I Nuclear size effect on isotope shift

[kelly0303]

Hello! The transition wavelength between 2 energy levels for an atom depends on the nuclear isotope through the mass of the isotope and the size of the nucleus. My question is only about the nuclear size effect. It can be shown that this effect can be written as (this is basically a taylor expansion):

$$\nu = \nu_0+F<r^2>+G(<r^2>)^2+...$$

where $\nu_0$ is the transition assuming a point nucleus, $<r^2>$ is the mean square nuclear radius and F and G are some parameters that depend on the electronic transition. Usually in literature (except for the super sensitive measurements), the higher order terms are dropped and only the F term is kept (this is how the so called King linearity of the isotope shift is obtained). I am not sure why we can drop the higher order terms. Let's say that $<r^2> = 5fm^2$, which is a reasonable value. Then $<r^2>^2 = 25fm^4$. I guess I am confused about the units. If we write this in terms of meters, the second term is smaller by a factor of $10^{-30}$, but can't we redefine units and treat 1 fm as the unit and in this case the $<r^2>^2$ term would be actually bigger? In QED, for example, the expansion is in terms of $\alpha$, which is unitless, so there is no confusion there. But here I am a bit confused. Can someone help me understand it? Thank you!

 

[hutchphd]

The natural units for the electron distance is the Bohr radius, so the expansion here is $r_{nuclear}/r_{bohr}$ so it is fermis vs angstroms

 

[kelly0303]

The natural units for the electron distance is the Bohr radius, so the expansion here is $r_{nuclear}/r_{bohr}$ so it is fermis vs angstroms

So the second term (assuming F and G are of similar order of magnitude) is about $10^{-10}$ smaller than the first one?

 

[hutchphd]

I honestly don't know the exact numbers, since I know precious little nuclear. There may be an "effective" radius in practice...I was justifying the expansion. In these units F and G may still be quite different in size. This depends upon the electronic wavefunction at the nucleus ( the s states get large there) and the coupling.
Whenever one does such an expansion, the expansion parameter is (at least tacitly) a dimensionless item, because your question is always relevant: "small compared to what?"

 

[Twigg]

I am not sure why we can drop the higher order terms.

I believe the higher order terms are related to higher order fluctuations in the magnetization density of the nucleus. I think dropping these terms is the same as saying "the nucleus is pretty much a dipole / it's uniformly magnetized". By magnetization density, I mean $\mathrm{M}(\mathrm{r}) = \frac{1}{2} \mathrm{r} \times \mathrm{J}(\mathrm{r})$ as defined in Jackson equation 5.53.

In the simplest case, you treat the nucleus as a magnetized sphere and ignore any variation of the magnetization over the nuclear radius, and that gives you the magnetic field $$\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi r^3} \left[ 3(\mathbf{\mu}\cdot \hat{r})\hat{r} - \mathbf{\mu}\right] + \frac{2\mu_0}{3} \mathbf{\mu} \delta^3 (\mathbf{r})$$ which I've taken from Griffiths quantum book equation 6.86 (see Jackson E&M book section 5.6 for a very comprehensive discussion). To next lowest order, I believe the nucleus is modeled as a sphere with magnetization that varies like $a + b r^2$ where $b \langle r^2 \rangle \ll a$. What I call $b$ here ends up being proportional to F. The terms of higher order than F are small because the nucleus acts very much like a uniformly magnetized sphere.

Edit: Oops, I was being a little lazy, just citing all my equations. Just filled in the ones that are short enough to copy.

 

[hutchphd]

I believe the higher order terms are related to higher order fluctuations in the magnetization density of the nucleus.

I think the primary issue is electrostatic effects caused by the finite nuclear size. Here is one treatment:
http://www.physics.drexel.edu/~bob/TermPapers/Joe Angelo Grad Quantum II Term Paper.pdf

The magnetic effects should be much smaller I believe. And the higher order terms are very much smaller because of the size of the expansion parameter.

 

[Twigg]

@hutchphd Nice catch, thanks! I never realized the nuclear size effect was outside the hyperfine term of the Hamiltonian. My bad

Just to re-cap, the reason the terms F,G,... fall off with higher orders is that the nuclear charge distribution isn't highly pathological. At 0th order, the nucleus behaves almost like a point charge ($\nu_0$). At 1st order, the nucleus behaves almost like a uniformly charged sphere with finite radius ($F$). At 2nd order, the nuclear charge density varies somewhat quadratically with radial position ($G$). And so on. Each successive model just adds slight corrections to the previous one.