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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

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  • $\begingroup$ en.wikipedia.org/wiki/Fine_structure $\endgroup$ – G. Smith Jun 21 at 0:58
  • $\begingroup$ 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). $\endgroup$ – Jitendra Jun 21 at 20:00
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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.

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