Particle Physics: Decomposition of a Helicity Spinor

Question

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 $e^{i\xi}$ ...

Answer 1

Consider a particle moving in the $\hat z$ direction, for simplicity. Define $\kappa= p/(E+m)$ and note it collapses to 1 for m =0. In this frame, $$ u_ \uparrow =\sqrt{E+m} \begin{pmatrix} 1\\ 0\\ \kappa\\ 0\end{pmatrix}, \qquad u_ \downarrow =\sqrt{E+m} \begin{pmatrix} 0\\ 1\\ 0\\ -\kappa \end{pmatrix}.\tag{4.65} $$

Up to normalizations, the eigenstates of $$ \gamma_5 = \begin{pmatrix} 0& 1\!\!1 \\ 1\!\!1 &0 \end{pmatrix} $$ have eigenvalues $\pm 1$, respectively. We can resolve the identity $I=P_R+P_L$, $$ P_R= \frac{1}{2} \begin{pmatrix} 1\!\!1 & 1\!\!1 \\ 1\!\!1 &1\!\!1 \end{pmatrix} \qquad P_L= \frac{1}{2} \begin{pmatrix} 1\!\!1 & -1\!\!1 \\- 1\!\!1 &1\!\!1 \tag{6.34}\end{pmatrix} $$ so that the R,L projections of the above helicity eigenstates are $$ u_ \uparrow =P_R u_ \uparrow +P_L u_ \uparrow \propto \left ( \tfrac{1+\kappa}{2}\right ) \begin{pmatrix} 1\\ 0\\ 1\\ 0\end{pmatrix}+ \left ( \tfrac{1-\kappa}{2}\right ) \begin{pmatrix} 1\\ 0\\ -1\\ 0\end{pmatrix} \tag{6.38}$$ and finally $$ u_ \downarrow = P_R u_ \downarrow +P_L u_ \downarrow \propto \left ( \tfrac{1-\kappa}{2}\right ) \begin{pmatrix} 0\\ 1\\ 0\\ 1\end{pmatrix}+ \left ( \tfrac{1+\kappa}{2}\right ) \begin{pmatrix} 0\\ 1\\ 0\\ -1\end{pmatrix} , $$ your resolution of the negative helicity spinor sought. Note its right and left chiral components are different than those of the above positive helicity given, even though they have the same eigenvalues, (and mismatched with your text's (6.32).... Probably sloppy notation).

Note that at $\kappa\to 1$, $u_\uparrow$ collapses to its right chiral projection, and $u_\downarrow$ to its own (different) left-chiral projection; which is why negative helicity is improperly/confusingly sometimes called "left", and positive helicity is misidentified as "right", to remind you of its connection to chirality seen here.

Apologies I failed imagining your text might lack consistent notation.