How would a free particle with known spin evolve? I searched a lot for a Hamiltonian of a pauli spinor with no potential energy but got no luck, so I tried deriving one my own.
I took an overkill shortcut and used pauli's equation:
$$i\hbar \frac{\partial}{\partial t}=\frac{1}{2m}[\vec{\sigma}\cdot (\hat{p}-q\vec{A})]^2+q\phi$$
And just set the magnetic and electric potential to zero:
$$i\hbar \frac{\partial}{\partial t}=\frac{1}{2m}[\vec{\sigma}\cdot \hat{p}]^2$$
But other than this, I don't know how to continue. I don't know what the dot product is, let alone its square.
All in all, is my approach correct? If so how to continue? In case everything I said was wrong, what's the reason? And finally, what is the solution to the differential equation?
 A: The operator $\hat{\sigma} \cdot \hat{\vec{p}}$ corresponds to the operator
$$
\sigma_x p_x + \sigma_y p_y + \sigma_z p_z.
$$
where the $\sigma_i$ stand for the Pauli matrices and the $p_i$ are the momentum operators (equal to $-i \hbar \frac{\partial}{\partial x_i}$ in position space.)  So if you wrote this thing out as a position-space operator, you would get
$$
\hat{\sigma} \cdot \hat{\vec{p}} = -i \hbar \begin{bmatrix} \partial_z & \partial_x - i \partial_y \\
\partial_x + i \partial_y & - \partial_z
\end{bmatrix}
$$
where $\partial_x \equiv \partial/\partial x$, etc.  This operator then acts on a two-component spinor, each of whose components is a function of $\vec{r}$.
In principle, this operator can then be squared.  However, this just works out to be
$$
\left(\hat{\sigma} \cdot \hat{\vec{p}} \right)^2 = -\hbar^2 \begin{bmatrix} \nabla^2 & 0 \\ 0 & \nabla^2 \end{bmatrix} 
$$
which just means that in the absence of a magnetic field, each of the components of the spinor satisfies the conventional Schrödinger equation, and can be solved using the typical techniques that you're already familiar with.
(The simplification of the squared operator, by the way, can be viewed as a special case of the immensely useful Pauli vector identity, which is
$$
(\vec{a} \cdot \hat{\sigma} )(\vec{b} \cdot \hat{\sigma}) = (\vec{a}\cdot\vec{b}) \mathbb{I} + i (\vec{a} \times \vec{b})\cdot \vec{\hat{\sigma}} 
$$
for any operators $\vec{a}$, $\vec{b}$ that commute with the Pauli matrices.)
