# Where can I find relativistic corrections to 2s and 2p levels of Hydrogen Atoms?

I am currently studying for a Quantum Mechanics test, and I want to calculate the 2p and 2s hydrogen atom corrections for the relativistic, spin-orbit and darwin corrections, using perturbation theory.

I know how to do this, but I would like to be able to check my work somehow, to know if what I have done is correct or not.

I would pressume that these values would be in a table somewhere, but so far I have been unable to find them. Does anybody know where I could find the results?

I don't care if there are no worked answers, as I know how to do the procedure, I just want to check my work

• en.wikipedia.org/wiki/Fine_structure – G. Smith Jun 21 at 0:58
• You may study about the fine-structure of hydrogen atom from Brandsden and Joachain. The hyperfine structure may also be relevant(depends on the nature of the problem). – Jitendra Jun 21 at 20:00

For reference, the approximate energy levels of the four lowest hydrogen states, including the fine structure corrections, are

\begin{align} \frac{E(1s_{1/2})}{m_ec^2}&=-\frac{1}{2}\alpha^2+\left(-\frac{5}{8}+0+\frac{1}{2}\right)\alpha^4=-\frac{1}{2}\alpha^2-\frac{1}{8}\alpha^4, \\ \\ \frac{E(2s_{1/2})}{m_ec^2}&=-\frac{1}{8}\alpha^2+\left(-\frac{13}{128}+0+\frac{1}{16}\right)\alpha^4=-\frac{1}{8}\alpha^2-\frac{5}{128}\alpha^4, \\ \\ \frac{E(2p_{1/2})}{m_ec^2}&=-\frac{1}{8}\alpha^2+\left(-\frac{7}{384}-\frac{1}{48}+0\right)\alpha^4=-\frac{1}{8}\alpha^2-\frac{5}{128}\alpha^4, \\ \\ \frac{E(2p_{3/2})}{m_ec^2}&=-\frac{1}{8}\alpha^2+\left(-\frac{7}{384}+\frac{1}{96}+0\right)\alpha^4=-\frac{1}{8}\alpha^2-\frac{1}{128}\alpha^4. \end{align}

Here $$\alpha=e^2/4\pi\epsilon_0\hbar c\approx 1/137$$ is the famous small dimensionless fine-structure constant used for perturbation expansions in QED.

The $$\alpha^2$$ term is from the standard Coulomb attraction while the $$\alpha^4$$ terms are the fine-structure corrections. Inside each set of parentheses, the first term comes from the $$O(v^4/c^2)$$ relativistic correction to the kinetic energy, the second term comes from the spin-orbit coupling, and the third term comes from the Darwin correction. Note that $$2s_{1/2}$$ and $$2p_{1/2}$$ end up with the same energy, but via different contributions of the three corrections.

The total values agree with the Dirac energy levels for hydrogen, when expanded through order $$\alpha^4$$.

If you want to include the effect of an atomic number $$Z$$, replace $$\alpha$$ with $$Z\alpha$$. If you want to include the effect of a non-infinite proton mass, replace $$m_e$$ with the reduced mass $$\mu=m_em_p/(m_e+m_p)$$.

The general formulas for an arbitrary state with quantum numbers $$n$$, $$l$$, and $$j$$ are

$$\frac{E_\text{ Coulomb}}{m_ec^2}=-\frac{1}{2n^2}\alpha^2,$$

$$\frac{E_\text{ Rel Kin}}{m_ec^2}=-\frac{1}{2n^4}\left(\frac{n}{l+1/2}-\frac{3}{4}\right)\alpha^4,$$

$$\frac{E_\text{ Spin-Orbit}}{m_ec^2}= \begin{cases} 0, & l=0 \\ \Large{-\frac{1}{2n^3}\left(\frac{3/4+l(l+1)-j(j+1)}{2l(l+1/2)(l+1)}\right)}\normalsize\alpha^4, & l\neq0 \end{cases},$$

$$\frac{E_\text{ Darwin}}{m_ec^2}= \begin{cases} \Large\frac{1}{2n^3}\normalsize\alpha^4, & l=0 \\ 0, & l\neq0 \end{cases},$$

and, for their total,

$$\frac{E_{n,j}}{m_ec^2}=-\frac{1}{2n^2}\alpha^2-\frac{1}{2n^4}\left(\frac{n}{j+1/2}-\frac{3}{4}\right)\alpha^4.$$

I'm not including any derivations because they are widely available and suitable for homework. I have tried to put the formulas into what I consider their cleanest forms, for reference purposes. For example, these dimensionless formulas seem nicer than those in Wikipedia.