# Implication of completely non relativistic Hamiltonian

I am studying the Application of perturbation theory to hydrogenic atoms subject to internal or external electromagnetic fields.

The very first equation it uses is the hamiltonian of hydrogenic atoms $$H_0=\frac{p^2}{2\mu}-\frac{Ze^2}{r}$$ This equation is said to have no spin and is completely non-relativistic.

• What does completely non-relativistic mean?
• These are two (generally) unrelated questions - you should ask them separately. In this case you can edit your question to restrict the scope, if you wish. Sep 1 '17 at 2:44
• @orthocresol removed the second question. Sep 1 '17 at 19:58

Your reduced equation is just the normal hydrogen Hamiltonian written in atomic units, that is with $\hbar=\frac{1}{4\pi\epsilon_{0}}=m_{e}=e=1$ (e is the charge of the electron, which they include for some reason despite it being 1 in atomic units). They also make the substitution $p=-i\hbar\nabla$ where, again to be consistent with atomic units, $\hbar=1$.

The form you have normally seen of the Hamiltonian is already spinless and non-relativistic (which are essentially synonymous for quantum mechanical operators). For something as light as Hydrogen, relativity (as in Einstein's theory of Special Relativity) is known to have very limited impact, and so it is neglected in forming the Hamiltonian.

There may be other questions on the Chem or Physics SE related to relativity in QM, but I would suggest unless you are particularly interested in relativity to simply focus on the QM side of things for now until you have that clear and can build on that knowledge.