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.

  • $\begingroup$ Bound states don't really have negative energy per se, they have less positive energy. $\endgroup$ – John Duffield Apr 18 '16 at 18:32
  • $\begingroup$ My pleasure. I can't answer your question I'm afraid. you could always ask Weinberg. $\endgroup$ – John Duffield Apr 18 '16 at 18:42
  • $\begingroup$ @JohnDuffield: I don't know what that is supposed to mean. How do I distinguish "less positive energy" from "it looks exactly like the same energy as in the "normal" particle"? $\endgroup$ – CuriousOne Apr 18 '16 at 18:47
  • $\begingroup$ @CuriousOne I believe he means that, say you have an electron and a positron. Then the energy of the system is less than $.511\ \mathrm{MeV}+.511\ \mathrm{MeV}$, but still greater than zero... $\endgroup$ – AccidentalFourierTransform Apr 18 '16 at 18:51
  • $\begingroup$ @CuriousOne : a hydrogen atom is a bound state. The mass is 13.6eV/c² less than the mass of the free proton and electron. If you consider the proton as fixed to simplify the situation, when you pull the electron away you add energy to it and increase its mass. We know of no negative-energy particles or objects. $\endgroup$ – John Duffield Apr 18 '16 at 18:54

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 $-m$.

  • $\begingroup$ Some relevant references come from the physicist V.R. Khalilov. $\endgroup$ – Chris Gerig Apr 21 '16 at 0:53
  • 1
    $\begingroup$ Thank you for your answer, it makes sense! So W. is talking about bound states with the external field and not QED bound states per se (e.g., a positronium), right? You say that the minimum energy is $-m$, which means that the Hamiltonian is bounded below, right? $\endgroup$ – AccidentalFourierTransform Apr 22 '16 at 10:39

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