So i am trying to understand the relaxation mechanism in NMR for nuclei with spin $I>\frac{1}{2}$. I know that nuclei with $I=\frac{1}{2}$ have spherical shape, because many paper have told me so. I guess the argument is that we calculate the quadrupolar moment for $I=\frac{1}{2}$ and the moment vanishes, consequently the nucleus has to be spherical, because only then the orientation of the nucleus relative to the electrical field gradient is irrelevant. But i never find the full argument in the texts. The most I found until now was "Application of the Wigner–Eckart theorem demonstrates that the quadrupolar interaction vanishes unless I ≥ 1", but this demonstration wasn't shown. Can anyone help me out with that, knowing a book or paper where the calculation is carried out? I am not very experienced with applying the Wignar-Eckart Theorem...
2 Answers
There are already many answers in David Bailey's linked questions. The intuitive way is already rigorous proof. The quadrupole moment operator $Q$ is a symmetric traceless matrix (rank two tensor). It is therefore a spin $2$ operator. You want to compute: $$ \langle I,m|Q_q|I,m'\rangle $$ Concretely, you can to convert to the more familiar cartesian components using spherical harmonics: $$ Q_{-2}=Q_{xy}\quad Q_{-1}=Q_{yz}\quad Q_0=2Q_{zz}-Q_{xx}-Q_{yy} \\ Q_1=Q_{zx} \quad Q_2=Q_{xx}-Q_{yy} $$ By Wigner-Ekhart's theorem: $$ \langle I,m|Q_q|I,m'\rangle=\langle I,m,I,m'|2,q\rangle Q_I $$ Where the first factor is the Clebsch-Gordon coefficient and $Q_I$ is independent of $m,m',q$. Without looking at the table, you already know that it is zero for $I<1$. This follows from spin composition: $$ I\oplus I=2I\oplus (2I-1)\oplus...\oplus 0 $$ Therefore for $2I<2$ you automatically have a zero coefficient. Geometrically, you can also view this as the triangle inequality for a triangle of sides $I,I,2$ which are the corresponding spins.
More generally, the $2^k$-pole moment operator $M^{(k)}$ is spin $k$ (previously $k=2$), so in general: $$ \langle I,m|M^{(k)}_q|I,m'\rangle=\langle I,m,I,m'|k,q\rangle M^{(k)}_{I,I} $$ so it is zero for $2I<k$. Similarly, if you are interested at transitions from $I$ to $I'$, the selection rule is $I+I'\geq k$.
Hope this helps.
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In extended GR the spatial configuration of the nucleus arises from relaxation of curvature and tension fields:
A spin-½ system lacks sufficient internal curvature tension to stabilize asymmetry.
So the nuclear configuration naturally relaxes into the lowest-energy geometry: a sphere.
This isn’t just "no quadrupole moment" — it’s that geometry can’t support one under those spin constraints.
Extending to Higher Spins For I≥1, extended GR interprets the emergence of non-spherical structure as the result of localized curvature bifurcations within the nuclear geometry.
These nuclei don’t just permit quadrupole interaction — they require tension asymmetry to stay stable.
That means: quadrupole moments aren't just possible — they’re geometrically necessary in many field configurations.
