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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.

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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).

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    $\begingroup$ Hey thanks for your response, Its been a while since I looked at this but I'll try and attack the problem again when I get a chance and see if I understand your response $\endgroup$ Feb 25, 2023 at 17:36
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    $\begingroup$ I understand now, I mustve been tired when i was attempting this. Thank you so much for your response, honestly wouldnt have got it without you! $\endgroup$ Feb 28, 2023 at 12:45

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