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When deriving the Lagrangian for Spin $\frac{1}{2}$ particles we are naturally led to using $\Psi$ and $\bar{\Psi}$. The Euler-Lagrange equations lead us to two wave equations: \begin{equation} (i\gamma_\mu \partial^\mu - m ) \Psi =0 \end{equation}

\begin{equation} (i \gamma_\mu \partial^\mu + m )\bar{\Psi} =0 \end{equation} which differ by a sign in front of the mass term. The same thing happens if we look at the electromagetic coupling of these $\frac{1}{2}$ fields. Again their coupling is different by a sign. This is interpreted as particle and anti-particle having opposite charge. Nevertheless it is unconventional to speak of the anti-particle having negative mass. Why is this the case?

John Rennie
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jak
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1 Answers1

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The second equation is actually incorrect. It should be written as follows:

$$ i\partial^{\mu}\overline{\Psi}\gamma_{\mu}+m\overline{\Psi}=0. $$ Here, $\overline{\Psi}$ is understood as a 4-component row vector (not in the sense of the vector rep. of the Lorentz group).

At any rate, $\overline{\Psi}$(or $\Psi^{\dagger}$) is not what you obtain from $\Psi$ by exchanging the roles of particles and antiparticles. The result of such operation is $\Psi^{C}\equiv-i\gamma^{2}\Psi^{\ast}$, and it satisfies the same Dirac equation as $\Psi$. Among other things, it means that antiparticles have the same mass as particles.

higgsss
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