The Necessary and Sufficient Conditions of Separability for Bipartite Pure States in Infinite Dimensional Hilbert Spaces

TL;DR

This paper establishes precise mathematical conditions for when bipartite pure states in infinite-dimensional Hilbert spaces are separable, linking quantum state properties to operator theory in functional analysis.

Contribution

It provides the necessary and sufficient conditions for separability of bipartite pure states in infinite dimensions, connecting quantum information to bounded linear operator theory.

Findings

Matrix of amplitudes is a compact operator.

Separable states correspond to rank-1 bounded linear operators.

Separable states characterized by one-dimensional image of the operator.

Abstract

In this paper, we present the necessary and sufficient conditions of separability for bipartite pure states in infinite dimensional Hilbert spaces. Let $M$ be the matrix of the amplitudes of $\ket\psi$, we prove $M$ is a compact operator. We also prove $\ket\psi$ is separable if and only if $M$ is a bounded linear operator with rank 1, that is the image of $M$ is a one dimensional Hilbert space. So we have related the separability for bipartite pure states in infinite dimensional Hilbert spaces to an important class of bounded linear operators in Functional analysis which has many interesting properties.

Equations

$$a_{ij}$$

$$|v⟩\otimes |w⟩$$

$$M$$

$$Mx=\left(\begin{array}[]{c}M_{1}^{T}\\ M_{2}^{T}\\ \vdots\\ M_{j}^{T}\\ \vdots\end{array}\right)x=\left(\begin{array}[]{c}M_{1}^{T}\cdot x\\ M_{2}^{T}\cdot x\\ \vdots\\ M_{j}^{T}\cdot x\\ \vdots\end{array}\right)=\left(\begin{array}[]{c}(M_{1},\bar{x})\\ (M_{2},\bar{x})\\ \vdots\\ (M_{j},\bar{x})\\ \vdots\end{array}\right)$$

$$M_{n}=\left(\begin{array}[]{ccccccc}a_{11}&a_{12}&\cdots&a_{1n}&0&0&\cdots\\ a_{21}&a_{22}&\cdots&a_{2n}&0&0&\cdots\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots\\ a_{n1}&a_{n2}&\cdots&a_{nn}&0&0&\cdots\\ 0&0&\cdots&0&0&0&\cdots\\ 0&0&\cdots&0&0&0&\cdots\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots\end{array}\right),$$

$$|s-s_n|\to 0(n\to \infty ).$$

$$M-M_{n}=\left(\begin{array}[]{ccccccc}0&0&\cdots&0&a_{1(n+1)}&a_{1(n+2)}&\cdots\\ 0&0&\cdots&0&a_{2(n+1)}&a_{2(n+2)}&\cdots\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots\\ 0&0&\cdots&0&a_{n(n+1)}&a_{n(n+2)}&\cdots\\ a_{(n+1)1}&a_{(n+1)2}&\cdots&a_{(n+1)n}&a_{(n+1)(n+1)}&a_{(n+1)(n+2)}&\cdots\\ a_{(n+2)1}&a_{(n+2)2}&\cdots&a_{(n+2)n}&a_{(n+2)(n+1)}&a_{(n+2)(n+2)}&\cdots\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots\end{array}\right)=\left(\begin{array}[]{c}B_{1}^{T}\\ B_{2}^{T}\\ \vdots\\ B_{n}^{T}\\ B_{n+1}^{T}\\ B_{n+2}^{T}\\ \vdots\end{array}\right).$$

$$\displaystyle(M-M_{n})\phi=\left(\begin{array}[]{c}B_{1}^{T}\\ B_{2}^{T}\\ \vdots\\ B_{j}^{T}\\ \vdots\end{array}\right)\phi$$

