# Are entanglement witnesses of this form optimal?

One can make an entanglement witness by taking the partial transpose of any pure entangled state.

Consider $$|\phi \rangle$$ as any pure entangled state. Then $$W = | \phi \rangle \langle \phi |^{T_2}$$ is an entanglement witness.

However, is there any way to prove that it is optimal? That is if $$\mathrm{Tr}(W \rho) = 0,$$ then $$\rho$$ must lie on the boundary of the separable state set.

Edit: This question is only in reference to 2 qubit systems - hence there are no PPT entangled states.

It's still not true for qubit systems. Consider these two unit vectors, both of which are entangled: $$|\phi\rangle = \frac{1}{\sqrt{2}} | 00\rangle + \frac{1}{\sqrt{2}} | 11\rangle,\\ |\psi\rangle = \frac{1}{\sqrt{2}} | 01\rangle + \frac{i}{\sqrt{2}} | 10\rangle.$$ Let $$\rho = |\psi\rangle\langle \psi |$$, which of course is not on the boundary of the set of separable states; it's as far away from that set as any state can be. However, we have $$\text{Tr}\Bigl( |\phi\rangle \langle \phi |^{T_2} \rho\Bigr) = 0.$$ To verify this equation, it may help to view the relevant operators as matrices: $$|\phi\rangle \langle \phi |^{T_2} = \frac{1}{2}\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}$$ and $$\rho = |\psi\rangle \langle \psi | = \frac{1}{2}\begin{pmatrix}0&0&0&0\\0&1&-i&0\\0&i&1&0\\0&0&0&0\end{pmatrix}.$$

For a counter-example, let $$\rho$$ be any entangled PPT state whose partial transpose has an entangled state $$|\phi\rangle$$ in its kernel. We then obtain $$\text{Tr}\Bigl( |\phi\rangle \langle \phi |^{T_2} \rho\Bigr) = \text{Tr}\Bigl( |\phi\rangle \langle \phi | \rho^{T_2}\Bigr) = 0,$$ but because $$\rho$$ is entangled it cannot be on the boundary of the set of separable states.
To construct an entangled PPT state $$\rho$$ whose partial transpose has an entangled state in its kernel, one may take any unextendable product basis $$|\phi_1\rangle |\psi_1\rangle,\ldots,|\phi_m\rangle |\psi_m\rangle \in \mathcal{A}\otimes\mathcal{B},$$ and then let $$\rho = \frac{1}{m} \sum_{k = 1}^m |\phi_k\rangle \langle \phi_k| \otimes |\psi_k\rangle \langle \psi_k|.$$ The partial transpose looks like this: $$\rho^{T_2} = \frac{1}{m} \sum_{k = 1}^m |\phi_k\rangle \langle \phi_k| \otimes |\overline{\psi_k}\rangle \langle \overline{\psi_k}|.$$ This operator has rank $$m<\text{dim}(\mathcal{A}\otimes\mathcal{B})$$ and every nonzero vector in its kernel is entangled.