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 \begin{equation} 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, \end{equation} 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.


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