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I see a lot of Venn kind of diagrams to depict the distinction between separable and entangled mixed states. Like this one (apologize my poor "paint"ings):

$\hskip2.0in$enter image description here

So there is an entangled state $\rho$. It is entangled because it is not inside the set of separable states. My question is inspired by the picture:

Is there a separable state that is furthest away from the entangled one?

I tried to indicate it in red, but can we calculate the separable state?

To be concrete: What is the separable state furthest away from a GHZ state:$(|00\cdots 0\rangle+|11\cdots 1\rangle)/\sqrt2$...

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  • $\begingroup$ Are you familiar with entanglement witnesses? I guess you could use the optimal entanglement witness to define a distance metric, which you can use to measure all separable states.... $\endgroup$ – DaftWullie May 13 at 15:26
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    $\begingroup$ This paper gives the distance to the closest separable state, arxiv.org/pdf/1006.3077.pdf, p.g 6 proposition 5, thus there must also be a state which can be defined at the 'furthest' $\endgroup$ – Sam Palmer May 13 at 16:02
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    $\begingroup$ In which distance measure? And yes, there is a farthest state (or several), since the set of separable states is closed. $\endgroup$ – Norbert Schuch May 13 at 16:21
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    $\begingroup$ @NorbertSchuch 's point is that there are many different distance measures that one could use (potentially with different answers). Do you have a context in which you want your question answered which might suggest a particular choice of distance measure? Or do you have a favourite measure? $\endgroup$ – DaftWullie May 14 at 7:22
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    $\begingroup$ @Yack I just looked at the abstract, but knowing the concept of robustness: Couldn't it be that the decomposition involves the states furthest away in a specific direction, which is determined by the closest separable state? $\endgroup$ – Norbert Schuch May 14 at 8:49
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Whenever under the distance measure you are using, the set of separable states is closed, there is a furthest point - this is a feature of closed sets. This is certainly the case whenever the distance measure is continuous relative to any other measure relative to which the separable states are closed, such as the trace norm distance.

In particular, there is always a furthest state as measured by the trace norm distance.

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  • $\begingroup$ Thanks, your answers ensures the existence. In order to make up an actual state, can you confirm that you meant this trace norm distance: en.wikipedia.org/wiki/Trace_distance ? $\endgroup$ – draks ... May 15 at 6:36
  • $\begingroup$ @draks... Yes. - $\endgroup$ – Norbert Schuch May 15 at 10:08
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Here's an example based on entanglement witnesses. Let $$ |\psi\rangle=\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle). $$ The aim is to define an operator $W$ such that $\text{Tr}(W\rho_\text{sep})\geq 0$ for all separable states $\rho_\text{sep}$. That means for any $\rho$ such that $\text{Tr}(W\rho)<0$, it must be entangled.

An easy way to define such an operator is $$ W=\alpha I-|\psi\rangle\langle\psi|, $$ where $$ \alpha=\max_{\rho_\text{sep}}\langle\psi|\rho_\text{sep}|\psi\rangle. $$ In this particular case, the maximisation is achieved with $\rho_\text{sep}=|0\rangle\langle 0|^{\otimes 3}$, I believe, so $\alpha=\frac12$.

Of course, $\text{Tr}(W|\psi\rangle\langle\psi|)=\alpha-1=-\frac12$, so we see that it's entangled.

Now consider any other state $\rho$. We can think of the value $\text{Tr}(W\rho)$ as measuring distance. The smallest possible value is $-\frac12$, and if you have that, you've got the state $|\psi\rangle$. The larger the value, there is a sense of getting further away from $|\psi\rangle$. Now, the maximum value of $\text{Tr}(W\rho)$ is $\alpha=\frac12$, which is achieved by $\langle\psi|\rho|\psi\rangle=0$. So this categorises the states that are furthest away according to this specific measure. There are separable states which are orthogonal to $|\psi\rangle$, such as $|001\rangle$, and so they can be considered the furthest separable states.

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  • $\begingroup$ Thanks, would a (mixed?) states composed of all orthogonal states also work? Picking one random example feels a little ... hmm... random. $\endgroup$ – draks ... May 15 at 6:38
  • $\begingroup$ Yes, of course. Any choice of separable state that is orthogonal to $|\psi\rangle$ is fine. That includes mixtures. That random feeling is why we were trying to push for more concrete details of what you were after to make it less random. For example, there are probably more sophisticated entanglement witnesses, or other forms of distance measure, which have better resolution, and might single out more states. $\endgroup$ – DaftWullie May 15 at 7:34
  • $\begingroup$ The "problem" with witnesses is that they give the furthest state in a specific direction. (Not that it has to be a problem, but it does make the distance measure somewhat special, e.g. it is not a metric.) $\endgroup$ – Norbert Schuch May 16 at 18:35
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A somewhat informal comment.

Some intuition can be developed from the fact that the pure states are rays in the complex projective space (subject to an additional normalization constraint). The most 'different' the two vectors in such a space can be is 'perpendicular' ⁠— if you keep $v_1$ moving away from $v_2$ in a certain plane, the distance will start increasing once you pass the perpendicular direction (think about the distance along the arc between two points on a circle, or about the maximum angle between any two lines passing through the origin). We also know that, because of the normalization condition, there is a maximum value for the scalar product between two vectors ⁠— given by the vectors' norm. Thus, in the sense of a particular norm, the two states can be as close as $0$, and as far as $1$ apart from each other.

Note that the trace norm distance between the states, which I was implicitly referring to, is precisely the one inherited from the scalar product of two vectors in the complex projective space (which, in turn, is inherited from the usual scalar product of vectors in $\mathbb{C}^n$): while for vectors we have $d(|\psi\rangle,|\psi'\rangle)=|\langle \psi | \psi' \rangle|$, for the corresponding pure density matrices we have $d(|\psi\rangle\langle\psi|, |\psi'\rangle\langle\psi'|) =\operatorname{tr} (|\psi\rangle\langle\psi||\psi'\rangle\langle\psi'| )$, which is obtained by squaring the previous expression.

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$\newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}\newcommand{\bke}[3]{\left<#1\middle|#2\middle|#3\right>}$

Let $\ket{GHZ}$ be a GHZ state with $n$ qubits. So thanks to Norbert's answer I found the following separable pure states that maximize the trace distance: $$ \ket\psi=\frac1{2^{n/2}}\otimes_{k=1}^n(\ket0 +\exp{\left(i\pi f(k)\right)} \ket1), $$ with $\sum_{k=1}^n f(k)=1$. By that we get the coefficient of $\ket{00\cdots 0}$ to be $1$ and the one of $\ket{11\cdots 1}$ to be $-1$. Therefore $\bk\psi{GHZ}=0$ holds. They all have trace distance: $$ \frac12\sum |\lambda_i|=1, $$ where $\lambda_i$ are the eigenvalues of $\left(\ket{GHZ}\bra{GHZ}-\ket\psi\bra\psi\right)$. Both states share the same permutation symmetry if $f(k)=1/n$...

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    $\begingroup$ You are supposed to edit your answer, not to delete and re-post it. $\endgroup$ – Norbert Schuch May 16 at 18:33

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