In the Bertlmann 2009 paper in the Annals of Physics (here), an optimal witness operator for an entangled state $\rho$, given that the closest separable state to it is $\rho_0$ is given by:

$$A_{\text{opt}} = \frac{\rho_0 - \rho - \langle \rho_0, \rho_0 - \rho \rangle I}{|| \rho_0 - \rho ||} $$

  1. How was this expression arrived at?
  2. Is this true only for the separable set, or any convex set?

1 Answer 1


The idea behind this expression is indeed a fairly general one.

$\newcommand{\bs}[1]{\boldsymbol{#1}}\newcommand{\calS}{\mathcal{S}}$An entanglement witness $\mathcal W$ is defined as an operator such that $\operatorname{Tr}(\mathcal W\rho)\ge0$ for all separable $\rho\in\mathcal S$, while for some entangled $\rho_{ent}$ we have $\operatorname{Tr}(\mathcal W\rho_{ent})<0$.

Geometrically, this definition is very easily understood as saying that $\mathcal W$ defines a hyperplane in the space of states that separates the separable states from the non-separable ones. Because of the convexity of the space of separable states, any non-separable state can be separated by such an operator (see e.g. (Horodecki 2007) or (Gühne 2008)).

Now forget for a second about states and quantum mechanics. Let $\calS$ denote a convex subset of $\mathbb R^n$ for some $n$, let $ v\notin \calS$, and let $ v_0\in\calS$ be the vector in $\calS$ that is the closest to $ v$ (in standard euclidean distance). This means that the line (or more generally hyperplane) that is orthogonal to $ v- v_0$ and touches $ v_0$ is tangent to $\calS$ at $ v_0$. Such a "line" is an "optimal" linear separation between $\calS$ and $ v$:

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We now simply need to define an operator that tells us on which side of such separation we are in. A natural candidate would be an operator which projects a candidate vector $ w$ on the line $ v- v_0$, or, more precisely, an operator $A$ defined via

$$A w\equiv\langle v- v_0, w- v_0\rangle.$$

Clearly, we then have $A v>0$, and $A v_0=0$, while all vectors in $\calS$ (together with the points on the $\calS$ side of the separation) correspond to $Aw<0$.

To obtain the given expression for the witness you now simply change the sign (because we conventionally define witnesses to be positive on the separable).

In conclusion, you have $A_{opt} w=\langle v- v_0, v_0- w\rangle$, that corresponds to $$A_{opt}=( v_0^*-v^*)-\langle v_0, v_0-v\rangle I.$$ where $v^*$ denotes the linear functional $v^*(w)\equiv \langle v,w\rangle$ (or, if you prefer, the bra $\langle v\rvert$).

To get to the expression given in the paper you now just add a normalisation factor, which I guess was added to make something simpler later on in the paper.

  • $\begingroup$ When we imagine them to be systems in $R^n$, we just take inner product, but since these are in matrix form, we take Trace(AB) which is like inner product of a vector made up of components of the matrix, but scaled in some dimensions. For example, inner product (using trace)between A = [a1 a2+ia3; a2 - ia3 a4] and B = [b1 b2+ib3; b2 - ib3 b4] is a1b1 + 2a2b2 - 2a3b3 + a4b4. Would it be okay to use trace as the inner product? Even to find the closest state? $\endgroup$ May 17, 2019 at 18:23
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    $\begingroup$ yes, the inner product here is defined as $\langle A,B\rangle\equiv\operatorname{Tr}(A^\dagger B)$, or $\mathrm{Tr}(AB)$ when $A$ is Hermitian $\endgroup$
    – glS
    May 17, 2019 at 18:39
  • $\begingroup$ @gIS, also, I can intuitively see that a line drawn orthogonal to the (line from closest point to point in consideration) at closest point will be a tangent, but is it possible to more rigorously prove this? Can I find the proof somewhere? This happens only due to the convexity of the class, right? $\endgroup$ Jun 2, 2019 at 10:30
  • $\begingroup$ @MahathiVempati yes it is due to the convexity of the space. For a rigorous proof I'm not sure. An argument that comes to mind is that if this was not true, then there would be a point $z$ belonging to the set but lying on the other side of the plane. But then by convexity of the set, the line from $z$ to $v_0$ would also be contained in the set, and therefore the plane would not be tangent to the space $\endgroup$
    – glS
    Jun 2, 2019 at 13:26
  • $\begingroup$ @gIS, This answer gave me extremely useful intuition, and I would like to include the last part in the paper I'm currently working on. I would also like to acknowledge you (by real name, if possible), or username, otherwise. Would that be okay? How can I go about doing that? $\endgroup$ Sep 29, 2019 at 6:55

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