How do we derive the minimal coupling Hamiltonian?

To derive the minimal coupling Hamiltonian, someone starts by calculating the Lagrangian $L$ of a particle in an electromagnetic field. The acting force is the Lorentz force $boldsymbol{F} = q(boldsymbol{E} + boldsymbol{v}/c times boldsymbol{B})$. Since $boldsymbol{F}$ depends on velocity, we have to find a generalized potential $U$ that satisfies the following equation: $$F_j = -frac{partial U}{partial x_j} + frac{mathrm{d}}{mathrm{d}t}left(frac{partial U}{partial dot{x}_j}right).$$ From Maxwell's equations $$boldsymbol{B} = nabla times boldsymbol{A}, qquad nabla times boldsymbol{E} = - frac{1}{c}frac{partial boldsymbol{B}}{partial t},$$ we get $$boldsymbol{E} = -nabla phi - frac{1}{c} frac{partial boldsymbol{A}}{partial t}.$$ Now by plugging our results into the Lorentz force equation and doing some vector calculus, we end up with $$boldsymbol{F} = qleft(-nabla phi - frac{1}{c}left(nabla (boldsymbol{v}cdot boldsymbol{A}) - frac{mathrm{d}boldsymbol{A}}{mathrm{d}t}right)right).$$ (Here we make use of the fact that $boldsymbol{v}times (nabla timesboldsymbol{A}) = nabla(boldsymbol{v}.boldsymbol{A})-(boldsymbol{v}.nabla)boldsymbol{A} $ and $d_{t}boldsymbol{A} = (boldsymbol{v}.nabla)boldsymbol{A}+partial_{t}boldsymbol{A}$.)

To obtain the generalized potential $U$, a final observation is needed, namely $$frac{mathrm{d}boldsymbol{A}}{mathrm{d}t} = frac{mathrm{d}}{mathrm{d}t}left(frac{partial}{partial boldsymbol{v}}(boldsymbol{v}cdot boldsymbol{A} - q phi)right),$$ which is true since the electrostatic potential $phi$ does not depend on the velocity. Comparing $$boldsymbol{F} = -nabla left(qphi - frac{q}{c}(boldsymbol{v}cdot boldsymbol{A})right) + frac{mathrm{d}}{mathrm{d}t}left(frac{partial}{partial boldsymbol{v}}(boldsymbol{v}cdot boldsymbol{A} - q phi)right)$$ with the equation for the generalized potential, we get $U = qphi - q/c(boldsymbol{v}cdot boldsymbol{A})$, which allows us to write down the Lagrangian $L = T - V = 1 / 2 m boldsymbol{v}cdot boldsymbol{v} - U$.

To derive the minimal coupling Hamiltonian, you have to transform the kinetic momentum (classically this would be $m boldsymbol{v}$) to the canonical momentum $boldsymbol{p} = partial L / partial boldsymbol{v} = m boldsymbol{v} + q / c boldsymbol{A}$. In quantum mechanics, the kinetic momentum corresponds to the momentum operator $hat{p}$, so the canonical momentum operator becomes $hat{p} - q/c A_j$. The Hamiltonian may be obtained by performing the Legendre transformation on $L$: $$H = boldsymbol{p} cdot boldsymbol{v} - L = frac{1}{2m}left(m boldsymbol{v} - frac{q}{c} boldsymbol{A}right)^2 - q phi quad Leftrightarrow quad hat{H} = frac{1}{2m}left(hat{p} - frac{q}{c} A_jright)^2 - q phi.$$