Constructing Separable States in Infinite-Dimensional Systems by Operator Matrices

Abstract

We introduce a class of states so-called semi-SSPPT (semi super strong positive partial transposition) states in infinite-dimensional bipartite systems by the Cholesky decomposition in terms of operator matrices and show that every semi-SSPPT state is separable. This gives a method of constructing separable states and generalizes the corresponding results in Chruściński et al. (Phys. Rev. A 77, 022113 2008) and Guo and Hou (J. Phys. A: Math. Theor. 45, 505303 2012). This criterion is specially convenient to be applied when one of the subsystem is a qubit system.

Appendices

Appendix A: SOT-Separability for Positive Operators and Radon-Nikodym Type Theorem for Spectral Measure

To prove theorem 1, we need to generalize the concept of separability from states to bounded positive operators and Radon-Nikodym theorem from vector mesasures in Hilbert space to the spectral measures.

Denote by $\mathcal {B}(H)$ and $\mathcal {B}_{+}(H)$ the set of all bounded linear operators and the set of all positive bounded operators acting on a complex Hilbert space H, respectively.

Definition A.1

A positive operator $T\in \mathcal {B}_{+}(H_{A}\otimes H_{B})$ is called SOT-separable if there exist positive operators $\{A_{k}\}\subset \mathcal {B}_{+}(H_{A})$ and $\{B_{k}\}\subset \mathcal {B}_{+}(H_{B})$ such that

$$ T=\sum\limits_{k} A_{k}\otimes B_{k} $$

(A.1)

or if T is the limit of the operators of the form as in (A.1) under the strong operator topology (briefly, SOT). Otherwise, T is said to be SOT-inseparable.

We remark that, if the sum in (A.1) is a series, we mean that the series is convergent under SOT. It is obvious that 0 is SOT-separable and, if dim H _A ⊗ H _B < ∞, then a nonzero positive operator T is SOT-separable if and only if $\frac {1}{\text {Tr}(T)}T$ is a separable quantum state.

Denote by $\mathcal {S}_{\text {SOT}}(H_{A}\otimes H_{B})$ the set of all SOT-separable operators, which is a SOT-closed convex cone in $\mathcal {B}(H_{A}\otimes H_{B})$.

The following is a SOT-separability criterion, which is similar to the entanglement witness criterion for states.

Proposition A.1

A positive operator $T\in \mathcal {B}(H_{A}\otimes H_{B})$ is SOT-inseparable if and only if there is a self-adjoint operator $W\in \mathcal {B}(H_{A}\otimes H_{B})$ of finite rank such that

• (1) Tr(W(A ⊗ B)) ≥ 0 holds for any $A\in \mathcal {B}_{+}(H_{A})$ and $B\in \mathcal {B}_{+}(H_{B})$;

• (2) Tr(WT) < 0.

Proof

Since $\mathcal {S}_{\text {SOT}}(H_{A}\otimes H_{B})$ is a SOT-closed convex subset of $\mathcal {B}(H_{A}\otimes H_{B})$, by Hahn-Banach theorem, $T\not \in \mathcal {S}_{\text {SOT}}(H_{A}\otimes H_{B})$ if and only if there exists a SOT-continuous linear functional ϕ on $\mathcal {B}(H_{A}\otimes H_{B})$ and a real number c such that Re(ϕ(S)) ≥ c for all $S\in \mathcal {S}_{\text {SOT}}(H_{A}\otimes H_{B})$ but Re(ϕ(T)) < c. As ϕ is SOT-continuous, there are vectors x ₁,…, x _r ; y ₁,…, y _r ∈ H _A ⊗ H _B with r < ∞ such that $\phi (X)={\sum }_{i=1}^{r} \langle y_{i}|X|x_{i}\rangle$ holds for all $X\in \mathcal {B}(H_{A}\otimes H_{B})$. Let $E={\sum }_{i=1}^{r} |x_{i}\rangle \langle y_{i}|$. Then $E\in \mathcal {B}(H_{A}\otimes H_{B})$ is a finite-rank operator which satisfies ϕ(X) = Tr(EX) for all X (Ref., for example, [24]). If X is self-adjoint, that is, if X ^† = X, then ϕ(X)^∗ = Tr(EX)^∗ = Tr((EX)^†) = Tr(E ^† X) and hence Re(ϕ(X)) = Tr(WX), where $W=\text {Re}(E)=\frac {1}{2} (E+E^{\dag })$ . It follows that Tr(WS) = Re(ϕ(S)) ≥ c for all $S\in \mathcal {S}_{\text {SOT}}(H_{A}\otimes H_{B})$ and Tr(WT) = Re(ϕ(T)) < c. Note that $0\in \mathcal {S}_{\text {SOT}}(H_{A}\otimes H_{B})$. So we must have c ≤ 0. We assert that Tr(WS) ≥ 0 holds for any $S\in \mathcal {S}_{\text {SOT}}(H_{A}\otimes H_{B})$. If, on the contrary, there is $S\in \mathcal {S}_{\text {SOT}}(H_{A}\otimes H_{B})$ so that Tr(WS) = a < 0. As $\mathcal {S}_{\text {SOT}}(H_{A}\otimes H_{B})$ is a convex cone, $tS\in \mathcal {S}_{\text {SOT}}(H_{A}\otimes H_{B})$ for any t > 0. Thus we have ta = Tr(W(tS)) ≥ c for all t > 0, which is a contradiction since ta →−∞ when t →∞. Therefore, we have found a self-adjoint operator W of finite rank such that (1) and (2) hold.

