- #1
kelly0303
- 579
- 33
Hello! 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 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!