The deuterium exists only with the proton and neutron of aligned spin, which suggests that the residual strong force is greated with aligned spins, i.e. the binding energy is greater if the spins are aligned.

On the other hand the mass of $\Delta^{+}$ is greater than the mass of proton $p$, even if they have the same quarks, since $\Delta^+$ has aligned spins of quarks. So it seems that here the binding energy decreases if spin are aligned (greater mass -> smaller binding energy).

Another fact in this sense is that in Weizsäcker formula the pairing term makes the binding energy of nucleus to increase if the spins are paired (i.e. opposite). (here the strong force is again residual as in deuterium)

So how are these two consistent? Does the strong force increase with aligned spins?

  • $\begingroup$ Can you please tell me what you mean by Δ+? $\endgroup$ Jun 25, 2018 at 23:03

2 Answers 2


First in order to get a bound state, you need to have a potential that is stronger then the kinetic energy of the nucleons, and to have that you need:

  1. the two nucleons don't repel charge wise, compared to nucleon interaction, the Coulomb force is not strong, but still worth considering

  2. the two nucleons being aligned in spin gives extra binding

Two nucleons in a zero orbital angular momentum state, that is the lowest energy state, can only align their spins if they are antialigned in isospin (Pauli exclusion). I think this is where you are getting confused. This is the lowest energy state that gives you a bound state.

But this is called an S state, a zero orbital angular momentum state. The alignment in spin gives them an extra binding energy, so the strong force itself does not increase with aligned spin, but the aligned spin itself will give them extra binding energy.

Without this extra binding energy (that is not extra strong force), they could not be bound. The alignment in spin gives them an extra binding energy, because of the dot product in the spin in the NN interaction.

This is why a proton and a neutron are of opposite isospin and can be aligned in spin and that gives them an extra energy (not strong force) to create a bound state.

I think where you are getting confused is you think the two nuclei are only bound by the strong force. But their bound state consists of multiple forces, as the EM force, strong force, residual strong force, that is the nuclear force, gravity (is not even measurable) and spin, and isospin (these two are not a forces, but the Pauli exclusion principle).

And you have to learn that the strong force is mediated by gluons. Gluons interact with each other, to form a flux tube. When two quarks are separated by distance, the strong force remains strong, until the energy is high enough to create more quarks. When the quarks are too close, the strong force is not that strong.

The strong force does pull quarks together, but it also gets weaker as the quarks get closer (i.e. it acts sort of like a spring), in a phenomenon known as "asymptotic freedom."

Please see here:



A question about deuterium is the isospin state. Is it singlet:

$$ \frac{1}{\sqrt 2}(pn - np),\ $$

or is it triplet:

$$ \frac{1}{\sqrt 2}(pn + np)\ ?$$

Well, if it were triplet, we would expect the other triplet isospin states to be bound:

$$ pp $$ $$ nn $$

(through the former may have Coulomb problems). Either way, they don't exist, so the isospin state is an antisymmetric singlet.

The total wave function is a product of space, spin, and isospin:

$$ \psi_{tot} = \psi(\vec r)\ \psi_{spin}\ \ \psi_{isospin} $$

With the spatial part in a symmetric S-state, we need the spin state to be symmetric for the overall wavefunction to be antisymmetric (Pauli Exclusion Principle)--and that has to $J=1$.

A deep dive into nucleon-nucleon potentials takes you to the Argonne V18 potential: https://journals.aps.org/prc/pdf/10.1103/PhysRevC.51.38

  • $\begingroup$ The diproton is unbound because two protons have the same isospin, and so their spin cannot be aligned (Pauli exclusion). Because their spin cannot be aligned, the strong force is not enough to compensate for the EM repulsion and for the Pauli principle. Same with the dineutron. But that does not have EM repulsion. $\endgroup$ Jun 27, 2018 at 1:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.