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I am confused about when the nuclear force is attractive and when not. Based on deuteron (the book I am following is Wong), we see that we can't have bound state with isospin T=1 (otherwise we would see, for example, a stable double neutron and no proton nucleus). Also, in the book I see mentioned in several places that one can't have a bound T=1 state. So based on this, I understand that nucleons repel each other in the T=1 state and attract for T=0 state. However, later, he introduces pairing interactions, which says that for a given nucleus, neutrons and proton prefer to pair together in a state with opposite spins and angular momenta (hence why the ground state of any nucleus with an even-even number of nucleons has a total spin of 0), but this implies that they pair in a state of T=1 (2 protons or 2 neutrons can only be in a T=1 state). So based on this it seems like T=1 is attractive. And I am really confused. Also, I see mentioned many times that the isospin dependence of the nuclear force is small, yet it seems like lots of things that decide whether is system is bound or not have to do with the isospin. Can someone help me understand this? Thank you!

JohnDoe122
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  • It is emphatically not true that "the ground state of any nucleus has a total spin of 0". This is only true for nuclei with an even number of protons and an even number of neutrons (an "even-even nucleus"). If either or both of these numbers are odd, the ground state spin is nonzero. – probably_someone Feb 21 '20 at 09:36
  • I also don't think it's necessarily wise to think of nucleon pairs inside the nucleus as having the same behavior as isolated nucleon pairs. Think about the neutron: a free neutron is unstable, but a neutron in a stable nucleus is stable. – probably_someone Feb 21 '20 at 09:39
  • Lastly, remember that the binding energy of deuterium is itself very small, so the small isospin dependence is just enough to make the isospin triplet states unbound. – probably_someone Feb 21 '20 at 09:40
  • @probably_someone thanks! Sorry for the even-even, corrected that. I am still not totally sure I understand. Is T=1 or T=0 attractive? Also in the classic plot for the nucleon-nucleon potential, there is a negative part between 1-2 fm. Why can't we have a bound neutron-neutron system, if the potential energy there is negative there? – JohnDoe122 Feb 21 '20 at 09:48
  • Also, why would the behavior of a force change when going from 2 to more particles? I imagine you add more terms in the Hamiltonian, but they would be the same terms as before, just for more particles. For example for Coulomb force, you just add more interaction terms between the new electrons, but the strength between 2 individual electrons is the same, whether you have 2 or 1000. Shouldn't any force/potential have this sort of superposition principle? Why is it different for the nuclear force? – JohnDoe122 Feb 21 '20 at 09:51
  • The fact that a potential is attractive doesn't mean that two interacting particles are guaranteed to form a bound state. Look at Newtonian gravity - any parabolic or hyperbolic orbit is unbound despite the potential being unambiguously attractive. Both T=0 and T=1 are attractive, but the difference between them is enough that the T=0 bound state is just slightly unbound in the T=1 case, for an isolated nucleon pair. – probably_someone Feb 21 '20 at 19:50
  • Likewise, the fact that the potential is negative doesn't really mean anything - the zero point of potential is arbitrary, and you can add any constant to a potential function and the physics will remain the same. What matters is the potential difference between the starting and ending configurations, relative to the initial kinetic energy of the particles. – probably_someone Feb 21 '20 at 20:08
  • The fact that the dynamics of a system can be completely different when you add more interacting bodies is a very well-known fact even in classical mechanics. Once again, look at Newtonian gravity - if you add a third body into a bound two-body system, the whole system becomes potentially unbound. This is, in fact, why we study emergent behavior as a phenomenon: to put it bluntly, in the words of Nobel laureate P.W. Anderson, "More is different." (https://science.sciencemag.org/content/177/4047/393) – probably_someone Feb 21 '20 at 20:19
  • One of the many interesting indicators that there is, in fact, something quite different going on when many nucleons are interacting is the EMC effect - essentially, the momentum distribution of the quarks inside an individual nucleon is different in deuterium than it is in iron. – probably_someone Feb 21 '20 at 20:31
  • Not identical but related: https://physics.stackexchange.com/questions/288357/why-is-nuclear-force-spin-dependent – Lewis Miller Feb 22 '20 at 01:27
  • @probably_someone thanks a lot for this! It's really helpful. One more thing I am confused about: I totally agree that adding more objects in a system changes the dynamics, but it doesn't changes the physics i.e. the gravitational force between any 2 objects is the same even if you have 2 or 3 objects in your system. From you second reply, I understood that you meant that the force between 2 nucleons changes inside of a nucleus (which is like saying that gravity changes between 2 objects, if you add a 3rd). What you meant there was that the dynamics of the system changes? – JohnDoe122 Feb 22 '20 at 02:09
  • @LewisMiller Thank you for that link! I have a question about that actually. From what I read, the nuclear force is a residual of the strong force. However a spin spin interaction is due to qed effects, isn't it? Similar to the hyperfine interactions, we have the magnetic field created by one dipole, interacting with the other dipole. Why do we add such a spin-spin interaction as part of the nuclear force and not as part of qed interaction between 2 nucleons? Like the force is there for sure, I am just not sure why we associate it with strong force and not EM. – JohnDoe122 Feb 22 '20 at 02:11
  • Spin dependant forces result from other than just QED forces. In fact in nuclei the spin dependance comes mostly from the strong forces. See my answer to this question: https://physics.stackexchange.com/questions/292913/why-is-the-density-of-the-fermi-gas-in-a-neutron-star-not-changing-the-potential/293523#293523 – Lewis Miller Feb 22 '20 at 15:56

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