Limits of Perturbation theory for hydrogenic atoms

1. May 4, 2009

Cheetox

1. The problem statement, all variables and given/known data
Why can't we use perturbation theory to calculate the effect of the spin orbit interaction in hydrogen like uranium?

2. Relevant equations

3. The attempt at a solution
Is it something to do with the fact that the perturbation must be small compared to the rest of the hamiltonian? But for hydrogen like uranium the ionization energy is still 28000 times that of the spin orbit shift? Or could it be to do with the fact that the bohr radius for uranium approaches the radius of the nucleus?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 4, 2009

millitiz

Well, I am a bit confused by your question, too.
So normally, we can obtain the EXACT result by using J operator and what not. And since we can get the exact wave function, why do we want to do it "approximately?"
The reason that Perturbation probably does not work is that, in order to get your energy, you need to know the wave function (or approximately the wave function) through the equation
E_n = <$$\Phi$$_n l H l $$\Phi$$_n>
We know H, it has many extra terms (especially the L dot S term)
This L dot S term made your previous wave function for Hydrogen invalid.
So without solving J and stuff like that, you actually don't really know what your phi look like (notice that this phi would not look like the original ones, ie, without the spin, because of the L dot S term). So essentially, you still need to go through solving J, and what not. And since you go through the whole thing, you can in fact get the exactly E. So I guess yes you could, it is probably just a bit redundant.