I liked Flip Tanedo's discussion. It is accurate as far as I can see. The key point is that mass mixes the chirality states. An electron might be produced as a pure left-chiral state in a weak interaction, but because it has mass it rapidly becomes a mixture of left and right chiral components. Another way of saying it is that the chirality operator (I've never seen it, but you could surely construct it - at least on the single particle sector of Fock space) does not commute with the mass term in the Hamiltonian.
All of this is independent of the helicity question. In pion decay the electron must be ejected with a certain helicity to conserve angular momentum, but it is the "wrong" helicity for the chirality state produced in the weak decay. Remember that for a massless Dirac particle helicity = chirality. The process is only allowed by the fact that the electron has a nonzero mass, so the helicity and chirality are mixed up slightly. I probably wasn't being particularly clear when I wrote that comment - it was very late, this this is all I really meant.
This section is a more mathematical way of describing the situation.
Everything comes from the representations of the Lorentz group, so let's examine that. A massive review of two component spinor techniques in four dimensions is
Dreiner, H. K., Haber, H. E., & Martin, S. P. (2010). Two-component spinor techniques and Feynman rules for quantum field theory and supersymmetry. Physics Reports, 494(1-2), 1–196. doi:10.1016/j.physrep.2010.05.002 (http://arxiv.org/abs/0812.1594)
I recommend following that to make sure you understand all the conventions etc. It also includes a useful appendix connecting all of this to the usual 4 component spinor notation.
In four dimensions the (double cover of the) Lorentz group is the product of two copies of SU(2): a left handed and right handed group. The generators of the two SU(2) are
$$\begin{array}{lcl}
\left[N_{i}^{\pm},N_{j}^{\pm}\right] &=& i\epsilon_{ijk}N_{k}^{\pm},\\
\left[N_{i}^{\pm},N_{j}^{\mp}\right] &=& 0,
\end{array}$$
and the generators of Lorentz rotations and boosts can be written
$$\begin{array}{lcl}
J_i &=& N_i^+ + N_i^-, \\
K_i &=& -i (N_i^+ - N_i^-).
\end{array}$$
Using the $N_i^\pm$ makes the representation theory simple, though I don't know any nice physical interpretation for them. Note that parity and conjugation both change $N_i^+ \leftrightarrow N_i^-$. By convention the $+$ generators correspond to the left hand group and the $-$ ones correspond to the right hand group. The irreducible representations are $(l,r)$ where $l,r$ are the spins under the left/right-handed SU(2)s respectively. You can see that chirality is not an observer-dependent thing - the separation of the Lorentz group into two pieces is invariant and the two pieces don't mix with each other.
Now you can write down left handed spinor fields which transform in the (1/2, 0) representation and right handed fields which transform in (0, 1/2), though you can also write these as the conjugates of left handed fields. From now on I'll use left handed fields exclusively, and following DHM will denote a right handed field by conjugation.
Now, given two left chiral spinor fields $\psi,\xi$ there are two types of mass terms you can construct:
- Majorana masses $ \frac{1}{2} m_\psi (\psi \psi + \psi^\dagger \psi^\dagger) + (\psi\rightarrow\xi) $
- Dirac mass $ m_D (\psi \xi + \xi^\dagger \psi^\dagger )$
There is some matrix structure hidden in the notation for field products. Check DHM if you want more detail. The upshot is that combinations like $\psi \psi$ etc. are Lorentz invariant.
Now the Majorana mass terms are ruled out for an electron because they violate charge conservation (there is no way to assign charges $Q_\psi,Q_\xi$ such that the terms are invariant under $U(1)_{EM}$ rotation). But, if we give $\psi$ and $\xi$ opposite charges the Dirac mass term is invariant. We interpret the fields as
- $\psi$ annihilates left handed electrons and creates right handed positrons
- $\xi$ annihilates left handed positrons and creates right handed electrons
- $\psi^\dagger$ annihilates right handed positrons and creates left handed electrons
- $\xi^\dagger$ annihilates right handed electrons and creates left handed positrons
We call the $\psi$ the field for left-chiral electrons and $\xi^\dagger$ the field for right-chiral electrons, though we have written it in terms of the left-chiral field $\xi$ which annihilates positrons. Confusing conventions. Oh well.
The weak interaction is chiral because only the $\psi$ component is charged under $SU(2)_L$. You have the operator $ \psi^\dagger \bar{\sigma}^\mu \nu W^-_\mu $ (among others) which annihilates a $W^-$ boson and creates an electron and corresponding anti-neutrino. But this state has zero overlap with the right-handed electron that has to appear in the final state. If this was the end of the story there would be no matrix element and the decay would be impossible.
But the Dirac mass is there. There is another field $\xi$ which is mixed up with the $\psi$ due to the mass term, and a propagating electron eventually becomes a mixture of these two fields. The mass term annihilates a left-handed electron and produces a right-handed electron. It is the $\xi^\dagger$ which eventually annihilates the state to give the matrix element.