How to calculate the Hamiltonian from the Lagrangian for a non-relativistic charged point particle in an EM field? I was given the equation of the Lagrangian:
\begin{equation}
L~=~\frac{1}{2}m \dot{x}^2+\frac{e}{c}\vec{\dot{x}}\cdot \vec{A}(\vec{x},t)-e\phi (\vec{x},t).
\end{equation}
I proceeded to use the equation:
\begin{equation}
H~=~\sum_{i} \dot{q_i} \frac{\partial L}{\partial \dot{q_i}} -L
\end{equation}
to get the Hamiltonian as:
\begin{equation}
H~=~\frac{1}{2}m \dot{x}^2+e\phi (\vec{x},t),
\end{equation}
but, in the text, the Hamiltonian is given as:
\begin{equation}
\hat{H}~=~\frac{1}{2m}(\frac{\hbar}{i}\nabla - \frac{e}{c}\vec{A})\cdot(\frac{\hbar}{i}\nabla - \frac{e}{c}\vec{A})+e\phi
\end{equation}
So,why and where did I go wrong?
 A: To make the correct answer clearer, allow me to introduce the canonical momentum $\vec{p}$, given by:
$$\vec{p}=\dfrac{\partial L}{\partial\dot{x}}$$
This way we can rewrite the Hamiltonian as:
$$H=\vec{p}\cdot\vec{\dot{x}}-L$$
Let's start by computing $\vec{p}$:
$$\vec{p}=\dfrac{\partial L}{\partial\dot{x}}=m\vec{\dot{x}}+\dfrac{e}{c}\vec{A}(\vec{x},t)$$
And you get:
\begin{align}H&=m\dot{x}^2+\dfrac{e}{c}\vec{x}\cdot\vec{A}-\dfrac{1}{2}m\dot{x}^2-\dfrac{e}{c}\vec{\dot{x}}\cdot\vec{A}+e\phi\\&=\dfrac{1}{2}m\dot{x}^2+e\phi\end{align}
But from the expression of the canonical momentum we found earlier, you can rewrite $\vec{\dot{x}}$ as:
$$\vec{\dot{x}}=\dfrac{1}{m}\left(\vec{p}-\dfrac{e}{c}\vec{A}\right)$$
Such that:
$$\dot{x}^2=\dfrac{1}{m^2}\left|\vec{p}-\dfrac{e}{c}\vec{A}\right|^2$$
Plugging this result into $H$:
$$H=\dfrac{1}{2m}\left|\vec{p}-\dfrac{e}{c}\vec{A}\right|^2+e\phi$$
Make the transition to quantum mechanics by promoting the classical momentum $\vec{p}$ to the operator $\hat{p}=-i\hbar\nabla$ and you're done.
