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.

  • $\begingroup$ Charge conjugation also inverts the spin of a single particle spinor and a creation/annihlation operater, $1/2\leftrightarrow-1/2$ $\endgroup$ – innisfree Oct 3 '13 at 15:42
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    $\begingroup$ There is a lot of terminology in physics that has historical roots and doesn't make much sense unless you understand that context. Learn to live with it, because it is not going away. $\endgroup$ – dmckee Oct 3 '13 at 23:47
  • $\begingroup$ It seems there were errors in my answer, so I deleted it. Sorry. $\endgroup$ – Trimok Oct 4 '13 at 16:50

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.

  • $\begingroup$ I check by myself, and it seems that my answer was not correct (at least, if we want that $\Psi^c = -i\gamma^2\Psi^*$). It seems strange, at first glance, that the charge conjugation changes the spin for the operators and not for the $u,v$. However, following your logic,I found $a_{p,1} \to b_{p,2}, \quad a_{p,2} \to -b_{p,1}\quad b_{p,1} \to - a_{p,2}, \quad b_{p,2} \to a_{p,1}$. There are several "typo" errors in your answer : at the second line, the coeff of $a_{p,2}$ is $e^{-ip.x}$ and the coeff of $b_{p,1}$ is $e^{i p.x}$. ....> $\endgroup$ – Trimok Oct 4 '13 at 17:12
  • $\begingroup$ ....> Your forget the "dagger" on the b. You have to correct the line $C\Psi(x)C$, where you have the same errors, plus a sign error (with my check) for the 2 last terms. $\endgroup$ – Trimok Oct 4 '13 at 17:13
  • $\begingroup$ @Trimok Thanks, I've fixed the dagger typos and the signs of the exponentials... are there any more left? There are subtleties that are ignored in some lecture notes... I read this stuff in Michele Maggiore's book, which is superb. $\endgroup$ – innisfree Oct 4 '13 at 17:53
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    $\begingroup$ In fact, it seems that there is no contradiction. For instance, if I begin with a state : $|X\rangle = b^\dagger_{p,\sigma}v^\sigma(p)e^{+ipx}|0\rangle$ (without sommation on the $\sigma$), the momentum / spin for this state are given by $\langle X|\gamma_0 (P,S_z)|X \rangle = \langle 0| b_{p,\sigma}b^\dagger_{p,\sigma}| 0 \rangle \bar v^\sigma(p) (P,S_z)v^\sigma(p) =\bar v^\sigma(p) (P,S_z)v^\sigma(p) =(p, \sigma)$. .....> $\endgroup$ – Trimok Oct 4 '13 at 18:39
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    $\begingroup$ .....If I make a change conjugation on this state, I change only the operator $b^\dagger_{p,\sigma}$ in $\pm ~d^\dagger_{p,-\sigma}$, so, we have : $\langle C(X)|\gamma_0 (P,S_z)|C(X) \rangle = \langle0| d_{p,-\sigma}d^\dagger_{p,-\sigma}| 0 \rangle \bar v^\sigma(p) (P,S_z)v^\sigma(p) =\bar v^\sigma(p) (P,S_z)v^\sigma(p) =(p, \sigma)$. So conjugation of charge does not change spin and momentum, as wished. $\endgroup$ – Trimok Oct 4 '13 at 18:40

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