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I asked a similar question about QED Lagrangian but I guess the question wasn't clear enough since I didn't get any correct answers. So, I'll try to ask the question in a different way: how does one write the QED Lagrangian using Weyl spinors for left handed electrons interacting with a photon (for example a left handed electron emitting a photon). If one considers an electron in terms of its Weyl spinors:

\begin{equation} \psi_{e} = \begin{pmatrix} \chi \\ \eta^{\dagger} \end{pmatrix} \end{equation}

Then one can write $\bar{\Psi_{e}}=(\eta \space \chi^{\dagger})$

Now, the QED Lagrangian is: $\bar{\Psi_{e}}\gamma^{\mu}A_{\mu}\Psi_{e}$ which would yield the following two terms:

$\chi^{\dagger}\bar{\sigma^{\mu}}A_{\mu}\chi - \eta^{\dagger}\bar{\sigma^{\mu}}A_{\mu}\eta$ $\space$ (1)

Assuming:

$\chi$: left handed electron

$\eta^{\dagger}$: right handed electron

$\eta$: left handed positron

$\chi^{\dagger}$: right handed positron

Then the Lagrangian in (1) implies the following 2 vertices:

(right-handed positron)---photon---(left-handed electron)

(right-handed electron)---photon---(left-handed positron)

So, where are the following type of vertices?

(left-handed electron)---photon---(left-handed electron)

(right-handed electron)---photon---(right-handed electron)

. . .

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You're not completely correct with your assumptions.

$\chi$ can describe both the anihiliation of a left-chiral electron or creation of a right-chiral positron

$\chi^\dagger$ can describe both the creation of a left-chiral electron or anihilation of a right-chiral positron

$\eta$ can describe both the creation of a right-chiral electron or anihilation of a left-chiral positron

$\eta^\dagger$ can describe both the anihiliation of a right-chiral electron or creation of a left-chiral positron

These allows to describe these interactions you're missing.

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  • $\begingroup$ Thanks! So, how does one write $\chi$ and $\eta$ explicitly in terms of creation and annihilation operators? I always thought each of $\chi$ and $\eta$ are only a two component spinors (where each element is a Grassmann number?) Also, based on what you wrote, shall one say this Dirac spinor: \begin{equation} \psi_{e} = \begin{pmatrix} \chi \\ \eta^{\dagger} \end{pmatrix} \end{equation} only describes annihilation of a left-handed electron and creation of a right-handed electron (ignoring the positron parts)? $\endgroup$ – physics_2015 May 27 at 23:38
  • $\begingroup$ You need $m=0$, or treat $m$ as perturbation and ignore it at the moment of quantization, or your creation/annihilation anihilators will mix left- and right-chiral fields. Then. for given $\vec{p}$ you can have $p_0=\pm|\vec p|$. For those two signs you have two equations $p_\mu\bar\sigma^\mu \chi = 0$, that have two different solutions $u(\vec p)$ and $v(\vec{p})$. Then you have $$ \chi(x) = \int\frac{d^3p}{\sqrt{2|\vec{p}|}} \big(b(\vec{p}) u(\vec{p}) e^{-i p x} + d^\dagger(\vec{p}) v(\vec{p}) e^{i p x} \big)$$ where $p^\mu=(|\vec{p}|,\vec{p})$ $\endgroup$ – Adam Latosiński May 28 at 0:09
  • $\begingroup$ As for the second question, it actuly made me notice a mistake I've made, because full $\psi_e$ should always describe annihilation of electrons (of both chirality) or creation of positrons (of both chirality). I've corrrected my answer. Also, we should be talking about chirality, not handedness=helicity. For massless particles they coincide, but not for massive ones. $\endgroup$ – Adam Latosiński May 28 at 0:22

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