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In Shankar's quantum mechanics book he says the spin of electron doesn't change the energy levels of hydrogen atom(page 397, 2nd edition). How doesn't spin(being a form of angular momentum) change the energy levels? The total Hamiltonian has a piece for spin.

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If we only account for the coulomb interaction which is the strongest one relevant to the hydrogen atom then the Hamiltonian won’t have any spin dependence. But that’s only an approximation, a very good one nonetheless.

However, if we look carefully enough we see that you are right. Energy does in fact depend on spin. For further details, look at spin-orbit interaction, a relativistic effect. But this is small compared to the coulomb interaction energy. To be precise, $\sim 10^5$ times smaller. This extra energy due to spin-orbit causes the spectral lines to spilt. But only when looked at with enough resolution. Hence it is also called as the fine structure.

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You can look at this lecture notes from MIT-OCW 8.06 or this hyperphysics page on the fine structure for further details.

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Electron spin does not appear in the Schroedinger equation. It does not affect it's single electron solutions. Many-electron solutions are affected only via Pauli exclusion. However, the Schroedinger equation is a non-relativistic approximation. In the Dirac equation spin impacts the solutions.

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  • $\begingroup$ The total Hamiltonian appears to be the sum two Hamiltonians: Orbital and spin. Shouldn't the total Hamiltonian enter the Schrödinger equation? $\endgroup$
    – Ben Stark
    Aug 23, 2020 at 10:48
  • $\begingroup$ It does not affect it's single electron solutions This is incorrect. The fine structure is modelled in QM via the spin-orbit interaction which in fact affects the single electron solutions of Schrodinger equation. $\endgroup$ Aug 23, 2020 at 12:34
  • $\begingroup$ @SuperfastJellyfish Spin-orbit coupling is not part of the Schroedinger equation. $\endgroup$
    – my2cts
    Aug 23, 2020 at 13:17
  • $\begingroup$ Schrodinger equation is just $H|\psi\rangle=i\partial_t|\psi\rangle$ and the Hamiltonian $H$ depends on how one models it. So I don’t know why you say Spin-orbit coupling is not part of the Schroedinger equation. you may look at the lecture notes attached in my answer that shows how one can incorporate it. $\endgroup$ Aug 23, 2020 at 13:28

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