$$=$$

$$\sum _{i=1}^{\infty }\parallel B_i{\parallel }^2$$

$$Mz=xy^{\dagger}z=x(\bar{y_{1}},\bar{y_{2}},\cdots,\bar{y_{n}},\cdots)\left(\begin{array}[]{c}z_{1}\\ z_{2}\\ \vdots\\ z_{n}\\ \vdots\end{array}\right)=x\left(\sum_{i=1}^{\infty}{\bar{y_{i}}z_{i}}\right),$$

$$M$$

Full text

A pure state is separable if and only if it can be written as a tensor product of states of different subsystems. It is also known that a state $|\psi\rangle$ of a bipartite system is separable if and only if it has Schmidt number 1 NC2000 . Peres presented a necessary and sufficient condition for the occurrence of Schmidt decomposition for a tripartite pure state Peres1995 and showed that the positivity of the partial transpose of a density matrix is a necessary condition for separability Peres1996 . Thapliyal showed that a multipartite pure state is Schmidt decomposable if and only if the density matrices obtained by tracing out any party are separable Thap1999 . In DafaLi2006 , Dafa Li obtained a necessary and sufficient conditions of separability for multipartite pure state in finite dimensional Hilbert spaces. In YuHu2007 , Yu and Hu gave other necessary and sufficient conditions, and simplified the proof of the main result in DafaLi2006 . They also obtained an algorithm to determine the separability of any multipartite pure state efficiently and quickly. But all of the previous works only aimed to determine the multipartite pure state in finite dimensional Hilbert spaces. In this letter, we gave the necessary and sufficient conditions of separability for bipartite pure state in infinite dimensional Hilbert spaces and relate this problem to an important class of bounded linear operators in Functional analysis.

Let $H_{A},H_{B}$ be two infinite dimensional Hilbert spaces. We define their tensor product as in Douglas179 . Let $H_{A}\stackrel{{\scriptstyle\displaystyle\otimes}}{{\scriptscriptstyle a}}H_{B}$ denote the algebraic tensor product of $H_{A}$ and $H_{B}$ consider as a linear space over $\mathbb{C}$. It is easy to see $(\sum_{i=1}^{m}{h_{i}\otimes k_{i}},\sum_{j=1}^{n}{h_{j}^{{}^{\prime}}\otimes k_{j}^{{}^{\prime}}})=\sum_{i=1}^{m}{\sum_{j=1}^{n}{(h_{i},h_{j}^{{}^{\prime}})(k_{i},k_{j}^{{}^{\prime}})}}$ defines an inner product of $H_{A}\stackrel{{\scriptstyle\scriptstyle{\bigotimes}}}{{\scriptstyle{a}}}H_{B}$. We should note that this space is not complete with this inner product. Pass to completion, we get a Hilbert space. As in Functional analysis, we call it the tensor product of $H_{A}$ and $H_{B}$ and denote it by $H_{A}\otimes H_{B}$. We can prove Douglas179 , if $\left\{e_{\alpha}\right\}_{\alpha\in A}$ and $\left\{f_{\beta}\right\}_{\beta\in B}$ are orthonormal basis for $H_{A}$ and $H_{B}$ respectively, then $\left\{e_{\alpha}\otimes f_{\beta}\right\}_{(\alpha,\beta)}\in H_{A}\otimes H_{B}$ is an orthonormal basis for $H_{A}\otimes H_{B}$.

Let us now consider two physical systems $A$ and $B$ represented by the Hilbert spaces $H_{A}$ and $H_{B}$ respectively. The joint system is represented by the Hilbert space $H_{A}\otimes H_{B}$. Let $|\psi\rangle\in H_{A}\otimes H_{B}$ be a pure state of a composite system $AB$. We also say $|\psi\rangle$ is separable if and only if it can be written as a tensor product of states of different subsystems. In this letter, we suppose $H_{A}$ and $H_{B}$ have countable number of dimensions, this is equivalent to say that $H_{A}$ and $H_{B}$ are separable topological spaces Conway . Let $|i\rangle$($|j\rangle$) be the orthonormal basis for Hilbert space $H_{A}$($H_{B}$). From above, we know $|i\rangle|j\rangle$ is an orthonormal basis for Hilbert space $H_{A}\otimes H_{B}$. Since $|\psi\rangle\in H_{A}\otimes H_{B}$, we can give the Fourier expansion of $|\psi\rangle$ under this basis. Then we can write $|\psi\rangle=\sum_{i,j}{a_{ij}|i\rangle|j\rangle}$, where $a_{ij}\in\mathbb{C}$ and $\sum_{i=1}^{\infty}{\sum_{j=1}^{\infty}{|a_{ij}|^{2}}}=1$. Let $M=(a_{ij})$ be the infinite (but countable) dimensional matrix of the amplitudes of $|\psi\rangle$. We will prove $M$ is a compact linear operator in Functional analysis and give the criterion for the separability.

