Let \(\lbrace \ket{0},\ket{1},\ket{2}\rbrace\) be the computational basis of a qutrit system, and consider the following quantum state on two qutrits defined in [1]:
\begin{align*} \rho &= \frac{2}{7} \ket{\psi}\bra{\psi} + \frac{1}{2} \sigma_+ + \frac{3}{14}\sigma_-,\\ \text{where}\quad \ket{\psi} &= \frac{1}{\sqrt{3}} (\ket{00}+\ket{11}+\ket{22})\\ \sigma_+ &= \frac{1}{3} (\ket{01}\bra{01}+\ket{12}\bra{12}+\ket{20}\bra{20})\\ \sigma_- &= \frac{1}{3} (\ket{10}\bra{10}+\ket{21}\bra{21}+\ket{02}\bra{02}), \end{align*}
(Click for a matrix form of ρ)
\begin{align*} \rho = \begin{pmatrix} 2/21 & . & . & . & 2/21 & . & . & . & 2/21\\ . & 1/6 & . & . & . & . & . & . & . \\ . & . & 1/14 & . & . & . & . & . & . \\ . & . & . & 1/14 & . & . & . & . & . \\ 2/21 & . & . & . & 2/21 & . & . & . & 2/21\\ . & . & . & . & . & 1/6 & . & . & . \\ . & . & . & . & . & . & 1/6 & . & . \\ . & . & . & . & . & . & . & 1/14 & . \\ 2/21 & . & . & . & 2/21 & . & . & . & 2/21 \end{pmatrix}. \end{align*}
The state \(\rho\) is PPT and entangled by the results of [1].
To construct a (symmetric) 2-extension of \(\rho\), consider the following vectors that are symmetric in the 2nd and 3rd system:
\begin{align*} \ket{\chi_0} &= \ket{0}\otimes\ket{00} + \ket{1}\otimes(\ket{01}+\ket{10})+\ket{2}\otimes(\ket{02}+\ket{20})\\ \ket{\chi_1} &= \ket{0}\otimes(\ket{01}+\ket{10}) + \ket{1}\otimes\ket{11} + \ket{2}\otimes(\ket{12}+\ket{21})\\ \ket{\chi_2} &= \ket{0}\otimes(\ket{02}+\ket{20})+\ket{1}\otimes(\ket{12}+\ket{21})+\ket{2}\otimes\ket{22}. \end{align*}
The following state \(\Omega\) is symmetric in the second and third system and satisfies \(\operatorname{tr}_2 \Omega = \operatorname{tr}_3 \Omega = \rho\), thus giving a symmetric 2-extension of \(\rho\):
\begin{align*} \Omega &= \frac{2}{63}(\ket{\chi_0}\bra{\chi_0} + \ket{\chi_1}\bra{\chi_1} + \ket{\chi_2}\bra{\chi_2})\\ & \quad {} + \frac{17}{126} (\ket{011}\bra{011} + \ket{122}\bra{122} + \ket{200}\bra{200})\\ & \quad {} + \frac{5}{126} (\ket{100}\bra{100} + \ket{211}\bra{211} + \ket{022}\bra{022}). \end{align*}
The marginal of \(\rho\) on the first system is maximally mixed, and thus \(\rho\) is a bona fide Choi state. Since a channel is antidegradable if and only if its Choi operator is 2-extendible [2], the operator \(\rho\) is in fact the Choi state of a channel that is PPT and antidegradable, but not entanglement-breaking.
References
[1] P. Horodecki, M. Horodecki, and R. Horodecki, “Bound Entanglement Can Be Activated,” Physical Review Letters, vol. 82, no. 5, pp. 1056–1059, Feb. 1999, doi: 10.1103/PhysRevLett.82.1056. [doi] [arXiv] [2] G. O. Myhr, “Symmetric extension of bipartite quantum states and its use in quantum key distribution with two-way postprocessing,” PhD thesis, Friedrich-Alexander-Universität Erlangen-Nürnberg, 2010. [arXiv]