Conversely, if there is some self-adjoint operator W of finite rank such that (1) and (2) hold, then Tr(WS) ≥ 0 holds for all $S\in \mathcal {S}_{\text {SOT}}(H_{A}\otimes H_{B})$ since ϕ : X↦Tr(WX) is a SOT-continuous linear functional. Thus ϕ separates strictly T and $\mathcal {S}_{\text {SOT}}(H_{A}\otimes H_{B})$. So $T\not \in \mathcal {S}_{\text {SOT}}(H_{A}\otimes H_{B})$. □

Next we discuss the relationship between separability and SOT-separability for a quantum state. Notice that, generally speaking, though a sequence {T _n } of positive trace-class operators converges to a state ρ under SOT, one can not assert that {T _n } converges to ρ under the trace-norm. For instance, let $\{a_{n}\}_{n=1}^{\infty }$ be a sequence of positive numbers so that ${\sum }_{n=1}^{\infty } a_{n} =a<\infty$. For any n, let T _n = diag(t ₁, t ₂,…, t _k …) with t _k = 0 if k ≤ n and t _{n + m} = a _m . Then $\{T_{n}\}_{n=1}^{\infty }$ is a sequence of positive trace-class operators and T _n → 0 in SOT. However, Tr(T _n ) = a which does not converge to 0. In addition, for a state ρ, let $T^{\prime }_{n}=\rho +T_{n}$. Then $T_{n}^{\prime }\to \rho$ under SOT but $\text {Tr}(T^{\prime }_{n})=1+a$ does not converge to Tr(ρ) = 1. Hence, $\{T^{\prime }_{n}\}$ does not converge to ρ in trace-norm. This suggests that a SOT-separable state may not be separable. However, the following result reveals surprisedly that this is not the case.

Proposition A.2

Let $\rho \in \mathcal {S}(H_{A}\otimes H_{B})$ be a state. Then ρ is separable if and only if ρ is SOT-separable.

Proof

Clearly, ρ is separable implies that ρ is SOT-separable by the definitions as the convergence in trace-class norm implies the convergence in SOT.

Conversely, assume that ρ is inseparable (i.e., entangled). We have to show that ρ is also SOT-inseparable. Let {|i _a 〉} and {|j _b 〉} be arbitrarily given orthonormal bases for H _A and H _B , respectively. Let P _k and Q _k be finite-rank projections on $H_{A}^{(k)}$ and $H_{B}^{(k)}$, the span of $\{|i_{a}\rangle \}_{i=1}^{k}$ and the span of $\{|j_{b}\rangle \}_{j=1}^{k}$, respectively. If k ≥ dim H _A (or k ≥ dim H _B ), let P _k = I _A (or Q _k = I _B ) with I _A the identity operator on H _A . Let

$$\rho_{k}=\frac{1}{\text{Tr}((P_{k}\otimes Q_{k})\rho(P_{k}\otimes Q_{k}))}(P_{k}\otimes Q_{k})\rho(P_{k}\otimes Q_{k}).$$

Obviously, ρ = ∥⋅∥_Tr-$\lim _{k\to \infty } \rho _{k}$. As ρ is inseparable, there exists infinitely many k so that ρ _k is inseparable (otherwise, ρ should be separable). Take a such k. Then ρ _k can be regarded as an inseparable state in the finite-dimensional system $H_{A}^{(k)}\otimes H_{B}^{(k)}$. Thus, by the entanglement witness criterion, there is a self-adjoint operator W _k on $H_{A}^{(k)}\otimes H_{B}^{(k)}$ such that Tr(W _k ρ _k ) < 0. Let W = (P _k ⊗ Q _k )W _k (P _k ⊗ Q _k ). W is a self-adjoint operator of rank ≤ k ² < ∞ and is an entanglement witness for ρ because for any pure states $P_{A}\in \mathcal {S}(H_{A})$ and $Q_{B}\in \mathcal {S}(H_{B})$,

$$\text{Tr}(W(P_{A}\otimes Q_{B}))=\text{Tr}(W_{k}((P_{k}P_{A}P_{k})\otimes (Q_{k}Q_{B}Q_{k})))\geq 0$$

and

$$\text{Tr}(W\rho)=\text{Tr}((P_{k}\otimes Q_{k})W_{k}(P_{k}\otimes Q_{k})\rho)=\text{Tr}(W_{k}\rho_{k})<0.$$

