Quadrupole deformation in nuclei

[kelly0303]

Hello! I am confused about the definition of the quadrupole moment in nuclei. One definition I found, in Wong, says that the quadrupole moment of a nucleus is given (ignoring some numerical constant) by: $$<J,M=J|r^2Y_{20}|J,M=J>$$ so the expectation value of a second order spherical harmonic (times $r^2$) in the state with maximum M of the nucleus. However, given that we have a second order tensor ($Y_{20}$), for a nucleus with $J=0$ we can't have a quadrupole moment (this follows from the addition rules of angular momentum). However, there are several nuclei that have a (pretty big) quadrupole deformation in the ground state (for example they show a clear rotational spectrum, which is possible only if the nucleus is deformed), yet their ground state has $J=0$ (for example Hf). How is this possible, shouldn't a $J=0$ state indicate a spherical nucleus? Can someone explain this to me? Thank you!

 

[mfb]

Which hafnium isotope are you looking at?

https://periodictable.com/Isotopes/072.176/index.p.html
https://periodictable.com/Isotopes/072.178/index.p.html
https://periodictable.com/Isotopes/072.180/index.p.html

The even/odd nuclei have a quadrupole moment and non-zero spin (obviously):

https://periodictable.com/Isotopes/072.177/index.p.html
https://periodictable.com/Isotopes/072.179/index.p.html

 

[kelly0303]

Which hafnium isotope are you looking at?

https://periodictable.com/Isotopes/072.176/index.p.html
https://periodictable.com/Isotopes/072.178/index.p.html
https://periodictable.com/Isotopes/072.180/index.p.html

The even/odd nuclei have a quadrupole moment and non-zero spin (obviously):

https://periodictable.com/Isotopes/072.177/index.p.html
https://periodictable.com/Isotopes/072.179/index.p.html

Hello! My example (the one from Wong) is Hf 170. But my question was general, not only about that one particular nucleus. In general, even-even nuclei have spin zero in the ground state. Yet nuclei in between filled shells are known to have quadrupole deformations. But as I mentioned in the post, this seems like a contradiction to me. So I am missing something.

 

[mfb]

https://periodictable.com/Isotopes/072.170/index.p.html
Still no quadrupole moment according to the same source.
Clicking around I don't find any even/even nucleus with quadrupole moment.

 

[kelly0303]

https://periodictable.com/Isotopes/072.170/index.p.html
Still no quadrupole moment according to the same source.
Clicking around I don't find any even/even nucleus with quadrupole moment.

That's weird. It has a quadrupole, as it has a spectrum specific to quadrupole deformation.

 

[mfb]

Aren't these excited states with $J\neq 0$?

 

[kelly0303]

Aren't these excited states with $J\neq 0$?

As far as I understand from Wong (I might be wrong tho) this spectrum is a rotational one (the energies increase as J(J+1)). In order to have that spectrum, the nucleus must be deformed (a spherical nucleus doesn't have rotational energy in QM). So the ground state must be deformed in order to develop this spectrum. Yet the ground state has spin 0, which means that the state is spherical. So there is a clear contradiction. Of course I am miss-understanding something, but I am not sure what.

 

[mfb]

But it's not in the ground state when it's rotating.

 

[kelly0303]

But it's not in the ground state when it's rotating.

That's true, but the ground state needs to be deformed in order to set it in motion i.e. give it a rotation quanta. Assuming that the shape of the nucleus doesn't change (which is what Wong is assuming in his derivations) the shape of the nucleus in the ground state and in any excited rotational state should be the same (at a point it might be more convenient to start vibrating and thus change its shape, rather than add one more rotational quanta, but that is another story). So in order to get the nucleus to rotate and keep it rotating, its shape in the ground state must have (at least) a quadrupole. What am I missing here?

 

[Meir Achuz]

I think 2nd order perturbation theory would work.