Definition 1. Conway $l^{2}$* denote the set of all infinite sequences $\{x_{n}\}_{n=1}^{\infty}$ such that $\sum_{i=1}^{\infty}{|x_{i}|^{2}}<\infty$. For $x,y$ in $l^{2}$ define the inner product by $(x,y)=\sum_{i=1}^{\infty}{x_{i}\bar{y_{j}}}$. It is easy to show $l^{2}$ is a Hilbert space with this inner product.*

For $x=\{x_{n}\}_{n=1}^{\infty}\in l^{2}$, denote $x=(x_{1},x_{2},\cdots,x_{n},\cdots)^{T},\bar{x}=(\bar{x_{1}},\bar{x_{2}},\cdots,\bar{x_{n}},\cdots)^{T}$.

[TABLE]

We set x=\left(\begin{array}[]{c}x_{1}\\ x_{2}\\ \vdots\\ x_{n}\\ \vdots\\ \end{array}\right),y=\left(\begin{array}[]{c}y_{1}\\ y_{2}\\ \vdots\\ y_{n}\\ \vdots\\ \end{array}\right). From $\sum_{i=1}^{\infty}{|x_{i}|^{2}}=1$ and $\sum_{j=1}^{\infty}{|y_{j}|^{2}}=1$, we see $x,y\in l^{2}$ and as above M=\left(a_{ij}\right)=\left(\begin{array}[]{cccc}a_{11}&\cdots&a_{1n}&\cdots\\ \vdots&\ddots&\vdots&\ddots\\ a_{n1}&\cdots&a_{nn}&\cdots\\ \vdots&\ddots&\vdots&\ddots\end{array}\right). From (1) and the multiplication law for infinite (countable) dimensional matrices, we see $M=xy^{\dagger}$.

$\Leftarrow$ Suppose that $M=xy^{\dagger},x,y\in l^{2}$. Because $M=(a_{ij})$ be the matrix of the amplitudes of $|\psi\rangle$, we know $\sum_{i=1}^{\infty}{\sum_{j=1}^{\infty}{|a_{ij}|^{2}}}=1$. From $M=xy^{\dagger}$ and the multiplication for infinite (countable) dimensional matrices, we see $a_{ij}=x_{i}\bar{y_{j}}$. From $\sum_{i=1}^{\infty}{\sum_{j=1}^{\infty}{|a_{ij}|^{2}}}=1$, we have $\sum_{j=1}^{\infty}{|y_{j}|^{2}}\sum_{i=1}^{\infty}{|x_{i}|^{2}}=1$. We denote $\parallel x\parallel=\sum_{i=1}^{\infty}{|x_{i}|^{2}}<+\infty,\parallel y\parallel=\sum_{j=1}^{\infty}{|y_{j}|^{2}}<+\infty$ (because $x,y\in l^{2}$). Under the suppose $\widetilde{x}=\frac{x}{\parallel x\parallel},\widetilde{y}=\frac{y}{\parallel y\parallel}$, we also have $M=\widetilde{x}\widetilde{y}$. So we can suppose $\parallel x\parallel=\parallel y\parallel=1$ and construct two states $|v\rangle=\sum_{i=1}^{\infty}{x_{i}|i\rangle},|w\rangle=\sum_{j=1}^{\infty}{\bar{y_{j}}|j\rangle}$. We have

