Energy Positivity of Classical QED Field Theory in Presence of Sources

Question

It's well known that the classical electromagnetic field has positive definite energy, simply because: $$\mathcal{H}=\frac{1}{2}\epsilon_0\vec{E}^2+\frac{1}{2\mu_0}\vec{B}^2.$$ However, this result only applies in the case of the free electromagnetic field. Is there a positive energy condition when considering a configuration consisting of source currents and charges? I would need a Lagrangian/Hamiltonian which includes the Lorentz force on the source current as well, but as far as I know there is only the Lagrangian for a single point particle in an electromagnetic field stated in literature, and not for a general current density $\vec{J}$.

However, we may examine QED as a classical field theory, which already has the sources included as the Dirac field. Here we find terms which may contribute negatively: $$H=\int d^{3} x\left[\frac{1}{2} \vec{A}^{2}+\frac{1}{2} \vec{B}^{2}+\bar{\psi}\left(-i \gamma^{i} \partial_{i}+m\right) \psi-e \vec{j} \cdot \vec{A}+\frac{e^{2}}{2} \int d^{3} x^{\prime} \frac{j^{0}(\vec{x}) j^{0}\left(\vec{x}^{\prime}\right)}{4 \pi\left|\vec{x}-\vec{x}^{\prime}\right|}\right]$$ namely the $-e \vec{j} \cdot \vec{A}$ term could be made arbitrarily large in either sign. Is there a Classical constraint on the Hamiltonian density? This is similar to a question I already asked, but here the emphasis is completely different. Just to clarify, my question is: in the presence of sources: charges and currents, and not just the free field contribution, does Classical Electromagnetism, or QED as a Classical Field Theory exhibit the locally positive energy density condition?

Answer 1

QED as a Classical Field Theory could mean a couple of different things. "Classical" means that all observables commute with each other, but it doesn't necessarily mean that all fields commute with each other. Here are two things that QED as a Classical Field Theory might mean:

1. It could mean a superclassical field theory in which the components $\psi_k$ of the spinor field $\psi$ are elements of a grassmann algebra (like in the integrand of a path integral in the quantum version of QED), but in that case the energy density is grassmann-valued instead of real-valued, so "positive" is undefined. In that context, the appropriate replacement for the positive-energy condition is reflection positivity, but that's only defined in the context of the path integral, so it doesn't really apply to classical QED.

2. On the other hand, if we take the components $\psi_k$ of the spinor field $\psi$ to be ordinary complex numbers, then the energy density is real-valued, so in this case the question makes sense — and the answer is no. The energy density doesn't have a lower bound even if we omit the coupling to the gauge field, because the kinetic term $\bar\psi(-i\gamma^k\partial_k+m)\psi$ itself doesn't have a lower bound.

So the question is undefined in version 1, and the answer is no in version 2, even if we omit the coupling to the gauge field. Here are two proofs of that statement about version 2:

• Consider the mass term, which in the usual representation is $\bar\psi\psi\equiv\psi^\dagger\gamma^0\psi$. The Dirac matrix $\gamma^0$ has eigenvalues $\pm 1$, and the quantity $\psi^\dagger(x)\psi(x)$ is manifestly nonnegative, so by considering a configuration of $\psi$ involving only negative eigenvalues of $\gamma^0$, we can make the energy density as negative as we want, with no limit, by taking the components of $\psi$ to have arbitrarily large magnitudes.

• Consider the derivative term $-i\bar\psi\gamma^k\partial_k\psi=-i\psi^\dagger\gamma^0\gamma^k\partial_k\psi$. For a plane wave, this becomes $\psi^\dagger\gamma^0\gamma^k p_k\psi$. The matrix $\gamma^0\gamma^k p_k$ has eigenvalues $\pm |p|$, so we can again make the energy density as negative as we want, even if we keep $\psi^\dagger(x)\psi(x)$ fixed, just by taking $|p|$ to be arbitrarily large.

These observations are related to the reason we take the components of a spinor field to be anticommuting operators in the quantum version of QED (which leads to the spin-statistics theorem), which in turn is related to why theorists often take the components to be grassmann-valued to define what they call the "classical" version of the model, the one I called version 1.

Answer 2

In classical (c-number) theory, it depends on the variant of EM theory, not on presence of sources.

In a theory that works only with single electromagnetic field, such as the macroscopic EM theory or Lorentz microscopic theory with single EM field, EM field energy density can be defined generally based on energy interpretation of the "mathematical" Poynting theorem, which is valid whether sources are present or not:

$$ w = \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0}B^2. $$

The interaction terms in Lagrangian/Hamiltonian like $-\rho \varphi + \mathbf j\cdot \mathbf A$, while they can be negative, are not part of EM field energy in the usual sense.

If you are asking whether total Hamiltonian including matter and interaction with matter can be negative, yes it can, because potentials $\varphi, \mathbf A$ can have arbitrarily negative value.

On the other hand, in a theory of point particles, such as the Coulomb/Newton theory of point masses with gravity or electrostatic interaction, or Frenkel's relativistic theory of charged particles, potential energy cannot be defined using Poynting's theorem (it breaks down mathematically at the point particles) but this is not a big problem, because energy can still be defined using particles' positions(in electrostatics) or individual fields in general. Then EM energy of the system can be negative.

But it is not usual for total energy to be negative because usually the EM interaction energy is much lower than sum of rest energies of the particles $mc^2$. Total energy of the system being negative would mean negative mass, a very strange concept.

For example, in electrostatics, Coulomb energy of a set of charges, which can be negative, can be expressed as integral of a density that is a function of individual particle fields $\mathbf E_i$:

$$ W = \frac{1}{2}\sum_i \sum_k' K\frac{q_i q_k}{r_{ik}} = \int \sum_i \sum_k' \mathbf E_i \cdot \mathbf E_k~ d^3\mathbf x. $$

This can be generalized (in Frenkel's theory) to the general relativistic case, and similar magnetic terms appear.

In the simplest case of two particles of opposite sign, the electromagnetic energy as defined above is negative. Hypothetically, the particles can be so close that magnitude of $W$ is higher than total rest mass of the particles and then total energy of the system is negative. But such a system has very strange behaviour (negative mass) and AFAIK hasn't been observed.