/quantum information – Are entanglement witnesses of this form optimal? (via Qpute.com) ## quantum information – Are entanglement witnesses of this form optimal? (via Qpute.com)

Answer to edited question:

It’s still not true for qubit systems. Consider these two unit vectors, both of which are entangled:
$$|phirangle = frac{1}{sqrt{2}} | 00rangle + frac{1}{sqrt{2}} | 11rangle,\ |psirangle = frac{1}{sqrt{2}} | 01rangle + frac{i}{sqrt{2}} | 10rangle.$$
Let $$rho = |psiranglelangle psi |$$, 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
$$text{Tr}Bigl( |phirangle langle phi |^{T_2} rhoBigr) = 0.$$
To verify this equation, it may help to view the relevant operators as matrices:
$$|phirangle langle phi |^{T_2} = frac{1}{2}begin{pmatrix}1&0&0&0\0&0&1&0\0&1&0&0\0&0&0&1end{pmatrix}$$
and
$$rho = |psirangle langle psi | = frac{1}{2}begin{pmatrix}0&0&0&0\0&1&i&0\0&-i&1&0\0&0&0&0end{pmatrix}.$$

Original answer:

If I understand it correctly, the statement suggested in the question is false.

For a counter-example, let $$rho$$ be any entangled PPT state whose partial transpose has an entangled state $$|phirangle$$ in its kernel. We then obtain
$$text{Tr}Bigl( |phirangle langle phi |^{T_2} rhoBigr) = text{Tr}Bigl( |phirangle langle phi | rho^{T_2}Bigr) = 0,$$
but because $$rho$$ is entangled it cannot be on the boundary of the set of separable states.

Incidentally, no entanglement witness of the form suggested in the question can ever detect entanglement in a PPT state.

To construct an entangled PPT state $$rho$$ whose partial transpose has an entangled state in its kernel, one may take any unextendable product basis
$$|phi_1rangle |psi_1rangle,ldots,|phi_mrangle |psi_mrangle in mathcal{A}otimesmathcal{B},$$
and then let
$$rho = frac{1}{m} sum_{k = 1}^m |phi_krangle langle phi_k| otimes |psi_krangle langle psi_k|.$$
The partial transpose looks like this:
$$rho^{T_2} = frac{1}{m} sum_{k = 1}^m |phi_krangle langle phi_k| otimes |overline{psi_k}rangle langle overline{psi_k}|.$$
This operator has rank $$m and every nonzero vector in its kernel is entangled.$$

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