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I have read the Levine Quantum Chemistry book and it says "Hamiltonian operator has no effect on the spin function" in chapter 10 (Electron Spin and the Spin-Statistics Theorem) and does the Hamiltonian operation like the following. Why the g(ms) state is taken out of the Hamiltonian?? Why is it constant with respect to the Hamiltonian?

$$\hat H[\psi(x,y,z)g(m_s)] = g(m_s)\hat H\psi(x,y,z) = E[\psi(x,y,z)g(m_s)]$$

Jan
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Subhadip Pal
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1 Answers1

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Because the operator $\mathrm d/\mathrm dx$ only acts on $x$ and not on $y$, we can write

$$\frac{\mathrm d}{\mathrm dx}[f(x)g(y)] = g(y) \left[\frac{\mathrm d}{\mathrm dx}f(x)\right]$$

Likewise, in this context, the Hamiltonian operator $\hat{H}$ only acts on $(x,y,z)$ and not $m_s$. This is explained by Levine1 in the paragraph immediately preceding this

To a very good approximation, the Hamiltonian operator for a system of electrons does not involve the spin variables but is a function only of spatial coordinates and derivatives with respect to spatial coordinates [...]

Is there something about this paragraph which you don't understand? This is pretty much all there is to be said on the topic.

  1. Levine, I. N. Quantum Chemistry (7th ed.), p 268
orthocresol
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  • 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. – Subhadip Pal Nov 12 '17 at 16:34
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    Spin doesn't correspond to a physical rotation of the electron! See, e.g. https://chemistry.stackexchange.com/questions/58020/do-electrons-really-spin 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 $m_s$). – orthocresol Nov 12 '17 at 16:35
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    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. – pentavalentcarbon Nov 12 '17 at 16:46