0
$\begingroup$

By the very definition of an antiparticle, we know that it has a negative energy $E$; which, in natural units, is: $$E = -\sqrt{m^2 + p^2}$$ where the symbols have their usual meanings. Thus, If we know were to describe the particle's 4-momentum vector, from an 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 writing above.

I am pretty sure my reasoning is somewhere wrong. My best guess is that I am making an assumption I shouldn't be making. Here are a few possible places the mistake might be:

  1. Even if the energy of a particle is negative, we represent it with a positive energy in its 4-momentum vector.
  2. The energy of the antiparticle is negative in some frame, but in the center of mass frame it turns out to be positive via some appropriate transformation.
$\endgroup$
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 Diracs Hole Theory is outdated and for multiple reasons, which can be seen e.g. on this wikipedia page, not a sensible describtion of nature.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.

| cite | improve this answer | |
$\endgroup$
  • 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 at 19:44
  • 1
    $\begingroup$ 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 at 20:30
  • 1
    $\begingroup$ 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 Sure, the Dirac Sea theory is outdated, but as that Wikipedia article mentions, "Dirac sea theory has been displaced by quantum field theory, though they are mathematically compatible".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 at 22:08

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.