Negative Energy in the 4-momentum of an Antiparticle?

Question

According to the Wikipedia article on Antiparticles:

Solutions of the Dirac equation contain negative energy quantum states. As a result, an electron could always radiate energy and fall into a negative energy state. Even worse, it could keep radiating infinite amounts of energy because there were infinitely many negative energy states available. To prevent this unphysical situation from happening, Dirac proposed that a "sea" of negative-energy electrons fills the universe, already occupying all of the lower-energy states so that, due to the Pauli exclusion principle, no other electron could fall into them. Sometimes, however, one of these negative-energy particles could be lifted out of this Dirac sea to become a positive-energy particle. But, when lifted out, it would leave behind a hole in the sea that would act exactly like a positive-energy electron with a reversed charge. These holes were interpreted as "negative-energy electrons" by Paul Dirac and mistakenly identified...However, these "negative-energy electrons" turned out to be positrons, and not protons.

Thus, by the very definition of an antiparticle, we know that it has negative energy $E$; which, in natural units $(\hbar = c =1)$, is: $$E = -\sqrt{m^2 + p^2}$$ where the symbols here have their usual meanings. Consequently, If we were to describe the antiparticle's 4-momentum vector, from some inertial reference frame, it would look like: $$\tilde{p} = \left(E, \textbf{p} \right) = \left(-\sqrt{m^2 + p^2}, \ \textbf{p} \right)$$ That is, the first entry of the 4-momentum vector ($p^{0}$) should be negative. However, in this link, the author asserts: $$E_{e^{+}} = E_{e^{-}} = 2m_ec^2 \ (> 0)$$ that is, the energy of the free positron $e_{+}$ (antiparticle of $e_{-}$) with respect to the center of mass lab frame is positive and thus, a direct contradiction with what I was stating above. Where exactly is the mistake here?

Answer 1

I guess you are referring to Diracs Hole Theory, within which the positron is a consequence of the negative energy solutions to the Dirac equation. Diracs Hole Theory is outdated and for multiple reasons, which can be seen e.g. on this wikipedia page, not a sensible describtion of nature. A single particle relativistic quantum theory does not make too much sense. The modern describtion of particle physics are quantum field theories, where the antiparticle and the particle are described by the same fermion field.

So when computing kinematics of an antiparticle its energy is not negative.

Answer 2

The frequency of positron solutions is negative of that of electron solutions. Via the formula $E=hf$ this may lead to the conclusion that their energy is negative, as we know electrons have positive energy. However, positrons have positive energy just like electrons. The conclusion must be that $E=h|f|$ and that the Dirac Hamiltonian gives the wrong sign of energy, as well as charge by the way, for positrons.