$^{208}$Pb is a spherical nucleus, so its transition to the first 3$^-$ state is an octupole vibration. Why does this state then have an associated quadrupole moment, as mentioned by several authors? E.g. this and this article.

Attaching two more articles, check this one and this one


The ground state of Pb-208 is spherical (with spin-parity $0^+$, and doubly-magic proton and neutron numbers). However there is nothing that says that its excited states must also be spherical. An octupole vibration may include dipole and quadrupole terms.

Edited: in the spirit of this related answer, a "rotational band" of states with spin-parity $\cdots \to 7^- \to 5^- \to 3^-$ could provide evidence of a nonzero moment of inertia (mass quadrupole moment) for whatever configuration of nucleons makes up the $3^-$ state. I haven't checked whether that sequence exists or whether that's the method used in the papers you cite, however.

  • $\begingroup$ Could you please elaborate? What the article say is that the vibration is purely octupolar, but after excitation, there exists a quadrupolar reorientation among the substates. I am not able to comprehend that. $\endgroup$ – Ana Mar 15 '16 at 5:16
  • $\begingroup$ Can you find me an open-access paper? My library doesn't have Physics Letters B going back that far. $\endgroup$ – rob Mar 15 '16 at 5:24
  • $\begingroup$ I couldn't locate an open access article for this problem. Attaching two other articles in the question above, see if there are accessible. $\endgroup$ – Ana Mar 15 '16 at 11:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.