Examining isotope shift in ions vs neutral atoms - why use ions?

Question

I'm currently deep-diving isotope shift (IS) spectroscopy literature. I've come across several papers that look at the isotope shifts of charged ions, and I want to try and understand why researchers would choose to use ions instead of neutral atoms in their experiments on this topic.

Does it have something to do with changing the possible transitions within the atom in such a way as to be more convenient for studying IS? If so, what makes having the outer electrons removed more convenient?

Secondly, how does an atom's IS spectroscopy compare to its ion in general?

I found a paper that states that highly charged ions have high resolution and signal-to-noise ratios. Also, that compared to neutral atoms, electron correlation effects are weak and mean that we can achieve highly accurate calculations of atomic properties. Why is this the case? I'd guess that with an electron or two missing, the charge distribution and coupling with the nucleus obviously changes, and is perhaps simpler to deal with..?

Or, could it be that researchers choose ions simply because it's easier to cool and trap them? (For instance, the trap depth achievable for charged particles is many orders of magnitude larger than for atoms and molecules. Laser cooling was also first developed for ions).

Any examples that might help with explaining would be so very appreciated!

Edit: another thought, is it to do with the fact that part of the isotopic shift (the specific mass part) is sensitive to electron correlation, and removing the outer electrons decreases the correlation effect?

Edit 2: here are some of the aforementioned papers looking at ions

• https://arxiv.org/abs/2102.02309
• https://link.springer.com/article/10.1134/S0030400X20080408
• https://iopscience.iop.org/article/10.1088/0256-307X/36/11/113101

As opposed to something like this, with a neutral atom

• https://journals.aps.org/prresearch/abstract/10.1103/PhysRevResearch.1.033113

Answer 1

Honestly, it's probably because single ions are heavily used in optical clocks, and maybe these papers were a spinoff from metrology research.

But it might well be that these people are interested only in the nucleus, and so they are making your life easier by removing all electrons apart from one. That one electron is the one that still gives you access to an optical transition to study the shift on.

E.g. equation 7 of this paper (the IOP science one you listed) has an expression for one of the correction terms, and it involved the wavefunction of the electron. If you had more electrons, you would have to do additional perturbation theory for the exchange interaction, the electron-electron repulsion, etc.
Also I see the othher papers are doing ab initio QED calculations, so removing all other electrons probably exponentially reduced the number of Feynman integrals to consider.