$^{208}$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.
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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.
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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. – Ana Mar 15 '16 at 05:16
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Can you find me an open-access paper? My library doesn't have Physics Letters B going back that far. – rob Mar 15 '16 at 05:24
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I couldn't locate an open access article for this problem. Attaching two other articles in the question above, see if there are accessible. – Ana Mar 15 '16 at 11:44