Correcting the energy spectrum of hydrogen for relativistic kinetic energy (to first order) obtains a first order correction in perturbation theory of
$$\Delta_{nl}^{(1)}=\frac{(E_{n}^{(0)})^{2}}{2m_{e}c^{2}}\left(3-\frac{4n}{l+\frac{1}{2}}\right)$$
similarly, correcting for the spin-orbit coupling, we have a first order correction in pertrubation theory of \begin{align}
\xi^{(1)}_{njl}&=\frac{(E_{n}^{(0)})^{2}}{m_{e}c^{2}}\left(\frac{n\left(j(j+1)-l(l+1)-\frac{3}{4}\right)}{l(l+\frac{1}{2})(l+1)}\right)
\end{align}
where $l$ is the orbital angular momentum of the electron in the hydrogen atom, and $j$ is the total angular momentum of the electron in the hydrogen atom, and $E_{n}^{(0)}$ is the uncorrected (Bohr) energy of an electron in hydrogen for energy $n$.
Adding the two together should provide the fine structure correction,
$$E_{FS}=\Delta_{nl}^{(1)}+\xi^{(1)}_{njl}=\frac{(E_{n}^{(0)})^{2}}{2m_{e}c^{2}}\left(3-\frac{4n}{j+\frac{1}{2}}\right)$$
Using the fact that $j=l\pm\frac{1}{2}$, $l=j\pm\frac{1}{2}$, it is possible to evaluate the sum seperately for each scenario. For $l=j-\frac{1}{2}$ I easily recover the formula for the fine structure correction, however for $j=l+\frac{1}{2}$ I obtain
$$\Delta_{n,l=j+1/2}^{(1)}=\frac{(E_{n}^{(0)})^{2}}{2m_{e}c^{2}}\left(3-\frac{4n}{j+1}\right)$$
and
\begin{align}
\xi^{(1)}_{njl}&=\frac{(E_{n}^{(0)})^{2}}{m_{e}c^{2}}\left(\frac{-n}{(j+\frac{1}{2})(j+1)}\right)
\end{align}
adding these two results together clearly doesn't recover the formula for the fine structure correction. I checked the solutions manual for Griffiths Introduction to quantum mechanics (https://physicaeducator.files.wordpress.com/2018/01/solutions-of-quantum-mechanics-by-griffith.pdf, pages 167-168, Q 6.17), and the results for the individual corrections are correct, however they seem to subtract the spin-orbit correction in the case where $l=j+\frac{1}{2}$, such that
$$E_{FS}=\Delta_{nl}^{(1)}-\xi^{(1)}_{njl}$$
whilst adding the two corrections together for $l=j-\frac{1}{2}$. Is there a reason why this is the case? I know that this formula should be obtained, since we observe it experimentally, and it can be recovered from the Dirac equation, however I cannot understand why the corrections are added together in one situation and subtracted in the other.
Best Answer
The calculation is indeed incorrect in the solutions. In both cases is necessary to add the terms, as you pointed out. That said, in the second one ($l=j+\frac{1}{2}$) there are two mistakes: (1) in the relativistic term there is a minus sign that wasn't deal with properly; (2) in the spin-orbit term they forgot the minus sign.
So, the fine structure correction for $l=j+\frac{1}{2}$ is: $$ E_{FS}=\Delta^{(1)}_{nl} + \xi^{(1)}_{njl} = \frac{(E^{(0)}_{n})^{2}}{2m_{e}c^{2}}\left(3−\frac{4n}{(j+1)}\right) + \frac{(E_{n}^{(0)})^{2}}{m_{e}c^{2}}\left(\frac{-n}{(j+\frac{1}{2})(j+1)}\right) \\ = \frac{(E^{(0)}_{n})^{2}}{2m_{e}c^{2}} \left(3-\frac{4n}{(j+1)} -\frac{2n}{(j+\frac{1}{2})(j+1)}\right) = \frac{(E^{(0)}_{n})^{2}}{2m_{e}c^{2}} \left(3+\frac{-4n(j+\frac{1}{2})-2n}{(j+\frac{1}{2})(j+1)}\right) \\ = \frac{(E^{(0)}_{n})^{2}}{2m_{e}c^{2}} \left(3+\frac{-4n}{(j+\frac{1}{2}) }\right) \;. $$ The terms you obtained are in agreement with the terms in the solutions, I think you can now spot the problem there (the minus signs).