Timeline for "Hamiltonian operator has no effect on the spin function" what does it mean?
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| Nov 12, 2017 at 16:46 | comment | added | pentavalentcarbon | To clarify, this is not the Hamiltonian, it is a Hamiltonian, specifically the one that acts on a non-relativistic stationary wavefunction (the time-independent Schrödinger equation, particle in a box/on a ring models, ...). Spin appears naturally in other Hamiltonians, such as Dirac's equation and its many reductions, and is "bolted on" for magnetic perturbations in the Schrödinger equation. | |
| Nov 12, 2017 at 16:35 | comment | added | orthocresol | Spin doesn't correspond to a physical rotation of the electron! See, e.g. chemistry.stackexchange.com/questions/58020/… If you consider both up spin and down spin to be degenerate, there will not be any term depending on $m_s$ that enters the Hamiltonian (depending on where you set your zero of energy, there might be a constant term, but this term doesn't depend on $m_s$). | |
| Nov 12, 2017 at 16:34 | comment | added | Subhadip Pal | the spin of an electron is due to the rotational motion about its own axis. Why it should not be there for the Hamiltonian operator?? I mean the electron should also have a rotational kinetic energy due to its motion about its own axis. | |
| Nov 12, 2017 at 16:18 | history | answered | orthocresol | CC BY-SA 3.0 |