[TABLE]

We see $|\psi\rangle$ is a separable pure state. ∎

$\Leftarrow$ Suppose $M=(M_{1},M_{2},\cdots,M_{j},\cdots)$ (we can suppose $M_{1}\neq 0$). If $M_{1},M_{j}$ are linearly independent for some $j>1$, then the $2\times 2$ submatrix is reversible, so $det(m)\neq 0$. This is a contradiction. So for each $j$ we have a constant $\lambda_{j}\in\mathbb{C}$, such that $M_{j}=\lambda_{j}M_{1}$. Then

[TABLE]

Set $x=M_{1},y=(1,\bar{\lambda_{2}},\cdots,\bar{\lambda_{j}},\cdots)^{T}$. Since $M$ is the matrix of the amplitudes of $|\psi\rangle$, it is obviously $x,y\in l^{2}$. From (2) we have $M=xy^{\dagger}$, as desired. ∎

Remark Theorem 2 generalize the corresponding result in DafaLi2006 to infinite dimensional Hilbert spaces.

Definition 2. Conway An operator $T$ on Hilbert space $H$ has finite rank if $Im(T)$(the image of $T$) is finite dimensional.

1. Denote M=\left(\begin{array}[]{c}M_{1}^{T}\\ M_{2}^{T}\\ \vdots\\ M_{j}^{T}\\ \vdots\end{array}\right). $\forall x\in l^{2}$ we have

[TABLE]

So $\parallel Mx\parallel^{2}=\sum_{i=1}^{\infty}{|(M_{i},\bar{x})|^{2}}\leq\sum_{i=1}^{\infty}{\parallel M_{i}\parallel^{2}\parallel x\parallel^{2}}=\sum_{i=1}^{\infty}{\sum_{j=1}^{\infty}{|a_{ij}|^{2}\parallel x\parallel^{2}}}=\parallel x\parallel^{2}$. We get $M$ is a well defined bounded linear operator on $l^{2}$ with the norm ${\parallel}M\parallel\leq 1$.

2. To prove $M$ is a compact operator, according to Conway we should only to show there is a sequence $\{T_{n}\}$ of operators of finite rank such that $\parallel M-T_{n}\parallel\rightarrow 0(n\rightarrow\infty)$. Because $M=(a_{ij})$ is the matrix of the amplitudes of $|\psi\rangle$, we have $\sum_{i=1}^{\infty}{\sum_{j=1}^{\infty}{|a_{ij}|^{2}}}=1$. We set

[TABLE]

obviously $M_{n}$ has finite rank.

Denote $s=\sum_{i=1}^{\infty}{\sum_{j=1}^{\infty}{|a_{ij}|^{2}}}=1,s_{n}=\sum_{i=1}^{n}{\sum_{j=1}^{n}{|a_{ij}|^{2}}}$, from the absolute convergence of $s$, we know

[TABLE]

We have

[TABLE]

From $\sum_{i,j=1}^{\infty}{|a_{ij}|^{2}}=1$, we see $B_{i}\in l^{2}(i\geq 1)$.

$(l^{2})_{1}$ denote the closed unit ball in $l^{2}$. $\forall\phi\in(l^{2})_{1}$, We have

[TABLE]

Then $\parallel(M-M_{n})\phi\parallel^{2}=\sum_{i=1}^{\infty}{|(B_{i},\bar{\phi})|^{2}}\leq\sum_{i=1}^{\infty}{\parallel B_{i}\parallel^{2}\parallel\phi\parallel^{2}}\leq\sum_{i=1}^{\infty}{\parallel B_{i}\parallel^{2}}$. From (4) we get

[TABLE]

