The Tech Basement

The gyromagnetic ratios determine how strongly the nuclear magnetic dipole moment interacts with the electromagnetic field, which directly affects the M1 transition strength. They show up in the $M1$ operator as

$$ \hat{M1} = g_L \hat{L} + g_S \hat{S}. $$

Protons and neutrons have different gyromagnetic ratios. The orbital motion of a charged particle produces a magnetic moment. Since only protons have charge, they alone contribute to the orbital part of the nuclear magnetic moment. The magnetic moment of a moving charge is given by

$$ \mu = \frac{q}{2m}L. $$

To compensate for various shortcomings of shell model calculations, like truncations and inert cores, it is possible to set the orbital g-factor for neutron to a non-zero number. This is is done in some cases for the sdpfsdg-mu interaction (like here where it is set to $-0.1$: https://doi.org/10.1103/csx6-6g5k). More commonly, the spin g-factors for protons and neutrons are often “quenched” in shell model calculations, in which case the free $g_s$ values of $5.585$ and $-3.826$ for protons and neutrons respectively, are multiplied with a factor that is typically between $0.7$ and $1$ depending on the interaction. See the nuclear shell model interaction overview for details.

What we can do is to artificially set the g-factors to zero to see what effect is has on the M1 strength function.

In dark red and in orange we have “only $g_l$” and “only $g_s$” where I have set $g_s = 0$ and $g_l = 0$ respectively. In the low energy region of approx $0-2$ MeV we see that the $L$ and $S$ term play almost exactly the same role. As the gamma energy increases past $2$ MeV however, we see that the “only $g_s$” part is increasingly doing the full duty of producing the M1 strength.

$L$ and $S$ are vector quantities which means that calculating the $L$ and $S$ parts separately and them adding them together does not work. That's because to get the GSF we have to calculate

$$ B(M1) = \frac{|(f|\hat{M1}|i)|^2}{2J_i + 1} $$

and expanding that square we get

$$ |(f|\hat{M1}|i)|^2 = (|(f|g_L\hat{L}|i) + (f|g_S\hat{S}|i)|)^2 $$

and dubbing the first term $M_L$ and the second term $M_S$ we get

$$ = M_L^2 + M_S^2 + 2M_L M_S $$

where the cross-term $2M_L M_S$ is the important part. Naively summing $M_L^2 + M_S^2$ is exactly the green line in the figure. From 0 to 5.2 MeV the naive sum underestimates the properly summed M1 strength meaning that there is constructive interference between the $L$ and $S$ terms that is not included, aka $2M_L M_S$ is positive. While beyond $5.2$ MeV it is the opposite.

In the figure above I have calculated the average interference angle between $L$ and $S$. If you visualise a simple xy-plane with two vectors in it, you can imagine that if the vectors are at 90 degrees their $2M_L M_S$ is equal to zero. If the angle is less than 90 then the cross-term is positive aka. constructive interference and vice versa for angles greater than 90.