For any $A\otimes B\in \mathcal {B}_{+}(H_{A}\otimes H_{B})$, (P _k ⊗ Q _k )(A ⊗ B)(P _k ⊗ Q _k ) = (P _k AP _k ) ⊗ (Q _k BQ _k ) is either zero or a positive multiple of a finite rank separable state. So, we still have

$$\text{Tr}(W(A\otimes B))\geq 0.$$

This implies by Proposition A.1 that ρ is SOT-inseparable, as desired. □

Now, let us turn to the question of establishing Radon-Nikodym type theorem for spectral measure. Let Ω be a nonempty set, $\mathcal B$ be a σ-algebra of subsets of Ω, H be a Hilbert space. Recall that a spectral measure for $({\Omega }, \mathcal {B}, H)$ is an operator-valued function $E: \mathcal {B}\to \mathcal {B}(H)$ such that

• (i) for each Δ in $\mathcal B$, E(Δ) is a projection;

• (ii) E(∅) = 0 and E(Ω) = I;

• (iii) for ${\Delta }_{1},{\Delta }_{2}\in \mathcal {B}$, E(Δ₁ ∩Δ₂) = E(Δ₁)E(Δ₂).

• (iv) if $\{{\Delta }_{i}\}\subset \mathcal {B}$ are pairwise disjoint sets, then $E(\cup _{i} {\Delta }_{i})={\sum }_{i} E({\Delta }_{i})$, here the sum converges in SOT (Ref. [24]).

Proposition A.3 (The Radon-Nikodym type theorem for spectral measure)

Let H be a complex Hilbert space, Ω be a nonempty set, $\mathcal B$ be a σ-algebra of subsets of Ω. Assume that E is a spectral measure for $({\Omega },\mathcal B, H)$ and μ is a positive measure on $({\Omega }, \mathcal {B})$ . If E ≪ μ, that is, if μ(Δ) = 0 ⇒ E(Δ) = 0, then there is an operator-valued function D on H such that 〈x|D(ω)|x〉 ≥ 0 a.e. μ and

$$E({\Delta})x=\mathrm{(B)}{\int}_{\Delta} D(\omega)x\mathrm{d} \mu_{\omega} $$

holds for every x ∈ H and ${\Delta }\in \mathcal {B}$ , where $\mathrm {(B)}{\int }_{\Delta }$ means the Bochner integral.

We remark that D(ω) may take a unbounded operator.

Proof

Recall that a Banach space X is said to have the Radon-Nikodym Property (RNP) if for any finite positive measure space $({\Omega }, \mathcal {F}, \mu )$ and vector-valued measure $F:\mathcal {F}\to X$, if F ≪ μ, then there exists a Bochner integrable vector-valued function f : Ω → X such that $F({\Delta })=\text {(B)}{\int }_{\Delta } f(\omega )\mathrm {d} \mu _{\omega }$ holds for any ${\Delta }\in \mathcal {F}$. Not every Banach space has RNP. However it is well-known that every Hilbert space has RNP (ref. [25, 26]).

Now let $({\Omega }, \mathcal {B}, \mu )$ be a finite positive measure space and $({\Omega }, \mathcal {B}, H,E)$ be a spectral measure space so that E ≪ μ. We remark here that we can not use the result in [26] because the spectral measure is not σ-bounded. For any vector x ∈ H, it is clear that $F_{x}: \mathcal {B}\to H$ defined by F _x (Δ) = E(Δ)x is a H-valued measure satisfying F _x ≪ μ and σ-boundedness. As H has RNP, there is a Bochner integrable vector-valued function D _x : Ω → H such that $E({\Delta })x=F_{x}({\Delta })=\text {(B)}{\int }_{\Delta } D_{x}(\omega )\mathrm {d}\mu _{\omega }$ holds for all ${\Delta }\in \mathcal {B}$. Note that, D _{αx + y} (ω) = αD _x (ω) + D _y (ω) a.e. μ. Hence there exists an operator-valued function D defined on Ω so that D(ω)x = D _x (ω) a.e. μ for each x ∈ H. And then $\text {(B)}{\int }_{\Delta } D(\omega )x\mathrm {d}\mu _{\omega }=\text {(B)}{\int }_{\Delta } D_{x}(\omega )\mathrm {d}\mu _{\omega }$ for any x ∈ H. Since, ${\int }_{\Delta } \langle x|D(\omega )|x\rangle \mathrm {d}\mu _{\omega } =\langle x|E({\Delta })|x\rangle \geq 0$ for any Borel set Δ, one sees that 〈x|D(ω)|x〉≥ 0 a.e. μ for each x ∈ H. So, almost all D(ω) are (may unbounded) operators with domain H satisfying 〈x|D(ω)|x〉≥ 0 and

