5
$\begingroup$

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)

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – user28400 Aug 18 '13 at 18:13
  • $\begingroup$ @user28400 : I have made an edit to the answer. Hope it helps. $\endgroup$ – Trimok Aug 19 '13 at 8:16
  • $\begingroup$ Do all the process (1abc, 2abc) gowith left projector when you compute the vertex for each one? $\endgroup$ – Vicky Mar 16 at 5:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.