0
$\begingroup$

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?

$\endgroup$
2
2
$\begingroup$

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 Diracs Hole Theory is outdated and for multiple reasons, which can be seen e.g. on this wikipedia page, not a sensible describtion of nature.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.

$\endgroup$
4
  • 3
    $\begingroup$ Positrons don't have negative energy in the Dirac sea model. The negative-energy states are occupied by electrons. A positron is the absence of one of those negative-energy electrons; it therefore has net positive charge and net positive energy. $\endgroup$ – benrg Oct 19 '20 at 19:44
  • 2
    $\begingroup$ You are totally right. I edited my answer and i am sorry for this sloppy statement (which is propably the reason for OP misconception). $\endgroup$ – AlmostClueless Oct 19 '20 at 20:30
  • 2
    $\begingroup$ What @benrg said. Sure, the Dirac Sea theory is outdated, but as that Wikipedia article mentions, "Dirac sea theory has been displaced by quantum field theory, though they are mathematically compatible". It's not exactly wrong, it's just inelegant, and it cannot be extended to describe bosons (eg, the $W^+$ and $W^-$ of the weak nuclear interaction). $\endgroup$ – PM 2Ring Oct 19 '20 at 22:08
  • $\begingroup$ @PM2Ring The Dirac Sea explanation of the positron is unphysical. $\endgroup$ – my2cts Mar 14 at 19:09
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.