Consider a weak current of the form
$
J^{\mu} = \bar{u}_{\nu}\gamma^{\mu}(1-\gamma^5)u_{e}
$
This describes the part of a weak process where a left-handed electron converts into a left-handed neutrino by emitting/absorbing a W boson. Equivalently, it should also describe the same process for a right-handed positron going to a right-handed anti-neutrino. How do you get this second part from the form of $J^{\mu}$, considering that $P_L = 1-\gamma^5$ is by definition the left handed projector? Whatever antiparticle states contained in $u$ and $\bar{u}$ should have eigenvalue $-1$ of $\gamma^5$ in order to be included in $J^{\mu}$, so, aren't they by definition left-handed?
(note: this is all in the massless approximation so that I can equate chirality and helicity/handedness)
Best Answer
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.