# Is there a two-qudit Choi entanglement witness $W^{(+)}$?

Example 2 in arXiv:1811.09896 states that the "Choi EW (entanglement witness) $$W^{(+)}$$ obtained from the Choi map in $$d=3$$ $$\ldots$$ is given by $$$$W^{(+)} = \frac{1}{6} \left( \sum_{i=0}^{2} [ 2| ii \rangle \langle ii | + | i,i-1 \rangle \langle i, i-1 | ] - 3 \mathrm{P}_{+} \right) \nonumber,$$$$ where $$\mathrm{P}_+ = |\phi^+\rangle \langle \phi^+|$$ with the Bell state $$|\phi^+\rangle = (|00\rangle + |11\rangle + |22\rangle) / \sqrt{3}$$." It is noted that this is applicable in the two-qutrit ($$9 \times 9$$ density matrix) setting.

I would like to know - if it exists - a two-qudit analogue, applicable to $$16 \times 16$$ density matrices. If it does, I presume the summation would run from 0 to 3, and the $$\sqrt{3}$$ in the new (obvious) Bell state formula be replaced by 2. Might the coefficient 2 be replaced by 3, and the $$3 P_+$$ by $$4 P_+$$? The $$\frac{1}{6}$$ does not seem of particular importance for testing purposes.

However, the last paragraph of arXiv:1105.4821 strongly suggests that there is no simple/direct two-qudit analogue.