TITLE: Are there negative energy states in QED?
QUESTION [8 upvotes]: I was reading Weinberg I, when I came upon the following statement$^1$ (slightly edited by me):

\begin{align}
(\not p+m)u=ie\not A\\
(\not p-m)v=ie\not A \tag{1}
\end{align}
The minus sign on the r.h.s. of the equation of $v$ shows that the $v$ are the famous "negative-energy" solutions of the Dirac equation. [...] Of course, for moderate external fields there are no negative-energy states in the theory.

What does Weinberg mean by this? For large enough $A$, is there some ket with $(H-E_0)|\varphi\rangle=-E_\varphi|\varphi\rangle$in the theory? (here, $E_0$ is the vacuum energy). Is this a flaw of the theory? Why moderate external field?
My thoughts

I believe this has nothing to do with negative norm states of QED, because these are gauge-dependent while $E_\varphi$ is not.
I believe this has nothing to do with bound states, as external fields have nothing to do with these, so the requirement for moderate external field wouldn't make sense.
Lastly, I think it might be related to $H$ being unbounded below. The QED interaction is $A\bar \psi\psi$, which is cubic in the fields. Therefore, "$H\to-\infty$ as $A\to-\infty$". But $H_\mathrm{QED}$ being unbounded would be something awful: the theory wouldn't have a ground state, so I don't think this is the answer either.


$^1$: The Quantum Theory of Fields, Volume 1: Foundations, page 567.

REPLY [3 votes]: It does involve bound states. For an electron in the Coulomb field $A(r)=-\frac{Ze}{r}$ that arises from a nucleus (of mass $m$), the lowest energy is $E_0=m\sqrt{1-(2Z\alpha)^2}$ where $\alpha$ is the fine structure constant. This is positive for $Z\le\frac{1}{2\alpha}$, which is what Weinberg means by "moderate" fields.
The assumption so far was that the nuclues is pointlike. When $Z$ is equal to or greater than this critical value $\frac{1}{2\alpha}$, we must instead view the nucleus as a finite-sized object. When this is done, you'll see that the energy levels become negative, up until the "continuum limit" value of $-m$.
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