# Magnetic moments of nucleons

I was comparing my notes of the nuclear physics class (undergraduate level) on magnetic moments of nucleons with the Krane's explanation.

In my notes I wrote that there are two types of magnetic moments:

1. The first one is the orbital one. It's written as $\mu_l=g_l l \mu_n$ where l is the orbital quantum number. I also wrote that $\vec{\mu_l}=g_l\vec{L}$ so that this vector is parallel to $\vec{L}$.

2. The second one is the spin one. It's written as $\mu_s=g_ss\mu_n$ where s is the spin quantum number, s=1/2 for nucleons. Its vectorial form is $\vec{\mu_s}=g_s\vec{S}$ so that this vector is parallel to $\vec{S}$

Then, the total magnetic moment is $\vec{\mu_j}=\vec{\mu_l}+\vec{\mu_s}$ where $\vec{\mu_j}=g_j\vec{J}$. The next step on the notes is about finding the value of $g_j$. I wrote that $|\vec{\mu_j}|=|\vec{\mu_l}|\cos{\theta}+|\vec{\mu_s}|\cos{\varphi}$ where $\varphi$ is the angle between $\vec{S}$ and $\vec{J}$ and $\theta$ is the angle between $\vec{L}$ and $\vec{J}$. In the next step I substitute $|\vec{\mu_l}|$ with $g_l\hbar (l(l+1))^{1/2}$, $|\vec{\mu_s}|$ with $g_s\hbar (s(s+1))^{1/2}$ and $|\vec{\mu_j}|$ with $g_j\hbar (j(j+1))^{1/2}$.

So here's my problem: why is $|\vec{\mu_l}|$ different from $\mu_l$? In fact the first one it's written like $g_l|\vec{L}|$ and the second one as $\mu_ng_ll$. The same happens with $|\vec{\mu_s}|$ and $\mu_s$.

Also: In my notes I wrote that $\vec{\mu_j}$ isn't parallel to $\vec{J}$ and it is, in fact, rotating about $\vec{J}$. So why $\vec{\mu_j}=g_j\vec{J}$? Shouldn't $\vec{\mu_j}$ and $\vec{J}$ be parallel this way?

Also: In my notes I wrote that μj→ isn't parallel to J⃗ and it is, in fact, rotating about J⃗ . So why μj→=gjJ⃗ ? Shouldn't μj→ and J⃗ be parallel this way?>

The angular momentum operators L^2 and L(z) commute with Hamiltonian and can be measured simultaneously giving the eigenvalues l(l+1) h_bar^2 and mh_bar ,

however the other components of L namely L(x) and L(y) do not commute along with **L(z)**so they can not be measured simultaneously...meaning thereby that direction of L remains indeterminate.

so one can not talk about the specific direction of orbital angular momentum L vector .

So when we describe the magnetic moment of a nucleous mue(j) = mue(l) + mue(s) (1) then

mue(l) = g(l). mue(N). sqrt (l(l+1)) ,

where mue(N) is magnetic moment of nucleon

For neutron as it is uncharged mue(l) will be zero and for proton g(l)=1

so for Proton

mue(lp)= mue(N). sqrt(l(l+1)) ....(2)

As nucleons are spin 1/2 particles the QM values of intrinsic magnetic moment can be written as

mue(s) = g(s). mue(N) . sqrt(s(s+1))..... (3)

So Total magnetic moment component in the j direction

mue(j) = mue(l) cos (l, j) + mue(s) .cos (s,j)

Those Cosine terms can be calculated in terms of l, s and j values.

Moreover the last nucleon in the extreme single particle model ( in odd A nucleus) and its state is to be considered which determines the magnetic moment . For even even nucleus the resultant spin is zero.

so classical description is not possible.though i have seen vector model drawing of coupling of angular momentum but

i think all its diagrams are not measurable. when one imposes the external magnetic field then the projections along z axis are measured.

For details see

Atomic and Nuclear Physics, Vol-II,S.N. Ghoshal,S. Chand & Co., New Delhi >India,Second Edition 1998