I would have a general question: If we consider the decay of the $W^{-}$ boson into $l^{-}\nu_{\bar{l}}$, how can we calculate the polarization of the $l^{-}$?For example, Mark Thomson has on page 299, Eq. (11.17), the following decomposition of a right-handed helicity spinor $u_{\uparrow}$:

$$u_{\uparrow} = \frac{1}{2}\left( 1 + \frac{p}{E + m}\right)u_{\text{R}} + \frac{1}{2}\left( 1 - \frac{p}{E + m}\right)u_{\text{L}} \qquad [1], $$ where $u_{\text{L}}$ and $u_{\text{R}}$ denote chiral states.

Question:

Is there a similar decomposition for a left-handed helicity spinor $u_{\downarrow}$ as in Eq. [1]? I coulnd't find it in the Thomson.

EDIT: Following Cosmas Zachos' comment, here is where I am stuck on proving [1] on my own. I think I might manage to prove for myself a representation for $u_{\downarrow}$ once I understand [1]. So: One line before Eq. (6.38) in Thomson, he has the following Eq.: $$u_{\uparrow}\left( p, \theta, \phi \right) = \frac{1}{2}\left( 1 + \kappa\right)N\begin{pmatrix} \cos\frac{\theta}{2} \\ \sin\frac{\theta}{2}e^{i\varphi} \\ \cos\frac{\theta}{2} \\ \sin\frac{\theta}{2}e^{i\varphi}\end{pmatrix} + \frac{1}{2}\left( 1-\kappa\right)N\underbrace{\begin{pmatrix} \cos\frac{\theta}{2} \\ \sin\frac{\theta}{2}e^{i\varphi} \\ -\cos\frac{\theta}{2} \\ -\sin\frac{\theta}{2}e^{i\varphi} \end{pmatrix}}_{\left(\star\right)}\qquad [2],$$ and then Eq. (6.38) (he also wrote somewhere that $s \equiv \sin\frac{\theta}{2}$ and $c\equiv \cos\frac{\theta}{2}$):

$$u_{\uparrow}\left( p, \theta, \phi \right) \propto \frac{1}{2}\left(1+\kappa\right)u_{\text{R}} + \frac{1}{2}\left( 1-\kappa\right)u_{\text{L}}.$$

I do not understand how $\left( \star \right)$ is supposed to be proportional to $u_{\text{L}}$. According to Thomson, "the above spinors all can be multiplied by an overall complex phase with no change in any physical predictions", page 108. According to Eq. (6.32), $$u_L = N\underbrace{\begin{pmatrix} -\sin\frac{\theta}{2} \\ \cos\frac{\theta}{2}e^{i\varphi} \\ \sin\frac{\theta}{2} \\ -\cos\frac{\theta}{2}e^{i\varphi} \end{pmatrix}}_{\left(\star\star\right)}.$$

Comparing $\left(\star\right)$ to $\left( \star\star\right)$ for the first component for now, and taking into account that we are allowed to habe a global phase factor of $e^{i\xi}$, I get:

$$e^{i\xi} \cdot \cos\frac{\theta}{2} = -\sin\frac{\theta}{2}$$

For me, this Equation is never satisfied, regardless of what I choose for $e^{i\xi}$ ...