The Hamiltonian of a system under only the effect of an electric field I have a maybe silly doubt: in quantum mechanics, we have the Hamiltonian as kinetic energy + potential energy. Now kinetic energy is obtained from the integral of force and displacement. Potential energy is also the integral of the electric field and the displacement it causes, whole multiplied by the charge (there is an electric field under consideration only). So if in the experiment, there's only an electric field, then kinetic energy is also the energy from work done by the force, and so is the potential energy. Aren't we then counting the same thing twice for the Hamiltonian?
 A: The question is unrelated to quantum mechanics, and even to classical mechanics (dealing with Lagrangians and Hamiltonians), but rather to the basic Newtonian mechanics: Indeed, when a particle is accelerated/decelerated by an electric field, both its kinetic energy and its potential energy change, and it is the same force that is responsible for both changes. Except that the sign of the energy change is different for the two contributions: the potential energy is converted in kinetic energy and vice versa, so that the net energy remains constant (is conserved):
$$
E=\frac{m\mathbf{v}^2}{2} + U(\mathbf{r})=\text{const}
$$
A: This answer is meant to address your comment to Roger Vadim's answer (which is clear and correct). Newton's 2nd law for a charge in a uniform electric field says that
\begin{align}
q \mathbf{E} = m \mathbf{a}
\end{align}
This equation is not telling us that $m \mathbf{a}$ is the same thing as the electric force $q\mathbf{E}$, it's telling us that by applying a net electric force we change the velocity of our particle. (In other words, $m\mathbf{a}$ is not a force, which is something a lot of people have trouble understanding).
When we integrate both sides of this equation over the trajectory of the charged particle, we end up with
\begin{align}
\int q\mathbf{E}\cdot d\mathbf{r} = \frac{1}{2}m(v_f^2 - v_i^2),
\end{align}
Again, the two sides here are not the same thing. The left side
\begin{align}
W_e = \int q\mathbf{E}\cdot d\mathbf{r}
\end{align}
is called the work done by the electric force, and we define the change in the electric potential energy to be $\Delta U_e = - W_e$. The right side is the change in the kinetic energy
\begin{align}
\Delta K = \frac{1}{2}m(v_f^2 - v_i^2)
\end{align}
The work tells us how much energy was transferred by applying the electric force, and the change in the kinetic energy tells us the effect of that energy transfer: it increases or decreases the speed of the moving charge.
If we rearrange the above work-energy equation, we get
\begin{align}
\Delta U_e + \Delta K = 0
\end{align}
We need both terms here because we need to keep track of both where the energy comes from and where it goes to, and those aren't the same. It's not really different than if you went to the bank and deposited \$1 into your account. You would have
\begin{align}
\Delta \text{money}_{\text{bank}} + \Delta \text{money}_{\text{you}} = 0
\end{align}
where $\Delta \text{money}_{\text{bank}} = \$1$ is the money the bank gains and $\Delta \text{money}_{\text{you}} = -\$1$ is the money you gave up. We're not double counting that \$1, we're just saying it went from one place to another, and overall the total amount of money in the world didn't change.
