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?

  • $\begingroup$ Possibly related: physics.stackexchange.com/q/118518/44126 $\endgroup$
    – rob
    Jun 26, 2014 at 1:45
  • $\begingroup$ @rob very interesting, thanks. So the Li-6 is a special case and I don't use the formula to determine the magnetic moment of the nucleus, isn't it? I can't understand why the only way to obtain the correct result is simply add the proton magnetic moment (2,79 $\mu_N$) and the neutron's one (-1,91 $\mu_N$). If I do it, I don't consider the Js particles... $\endgroup$
    – sunrise
    Jun 26, 2014 at 6:24
  • 1
    $\begingroup$ I'm actually not sure. It doesn't make sense for the shell model to fail on light nuclei. One interpretation is that the $2s$ nuclear shell fills before the $1p$ shell, but that gives the wrong parity for lithium-7. I'm looking forward to a real answer to your question. $\endgroup$
    – rob
    Jun 26, 2014 at 12:28
  • $\begingroup$ Oops, after offering the bounty I realized that this duplicated another question from a week earlier in 2014 physics.stackexchange.com/questions/118518/… , and I also realized that I think I know the part of the answer about the spin. So I'll add an answer to that question. $\endgroup$
    – user4552
    May 26, 2019 at 18:22

2 Answers 2


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.

  • $\begingroup$ Unless I'm misunderstanding, this answer doesn't explain why the wavefunction is what it is. It just assumes that it is a certain wavefunction (L=0, spin triplet) and goes on to explain the magnetic moment based on that. $\endgroup$
    – user4552
    May 25, 2019 at 23:47
  • $\begingroup$ @BenCrowell I think the way to see it is that from the experimental measure of the parity, $P$, you get $P = +1$ and therefore $L$ even. But the higher $L$ is, more energy your nuclous has. So if taking $L = 0$ you can find a solution to the problem, then this is the most likely value for $L$. If you wouldn't have obtained the correct result for $L = 0$, you would have proposed $L = 2$ or higher $\endgroup$
    – Vicky
    May 26, 2019 at 0:29
  • $\begingroup$ @Vicky: But the higher L is, more energy your nuclous has. No, this is not true in general. $\endgroup$
    – user4552
    May 26, 2019 at 18:23
  • $\begingroup$ @BenCrowell So more $L$, which means more rotational energy, suppose no more energy? How is that? $\endgroup$
    – Vicky
    May 26, 2019 at 23:24

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.


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