Problem with shell model and magnetic moment of Lithium-6 I have a problem with the calculus of magnetic moment of Li-6.
The configuration of protons is $1p_{3/2}$, and the neutrons' one is the same. 
I have to add the magnetic moment of uncoupled proton and uncoupled neutron.
I use the following formula for $J=l+\frac{1}{2}$ (J is the particle spin):
$$ \frac{\mu}{\mu_N}=g_lJ+\frac{g_s-g_l}{2}$$
For the proton I have: $g_l=1; g_s=5.58 \rightarrow  \frac{\mu}{\mu_N}=J+2.29=3.79$
For the neutron I have: $g_l=0; g_s=-3.82 \rightarrow  \frac{\mu}{\mu_N}=-1.91$
So the total $\frac{\mu}{\mu_N}=3.79-1.91=1.88$, exactly 1 more than the correct value, 0.88!
What's wrong?
 A: The static magnetic moment of Li-6 
$$\mu_{6Li} = 0.822 \mu_N$$
comes from its nuclear spin $I^\pi = 1^+$, with positive parity $\pi$, so in the ground state of Li-6, only even values of $l = 0, 2, ..$ would be allowed,  neglecting the paired $2p$ plus $2n$ in the $s_{1/2}$-state core with net $I=0$. 
The nuclear spin then comes from $L$-$S$ coupling of the two unpaired $p$ and $n$, which have to be in a spin triplet ($S = 1$) state, since $I=1$ requires the combined $p$ + $n$ orbital $L=0$ (there is a small admixture of $L=2$). For the $L=0$ level, for each particle outside the closed shell, $I = 0 +1/2$ in the formula (using $I$ for nuclear spin, instead of the atomic notation $J$)
$$ \frac{\mu}{\mu_N}=g_lI+\frac{g_s-g_l}{2}$$
$p$: $1 \frac{1}{2} + \frac{5.58 - 1}{2} = 2.79$
$n$: $\frac{-3.82}{2} = -1.91$
$p + n = 0.88$, close to $0.822$ (most of the difference comes from the $L=2$ level that was ignored above).
The value $1.88$ is the Schmidt line assuming $i$-$i$ coupling (independent combination of each particle's $l$ and $s$). But parity and the measured moment rule out $i$-$i$ coupling. The Schmidt lines just give the magnetic moments in the limit of the extreme shell model.
A: You can't determine the magnetic moment of the Li-6 since there are two un-coupled nucleons in the nucleus. The answer from @OscarRondon seems incorrect to me. Of course, if the experimental value is $I=1^+$, the total spin of both particles $S$ has to be $S=0$, but you can't theoretically exclude the $L=2$. Even if you neglect it and take $L=0$, you can't say that this is the angular momentum of the separate particles as his eq. ($I=0+\frac{1}{2}$) implies. That is in complete disagreement with the fact that the particles are both in $1p_\frac{3}{2}$ state which, since their spin is $s=1/2$, allows only for $l=1, 2$. On top of that, the equation that both the question and the answer from @OscarRondon use to calculate the magnetic moment is wrong, because it can only be used for a nucleus with only one un-coupled nucleon.
Take what I said with a grain of salt; I'm only a undergraduate student.
