# Shouldn't Charge Conjugation be known as "positive/negative frequency symmetry"?

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.

• Charge conjugation also inverts the spin of a single particle spinor and a creation/annihlation operater, $1/2\leftrightarrow-1/2$ Oct 3, 2013 at 15:42
• 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. Oct 3, 2013 at 23:47
• It seems there were errors in my answer, so I deleted it. Sorry. Oct 4, 2013 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.
• 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 $e^{i p.x}$. ....> Oct 4, 2013 at 17:12
• ....> 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. Oct 4, 2013 at 17:13
• 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 $\langle X|\gamma_0 (P,S_z)|X \rangle = \langle 0| b_{p,\sigma}b^\dagger_{p,\sigma}| 0 \rangle \bar v^\sigma(p) (P,S_z)v^\sigma(p) =\bar v^\sigma(p) (P,S_z)v^\sigma(p) =(p, \sigma)$. .....> Oct 4, 2013 at 18:39
• .....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. Oct 4, 2013 at 18:40