Weak interaction and the Chirality of anti-particles

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)

1 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.

• I understand why the anti-particles appear, but not why they are necessarily right-handed. You are right that we cannot separate the particle from the antiparticle, but the left projection operator $1-\gamma^5$ stays the same - so why is the anti particle that is involved right handed? – user28400 Aug 18 '13 at 18:13
• @user28400 : I have made an edit to the answer. Hope it helps. – Trimok Aug 19 '13 at 8:16
• Do all the process (1abc, 2abc) gowith left projector when you compute the vertex for each one? – Vicky Mar 16 at 5:15