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I have a silly confusion. The Pairing term of the semi-empirical mass formula tells that nucleons tend to pair up. I believe that pairing up means forming a state in which two nucleons combine to give a total $S=0$ (or total $J=0?$). Pairing increases the binding energy.

On the other hand, we know that nuclear forces favour the alignment of spin i.e. $S=1$ is favoured over $S=0$.

Don't these two statements contradict each other and say completely opposite things?

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Because of the Pauli principle the $s$-wave interaction between two nucleons involves either isospin $I=1$ and spin $S=0$, or $I=0$ and $S=1$. Both of these channels are attractive, but the $(I,S)=(0,1)$ is slightly more so. Indeed, only $(I,S)=(0,1)$ forms a bound state, the deuteron, whereas the $I=1$ states (in particular $nn$ and $pp$) are unbound.

For finite nuclei the important observation is that the pairing interaction can act coherently over the entire Fermi surface. This is the phenomenon that leads to Cooper pairing and superfluidity. Finite nuclei have more neutrons than protons (because of the Coulomb interaction), the proton and neutron Fermi surface are shifted with respect to one another, and the $np$ interaction cannot act coherently. The main pairing interaction is therefore $nn$ and $pp$, which acts in the $S=0$ channel.

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  • $\begingroup$ What does act coherently over the Fermi surface mean exactly, and why does it depend on the surfaces being shifted? $\endgroup$ Nov 22, 2023 at 15:04
  • $\begingroup$ Coherent means that all pairs with back-to-back momenta $|p_F,-p_F>$ can mix. This does not work if the Fermi surfaces are shifted $|p_F,-p'_F>$ because now the pair momentum is not zero, and different pairs cannot mix by momentum conservation. $\endgroup$
    – Thomas
    Nov 27, 2023 at 16:46
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The semi-empirical mass formula's pairing term is attractive only for even-even nuclei, see for example, Miriam Hein et al. $2022$ Eur. J. Phys. $43$ $035801$. That is, it is pairing between protons or pairing between neutrons. This is the isospin $T=1$ channel, where the $S=0$ interaction is most attractive. Your statement that $S=1$ is most attractive is true for $T=0$ i.e. the deuteron channel, but this is not the pairing contained in the semi-empirical mass formula term.

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