# Does Hamiltonian including spin involves tensor products?

I'm trying to learn introductory QM from a book and I'm confused about how spin is incorporated into the formalism.

From what I have gathered so far, the state of a particle can be fully described by two parts:

(1) the wave function of the particle as a solution to the time independent Schrodinger equation

(2) the spin of the particle

I keep reading that the "whole system" is described by the Hamiltonian, so I was wondering if in more advanced QM treatments, they consider the "whole system" to be a state in the tensor product spanned by states of the form $$\psi_n \otimes | s, m_s \rangle$$ where $$\psi_n$$ is an eigenstate of $$H$$, equipped with the Hamiltonian $$H \otimes I$$ and time independent Schrodinger equation: $$(H \otimes I) (\psi \otimes | \sigma \rangle) = E\psi \otimes | \sigma \rangle$$

Yes. Of course, if the Hamiltonian can be written in the form $$H\otimes I$$, then the addition of the spin degree of freedom serves only to give every state a two-fold degeneracy, and is therefore a bit boring. Typically, states with different spin have different energies (see e.g. spin-orbit coupling, Zeeman splitting, etc), in which case your Hamiltonian will not be so trivial.