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I know that charge conjugation exchanges the creation (or annihilation) operators of the particles with those of the anti-particles and therefore merits the name charge conjugation.

However, if operated on the single electron Dirac plane wave $u(p)$ it results in v(p) and vice-versa. For me, however, $v(p)$ is not the single positron plane wave. For me it is the negative frequency solution. So for the single particles solutions of the Dirac equation it is more like a symmetry between positive and negative solutions.

For a charge conjugation operator I would expect that it changes a in-going single electron plane wave to a in-going single positron wave. But $v(p)$ represents a out-going plane wave in Feynman diagrams.

It is also said that $C$ changes the negative frequency wave $v(p)$ to a positive frequency wave solution $u(p)$ which finally represents the positron. Okay, but again then $C$ should not be called a charge conjugation, but symmetry between positive and negative frequency solutions. I would be grateful to get an explanation on that.

Emilio Pisanty
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Lagrangian
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1 Answers1

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When investigating spinorial representations of the Lorentz group, one finds that if $\Psi$ is a left-handed Dirac spinor, then $\Psi^c = -i\gamma^2\Psi^*$ is a right-handed Dirac spinor. At that moment, however, the physical meaning of the operation is latent.

Having quantized the Dirac spinor, $$ \Psi(x) = \int \frac{d^3p}{(2\pi)^3\sqrt{2E_p}} \sum_{s=1,2} \left(a_{p,s}u^s(p)e^{-ipx} + b^\dagger_{p,s}v^s(p)e^{+ipx} \right)\\ \propto \left( a_{p,1}u^1(p)e^{-ipx} +a_{p,2}u^2(p)e^{-ipx} +b^\dagger_{p,1}v^1(p)e^{+ipx} +b^\dagger_{p,2}v^2(p)e^{+ipx} \right) $$ we reconsider the meaning of $\Psi^c$. By brute force we find that the one-particle spinors obey, $$ [-i\gamma^2u^1(p)]^* = v^2(p),\\ [-i\gamma^2u^2(p)]^* = -v^1(p),\\ [-i\gamma^2v^1(p)]^* = -u^2(p),\\ [-i\gamma^2v^2(p)]^* = u^1(p). $$ Because of these transformations properties, we guess that the creation and annihilation operators transform in an analogous fashion, e.g. $$ Ca_{p,1}C = \eta_c b_{p,2}, $$ note well, however, that $$ Cu^s(p)C = u^s(p), $$ etc. With all these formula, you can show that, $$ C\Psi(x)C \propto \eta_c \left( b_{p,2}u^1(p)e^{-ipx} -b_{p,1}u^2(p)e^{-ipx} -a^\dagger_{p,2}v^1(p)e^{+ipx} +a^\dagger_{p,1}v^2(p)e^{+ipx} \right)\\ = -i\eta_c\gamma^2\Psi^* $$ Justifying our guesses.

Lastly, we check the consequences of the transformation properties of the creation and annihlation operators. Looking at e.g. a $U(1)$ charge, $$ Q \propto a^\dagger a - b^\dagger b $$ it's clear that $CQC=-Q$, justifying the name charge-conjugation.

innisfree
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