# Can the ground state of an even-even nucleus have a quadrupole moment?

The question I'm stuck with is the following:

The quadrupole moment in a way describes the shape of the electric charge distribution of a nucleus right? Oblate if negative, prolate if positive and 0 if spherical.

Furthermore, things that add to it are the rotation of the nucleus and which excited state it is in...etc.

But can an even-even nucleus in the ground state have a quadrupole moment? On one side I would think yes since a nucleus in the ground state can still be deformed, so it must have a non spherical charge distribution which means it needs to have a quadrupole moment. On the other hand every term I know to add to the quadrupole moment should be 0.

I would be grateful to anyone who could shed some light on the issue.

• Can you clarify what you mean when you say "every term I know to add to the quadrupole moment should be 0"? Commented May 29, 2020 at 17:08
• That there is no contribution from an excited state, no contribution from rotation...etc. I do fail to see how this changes the nature of the question. Commented May 31, 2020 at 7:16
• Ok, so you're presumably starting off with some model for the nucleus, that makes some kind of prediction for the quadrupole moment based on various terms. What is the specific model that you're using, and is it even compatible with non-spherical nuclei in the first place? (In other words, does the model that you're getting these contributions from assume that the nucleus is spherical?) Commented May 31, 2020 at 7:55

The matrix element $$\langle \alpha L\vert Q_\nu\vert \alpha L\rangle$$, where $$\alpha$$ denotes all quantum numbers other than angular momentum needed to specify your state, is $$0$$ when $$L=0$$ by the Wigner-Eckart theorem.