Is Fine Structure and Spin-Orbit Coupling Observed at all Energy Levels of Hydrogen, or just One

Question

I have been trying to figure out how to gather some intuition regarding the origin of spin. I thought a good place to start was the spectra of hydrogen, which was first interpreted as evidence of an "intrinsic angular momentum" or spin by means of fine structure. Now my question here is more experimental. How exactly does fine structure manifest itself in spectral data? From my understanding, it occurs twice within each energy level (only for hydrogen and alkali metals).

However, the examples I have seen thus far on websites (hyperphysics) only mentions the 2P transition.

Can anyone enlighten me on the details, is doublet splitting observed for all spectral lines of hydrogen, or just red line. Thanks.

Answer 1

In hydrogen, every level with $l > 0$ is split into two. For example, what would be simply $2p$ without spin becomes the two levels $2P_{3/2}$ and $2P_{1/2}$ as in your diagram. The $2s$ state is not split because when $l=0$ there is only one value for the total angular momentum: it is then $j=1/2$ and the notation is $S_{1/2}$.

Here are two more examples:

$3p$ gives $3P_{3/2}$ and $3P_{1/2}$

$3d$ gives $3D_{5/2}$ and $3D_{3/2}$

When the atom changes state in a transition, in principle all the changes can happen, but some of them are very much more likely than others. The most likely ones involve the electric dipole moment of the atom, and for these transitions $l$ has to change by either $+1$ or $-1$, and $j$ may change only by $0$ or $+1$ or $-1$. As a result, although there are 4 possible transitions from $3d$ to $3p$ only three of them are seen in the emission spectrum (the transition from $j=5/2$ to $j=1/2$ being not possible by electric dipole radiation), and the transition from $3d$ to $1s$ is very weak so not normally considered.

The result of all this is that the transitions from $P$ to $S$ states appear as doublets, and the transitions from $D$ to $P$ states appear as triplets. However, when the splitting of the doublet or triplet is small, it may be too small to be resolved by a given instrument.

Answer 2

Fine structure arrises from relativistic corrections to the non-relativistic Coulomb hamiltonian of the atom. An absolutely fantastic introduction to what you need to understand atomic physics and laser spectroscopy is in MIT's open courseware, two series of lectures by Wolfgang Ketterle (https://www.youtube.com/playlist?list=PLUl4u3cNGP62FPGcyFJkzhqq9c5cHCK32).

There are 3 parts to fine structure:

1: The relativistic correction to the kinetic energy v. momentum relation.

2: The Darwin term, due to the smearing of the electron wave function near the nucleus.

3: Spin-Orbit Coupling.

The spin-orbit coupling hamiltonian is:

$$ \hat H_{SO}= \frac 1 2 \big(\frac{Ze^2}{4\pi\epsilon_0}\big) \big(\frac g {2m_e^2c^2}\big)\frac{\vec L\cdot\vec S}{r^3}$$

Squaring $\vec J = \vec L + \vec S$ gives:

$$ J^2 = L^2 + S^2 + 2\vec L\cdot S $$

which, for eigenstates of the unperturbed hamiltonian and $s=\frac 1 2$:

$$ j(j+1) = l(l+1) +\frac 3 4 +2\vec L\cdot S$$

With that and:

$$\big{\langle}\frac 1 {r^3}\big{\rangle}=\frac{Z^3}{n^3a_0^3} \frac 1 {l(l+\frac 1 2)(l+1)}$$

you get:

$$\big{\langle}\hat H_{SO}\big{\rangle}=\frac{E_n^2}{m_ec^2} \times n\frac{j(j+1)-l(l+1)-\frac 3 4}{l(l+\frac 1 2)(l+1)}$$

which shows that it's not just $l=1$.