# Problem with shell model and magnetic moment of Lithium-6

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.