What are examples of weakly optimal witnesses?

Question

While discussing witnesses, in https://arxiv.org/abs/0811.2803 the authors mention (page 16 of the arxiv version, below Eq. (32)) that a necessary condition for a witness $W$ to be optimal is that it touches the set of separable states, meaning there's some separable $\rho_s$ such that $\operatorname{tr}(\rho_s W)=0$. They then mention that this condition is not sufficient for optimality, and a witness satisfying this condition is sometimes called "weakly optimal".

From a geometric perspective, this seems weird to me: I imagine witnesses as hyperplanes in state space separating separable states $\mathcal C$ from the rest, as per the hyperplane separation theorem. So if a witness corresponds to a hyperplane touching $\mathcal C$, it seems intuitive that it should be optimal, meaning the plane can't be "made closer" to $\mathcal C$. On the other hand, it might just be possible that one can move the plane in a direction "orthogonal" to the one touching $\mathcal C$. It's just hard to imagine things in such high dimensions.

What are examples of witnesses that are "weakly optimal" in the above sense?

Answer 1

Consider the optimal witness $$W = I-2|\psi^-\rangle \langle \psi^-|,$$ and now add some amount of $|01\rangle\langle01|$ to it, i.e., let $$W' = W + \alpha |01\rangle\langle01|,$$ for some $\alpha \in (0,2)$. Now this is clearly still a witness, since we just added a positive operator to it, and it still touches the set of separable states at $|10\rangle$, so it is still weakly optimal, but it is no longer optimal, as $\langle 01 |W' |01\rangle = \alpha > 0$.

Geometrically, we can think of this addition as tilting $W$ away from the set of separable states, while keeping fixed the contact point with $|10\rangle$.

Answer 2

The following seems to work: consider in $\mathbb{C}^3\otimes\mathbb{C}^3$ the operator: $$W \equiv \frac{I - 2\mathbb{P}(|\Psi^-\rangle)}{2} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix},$$ where $\{|0\rangle,|1\rangle,|2\rangle\}$ span each of the two qutrit spaces. This $W$ is a witness because it has the form $W=\alpha I-\mathbb{P}(|\Psi^-\rangle)$ with $\mathbb{P}(|\Psi^-\rangle)$ projection onto $|\Psi^-\rangle\equiv\frac{|01\rangle-|10\rangle}{\sqrt2}$, and $\alpha=\frac12$ the largest (squared) Schmidt coefficient of $|\Psi^-\rangle$. Its action is trivial on all computational basis states except for $$W|01\rangle=|10\rangle, \qquad W|10\rangle=|01\rangle.$$

It's also (at least) weakly optimal because $\langle \mathbb{P}_{01},W\rangle=\langle \mathbb{P}_{10},W\rangle=0$, and it's a proper witness because it detects $|\Psi^-\rangle$ as entangled.

However, it's not optimal, because denoting with $P_W$ the set of product states $|e,f\rangle$ such that $\langle \mathbb{P}_e\otimes\mathbb{P}_f,W\rangle=0$, as per the results mentioned here, one can observe that $|2,2\rangle$ isn't in the span of $P_W$. That's because for $|2,2\rangle$ to be spanned by vectors in $P_W$ one would need vectors in there of the form $|e\rangle\otimes|f\rangle$ where both $|e\rangle$ and $|f\rangle$ have nonzero overlap with $|2\rangle$. But then, expanding the expectation value, that would produce terms of the form $|\langle 2,f\rangle\langle 2,e\rangle|^2$, which can't be zero. Thus $|2,2\rangle$ is not spanned by $P_W$.

The projection $P\equiv \mathbb{P}_{2,2}\equiv \mathbb{P}_2\otimes\mathbb{P}_2$ is then a positive semidefinite nonzero operator such that $$W-P = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix}$$ is still a witness. Hence $W$ isn't optimal.

Answer 3

Here's another simple 2-qubit example based on the flip/swap operator $\mathbb F$—which is the Choi matrix of the transpose map, hence a witness: $$ W=\mathbb F+\alpha|00\rangle\langle 00|=\begin{pmatrix} 1+\alpha&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1 \end{pmatrix} $$ for any $\alpha>0$. This operator is

• block-positive because it is the sum of a witness and a positive operator.
• a proper witness because it detects the entanglement of $|\psi\rangle\langle\psi|$ with $|\psi\rangle=\frac1{\sqrt2}(|01\rangle-|10\rangle)$ (because $\langle\psi|W|\psi\rangle=-1$).
• not optimal because subtracting $\alpha|00\rangle\langle 00|$ from $W$ still is a witness, cf. Thm. 1 in "Optimization of entanglement witnesses" by Lewenstein et al. (arXiv).
• weakly optimal because $\langle 01|W|01\rangle=0$ (or $\langle 10|W|10\rangle=0$).