Comments on calculations of nuclear spin-spin coupling constants using the blinder operator

It is pointed out that the finite second-order hyperfine self-coupling energies, obtained by the Blinder operator, are unphysical and 10² times too large. The additional terms in the first-order matrix element are also unphysical. Therefore the Blinder operator gives no improvement over the simple delta function. The consequences on the recent calculations by Paviot and Hoarau and by Sänger and Voitländer are discussed.