# Negative Energy in the 4-momentum of an Antiparticle?

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.
• where did you find, that the definition of antiparticle gives you negativ energie? – trula Oct 19 at 18:04
• – PM 2Ring Oct 19 at 22:09

• 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 Sure, the Dirac Sea theory is outdated, but as that Wikipedia article mentions, "Dirac sea theory has been displaced by quantum field theory, though they are mathematically compatible".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). – PM 2Ring Oct 19 at 22:08