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

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