Are all pure entangled states `robust'?

Let $$\mathcal{H}_A \otimes \mathcal{H}_B$$ be the tensor product of two finite dimensional Hilbert spaces, let $$d = \operatorname{dim}(\mathcal{H}_A \otimes \mathcal{H}_B)$$ and let $$| \psi \rangle \in \mathcal{H}_A \otimes \mathcal{H}_B$$ be a pure entangled state.

We say the entanglement in $$| \psi \rangle$$ is $$\epsilon_0$$-robust, for some $$\epsilon_0 \in [0,1]$$, if $$(1-\epsilon) | \psi \rangle \langle \psi | + \epsilon \, \mathbb{I}/d$$ is entangled for all $$\epsilon \in [0, \epsilon_0]$$. More generally we say the entanglement in $$| \psi \rangle$$ is completely $$\epsilon_0$$-robust if $$(1-\epsilon) | \psi \rangle \langle \psi | + \epsilon \, \tau$$ is entangled for all $$\epsilon \in [0, \epsilon_0]$$ and all states $$\tau$$ on $$\mathcal{H}_A \otimes \mathcal{H}_B$$.

Are there any pure entangled states that are not $$\epsilon_0$$-robust (or completely $$\epsilon_0$$-robust) for all $$\epsilon_0 > 0$$?

The set of separable states is closed.

Thus, around any entangled state - not necessarily pure - there is an $$\epsilon$$-ball which lies entirely within the entangled states.

Or, in the language of your question: All entangled states are "robust".

(As illustrated by DaftWullie's answer, the size of this ball can depend on the state: There are pure entangled states arbitrarily close to separable ones.)

• Thanks! Interestingly, things seem to break down if we move to infinite dimensional Hilbert spaces Is the set of separable quantum states closed? Aug 4, 2020 at 15:53
• @Rammus You also can't add the identity in infinite dimensions. Aug 4, 2020 at 20:33
• Yes, but one could still consider epsilon balls around the state. It seems like it is not the case though, apparently the set of entangled states is dense (w.r.t. the trace norm) in the set of quantum states on infinite dimensional bipartite systems -- Bipartite Mixed States of Infinite-Dimensional Systems are Generically Nonseparable. Thanks for the help! Aug 4, 2020 at 21:56

For a fixed $$\epsilon_0$$, why not simply consider $$|\psi\rangle=\cos\theta|00\rangle+\sin\theta|11\rangle?$$ Since it's a two-qubit state, entanglement can be determined using the PPT criterion. Hence, $$\rho=(1-\epsilon)|\psi\rangle\langle\psi|+\epsilon I/4$$ is entangled if $$\epsilon<2\sin(2\theta)/(1+2\sin(2\theta))$$. Any $$\epsilon$$ you give me, and I just pick $$0<\theta<\arcsin(\epsilon_0/2)/2\approx\epsilon_0/4$$, and the state is not $$\epsilon_0$$-robust. Given there exists a state that is not $$\epsilon_0$$-robust, entanglement is not completely $$\epsilon_0$$-robust.

To prove things the other way around (for fixed $$|\psi\rangle$$, is there always a non-zero $$\epsilon_0$$ such that for all $$\epsilon<\epsilon_0$$, the mixed state is entangled?), we can consider entanglement witnesses. Let $$W$$ be an entanglement witness for $$|\psi\rangle$$. We have $$\text{Tr}(W\rho)\geq 0$$ for all separable states $$\rho$$ and $$\text{Tr}(W|\psi\rangle\langle \psi|)=-\sigma<0$$.

Now, the trace of $$W$$ will be some specific value $$\text{Tr}(W)=k\geq 0$$ (this is positive since the maximally mixed state is separable). Consider $$\text{Tr}(W(\epsilon I+(1-\epsilon)|\psi\rangle\langle\psi|))=\epsilon k-(1-\epsilon)\sigma.$$ For any $$\epsilon<\frac{\sigma}{k+\sigma},$$ the trace is negative and hence the state is entangled.

• Thanks, but I didn't mean to ask if for a fixed $\epsilon_0$ whether there exists a state that is not $\epsilon_0$-robust. Rather I meant to ask whether there exists a state which has no non-zero robustness, i.e. $| \psi \rangle$ is entangled but $(1-\epsilon) | \psi \rangle + \epsilon \, \mathbb{I}/d$ is separable for all $\epsilon > 0$. Aug 4, 2020 at 15:16