# How is the expression for the optimal entanglement witness derived?

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:

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

2. Is this true only for the separable set, or any convex set?

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


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 $$v$$:

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.

• 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? – Mahathi Vempati May 17 at 18:23
• 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 – glS May 17 at 18:39
• @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? – Mahathi Vempati Jun 2 at 10:30
• @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 – glS Jun 2 at 13:26