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.


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

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    $\begingroup$ Thank you. I edited my question to exactly show where I am failing to prove $[1]$, because I was stumbling upon this when trying to derive a decomposition of $u_{\downarrow}$ for myself. $\endgroup$
    – user248824
    Jan 23, 2021 at 21:49
  • $\begingroup$ Linked. $\endgroup$ Jan 23, 2021 at 22:19
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    $\begingroup$ Let me give it a try tomorrow then. :) $\endgroup$
    – user248824
    Jan 23, 2021 at 22:25
  • $\begingroup$ You might have a point: there is a break in conventions between (6.32) and (6.38)! Have not looked into it, but he might be switching Dirac matrix to Weyl conventions. $\endgroup$ Jan 23, 2021 at 23:30

1 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.

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    $\begingroup$ Hi Cosmas, I see what you mean with your last sentence! I thought I must be missing sth, but I think you're right that there is an inconsistency in the text. But I like the following argument of yours most: In the ultra-relativistic limit, i. e. $\kappa \rightarrow 1$, we know that $u_{\downarrow}$ has to collapse to $u_{\text{L}}$ and thus we could already infer from that that it is has to be $u_{\downarrow} \propto \frac{1-\kappa}{2}u_{\text{R}} + \frac{1+\kappa}{2}u_{\text{L}}$. :) $\endgroup$
    – user248824
    Jan 24, 2021 at 10:15
  • $\begingroup$ I reversed your edit, as it is unsound! You already used $u_R,u_L$ for the positive helicity chiral projections above. That's why I wrote the 4-spinors explicitly. $\endgroup$ Jan 29, 2021 at 18:48

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