  We assume all interaction to be Coulombic and solve the modified Faddeev
equation for energies between the $Ps(n=2)$ and $\bar{H}(n=3)$, which involve
six and eight open channels. We find that 99% of the antihydrogen are formed in
$\bar{H}(n=2)$. Just above the $Ps(n=2)$ threshold the S, P, and D partial
waves contribute more than 4000 square Bohr radii near the maximum. Evidences
indicate that the induced long range dipole potential from the degenerate
$Ps(n=2)$ targets is responsible for such a large antihydrogen formation cross
section.
