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?


1 Answer 1


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.

  • $\begingroup$ Thanks for you answer, you are of course right, that $\bar \Psi = \Psi^{\dagger} \gamma_0 $ has to stand on the left-hand side of $\gamma_\mu$. Do you have of any tip about interpretation of the second equation, or $\bar \Psi$ or where i can read about it? $\endgroup$
    – jak
    Jan 28, 2014 at 8:15
  • $\begingroup$ I read about charge conjugation and as far as i understand $\Psi^{C}\equiv-i\gamma^{2}\Psi^{\ast}$ is introduced because it does the job, i.e. changing the sign in front of the coupling to the EM-Field and conventionally this is interpreted as wave-function of the anti-particle. Nevertheless this doesn't seem like something natural to do to me, nor does it prevent us from interpreting $\overline{\Psi}$ as wave function for the anti-particle instead, or am i missing something? $\bar{\Psi}$ has the EM-coupling sign changed, too and additional the mass sign. $\endgroup$
    – jak
    Jan 28, 2014 at 8:39
  • $\begingroup$ @JakobH Consider the mode expansion $\Psi(x) = \int\frac{d^{3}p}{(2\pi)^{3}\sqrt{2E_{\textbf{p}}}} \sum_{\sigma}\left(a_{\textbf{p}\sigma}u_{\sigma}(p)e^{-ipx}+b_{\textbf{p}\sigma}^{\dagger}v_{\sigma}(p)e^{ipx}\right)$. Then, the natural "antiparticle wave function" is obtained by exchanging particles and antiparticles (i.e., $a$ and $b$) in $\Psi$. It turns out that $\Psi^{C} = -i\gamma^{2}\Psi^{\ast}$ does the job, as you pointed out. $\endgroup$
    – higgsss
    Jan 28, 2014 at 9:00
  • $\begingroup$ i looked it up, and I find that $\bar \Psi(x) = \int\frac{d^{3}p}{(2\pi)^{3}\sqrt{2E_{\textbf{p}}}} \sum_{\sigma}\left(a_{\textbf{p}\sigma}^{\dagger}u_{\sigma}(p)e^{ipx}+b_{\textbf{p}\sigma‌​}v_{\sigma}(p)e^{-ipx}\right)$ which would mean, that in field theory $\bar \Psi$ creates anti particles, destroying particles as opposite to $\Psi$ $\endgroup$
    – jak
    Jan 29, 2014 at 7:20
  • $\begingroup$ @JakobH $\overline{\Psi}$ as you defined certainly creates a particle and destroys an antiparticle at the same time, but it isn't something that transforms like a Dirac field under a Lorentz transformation. To make it an object that transforms like a Dirac field, one should also exchange $u$ and $v$ from $\overline{\Psi}$. $\endgroup$
    – higgsss
    Jan 29, 2014 at 21:52

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