nuclear physics - How is the spin of the odd-odd nucleus 6Li explained?

Spins of odd-odd nuclei is difficult to predict. But $^{6}Li$ is light - only 6 nucleons. $^{6}Li$ should have spin $(\frac{3}{2} + \frac{3}{2}) = 3$ by shell model, as should have one proton on $p_{3/2}$ level and one neutron on $p_{3/2}$ level. How it is explained that it has spin 1?

In the answers to question like that we see frequently formalism from some theory like $πp_{3/2}⊗νp_{3/2}$ and $3/2⊗3/2$. Please, explain to what theory that formalism belongs and explain how to understand it and where to read about.

Answer

The way to understand why 6-Li has $J^P = 1^+$ ($J$ the toal angular momentum and $P$ the parity) is through the measure of its magnetic momentum,

$$ \mu^{exp} \simeq 0.88\mu_N $$

Where $\mu_N$ is the Bohr's magnetic momentum.

This result can be understood assuming that 6-Li behaves as an alpha particle plus a deuteron. The alpha particle has $J^P = 0^+$ so taking $\mu^{exp}$ as the mean value of the magnetic momentum operator of the remnant deuteron one has $$ \mu^{exp} = \langle \mu \rangle = \langle \sum_{p, n}g\mu_N J_z \rangle = \sum_{p, n}g\mu_N m_J = (2·2.79 - 2·1.91)m_J\mu_N \equiv 0.88\mu_N \tag1$$

$m_J$ is the eigenvalue of $J_z$, $\sum_{p, n}$ is the sum over the deuteron's proton and neutron, $g[p] = 2·2.79$ and $g[n] = -2·1.91$. Since from the point of view of Nuclear Physics -nuclear shell model- there is no difference between $n$ and $p$, we can assume that $p$ and $n$ contributes equally as $m_J = +1/2$ and with this value you can see that Eq. (1) is fullfilled. Now, only with this, you could say that $J$ is zero or 1.

Nevertheless, you know that isospin of deuteron is null (see below) wich means that this part of the wave function is antisymmetric. The rest of it must be, therefore, symmetric: spatial plus $J$ parts. For the spatial one we can select $L = 0$ wich implies symmetric spatial state and positive parity

$$P = (-1)^L = +1$$

Finally, $J$ part must be symmetric, i.e., $J = 1$

Therefore,

$$ ^{6}Li \sim \alpha +\ ^{2}_1H $$

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Deuteron's isospin equals to zero, $T = 0$

For $T = 1$ you have the triplet while for $T = 0$ the singlet isospin state,

$$ \cases{ T = 1: \ pp, nn, (pn + np)/\sqrt{2}\ \leftarrow\ symmetric\ states\\ T = 0: \ (pn - np)/\sqrt{2}\ \leftarrow\ anti-symmetric\ state } $$

Since we are considering isospin symmetry, all triplet states have (up to electromagnetic corrections) the same energy and therefore they are equally probable to exist in nature. But we have not seen $pp, nn$ nucleous so for deuteron the isospin state must be the singlet, which is anti-symmetric.