Nuclear Force Behavior: Confusing Isospin Dependence

[kelly0303]

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!

 

[mfb]

The energy differences between different states are much larger than the energy differences due to spin-spin interactions. Two neutrons with opposite spin in tritium might be not optimal in terms of spin, but it's optimal in terms of having both neutrons in the lowest energy state. Similar for the protons in helium-3.

 

[kelly0303]

The energy differences between different states are much larger than the energy differences due to spin-spin interactions. Two neutrons with opposite spin in tritium might be not optimal in terms of spin, but it's optimal in terms of having both neutrons in the lowest energy state. Similar for the protons in helium-3.

Thank you for this! Why is 2 neutrons with opposite spin optimal? I thought that opposite spin is attractive. Also I am still confused. When is the nuclear force (between 2 nucleons) attractive and when it is not? Based on the classical plot of nuclear potential, between 1-2 fm you have a negative potential for any 2 nucleons, so any 2 nucleons should have a bound state. Yet nn doesn't exist. I am not sure I understand why.

 

[mfb]

If you have more than two nucleons it's not helpful to find the force between two of them. You need to consider the energy of the whole system. Two neutrons (or two protons) in the lowest energy state is the lowest-energy configuration, independent of what happens with spins. Spin is very important in the two-nucleon system.

 

[kelly0303]

If you have more than two nucleons it's not helpful to find the force between two of them. You need to consider the energy of the whole system. Two neutrons (or two protons) in the lowest energy state is the lowest-energy configuration, independent of what happens with spins. Spin is very important in the two-nucleon system.

Why can't I get the energy of the system, from the energy between 2 nucleons at a time? Shouldn't any potential work like that (might not be solvable analytically, but the Hamiltonian of N particles should follow directly from the hamiltonian of 2)? So spin matters when I have just 2 nucleons, and for some reason the spin doesn't allow nn and pp to be bound (again the binding energy looks negative in the 1-2 fm region, so I am not sure I understand why).. Why doesn't the spin matter anymore when I have more nucleons? For example for He, I have just 2 more nucleons, yet the protons and the neutrons spin among themselves (i.e. nn and pp) to give an overall spin of zero. How come that only 2 extra nucleons removed the important effect of the spin for the case of deuteron? Also I am not sure I see how adding more particles changes the relative strengths of the particles involved i.e. the terms in the hamiltonian should be the same, no matter if I have 2 or 100 particles, right? Is there an actual (at least approximate) formula for the nuclear force? Maybe that would clarify the things. (Sorry for asking so many questions).

 

[Vanadium 50]

You're not thinking about the answers you are getting. You are reacting to them.

How do I know? Because only a few minutes pass between an answer and your reaction that you don't understand. There hasn't been enough time for you to think things through. You would be much better served by thinking about the responses you are getting and if you need a follow up to ask something focused and well-posed rather than just shotgunning questions back as fast as you can. You are much, much more likely to reach understanding that way.

 

[kelly0303]

You're not thinking about the answers you are getting. You are reacting to them.

How do I know? Because only a few minutes pass between an answer and your reaction that you don't understand. There hasn't been enough time for you to think things through. You would be much better served by thinking about the responses you are getting and if you need a follow up to ask something focused and well-posed rather than just shotgunning questions back as fast as you can. You are much, much more likely to reach understanding that way.

I am not sure what you mean. I understood the answer I received. I just don't see how it clarifies my questions, so I added more details about what I want to know. For example, why does the nuclear force potential (this) has a negative value between 0.5 and 1.5 fm, for any 2 nucleons, yet we only see np (as deuterium) but no nn. The answer I got clearly doesn't touch on that at all, so I don't need much time to clarify my question from that angle.

 

[mfb]

The answer I got clearly doesn't touch on that at all

... or you didn't understand how it does. I'm with V50 here.

 

[snorkack]

I am not sure what you mean. I understood the answer I received. I just don't see how it clarifies my questions, so I added more details about what I want to know. For example, why does the nuclear force potential (this) has a negative value between 0.5 and 1.5 fm, for any 2 nucleons, yet we only see np (as deuterium) but no nn. The answer I got clearly doesn't touch on that at all, so I don't need much time to clarify my question from that angle.

For one thing, at long distances nuclear potential at long distances drops off faster than 1/r, and importantly faster than 1/r², too.
This means that the number of bound states in a nuclear potential hole is finite, and may be zero. A too narrow and shallow potential, though attractive, cannot bind a particle.
Now, the reason why there is no nn is closely tied to the reason why deuterium has spin 1.
The nuclear attractive force between proton and neutron of opposite spin is weaker than the attractive force between proton and neutron of same spin. Just weak enough that proton and neutron of opposite spin are not bound at all despite being attracted, while proton and neutron of same spin are attracted strongly enough to be bound.
Two neutrons of same spin would be attracted by a nuclear force, but they are also repelled by exchange force. So nn cannot be bound either way - not if same spin (because identical fermions) and not if opposite spins (because, as with np, attraction barely not strong enough to bind).

 

[kelly0303]

... or you didn't understand how it does. I'm with V50 here.

Well, whether it does and I didn't understand or it doesn't at all, I still need further explanations in either cases (hence why my follow up question). Also, if it does and I didn't understand it, I won't be able to realize that it does... so again I would like more details (or an explanation from a different perspective).

 

[kelly0303]

For one thing, at long distances nuclear potential at long distances drops off faster than 1/r, and importantly faster than 1/r², too.
This means that the number of bound states in a nuclear potential hole is finite, and may be zero. A too narrow and shallow potential, though attractive, cannot bind a particle.
Now, the reason why there is no nn is closely tied to the reason why deuterium has spin 1.
The nuclear attractive force between proton and neutron of opposite spin is weaker than the attractive force between proton and neutron of same spin. Just weak enough that proton and neutron of opposite spin are not bound at all despite being attracted, while proton and neutron of same spin are attracted strongly enough to be bound.
Two neutrons of same spin would be attracted by a nuclear force, but they are also repelled by exchange force. So nn cannot be bound either way - not if same spin (because identical fermions) and not if opposite spins (because, as with np, attraction barely not strong enough to bind).

Thank you for your reply! I see, so same spins create more binding. What exactly is the exchange force? I see that you mention it as a different thing from the nuclear force. What creates it? Also, why does inside nuclei with more nucleons, we have the pairing effect i.e. 2 protons or 2 neutrons will pair together with opposite spins (leading to overall zero spin in even-even nuclei for example)? If the force creates more binding in the same spin case, why do we get this opposite effect now? Thank you!

 

[snorkack]

Thank you for your reply! I see, so same spins create more binding. What exactly is the exchange force? I see that you mention it as a different thing from the nuclear force. What creates it?

Interaction between identical fermions. Even neutrinos should suffer from exchange force.

Also, why does inside nuclei with more nucleons, we have the pairing effect i.e. 2 protons or 2 neutrons will pair together with opposite spins (leading to overall zero spin in even-even nuclei for example)? If the force creates more binding in the same spin case, why do we get this opposite effect now? Thank you!

The reason is that 2 neutrons with opposite spins have nuclear attraction which is barely too weak to bind them to a dineutron nucleus. Add a proton, and it adds enough nuclear attraction to bind all three.