The charged current part of the Lagrangian of the electoweak interaction, for the first generation of leptons, is :
$$L_c = \frac{g}{\sqrt{2}}(\bar \nu_L \gamma^\mu e_L W^+_\mu + \bar e_L \gamma^\mu \nu_L W^-_\mu )$$
The first part corresponds to different versions of the same vertex :
$e_L + W^+ \leftrightarrow \nu_L \tag{1a}$
$(\bar\nu)_R + W^+ \leftrightarrow(\bar e)_R \tag{1b}$
$W^+ \leftrightarrow (\bar e)_R +\nu_L \tag{1c}$
The second part corresponds to different versions of the hermitian congugate vertex :
$\nu_L + W^- \leftrightarrow e_L \tag{2a}$
$ (\bar e)_R + W^- \leftrightarrow(\bar \nu)_R \tag{2b}$
$W^- \leftrightarrow e_L +(\bar \nu)_R \tag{2c}$
Here, $(\bar e)_R$ and $(\bar\nu)_R$ are the anti-particle of $e_L$ and $\nu_L$
Roughly speaking, you can change the side of a particle relatively to the $\leftrightarrow$, if you take the anti-particle.
Why the right-handed particles appear ? The fundamental reason is that we cannot separate particles and anti-particles, for instance, we cannot separate the creation of a particle and the destruction of an anti-particle.
[EDIT]
(Precisions due to OP comments)
The quantized Dirac field may be written :
$$\psi(x) = \int \frac{d^3p}{(2\pi)^\frac{3}{2} (\frac{E_p}{m})^\frac{1}{2}}~\sum_s(b(p,s) u(p,s)e^{-ip.x} + d^+(p,s) v(p,s)e^{+ip.x} )$$
$$\psi^*(x) = \int \frac{d^3p}{(2\pi)^\frac{3}{2} (\frac{E_p}{m})^\frac{1}{2}}~\sum_s(b^+(p,s) \bar u(p,s)e^{+ip.x} + d(p,s) \bar v(p,s)e^{-ip.x} )$$
Here, the $u$ and $v$ are spinors corresponding to particle and anti-particle, the $b$ and $b^+$ are particle creation and anihilation operators, the $d$ and $d^+$ are anti-particle creation and anihilation operators.
We see, that in Fourier modes of the Dirac quantized field, the elementary freedom degree is (below $p$ and $s$ are fixed):
$$b(p,s) u(p,s)e^{-ip.x} + d^+(p,s) v(p,s)e^{+ip.x}$$
Now, suppose we are considering massless particles, so that helicity and chirality are the same thing. Suppose that, for the particle (spinor $u(p,s)$) the couple $s,p$ corresponds to some helicity. We see, that, for the anti-particle ($v$), there is a term $e^{+ip.x}$ instead of $e^{-ip.x}$ for the particle. That means that the considered momentum is $-p$ for the anti-particle, while the considered momentum is $p$ for the particle. The momenta are opposed for a same $s$, so it means that the helicities are opposed.
It is important to differentiate between helicity and chirality. Helicity is the spin angular momentum of a particle projected onto its direction of motion. For a massive particle this quantity is frame dependent. Furthermore, since angular momentum is conserved, as a particle propagates helicity is conserved.
On the other hand, chirality is an innate property of a particle and doesn't change with frame. However, the mass term for a Dirac particle is,
\begin{equation}
-m(\psi_L^\dagger \psi_R + \psi_R^\dagger\psi_L)
\end{equation}
(in this notation the Dirac spinor is $\Psi= (\psi_L , \psi_R) ^T $). This term can be thought of as an interaction term in the Lagrangian which switches the chirality of a particle (e.g. a left chiral particle can spontaneously turn into a right chiral particle)
For a massless particle, chirality is equal to helicity.
With that background we can finally address your questions.
- Both helicity and chirality definitely make sense for a massive
Dirac spinor. However, that doesn't mean that a Dirac spinor is a
helicity and chirality eigenstate. In the same sense that energy
makes sense for a particle, but it may not be an energy eigenstate.
- As you mention the left chiral and right chiral fields can't be
decoupled from each other due to the mass term. The mass term can
always switch a right handed field to a left handed field and vice
versa.
- As I said above, the helicity of an electron is indeed frame
dependent. So it may look like a left or right helicity electron
depending on the frame, however its chirality is not frame
dependent.
- If we write the Dirac Lagrangian in terms of chirality eignestates
then we have, \begin{equation} {\cal L} _D = i \psi _L ^\dagger
\sigma ^\mu \partial _\mu \psi _L + i \psi _R ^\dagger \bar{\sigma}
^\mu \partial _\mu \psi _R - m \psi _L ^\dagger \psi _R - m \psi _R
^\dagger \psi _L \end{equation} Then we can think of $\psi_L$ (left
chiral particle) and $\psi_R$ (right chiral particle) as two
different particles that can turn into each other spontaneously
through a mass term. Putting them together, into a Dirac spinor
masks this property. However, they are still well defined
separately.
Best Answer
You are correct that for a massive spinor, helicity is not Lorentz invariant. For a massless spinor, helicity is Lorentz invariant and coincides with chirality. Chirality is always Lorentz invariant.
Helicity defined $$ \hat h = \vec\Sigma \cdot \hat p, $$ commutes with the Hamiltonian, $$ [\hat h, H] = 0, $$ but is clearly not Lorentz invariant, because it contains a dot product of a three-momentum.
Chirality defined $$ \gamma_5 = i\gamma_0 \ldots \gamma_3, $$ is Lorentz invariant, but does not commute with the Hamiltonian, $$ [\gamma_5, H] \propto m $$ because a mass term mixes chirality, $m\bar\psi_L\psi_R$. If $m=0$, you can show from the massless Dirac equation that $\gamma_5 = \hat h$ when acting on a spinor.
Your second answer is closest to the truth:
A left chiral spinor can be written $$ \psi_L = \frac12 (1+\gamma_5) \psi. $$ If $m=0$, the left and right chiral parts of a spinor are independent. They obey separate Dirac equations.
If $m\neq0$, the mass states $\psi$, $$ m(\bar\psi_R \psi_L + \bar\psi_L \psi_R) = m\bar\psi\psi\\ \psi = \psi_L + \psi_R $$ are not equal to the interaction states, $\psi_L$ and $\psi_R$. There is a single Dirac equation for $\psi$ that is not separable into two equations of motion (one for $\psi_R$ and one for $\psi_L$).
If an electron, say, is propagating freely, it is a mass eigenstate, with both left and right chiral parts propagating.