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The Wikipedia page on J-couplings states that

Scalar or J-couplings (also called indirect dipole dipole coupling) are mediated through chemical bonds connecting two spins. It is an indirect interaction between two nuclear spins which arises from hyperfine interactions between the nuclei and local electrons.

My question is: provided that J-couplings are some metric for hyperfine interactions between electron spins and nuclear spins, is there something special we can say regarding J-couplings between two atoms in a lattice with even integer mass numbers, i.e. where the nuclei of both atoms have a spin quantum number of zero? Shouldn't the lack of a nuclear spin on both bonded atoms abrogate any j-coupling term?

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Yes. If any of the two nuclei has spin $j=0$, then there will be no J-coupling between them. This is made clear by the J-coupling hamiltonian, $$H\propto\sum_{ik}I^{(1)}_i J_{ik} I_k^{(2)}.$$

If either spin is zero then all the spin components $I_i$ will be zero as operators, and $H$ will be zero.

You should note that atoms with even mass numbers need not have spin zero. For instance, 60Co has spin 5.

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  • $\begingroup$ Thank you very much for your clear answer. However, may I ask - is there an intuitive reason why the nucleus of 60Co has such a high angular momentum? $\endgroup$ – Bill Boson Oct 22 '14 at 13:13
  • $\begingroup$ There is no requirement that, as you add nucleons to a potential well, they will pair up neatly and leave spin zero. It doesn't happen to electrons in atoms, either (this is known as Hund's rules, and in fact electrons will try and maximize their spin in many cases). Why would you expect it to hold in nuclei? $\endgroup$ – Emilio Pisanty Oct 22 '14 at 13:27

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