Why do spin-$\frac{1}{2}$ nuclei have zero electric quadrupole moment? How does this come about, and how can one tell in general whether a spin-$j$ nucleus can have a nonzero quadrupole (or higher multipole) moment?
[Physics] Why do spin-$\frac{1}{2}$ nuclei have zero electric quadrupole moment
electromagnetismnuclear-physicsquantum mechanicsquantum-spin
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I think I can clear up most of this. Maybe someone whose qm chops are better than mine could help with the parts I'm fuzzy on.
Suppose an even-even nucleus has a prolate deformation (like an American football). This is very common, and basically occurs for any nucleus whose N and Z are both far from any magic numbers. What we really mean when we say that it's deformed is that there are correlations between the different nucleons (neutrons and protons), and the correlations have a certain spatial pattern or organization.
But in its ground state, this nucleus has spin 0. A spin $I=0$ has only a single $I_z$ state, which is $I_z=0$. That means that a zero spin has no orientation degree of freedom. Now how can this be when the thing is supposed to be shaped like a football? Obviously it's possible to give a football different orientations, and those orientations are distinguishable. Well, the general idea is to think of the spin-0 ground state as a superposition of every possible orientation. (I don't know if this description is really rigorous, but I think it's good enough for the present purpose. We have issues like these, and also states with similar orientations can have nonvanishing inner products with each other.)
So in the ground state, the neutrons and protons have this quadrupole-type pattern of correlations with each other, but they have no such correlation with anything external.
I think the way this shows up in the formula you posted is that for $I=0$, it misbehaves. (The numerator and denominator are both zero.)
The formula also misbehaves if you plug in, e.g., $I=1/2$, $I_z=1/2$, $\eta=0$, and $I^2\rightarrow I(I+1)=3/4 $. Here I don't know if there's any geometrical explanation as simple as the one I gave above. This may be where you can't avoid the Wigner-Eckart theorem.
Let's come back to the deformed even-even nucleus with a spin 0 ground state. Although you can't orient the ground state, you can take this nuclear shape and excite it into a state of end-over-end rotation. If you do this, you get a rotational band with spins 0, 2, 4, ... and energies that go approximately like $E\propto I(I+1)$, which is basically the classical $I^2$ result for a rotor, with a quantum correction term added on. The existence of a set of states with this pattern of spins and energies is one of the classic ways of verifying that the nucleus really is deformed. (A spherical nucleus can't rotate collectively.) But in NMR or NQR you would never see the excited states.
In such a rotational band, we also observe anomalously fast electromagnetic transitions such as $4\rightarrow2$ and $2\rightarrow0$. These transitions are fast because they arise from the collective rotation of the whole nucleus, which makes it radiate coherently like an antenna. The speed of these transitions can be described using a transition quadrupole moment, which is different from, but related to, the static quadrupole moment of the nuclear shape. In the excited states, the nucleons have quadrupole correlations not just with each other, but also with the outside world. We can see this because the gamma radiation emitted in the transitions is externally detectable and also has a radiation pattern that is asymmetric as measured in the lab.
When nuclear physicists say that the spin 0 ground state of a rotational band "has" a certain quadrupole moment, what we really mean is that we make a somewhat unrealistic model in which we break rotational symmetry (and therefore slightly violate angular momentum conservation) by modeling the nucleus as if it had a fixed orientation in the lab frame. In this model, the nucleus has a quadrupole moment.
The interaction between nucleons in a nucleus is very strong so the energy changes associated with flipping a spin are very high. The first few energy levels of a carbon nucleus can be found in this document. The energy spacing between the ground state $J=0$ and the first excited state $J=2$ is 4.44 MeV. This isn't much by the standards of modern accelerators, but it is far, far too big for the transition to be caused by an external magnetic field.
So as far as NMR is concerned the carbon atom behaves as an elementary particle of spin zero with no internal excitations.
Best Answer
The answer is: The Wigner-Eckart theorem.
In plain old angular momentum coupling (see Clebsch-Gordan coefficients) we learn that if a system consists of a spin-1/2 subsystem and a spin-2 subsystem, then the total system altogether can be spin-5/2 or spin-3/2, no other value. The rules are
Well, the Wigner-Eckart theorem teaches us that similar rules apply involving operators. The electric quadrupole operator $Q$ can be expressed as a rank-2 spherical tensor operator. When it acts on a spin-1/2 system, the resulting spin can only be $\frac{5}{2}$ or $\frac{3}{2}$. So if you're calculating $\langle\frac{1}{2}| Q|\frac{1}{2}\rangle$, you can group it like $\langle \frac{1}{2}| \;\;\; Q |\frac{1}{2}\rangle$. In other words, this is the inner product of a spin-$\frac{1}{2}$ system with $Q|\frac{1}{2}\rangle$ (which is spin $\frac{3}{2}$ or $\frac{5}{2}$). So they're orthogonal; the inner product is zero. $\langle\frac{1}{2}| Q |\frac{1}{2}\rangle=0$.
By similar logic, dipole moments (electric or magnetic) require spin ≥ 1/2, quadrupole moments require spin ≥ 1, octupole moments require spin ≥ 3/2, etc. (Dipole operators are rank 1, quadrupole are rank 2, octupole are rank 3, etc.)