(Why/how) are antiparticles and charge-conjugates different things?
I am trying to understand the effect of discrete symmetries on spinor fields (neutrinos in particular). In the article, Dirac, Majorana and Weyl fermions (section 7, pg 25), the author stresses the distinction between
a) the 'Lorentz-covariant conjugate',(LCC) operation, given by $\widehat{\psi} = \gamma_{0} C \psi^*$
b) the charge conjugation operation, given by $\mathcal{C}\psi \mathcal{C}^{-1} = \eta_c \widehat{\psi}$.
(Here $\psi(x)$ describes a fermion field, $\eta_c$ is a complex phase and $C$ satisfies $C^{-1} \gamma_\mu C = – \gamma_\mu^T$.)
As far as I can tell, Lorentz-covariant conjugation replaces all particles with their antiparticles, whereas charge conjugation changes the sign of all the charges associated with the particle. However I'm not sure I understand the difference.
The distinction is important in the case of chiral fields ($P_{L/R} \psi = 1 \pm \gamma_5 \psi = \psi_{L/R}$) because we have:
$$ \mathcal{C} \psi_L \mathcal{C}^{-1} = \eta_C \widehat{\psi}_L$$
but
$$ \widehat{\psi_L} = \widehat{\psi}_R.$$
In other words, the anti-particle state of a left-chiral state is right-chiral but the charge conjugate state of a left-chiral state is left-chiral. There is a supporting statement on pg 26:
…can be best seen with Weyl fields for which chirality is the same as helicity. Helicity involves spin and momentum, none of which changes
under charge conjugation. Thus helicity is unaffected by charge
conjugation, and so must be chirality.
All the operator manipulation seems reasonable but I am confused because the original definitions only differ by a complex phase so it's hard to understand where the difference comes from. Why can't I write
$$\mathcal{C} \psi_L \mathcal{C}^{-1} = \eta_c \widehat{\psi_L} = \eta_c \widehat{\psi}_R$$
implying that charge conjugation does change chirality?
I can't find many examples of these concepts being presented separately in this way.
I think this question addresses a similar issue but has no answer. This answer suggests that my problem might be linked to the definition of anti-particle.
Best Answer
Frankly I find this so-called pedagogical article quite unintelligible and fail to see what the author wanted to say about these two operations. I also can't make any sense of the "derivation" of 7.3 based on the chiral projection being a "numerical matrix" and therefore commuting with charge conjugation operator.
Moreso the remark:
is misleading too. It's right that the left-handed Weyl spinor corresponds to the left helicity particle. But it also describes right helicity antiparticle (that the author notes himself earlier!) E.g. for a long time we didn't know that the neutrino has a mass and therefore we could in principle describe it by a single Weyl spinor. But in addition to the left helicity neutrino the same spinor describes right helicity anti-neutrino.
You can't redefine it away because helicity corresponds to a specific value of the angular momentum and therefore is observable.
The charge conjugation transforms the particle into the antiparticle with the same helicity. You can't get the left helicity antiparticle from purely left-handed Weyl spinor! So it's no surprise that the charge conjugation converts left-handed Weyl spinor into the right-handed and vice versa.
This flip is also why the weak interaction term $i\psi_L^\dagger\gamma^0\gamma_\mu W^\mu\psi_L$ is not invariant under $\mathcal{C}$ transform. The spatial reflection also changes the chirality and thus it happens to be invariant under the combined $\mathcal{CP}$.
For the Dirac fermion all works perfectly for a straightforward definition of the charge conjugation that you denote as $\hat{\psi}$. In the Fock space this leads to $\mathcal{C}a_s^\dagger(p)\mathcal{C}^{-1}=b_s^\dagger(p)$ where $a_s^\dagger$ and $b_s^\dagger$ are creation operators for particle and antiparticle respectively with a helicity $s$. As expected in the massless limit the $a_s^\dagger$ and $b_s^\dagger$ act on different chiral components separately while $a_s^\dagger$ and $b_{-s}^\dagger$ on the same one. So as I said the charge conjugation changes chirality.
It also works perfectly for a Majorana fermion understood as a Dirac spinor with a reality condition $\psi=\hat{\psi}$. The reality condition leads to the identification $a_s^\dagger(p)=b_s^\dagger(p)$ which is exactly that the Majorana particle is the same as an antiparticle with the same helicity. You can't talk about chirality anymore even in the massless case because you tie the right component to the left by definition.
In summary I have no idea what the author of this article tried to say. All works fine for straightforward definition that is used by everyone and it does change chirality.
UPDATE
Note that before we discussed only the chirality of fields but not the states.
The common way to derive helicity=chirality in the massless case is done simply for the Dirac equation. The helicity operator is given by $\frac{\vec{p}}{E}\cdot\vec{\sigma}$. The Weyl equation takes the form, \begin{equation*} \vec{p}\cdot\vec{\sigma}\psi_{L,R}=\pm E\psi_{L,R} \end{equation*} Dividing it on $E$ we get that the helicity eigenstates exactly coincide with chirality eigenstates. We may conclude that the helicity operator equals $\gamma_5$ i.e. chirality.
Any solution restricted to the left component will get left helicity. However the solutions associated with antiparticles have negative energies. They are reinterpreted in such a way that their physical energy, momentum and spin (and thus helicity) change the sign. In the QFT this is achieved by stating that $\psi$ operator contains the annihilation operator for the particle but the creation operator for the antiparticle. Because in the case of fermions the operators anticommute e.g. in the spin operator, \begin{equation*} \hat{S}_k=\frac{1}{2}\int d^4x :\psi^\dagger \begin{pmatrix}\sigma_k&0\\0&\sigma_k\end{pmatrix} \psi: \end{equation*} we get that the antiparticle contribution has the right (i.e. opposite to the naive) sign.
Considering this if we define the chirality operator as $\int d^4x :\psi^\dagger \gamma_5 \psi:$ it will coincide with helicity operator also in QFT. So we should assign right chirality to the antiparticle described by left-handed Weyl spinor and it will not change under $\mathcal{C}$
In summary. The chirality of the field operator is changing. The chirality of the state is not.