# "Hamiltonian operator has no effect on the spin function" what does it mean?

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?

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

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
• 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 '17 at 16:34
• 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 '17 at 16:35
• 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 '17 at 16:46