# Why are atomic quadrupole moments calculated using nuclear spin?

It's my understanding that electric quadrupoles interact with the gradient of an electric field, and I understand roughly how this works. I am trying to calculate the interaction between an atomic nucleus and a field gradient using this understanding. However, this does not seem trivial and from what I've read (in textbooks and online) an atom's nuclear quadrupole moment is directly proportional to its total spin. Why is this?

Sources:

According to Wikipedia, nuclei have electric quadrupole moments if their total spin is 1 or greater.

(Most of my sources so far are textbooks, which are not public-accessible, but the above source explains things fairly well.)

Edit: As far as I can tell, the formula for the energy of a quadrupole in a field gradient is very close to this one (but potentially different, which is why I'm trying to derive it):

$E_Q = \frac{e V_{zz} Q} {(4 I) (2 I - 1) {\hbar}} [(3 I_z^2 - I^2) + {\eta} * (I_x^2 - I_y^2)]$

where:

• $Q$ is the atom's quadrupole moment

• $I_x$ its x spin (etc)

• $V_{zz}$ the z field gradient

• ${\eta} = \frac{V_{xx}-V_{yy}}{V_{zz}}$ the "asymmetry parameter".

As you can see, it is heavily dependent on spin.

• For fundamental particles you can't have a quadrupole moment with spin < 1 due to the Wigner-Eckart theorem, see this answer. But for nuclei it seems to be more complicated: on the chart of nuclides you can color by quadrupole deformation parameter $\beta_2$, and only even-even nuclei (which all have ground state $J^\pi = 0^+$) have data. Famously, U-238 is cigar-shaped. So I'm confused myself, and looking forward to an answer to your question.
– rob
Aug 4, 2014 at 7:36
• By the way, the WP link doesn't seem to have been written by a competent person, so I wouldn't depend on it. They have this: "whereas the dipole moment is a measure of the degree to which the nucleus is stretched out in a line, as is a prolate spheroid or an American football, the quadrupole moment is a measure of flatness or oblateness." This is completely wrong. Both prolate and oblate shapes have quadrupole moments.
– user4552
Aug 4, 2014 at 17:36
• @BenCrowell You're right, the dipole line is completely wrong... but I figure the article might still come in handy, and the fact that I cited it for is still true. (In the meantime, I'm going to fix the article. So goes the magic of Wikipedia) Aug 4, 2014 at 18:04

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.