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 $\epsilon_0 > 0$?


2 Answers 2


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.

  • $\begingroup$ 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 $\epsilon > 0$. $\endgroup$
    – Rammus
    Aug 4, 2020 at 15:16

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