$$ E({\Delta})x=\text{(B)}{\int}_{\Delta} D(\omega)x\mathrm{d}\mu_{\omega} $$

(A.2)

holds for all x ∈ H and ${\Delta }\in \mathcal B$. □

Some times we denote the relation in (A.2) by

$$ E({\Delta})=\mathrm{(SOT)}{\int}_{\Delta} D(\omega)\mathrm{d}\mu_{\omega} $$

(A.3)

holds for any ${\Delta }\in \mathcal B$.

Appendix B: Proof of the Main Result

Now we are at a position to give a proof of the main result theorem 1.

Proof of Theorem 1.

Assume that $\rho \in \mathcal {S}(H_{A}\otimes H_{B})$ is a semi-SSPPT state. We have to show that ρ is separable. We only need to check the case that ρ is semi-SSPPT up to part B since the proof for the case of semi-SSPPT up to A is similar.

As ρ is a semi-SSPPT state up to part B, we may write ρ = X ^† X, where X is the upper triangular operator matrix of the form mentioned in (4) with respect to an orthonormal basis {|i _a 〉} of H _A . Let C _k be the operator matrix with the same size as that of X, which is induced from X by replacing all entries by zero except for the kth row of X, i.e.,

k = 1, 2,…. Then C _k is a Hilbert-Schmidt operator and

$$\begin{array}{@{}rcl@{}} \rho=\sum\limits_{k}C_{k}^{\dag} C_{k}, \end{array} $$

(8)

Here the series converges in the trace-norm. If C _k ≠0, write $C_{k}^{\dag } C_{k}=p_{k}\rho _{k}$ where $p_{k}=\text {Tr}(C_{k}^{\dag } C_{k})$. We will show that ρ _k is separable for any k whenever C _k ≠0, and then, $\rho ={\sum }_{k} p_{k}\rho _{k}$ is separable, too.

Consider the case when k = 1. We have

$$ p_{1}\rho_{1}=(X_{1}^{\dag} S_{1i}^{\dag} S_{1j}X_{1})=\sum\limits_{i,j}(|i_{a}\rangle\langle j_{a}|)\otimes (X_{1}^{\dag} S_{1i}^{\dag} S_{1j}X_{1}) $$

(B.1)

with S ₁₁ = I _B . Since ρ is semi-SSPPT up to part B, {S _1i } is a commutative set of normal operators. Then there exists a normal operator $N_{1}\in \mathcal {B}(H_{B})$ and bounded Borel functions {f _1i } such that S _1i = f _1i (N ₁). Let $N_{1}={\int }_{\sigma (N_{1})} \omega \mathrm {d}E_{\omega }$ be the spectral decomposition, where $(\sigma (N_{1}), \mathcal {B}, H,E)$ is the spectral measure of N ₁, $\mathcal B$ is the σ-algebra of all Borel subsets of σ(N ₁), the spectrum of N ₁. Thus, we have $S_{1i}={\int }_{\sigma (N_{1})} f_{1i}(\omega )\mathrm {d} E_{\omega }$. Because H _B is separable, there exists a probability measure, that is, the scalar spectral measure, $(\sigma (N_{1}), \mathcal {B}, \mu )$ so that, for any ${\Delta }\in \mathcal {B}$, E(Δ) = 0 if and only if μ(Δ) = 0 (Ref. [24]). Then, by Proposition A.3, there exists an operator-valued function D such that 〈x|D(ω)|x〉≥ 0 a.e. μ for each x ∈ H and

$$\begin{array}{@{}rcl@{}} E({\Delta})=\text{(SOT)}{\int}_{\Delta} D(\omega) \mathrm{d}\mu_{\omega} \end{array} $$

holds for all ${\Delta }\in \mathcal {B}$. By (B.1) one gets, for each product vector x _A ⊗ x _B ∈ H _A ⊗ H _B ,

