Decomposition of completely symmetric states

Abstract

Symmetry is a fundamental milestone of quantum physics, and the relation between entanglement is one of the central mysteries of quantum mechanics. In this paper, we consider a subclass of symmetric quantum states in the bipartite system, namely the completely symmetric states, which is invariant under the index permutation. We investigate the separability of these states. After studying some examples, we conjecture that the completely symmetric state is separable if and only if it is S-separable, i.e., each term in this decomposition is a symmetric pure product state ${|x,x\rangle }{\langle x,x|}$. It was proved to be true when the rank does not exceed $\max \{4,N+1\}$. After studying the properties of these state, we propose a numerical algorithm which is able to detect S-separability. This algorithm is based on the best separable approximation, which furthermore turns out to be applicable to test the separability of quantum states in bosonic system. Besides, we analyse the convergence behaviour of this algorithm. Some numerical examples are tested to show the effectiveness of the algorithm.

Appendices

Proof of Eq.  (104)

Fig. 9

$x = \cos (\theta )x_{*}+ \sin (\theta )x_{\bot }$

In this appendix, we prove Eq. (104). Let us describe this question formally with a lemma.

Lemma 29

Let $x_{*}$ and $x_{\bot }$ are two orthogonal unit vectors in the $\mathbb {R}^{N}$ space and

$$\begin{aligned} x = \cos (\theta )x_{*}+ \sin (\theta )x_{\bot },\theta \in [0,2\pi ]. \end{aligned}$$

(175)

Then we have

$$\begin{aligned} \left| \sin (\theta )\right| = \sqrt{\left\| x-x_{*}\right\| ^{2} - \frac{\left\| x-x_{*}\right\| ^4}{4}}. \end{aligned}$$

(176)

Proof

The following graph shows the relations of x,$x_{*}$, and $x_{\bot }$ when $\langle x,x_{*}\rangle >0$ (left one) and $\langle x,x_{*}\rangle <0$ (right one):

From the above graphs (Fig.  9), we have

$$\begin{aligned} \left\| x-x_{*}\right\| = \left| AB\right| . \end{aligned}$$

(177)

And

$$\begin{aligned} \begin{aligned} \left| OC\right|&= \left| \cos (\theta )\right| ,\\ \left| BC\right|&= 1 - \cos (\theta ),\\ \left| AC\right|&= \sin (\theta ). \end{aligned} \end{aligned}$$

(178)

Moreover,

$$\begin{aligned} \begin{aligned} \left| AB\right|&= \sqrt{\left| AC\right| ^2+\left| BC\right| ^2}\\&= \sqrt{(1-\cos (\theta ))^2+\sin ^2(\theta )}\\&= \sqrt{2-2\cos (\theta )}. \end{aligned} \end{aligned}$$

(179)

Therefore,

$$\begin{aligned} \cos (\theta ) = 1 - \frac{\left| AB\right| ^2}{2}. \end{aligned}$$

(180)

Forward,

$$\begin{aligned} \begin{aligned} \left| \sin (\theta )\right|&= \sqrt{1-\cos ^2(\theta )}\\&= \sqrt{1 - \left( 1-\frac{\left| AB\right| ^2}{2}\right) ^2}\\&= \sqrt{\left| AB\right| ^2-\frac{\left| AB\right| ^2}{4}}\\&= \sqrt{\left\| x-x_{*}\right\| ^{2} - \frac{\left\| x-x_{*}\right\| ^4}{4}}. \end{aligned} \end{aligned}$$

(181)

$\square$

Fig. 10

Relation between $x_{\mathrm{SQP}}$ and $x_{\mathrm{NT}}$

Proof of Eq. (162)

In this appendix, we prove Eq. (162), that is to prove

$$\begin{aligned} \left\| x_{\mathrm{SQP}}-x_{*}\right\| \leqslant \left\| x_{\mathrm{NT}}-x_*\right\| , \end{aligned}$$

(182)

where $\left\| x_{\mathrm{SQP}}\right\| \leqslant \left\| x_{\mathrm{NT}}\right\|$ and $x_{*},x_{\mathrm{SQP}}$ are unit vectors. The following Fig. 10 shows the relationship of $x_{\mathrm{SQP}}$ and $x_{\mathrm{NT}}$.

Note that

$$\begin{aligned} \angle OAC = \frac{1}{2}(\pi - \angle AOC). \end{aligned}$$

(183)

Hence,

$$\begin{aligned} \begin{aligned} \angle BAC&= \pi -\angle OAC\\&= \pi - \frac{1}{2}(\pi - \angle AOC)\\&= \frac{\pi }{2}+\frac{1}{2}\angle AOC\\&> \frac{\pi }{2}. \end{aligned} \end{aligned}$$

(184)

Moreover,

$$\begin{aligned} \sin (\angle ABC) \leqslant \sin (\pi - \angle BAC) = \sin (\angle BAC). \end{aligned}$$

(185)

Forward,

$$\begin{aligned} \frac{\left\| x_{\mathrm{SQP}}-x_{*}\right\| }{\left\| x_{NT}-x_{*}\right\| }= \frac{\sin (\angle ABC)}{\sin (\angle BAC)}. \end{aligned}$$

(186)

By Eq. 185, we have

$$\begin{aligned} \frac{\left\| x_{\mathrm{SQP}}-x_{*}\right\| }{\left\| x_{NT}-x_{*}\right\| }\leqslant 1, \end{aligned}$$

(187)

which completes our proof.