Sinkhorn–Knopp theorem for PPT states

Abstract

Given a PPT state $A=\sum _{i=1}^nA_i\otimes B_i \in M_k\otimes M_k$ and a rank k tensor v within the image of A, we provide an algorithm that checks whether the positive map $G_A:M_k\rightarrow M_k$, $G_A(X)=\sum _{i=1}^n tr(A_iX)B_i$, is equivalent to a doubly stochastic map. This procedure is based on the search for Perron eigenvectors of completely positive maps and unique solutions of, at most, k unconstrained quadratic minimization problems. As a corollary, we can check whether this state can be put in the filter normal form. This normal form is an important tool for studying quantum entanglement. An extension of this procedure to PPT states in $M_k\otimes M_m$ is also presented.

Appendix

Proof of Lemma 1

By Definition 2, $T(X)=\sum _{i=1}^mA_iXA_i^*.$

Now, if $T(V)=\sum _{i=1}^mA_iVA_i^*\in VM_kV$, then $A_iVA_i^*\in VM_kV$, for every i.

Since $A_iVA_i^*=A_iV(A_iV)^*$, then $A_iV\in VM_k$. Thus, $VA_i^*\in M_kV$, for every i.

Hence, $A_i(VX+YV)A_i^*\in VM_k+M_kV$, for every $X,Y\in M_k$. $\square$

Proof of Lemma 2

(a) Note that

$$\begin{aligned} 0= & {} tr(T(V_1)(V-V_1))=tr(V_1T^*(V-V_1))=tr(VV_1VT^*(V-V_1))\\= & {} tr(V_1VT^*(V-V_1)V). \end{aligned}$$

Thus, $\mathfrak {I}(VT^*(V-V_1)V)\subset \mathfrak {I}(V-V_1)$ and

$$\begin{aligned} VT^*((V-V_1)M_k(V-V_1))V\subset (V-V_1)M_k(V-V_1). \end{aligned}$$

(b) If $T|_{VM_kV}$ is not irreducible, neither is $VT^*(\cdot )V|_{VM_kV}$ by item (a).

Now, if $VT^*(\cdot )V:VM_kV\rightarrow VM_kV$ is not irreducible, then there is a proper subalgebra $V_1M_kV_1\subset VM_kV$ such that $VT^*(V_1M_kV_1)V\subset V_1M_kV_1$ and $V_1\ne 0$.

Hence, $0=tr(VT^*(V_1)V(V-V_1))=tr(T^*(V_1)V(V-V_1)V)=tr(V_1T(V-V_1))$. Thus,

$$\begin{aligned} T((V-V_1)M_k(V-V_1))\subset (V-V_1)M_k(V-V_1). \end{aligned}$$

(c) Let $\gamma \in P_k\cap VM_kV$ be such that $\gamma ^2=\delta$. Denote by $\gamma ^{+}$ the Hermitian pseudo-inverse of $\gamma$. So $\gamma ^{+}\gamma =\gamma \gamma ^{+}=V$.

Note that $\frac{1}{\lambda }\gamma ^{+}T(\gamma V\gamma )\gamma ^{+}=V$. Therefore, the operator norm of

$$\begin{aligned} \frac{1}{\lambda }\gamma ^{+}T(\gamma (\cdot ) \gamma )\gamma ^{+}:VM_kV\rightarrow VM_kV \end{aligned}$$

induced by the spectral norm on $M_k$ is 1 [2, Theorem 2.3.7].

Let $Y\in VM_kV$ be an eigenvector of $T:VM_kV\rightarrow VM_kV$ associated with $\alpha$ such that the spectral norm of $\gamma ^{+}Y\gamma ^{+}$ is 1. Then, the spectral norm of $\frac{1}{\lambda }\gamma ^{+}T(\gamma \gamma ^{+}Y\gamma ^{+} \gamma )\gamma ^{+}=\frac{\alpha }{\lambda }\gamma ^{+}Y\gamma ^{+}$ is smaller or equal to the operator norm of $\frac{1}{\lambda }\gamma ^{+}T(\gamma (\cdot ) \gamma )\gamma ^{+}:VM_kV\rightarrow VM_kV$. Thus, $\frac{|\alpha |}{\lambda }\le 1$. $\square$

Proof of Lemma 3

Since $VT^*(\cdot )V|_{VM_kV}$ and $T|_{VM_kV}$ are adjoint with respect to the trace inner product and $\lambda \in {\mathbb {R}}$, then $\text {rank}(VT^*(\cdot )V-\lambda Id|_{VM_kV})= \text {rank}(T-\lambda Id|_{VM_kV})$. So the geometric multiplicity of $\lambda$ is the same for both maps.

Moreover, $T|_{VM_kV}$ is irreducible if and only if $VT^*(\cdot )V|_{VM_kV}$ is irreducible by Lemma 2. item (b). So conditions (1) and (2) are necessary for irreducibility by Perron–Frobenius theory [11, Theorem 2.3].

Next, let us assume by contradiction that conditions (1) and (2) hold and $T|_{VM_kV}$ is not irreducible.

Thus, there is an orthogonal projection $V_1\in M_k$ such that $T(V_1M_kV_1)\subset V_1M_kV_1$, $V_1V=VV_1=V_1$ and $V-V_1\ne 0$.

By item (a) of Lemma 2, $VT^*((V-V_1)M_k(V-V_1))V\subset (V-V_1)M_k(V-V_1)$. Since $T^*:M_k\rightarrow M_k$ is completely positive, then $VT^*(\cdot )V:M_k\rightarrow M_k$ is too.

Hence, by Lemma 1,

$$\begin{aligned} VT^*((V-V_1)M_k+M_k(V-V_1))V\subset (V-V_1)M_k+M_k(V-V_1). \end{aligned}$$

(5)

Now, since $V_1V=V_1$ and $\lambda \delta =VT^*(\delta )V=VT^*((V-V_1)\delta +V_1\delta (V-V_1))V+VT^*(V_1\delta V_1)V$, then, by Eq. 5, $\lambda V_1\delta V_1=V_1T^*(V_1\delta V_1)V_1.$

Note that $V_1\delta V_1\ne 0$, since $\mathfrak {I}(V_1)\subset \mathfrak {I}(V)=\mathfrak {I}(\delta )$. Hence, $\lambda$ is an eigenvalue of $V_1T^*(\cdot )V_1: V_1M_kV_1\rightarrow V_1M_kV_1$. Therefore, it is also an eigenvalue of its adjoint $T:V_1M_kV_1\rightarrow V_1M_kV_1$.

Actually, $\lambda$ is the spectral radius of $T|_{V_1M_kV_1}$, since $V_1M_kV_1\subset VM_kV$ and by hypothesis. So, by Perron–Frobenius theory, there is $\gamma '\in (V_1M_kV_1 \cap P_k){\setminus } \{0\}$ such that $T(\gamma ')=\lambda \gamma '$.

Note that $\gamma$ and $\gamma '$ are linearly independent, since $\mathfrak {I}(\gamma ')\subset \mathfrak {I}(V_1)\ne \mathfrak {I}(V)=\mathfrak {I}(\gamma )$. Thus, the geometric multiplicity of $\lambda$ is not 1. Absurd! $\square$