What's the Lagrangian for the Pauli equation? The Pauli Hamiltonian is
$$H=  \frac{1}{2m}(\boldsymbol{\sigma}\cdot(\mathbf{p} - q \mathbf{A}))^2 + q \phi $$
For the 1-component Schrodinger equation in absence of applied field we have the Lagrangian
$$\mathcal{L} = i\hbar\psi^*\dot \psi - \frac{\hbar^2}{2m}\Big( \boldsymbol{\nabla}\psi^*\cdot\boldsymbol{\nabla}\psi \Big)- \psi^*V\psi $$
What is the 2-component analog of this which corresponds to the Pauli Hamiltonian?
 A: Well, it is a straightforward transcription, if you appreciate the language.
For the 1-component (classical) Schroedinger field in absence of applied external EM fields, the Lagrangian density is actually 
$$\mathcal{L} = i\hbar\psi^*\dot \psi - \frac{1}{2m}\Big(  \psi^*\boldsymbol{p}^2\psi \Big)- \psi^*V\psi ~~;$$
(I have integrated your expression, corrected,  by parts and supplanted ${\boldsymbol p}=-i\hbar {\boldsymbol \nabla}$.)
The usual degenerate canonical momentum for ψ is but $\Pi_\psi=i\hbar \psi^* $, which eliminates the time derivative term in the Legendre transform to the hamiltonian, since $\psi^* \Pi_\psi-i\hbar \psi^* \partial_t\psi=0$.
The ensuing field Hamiltonian density is then also trivial,
$$
{\cal H}= \psi^* \Big(  \frac{1}{2m}   \boldsymbol{p}^2      +V\Big) \psi,
$$ 
netting you the standard Schroedinger hamiltonian.
To generalize to Pauli's, promote the wave-fields/functions to two-spinors, insert the requisite vector potentials (in the non-quantized classical potential part), and you have 
 $$\mathcal{L} = i\hbar\psi^\dagger\cdot  \dot \psi - \psi^\dagger \cdot \Big(   \frac{1}{2m}(\boldsymbol{\sigma}\cdot(\mathbf{p} - q \mathbf{A}))^2 + q \phi \Big) \psi.$$
Recall the Pauli matrices are Hermitian. The action is the space & time integral of this.
Again, variation w.r.t. the classical two-spinor fields/variables (like the qs and ps of classical mechanics) yields the corresponding Schroedinger/Pauli equation.
