# 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.)

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 $\epsilon > 0$. – Rammus Aug 4 at 15:16