Physics Asked by Carson James on June 7, 2021
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: $$(H otimes I) (psi otimes | sigma rangle) = Epsi otimes | sigma rangle $$
Yes. Of course, if the Hamiltonian can be written in the form $Hotimes I$, 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.
Correct answer by J. Murray on June 7, 2021
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP