My particle physics lecture notes seem to have contradictory statements.

Firstly they argue that p-n is stable while p-p and n-n are not because the nuclear interaction favours spin alignment (and by Pauli Exclusion principle p-p and n-n have to have their spins anti-aligned).

Also for the asymmetry term explanation in the semi-empirical mass formula they say that the p-n force in the nucleus is slightly ore attractive than p-p or n-n as the state can be symmetric and antiymmstric. (Again favouring spin alignment)

However for the pairing term explanation in the semi-empirical mass formula they say that n-n and p-p pairwise configurations are energetically favoured since the spins can be anti-aligned so that the patial states are the same and therefore have a high degree of overlap and form a strong bond.

So how can p-p or n-n bonds be stronger within a large nucleus, but p-n bonds be stronger if that is the total nucleus? And what would be preferred, an odd-odd nucleus (e.g. nnn ppp) with more p-n's or an even-even nucleus (e.g. nn pppp) with more p-p's?

(This question is similar but doesn't seem to have helpful answers Does the strong force increase or decrease with aligned spins?)

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    $\begingroup$ Do any long-lived odd-odd nuclei have ground states with zero spin? (I haven’t looked recently, but the low-mass ones don’t.) $\endgroup$
    – rob
    May 11, 2021 at 23:26
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    $\begingroup$ btw: the deuteron is an odd-ball nucleus. The nucleons are far apart, and loosely bound (2.2MeV), it's not like a 'normal' nucleus, where everything is crammed in. $\endgroup$
    – JEB
    May 12, 2021 at 5:43
  • $\begingroup$ FWIW, Wikipedia has tables of nuclides which are handy for quick comparisons of neighbouring nuclides. For more detailed info, there are pages dedicated to the isotopes of each element, eg Isotopes of lithium, which are linked from those tables. $\endgroup$
    – PM 2Ring
    May 13, 2021 at 1:46
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    $\begingroup$ @rob Wiki says no. At least, none of the 9 primordial odd-odd nuclides (5 stable, 4 long-lived) have zero spin in the ground state. $\endgroup$
    – PM 2Ring
    May 13, 2021 at 1:56

1 Answer 1


Regarding the deuteron ($np$), the concern is isospin. In the strong nuclear force, neutrons and protons are the isospin down and up states of a single particle: the nucleon. Thus: $n$ and $p$ are identical particles, and need to be in an antisymmetric state:

$$ \Psi(n_1, n_2) = \psi(\vec r)\chi_S\tau_I^{I_3} $$


$$\psi(\vec r) = f(r)Y_0^0(\theta, \phi) $$

is symmetric S-wave. If the spin wave function is an antisymmetric $S=0$ singlet:

$$\chi_0 = \frac 1{\sqrt 2}(|\uparrow\downarrow\rangle- |\downarrow\uparrow\rangle)$$

then the Pauli exclusion principle says the isospin wave function is a symmetric $I=1$ triplet:

$$ \tau_1^1 = |pp\rangle$$ $$ \tau_1^0 = \frac 1{\sqrt 2}(|pn\rangle+ |np\rangle)$$ $$ \tau_1^{-1} = |nn\rangle$$

Since $ \tau_1^1$ ($^2$He) and $\tau_1^{-1}$ (the di-neutron) are not observed, we assume the deuteron ($^2$H) is iso-singlet ($I=0$):

$$ \tau_0^0 = \frac 1{\sqrt 2}(|pn\rangle- |np\rangle)$$

and hence the spin state is $S=1$, and indeed, the deuteron is spin 1.

The two-nucleon potential is complicated. Perhaps the most famous version is the Argonne V-18 potential (https://www.phy.anl.gov/theory/research/av18/), named because it was developed at Argonne National Lab, it is a formula for $V(n_1, n_2)$, and it has 18 terms.

The various terms can be attributed to effective field theory processes such as one-pion exchange (the pion acts like a pseudoscalar exchange boson). This one term looks like:

$$ V_{\pi}=V_0(\vec \tau_1\cdot \vec \tau_2)\big[ \vec \sigma_1\cdot \vec \sigma_2+S_{12}\big( 1+\vec 3m_{\pi}r+\vec 3(m_{\pi}r)^2 \big) \big] \frac{e^{-m_{\pi}r}}{m_{\pi}r} $$

where $\vec \sigma_1\cdot \vec \sigma_2$ is the dot product of spin operators (a scalar operator) and

$$S_{12}=2\big[3\frac{(\vec S\cdot \vec r)^2}{r^2}-\vec S^2\big]$$

is a (non central) tensor spin operator. ($\vec S=\vec \sigma_1 + \vec \sigma_2$ is the total spin).

The isoscalar operator is the dot product (in isospin space) of the iso-spin operators:

$$\vec \tau_1\cdot \vec \tau_2$$

(It's called isospin because the math describing it is identical to spin 1/2 particles.) Note also, that the pion is an isovector triplet:

$$ \pi^+ = |u\bar u\rangle$$ $$ \pi^0 = (|u\bar d\rangle - |d\bar u\rangle)/\sqrt 2$$ $$ \pi^- = |d\bar d\rangle$$

(The minus sign is because of antiquarks, not because the state is antisymmetric).

...and that's just the pion. There is also $\omega$, $\eta$, $\eta'$, $f_0$, $\rho$, $\sigma$, kaon exchange (that includes strangeness), and more. Other terms include spin-orbit ($\vec L\cdot \vec S)$, sigma minus p ($(\vec \sigma_1\cdot \vec p_1)(\vec \sigma_2\cdot \vec p_2)$), exchange forces $[(\vec \sigma_1\cdot \vec L)(\vec \sigma_2\cdot \vec L) + (\vec \sigma_2\cdot \vec L)(\vec \sigma_1\cdot \vec L)]$, for example.

The takeaway here is that the spin and isospin nature of the two nucleon force cannot be simply described as "the spins like to be aligned". Another takeaway that may not be obvious to non-experts (esp. b/c of the pictures we see of nuclei) is that protons and neutrons don't retain their identity. In a deuteron in the $S_z=0$ state, not only is one of the particle in a mix of spin up and down, it's also in a mixture of proton and neutron.

When you start looking at nuclei, you now have to consider three and four nucleon forces, which are novel for many physicists. The three nucleon force, for example, is not the standard 3 body problem with two-nucleon forces between all particles. It's that, plus a new forces that can only be considered as a 3-body interaction (https://en.wikipedia.org/wiki/Three-body_force).

There is also the EMC effect (https://en.wikipedia.org/wiki/EMC_effect), which is observation that quark structure functions of the proton and neutron are modified in the nuclear environment...that is, protons in nuclear environments may be different from free protons.

And that's all effective field theory. A fundamental QCD based description is a long way off.

  • $\begingroup$ Thanks that has been very helpful. One last thing. Would I be right in thinking within a large nucleus the asymmetry term favours p-n pairs as they are energetically favourable from the spin dependent parts of the strong interaction, whereas the pairing term favours p-p and n-n pairs because their spins must be anti aligned which indirectly increases their spatial overlap. Since these are different effects it explains why we need both terms in the SEMF? $\endgroup$
    – Alex Gower
    May 12, 2021 at 9:49
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    $\begingroup$ idk. I don't go past $^3$He. $\endgroup$
    – JEB
    May 12, 2021 at 15:02

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