Is there a way to get the spin naturally in nonrelativistic theories? We all know how spin is added in a rather ad-hoc way in quantum mechanics. In the other hand, in relativistic quantum field theories the spin structure arises quite naturally from the fields. 
Is there not a way to construct a theory (perhaps a nonreltivistic field theory) in which the spin structure arises naturally?
 A: The Pauli equation, which describe non-relativistic, spin-1/2 electrons and is the non-relativistic limit of the Dirac equation, can be obtained from the Schrödinger equation throught minimal coupling in a rather similar way of its relativistic counterpart.
Consider the Schrödinger equation for a particle with charge $q$ in an external electrostatic potential $\Phi$:
$$ \tag{A} i \partial_t | \psi(t) \rangle = \left( \frac{\textbf{p}^2}{2m} + q \Phi \right) | \psi(t) \rangle$$
now remember from the properties of the Pauli matrices the Pauli vector identity:
$$ \tag{B} (\boldsymbol \sigma \cdot \textbf a) (\boldsymbol \sigma \cdot \textbf b)
= 
\textbf a \cdot \textbf b + i \boldsymbol \sigma \cdot (\textbf a \times \textbf b),  $$
which gives in particular for $\textbf p $ the identity
$$ \tag{C} (\boldsymbol\sigma \cdot \textbf p)^2 = \textbf p \cdot \textbf p \equiv \textbf p^2, $$
which allows us to rewrite the Schrödinger equation (A) as
$$ \tag{C} i \partial_t | \psi(t) \rangle
= \left( \frac{(\boldsymbol \sigma \cdot \textbf p)^2}{2m} + q \Phi \right) |\psi(t)\rangle. $$
Now use the minimal coupling
$$ \tag{D} \textbf p \rightarrow \textbf p - q \textbf A, $$
obtaining from (C) the Pauli-Schrödinger equation:
$$ i \partial_t |\psi(t)\rangle =
\left[ \frac{1}{2m} \left( \boldsymbol \sigma \cdot (\textbf p- q \textbf A) \right)^2 + q \Phi \right] | \psi(t) \rangle $$
and squaring the expression we can explicit the interaction term between spin and magnetic field:
$$ i \partial_t |\psi(t)\rangle =
\left[ \frac{1}{2m} \left( (\text p - q \textbf A)^2 - q \boldsymbol \sigma \cdot \textbf B \right) + q \Phi\right] | \psi(t) \rangle, $$
with $ \textbf B = \nabla \times \textbf A.$
