Hamiltonian of charged particle in an EM field and magnetic field does no work on charged particles I am trying to understand a part of I.E.P.'s answer here.
I.E.P. states that one can see from the following Hamiltonian,
$$
H = \frac{1}{2m}|{\bf p}+q{\bf A}|^2 +q \phi \tag{8.35}
$$
that the magnetic field does not work on the charged particle, and thus does not contribute to the energy.

*

*How can one see that directly from this Hamiltonian?


*The corresponding Lagrangian for this system is
\begin{equation}
    L = \frac{1}{2} m\dot{\bf r}^2 - q \phi + q \dot{\bf r}\cdot {\bf A}. \tag{8.32}
\end{equation}
by Goldstein's matrix formulation to the Hamiltonian formalism, since $L$ is not a homogeneous function of degree 2, $H$ is not equal to kinetic energy + potential energy. HOWEVER, Goldstein does state that

There is now a linear term in the generalized velocities such that the matrix $\mathbf{a}$ has the elements $q A_i$. Because of this linear term in $V$, the Hamiltonian is not $T + V$. However, it is still in this case the total energy since the “potential” energy in an electromagnetic field is determined by $\phi$ alone.

Can I have some help also in how to reconcile the above quote of Goldstein's with his comment about $H\neq E$ unless $L$ is a homogeneous function of the velocities squared?
 A: *

*Consider the (Lagrangian) energy function
$$ h(q,\dot{q},t)~=~\left(\sum_j\dot{q}^j\frac{\partial }{\partial \dot{q}^j}-1 \right)L(q,\dot{q},t), \tag{2.53} $$
which should not be confused with the Hamiltonian function $H(q,p,t)$. They are different functions, although their values agree.


*OP's quote from Ref. 1 is closely related to the following fact. If $L_n$ denotes the part of the Lagrangian $L$ that is a homogeneous polynomial of $n$'th degree in the generalized velocities $\dot{q}^j$, and if the Lagrangian is of the form $L=L_2+L_1+L_0$, then the energy is $$h~=~L_2-L_0.\tag{2.57}$$


*In particular, for a non-relativistic charge in an E&M background, Ref. 1 denotes $L_2=T$ and $L_1+L_0=-V$. The energy function is
$$ h({\bf r},\dot{\bf r},t)~=~ \frac{m}{2}\dot{\bf r}^2 +q \phi({\bf r}) $$
is different from $T+V$. Note that the energy $h$ is independent of the magnetic potential ${\bf A}$, i.e. the magnetic force produces no work, cf. the work-energy theorem.


*Concerning the relationship between Hamiltonian and energy, see also e.g. this Phys.SE posts and links therein.
References:

*

*H. Goldstein, Classical Mechanics, 3rd edition; Chapter 2 + 8.