$$\begin{array}{rl} & \rho_{1}(x_{A}\otimes x_{B}) \\=& p_{1}^{-1} {\sum}_{i,j}(|i_{a}\rangle\langle j_{a}|)\otimes (X_{1}^{\dag} {\int}_{\sigma(N_{1})} f_{1i}(\omega)^{*} f_{1j}(\omega)\mathrm{d}E_{\omega} X_{1})(x_{A}\otimes x_{B})\\ =& p_{1}^{-1} {\sum}_{i,j}(|i_{a}\rangle\langle j_{a}|)x_{A}\otimes (\mathrm{(B)}{\int}_{\sigma(N_{1})} f_{1i}(\omega)^{*} f_{1j}(\omega)X_{1}^{\dag} D(\omega)X_{1}x_{B}\mathrm{d}\mu_{\omega})\\ =& {\sum}_{i,j}\{\text{(B)}{\int}_{\sigma(N_{1})} [ p_{1}^{-1}f_{1i}(\omega)^{*}f_{1j}(\omega)|i_{a}\rangle\langle j_{a}\rangle]x_{A}\otimes [X_{1}^{\dag} D(\omega)X_{1}]x_{B}\mathrm{d}\mu_{\omega}\}. \end{array} $$

Take an orthonormal basis {|j _b 〉} of H _B . For any n, let P _n be the n-rank projection onto span$\{|i_{a}\rangle \}_{i=1}^{n}$ and Q _n be the n-rank projection onto span$\{|j_{b}\rangle \}_{j=1}^{n}$. Then

$$\begin{array}{rl} & (P_{n}\otimes Q_{n})\rho_{1}(P_{n}\otimes Q_{n})(x_{A}\otimes x_{B})\\ = &\text{(B)}{\int}_{\sigma(N_{1})} [ p_{1}^{-1} {\sum}_{i,j=1}^{n}f_{1i}(\omega)^{*}f_{1j}(\omega)|i_{a}\rangle\langle j_{a}\rangle]\otimes [Q_{n}X_{1}^{\dag} D(\omega)X_{1}Q_{n}]\mathrm{d}\mu_{\omega}(x_{A}\otimes x_{B}) \end{array} $$

holds for any x _A ⊗ x _B ∈ H _A ⊗ H _B , which entails that

$$\begin{array}{rl} &(P_{n}\otimes Q_{n})\rho_{1}(P_{n}\otimes Q_{n})\\ =& \mathrm{(B)}{\int}_{\sigma(N_{1})} [ p_{1}^{-1} {\sum}_{i,j=1}^{n}f_{1i}(\omega)^{*}f_{1j}(\omega)|i_{a}\rangle\langle j_{a}\rangle]\otimes [Q_{n}X_{1}^{\dag} D(\omega)X_{1}Q_{n}]\mathrm{d}\mu_{\omega}.\end{array} $$

Let $A_{n}(\omega )={\sum }_{i,j=1}^{n} p_{1}^{-1}f_{1i}(\omega )^{*}f_{1j}(\omega )|i_{a}\rangle \langle j_{a}\rangle ]$ and $B_{n}(\omega )=Q_{n}X_{1}^{\dag } D(\omega )X_{1}Q_{n}$. It is easily seen that A _n (ω) is a rank one positive operator on H _A for each ω ∈ σ(N ₁). As B _n (ω) is bounded and satisfies 〈x _B |B _n (ω)|x _B 〉≥ 0 for any x _B ∈ H _n , B _n (ω) must be a positive operator. Therefore,

$$\begin{array}{rl}\sigma_{n}=&\frac{1}{\text{Tr}((P_{n}\otimes Q_{n})\rho_{1}(P_{n}\otimes Q_{n}))}(P_{n}\otimes Q_{n})\rho_{1}(P_{n}\otimes Q_{n})\\ =& \frac{1}{\text{Tr}((P_{n}\otimes Q_{n})\rho_{1}(P_{n}\otimes Q_{n}))}\mathrm{(B)}{\int}_{\sigma(N_{1})}A_{n}(\omega)\otimes B_{n}(\omega)\ \mathrm{d}\mu_{\omega}\end{array}$$

is a separable state. Since $\rho _{1}=\text {SOT-}\lim _{n\to \infty }(P_{n}\otimes Q_{n})\rho _{1}(P_{n}\otimes Q_{n})$, we see that ρ ₁ is SOT-separable. Then the proposition A.2 ensures that ρ ₁ is a separable state, as desired.

Similarly, one can check that ρ _k is separable for each k, k ≥ 1. Hence, we see that ρ is a separable state, as desired. □