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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.

Thomas Fritsch
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coodan
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  • related: https://physics.stackexchange.com/questions/454246/how-do-you-go-about-guessing-the-ground-state-spin-and-parity-of-a-nucleus –  May 25 '19 at 23:13
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    You don't explain why you think it should have spin 3. The predominant configuration in the wavefunction for the valence particles is presumably $\pi p_{3/2} \otimes \nu p_{3/2}$. You can couple spins $3/2\otimes 3/2$ to anything from 0 to 3. Unless you do a shell-model calculation that handles the proton-neutron residual interaction, I don't see why there would be any expectation that you could guess the spin of the ground state, or why 3 would be any more likely than 0, 1, or 2. It's interesting that the spin-0, 1, and 2 are just resonances but can undergo gamma decay. –  May 25 '19 at 23:36
  • Thank you, Ben Crowell, and this question was in fact, to you, as after the reading of that answers it would be interesting to see 3/2⊗3/2 conception fully presented by specialist. I will edit my question. Please, explain about πp3/2 and νp3/2 detally. – coodan May 26 '19 at 08:11
  • @coodan: I've written an answer for the question that this one duplicates. –  May 26 '19 at 18:47
  • Thank you, Ben. Sorry for being not possible to comment your answer there yet. But it get more obvious that this idea allows any combinations of spins for neutron and proton, de facto unifying nucleons to single type, is not it? Why it does not interrupt Pauli principle? Please, explain - neutron and proton are interacting in some special way? Zero spin case for odd-odd nuclei (usual for heavy odd-odd) is especially interesting. – coodan May 26 '19 at 20:00
  • The motivation to my question about lithium was exactly that freedom and unity. Ben, what is your personal feeling after revisioning a lot of nuclei spins - is it like it was quads of fermions of one type with some unclear rules for filling or it is system constructed somehow else? – coodan May 26 '19 at 20:11

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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,

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

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.

Vicky
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  • IMO clustering explanations like this should always be considered guilty until proven innocent, because they violate the Pauli exclusion principle. A deuteron has both particles in the 1s1/2 state, but that state is also occupied by the neutrons and protons that are in the alpha. –  May 25 '19 at 23:26
  • @BenCrowell The Mayer's shell nucleous model is not always accurate as happens with the atomic shell model. So when it fails you have to look for another explanation and as far as I know this is a good one and the one explained in bachelor studies. Plus I have upgraded my answer in order to prove it is not against Pauli's principle – Vicky May 25 '19 at 23:39
  • Since the question is a duplicate of https://physics.stackexchange.com/questions/118518/shell-model-of-an-odd-odd-nucleus-6li , you might want to delete this answer and copy it over as an answer to that question. –  May 25 '19 at 23:44
  • @BenCrowell With all due respect to rob, I think mine is more nuance and therefore this solution is still useful – Vicky May 25 '19 at 23:49
  • But the magnetic moment is a reasult of spin, and not spin is a reasult of magnetic moment. Measuring it, you just prove, that spin is 1. – coodan May 26 '19 at 08:20
  • @BenCrowell, this is frequent question, I thing, but, please, give explanation here, as it is not possible for newbie to comment in other answers. – coodan May 26 '19 at 08:31
  • @coodan I don't get your point. What do you want to be explained? Magnetic moment is the result of $S + L \equiv J$ not just $S$. Due to the measurement of parity you know this is positive, which implies $L$ even. You can start proving models with $L = 0$, so if you get the correct answer then you have finished. If you don't, then you prove $L = 2$ and so on. In this case, taking $L = 0$ gives you the right answer so you can say that assuming 6-Li is like an alpha plus a deuteron with $J = S = 1$ is a good idea. Again, Mayer's model is not always accurate – Vicky May 26 '19 at 14:46
  • The problem is that it is not alpa plus deuteron. But deuteron first, and than alpha. To fill p3/2 get more energetically profitable than previous s1/2? Why? What is modern explanation for that? – coodan May 26 '19 at 15:53
  • @coodan I'm not sure if I'm understanding your point. When I say $^{6}Li \sim \alpha +\ ^{2}1H $ I'm meaning that you can use Mayer's model for alpha and deuteron separatedly, so for alpha you get $J^P = 0^+$ and for deuteron, as I proved in the answer, $J^P = 1^+$ since you will find the $p$ and $n$ of this one in $1s{1/2}$ level. Now what is more important is that this is not against experimental data (e.g. magnetic momentum). Is this more clear? – Vicky May 29 '19 at 03:51
  • Yes, but this suggest that we have 1s levels filled by 1 proton and 1 neutron and then 1p level filled by 2 protons and 2 neutrons. Deuteron filled, then helium filled. Inside out filling of levels. – coodan May 29 '19 at 08:43
  • @coodan You are misunderstanding me. When I say use Mayer's model separatedly for alpha and deuteron i mean that the $2p$ and $2n$ from alpha are in 1s and the $p$ and $n$ from deuteron are in 1s level as well. You use Mayer's model separatedly for both, not filling shells for deuteron and after that filling from the next up level the rest shells for alpha. Two separated fillings for $\alpha$ and $^{2}_1H$ – Vicky May 29 '19 at 14:30