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## Main Question or Discussion Point

Hello! I don't know much about this, so maybe the answer to my questions follows directly from the math of it, but I was wondering if there is an answer providing more physics intuition to this, not just math: Why can a nucleus have an octupole deformation, as a ground state stationary state (https://www.nature.com/articles/nature12073), but no nucleus so far was found to have dipole deformation. I understand that both type of deformations would have to vanish in a stationary state if parity would not be violated. Given that parity is actually violated by the weak interaction (ignore beyond the SM physics for now), we expect to have (a small) octupole and dipole deformation in some nuclei (probably in all nuclei in principle, but for most of them it is too small to be measured). In all cases I encountered so far in physics, when one makes a multipole expansion, the higher the multipole the lower the given effect. So based on that logic I would expect that a dipole deformation to be bigger than an octupole one. Yet, the octupole one was measured, but no dipole one. Why is this the case? Why can the weak interaction lead to a measurable octupole deformation, but not to a dipole one? Thank you!