From (5) and (6), $\parallel(M-M_{n})\phi\parallel^{2}\leq|s-s_{n}|,\forall\phi\in(l^{2})_{1}$. From the definition of the norm of the operators on $l^{2}$, we have $\parallel M-M_{n}\parallel^{2}=\sum\limits_{\phi\in(l^{2})_{1}}{\parallel(M-M_{n})\phi\parallel^{2}}\leq|s-s_{n}|$. From (3), we get $\parallel M-M_{n}\parallel\rightarrow 0(n\rightarrow\infty)$. So we see $M$ is a compact operator on $l^{2}$. ∎

Remark: We know all the compact operators form a closed two sided ideal in operator algebra and they have many interesting properties. For example Conway

(1) “If $A$ is a compact linear operator on $H,\lambda\in\mathbb{C}$ and $\lambda\neq 0$, then the image of $A$ is closed and $\dim\ker(A-\lambda I)=\dim\ker(A-\lambda I)^{\dagger}<\infty$”, this is a famous theorem named “The Fredholm Alternative” in Functional analysis. Someone call this is “the linear algebra of infinite dimensional spaces”.

(2) “$A$ is compact if and only if $A^{\dagger}$ is compact.” This is a theorem of Schaduer.

We will see if $|\psi\rangle$ is separable, $M$ is not only a compact linear operator, but also an operator with rank 1. We know from Analysis the operators which have finite rank must be a compact operator Conway .

We have

[TABLE]

so $Im{M}\subset\textrm{span}{(x)}$. But $M\neq 0$, we get $\dim Im{M}=1$.

$\Leftarrow$ Suppose $\dim Im{M}=1$. Denote $M=(M_{1},M_{2},\cdots,M_{i},\cdots)$( we can suppose $M_{1}\neq 0$). Suppose the vector $e_{i}\in l^{2}$ is the vector with all [math]s except for a 1 in the $i$th coordinate. We have $Me_{1}=M_{1}$ and $Me_{i}=M_{i}$. Then $M_{1},M_{i}\in Im{M}$, but $\dim Im{M}=1$, so there exists $\lambda_{i}\in\mathbb{C}$ such that $M_{i}=\lambda_{i}M_{1}$. We have

[TABLE]

Denote $x=M_{1},y=(1,\lambda_{2},\cdots,\lambda_{i},\cdots)^{\dagger}$. $M$ is the matrix of the amplitudes of $|\psi\rangle$, we see that $x,y\in l^{2}$. Finally, from (7) we get $M=xy^{\dagger}$, as desired. ∎

1. $Im{M^{\dagger}}$ is also closed;
2. $Im{M}=\ker{M^{\dagger}}^{\perp}$;
3. $Im{M^{\dagger}}=\ker{M^{\perp}}$;
4. $M$ is an open mapping;
5. $M^{\dagger}$ is an open mapping.*

Denote $\mathscr{L}(l^{2})$ to be the algebra of bounded linear operators on $l^{2}$ and $\mathscr{B}^{0}(l^{2})$ to be the ideal of compact operators on $l^{2}$. We also define two subsets of $\mathscr{L}(l^{2})$, $\mathscr{F}=\{M\in\mathscr{L}(l^{2}):M\quad\textrm{is a bounded linear operator with rank 1}\}$. $\mathscr{T}=\{M\in\mathscr{L}(l^{2}):M\quad\textrm{is the matrix of a}\textrm{ bipatite pure state}\newline \textrm{ of a composite}{\textrm{system AB}}\}$.

From theorem 3 and 5, we get $\mathscr{T}\subset\mathscr{F}\subset\mathscr{B}^{0}(l^{2})\subset\mathscr{L}(l^{2})$, and relate the separability for bipartite pure states to an important class of bounded linear operators in Functional analysis.

So, given a pure state $|\psi\rangle$ in $H_{A}\otimes H_{B}$, $M$ is the matrix of the amplitudes of $|\psi\rangle$, if $M$ is not a rank 1 operator on $l^{2}$, we can conclude that $|\psi\rangle$ is not separable. But the rank 1 operators in an infinite dimensional Hilbert space are rare, so the separable pure states in $H_{A}\otimes H_{B}$ are also rare.