Hamiltonian and Energy of a charged particle in an Electromagnetic field The Lagrangian of a charged particle of charge $e$ moving in an electromagnetic field is given by $$L=\frac{1}{2}m\dot{\textbf{r}}^2-e\phi-e\textbf{A}\cdot \textbf{v}$$ where $\phi(\textbf{r},t)$ is the scalar potential and $\textbf{A}(\textbf{r},t)$ is the vector potential. Here the "potential" $U=e(\phi-\textbf{A}\cdot \textbf{v})$ is velocity dependent. The corresponding Hamiltonian is given by $$H=\frac{(\textbf{p}-e\textbf{A})^2}{2m}+e\phi$$ 


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*Is it possible to define a total energy for the the charged particle? If yes, what is the expression for total energy and is that a constant of motion? 

*Does the expression for the Hamiltonian coincide with that of the total energy? My guess is that the Hamiltonian cannot represent the total energy because it is not gauge invariant.
 A: *

*For a particle-field system the only way to define a gauge invariant energy is to consider the energy carried by the field as well, in the form of the energy momentum tensor $T^{\mu\nu}$ in the presence of charges. $T^{\mu\nu}$ is a manifestly gauge invariant quantity. To derive this use Noether's theorem and the maxwell equations in the presence of charges. This gives a conserved energy for the system 

*So the answer to this would be no, it does not correspond to the total energy as you must also consider the field energy to attain a conserved quantity. Intuitively, if we choose some gauge where $\phi = 0$, then the change in the energy of particle is compensated for by a change in the energy carried by the fields. Thus resulting in the same total energy for the system.
A: This question is studied in detail in http://aapt.scitation.org/doi/abs/10.1119/1.12463. The punch line is that there is a natural way to define the total energy of the electromagnetic field and particles together, but there is no single natural way to divide it up between the two contributions.
A: That Hamiltonian already represents the total energy of the particle but it is not conserved because the particle is an open system. Adding the energy of the field and of the particles associated to the Coulomb interaction gives a conserved energy for the total closed system of particles plus field
$$H_\mathrm{tot}= \sum_i \frac{(\boldsymbol{p}_i - e_i \boldsymbol{A})^2}{2m_i^\mathrm{bare}} + \frac{1}{2} \sum_i\sum_{j\neq i} \frac{e_i e_j}{4\pi\epsilon_0 | \boldsymbol{r}_i - \boldsymbol{r}_j |} + \frac{1}{8\pi} \int d^3x \> ( \boldsymbol{E}_\mathrm{T}^2 + \boldsymbol{B}^2) . $$
where $m_i^\mathrm{bare}$ is the bare mass, an infinite quantity that has to be renormalized using the "electromagnetic mass" associated to the divergent vector potential and $\boldsymbol{E}_\mathrm{T}$ is the transverse component of the electric field.
