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, $\hbar=1$.