right) }}} \right. \Bigg )\\&+\left. \Bigg ( \sum \limits _{{{i}_{V\backslash \{s,t\}}}\in I}{{{r}_{\left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) }}} \right. \Bigg )\,{{r}_{\left( {{{{i}'}}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{{{i}'}}_{N}} \right) \left( {{{{i}'}}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{{{i}'}}_{N}} \right) }}. \end{aligned} \end{aligned}$$

(20)

By the property (iii) of positive semidefinite matrices, it follows that

$$\begin{aligned} \begin{aligned}&\left| {{r_{\left( {{i_1} \cdots {0_s} \cdots {i_t} \cdots {i_N}} \right) \left( {{i_1} \cdots {j_s} \cdots {0_t} \cdots {i_N}} \right) }}} \right| \\ \le&\sqrt{{r_{\left( {{i_1} \cdots {0_s} \cdots {i_t} \cdots {i_N}} \right) \left( {{i_1} \cdots {0_s} \cdots {i_t} \cdots {i_N}} \right) }}{r_{\left( {{i_1} \cdots {j_s} \cdots {0_t} \cdots {i_N}} \right) \left( {{i_1} \cdots {j_s} \cdots {0_t} \cdots {i_N}} \right) }}} \end{aligned} \end{aligned}$$

So we can get

$$\begin{aligned} \begin{aligned}&\left| \sum \limits _{{{i}_{V\backslash \{s,t\}}}\in I}{{{r}_{\left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i}_{N}} \right) }}} \right| \\ \le&\sum \limits _{{{i}_{V\backslash \{s,t\}}}\in I}{\left| {{r}_{\left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i}_{N}} \right) }} \right| } \\ \le&\sum \limits _{{{i}_{V\backslash \{s,t\}}}\in I}{\sqrt{{{r}_{\left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) }}{{r}_{\left( {{i}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i}_{N}} \right) }}}} \\ \le&\sqrt{\left( \sum \limits _{{{i}_{V\backslash \{s,t\}}}\in I}{{{r}_{\left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) }}})(\sum \limits _{{{i}_{V\backslash \{s,t\}}}\in I}{{{r}_{\left( {{i}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i}_{N}} \right) }}}\right) }.\\ \end{aligned} \end{aligned}$$

(21)

After both sides of this Formula (21) is squared and subtracted, we obtain

$$\begin{aligned} \begin{aligned}&{{\left| \sum \limits _{{{i}_{V\backslash \{s,t\}}}\in I}{{{r}_{\left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i}_{N}} \right) }}} \right| }^{2}} \\&\quad - \left( \sum \limits _{{{i}_{V\backslash \{s,t\}}}\in I}{{{r}_{\left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) }}} \right) \left( \sum \limits _{{{i}_{V\backslash \{s,t\}}}\in I}{{{r}_{\left( {{i}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{i}_{N}} \right) }}} \right) \\&\le 0 .\\ \end{aligned} \end{aligned}$$

(22)

But from Eq. (20), we obtain that the difference between the two expressions in left side of Formula (22) is $(\sum \limits _{{{i}_{V\backslash \{s,t\}}}\in I}{{{r}_{\left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) }}} ){{r}_{\left( {{{{i}'}}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{{{i}'}}_{N}} \right) \left( {{{{i}'}}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{{{i}'}}_{N}} \right) }}$, which is greater than or equal to 0, so we have

$$\begin{aligned} \begin{aligned}&\left( \sum \limits _{{{i}_{V\backslash \{s,t\}}}\in I}{{{r}_{\left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) \left( {{i}_{1}}\cdots {{0}_{s}}\cdots {{i}_{t}}\cdots {{i}_{N}} \right) }}} \right) {{r}_{\left( {{{{i}'}}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{{{i}'}}_{N}} \right) \left( {{{{i}'}}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{{{i}'}}_{N}} \right) }} \\&= |{{c}_{ti}}{{|}^{2}}{{r}_{\left( {{{{i}'}}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{{{i}'}}_{N}} \right) \left( {{{{i}'}}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{{{i}'}}_{N}} \right) }}\\&= 0 .\\ \end{aligned} \end{aligned}$$

As $|c_{ti}|^2 \ne 0$, we can get ${{r}_{\left( {{{{i}'}}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{{{i}'}}_{N}} \right) \left( {{{{i}'}}_{1}}\cdots {{j}_{s}}\cdots {{0}_{t}}\cdots {{{{i}'}}_{N}} \right) }}=0$.

This end the proof. $\square$