# A confusing point of the Hamiltonian for a particle interacting with electromagnetic fields

In non-relativistic quantum theory the Hamiltonian for a particle interacting with electromagnetic fields is $$H=\frac{(\mathbf{p}-\mathbf{A}*e/c)^2}{2m}+e\phi+\int\,d^3x \frac{\mathbf{E^2}+\mathbf{B^2}}{8\pi}.\tag{1}$$

According to Hamiltonian's equations, $$\dot{r_i}=\frac{\partial H}{\partial p_i}, \tag{2}$$ $$\dot{p_i}=-\frac{\partial H}{\partial r_i}.\tag{3}$$ They certainly can not produce the equations of motion of the particle as well as the electromagnetic fields. Where am I wrong? What are the coordinates and the canonical momentum for the fields?

• Note，by the way, the Hamiltonian (1) shall not be wrong since you can find it in many books about quantum theory of EM radiation such as Greiner's book "Quantum Mechannics: Special Chapters". – yztsz Jul 27 '18 at 8:19
• Just because something is in many books, it does not mean it is free of flaws. However, you should clarify what you mean, when you write "certainly can not produce the equations of motion of the particle as well as the electromagnetic fields". Why do you think there is a problem? – Ján Lalinský Jul 27 '18 at 17:11

I) The Hamiltonian for point charges and EM fields can certainly produce the EOMs of the particle(s) as well as the EM fields.

A full explanation is quite a long story. For pedagogical reasons, to see how this works, it is best to:

1. Firstly, understand the corresponding Lagrangian formulation.

2. Secondly, understand how the Hamiltonian formulations work for point charges and EM fields separately, see e.g. this & this Phys.SE posts.

3. Thirdly, try to construct a Hamiltonian formulation for both point charges and EM fields together.

II) One correction: OP's Hamiltonian (1) yields the correct total energy, but OP asks how to produce Maxwell's equations. For the latter purpose, OP's Hamiltonian (1) is missing a Lagrange multiplier term that imposes Gauss' law.

III) Concretely, the minimal phase space is as follows:

1. Particle position ${\bf r}(t)$ and particle momentum ${\bf p}(t)$: $$\{r^k(t), p_{\ell}(t)\}= \delta_{\ell}^k.$$

2. (Minus$^1$) the electric field ${\bf E}(x)$ is the canonical conjugate variable to the magnetic gauge potential ${\bf A}(x)$: $$\{A_i({\bf x},t), E^j({\bf x}^{\prime},t)\}~=~ -\delta_i^j~\delta^3({\bf x}\!-\!{\bf x}^{\prime}).$$

3. Lagrange multiplier $A^0(x)\equiv \phi(x)$.

IV) The equations come about as follows:

1. The magnetic field ${\bf B}\equiv{\bf \nabla}\times {\bf A}$ is defined as the curl of the magnetic gauge potential ${\bf A}$.

2. The Hamilton's equations for ${\bf r}$ and ${\bf p}$ yield (i) the Newton's 2nd law with a Lorentz force, and (ii) the relation between velocity $\dot{\bf r}$ and momentum ${\bf p}$.

3. The Hamilton's equations for ${\bf A}$ and ${\bf E}$ yield (i) the Maxwell–Ampere's law, and (ii) the relation between the electric field ${\bf E}$ and the gauge potential $A_{\mu}$.

4. The Lagrange multiplier $A^0\equiv\phi$ imposes Gauss' law.

5. The source-free Maxwell equations follows from the existence of the gauge potential $A_{\mu}$.

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$^1$ We use $(-,+,+,+)$ Minkowski sign convention with $c=1$.

• But what are the coordinates and the canonical momentum for the fields since $\mathbf{A}$ and $\phi$ are the function of the particle's coordinate $\mathbf{r}$ while $\mathbf{r}$ has been the generalized coordinate – yztsz Jul 27 '18 at 8:47
• I update the answer. – Qmechanic Jul 27 '18 at 9:22
• Re 3., II. -- unfortunately, this cannot work, because for point particles and total field $\mathbf A$, the usual interaction term $q\mathbf v \cdot \mathbf A$ is mathematically undefined and Poynting's expression for EM energy is infinite. One really has to decide one of the two:1) particles are points, but then field energy is not given by Poynting's formulae; but, for example, by terms bi-linear in particle adjunct fields (Frenkel's formulae) or 2) the field energy is given by Poynting's formulae, but then the particles are not points, they have internal dof and the model gets complicated. – Ján Lalinský Jul 27 '18 at 17:31

Normally the EMF Hamiltonian must be written in terms of $Q$s and $P$s of the electromagnetic field which is done by representing it via harmonic oscillators. You may find decomposition of EMF into a set of harmonic oscillators in many textbooks.

Remove the second term in the hamiltonian altogether. In the first term only consider the field of other particles.

• the integrand is the energy density due to EM field, so the integral is energy and dimensionally consistent with other terms in the Hsmiltonian, so I don't see what is wrong. – wcc Jul 27 '18 at 7:27
• Check my comment. – yztsz Jul 27 '18 at 8:25