Dirac equation solution - four-component spinors - left-/right-handed in ultrarelativistic limit

Question

I am confused about the solution to the Dirac equation and how it corresponds to left-/right-handed Weyl spinors. In Srednicki, page 242, it is stated that taking the ultrarelativistic limit ($|\bar{p}|>>m$), projects the solutions of the Dirac equation to purely left- or right-handed spinors.

From the Dirac equation: $$(p_\mu \gamma^\mu + m)u_s(\bar{p})=0$$

we get the solutions (for simplicity, I set p_x=p_y=0):

$$ u_+ = \begin{pmatrix}1 \\ 0 \\ \frac{p_z}{E+m} \\ 0 \end{pmatrix} , ~~u_- = \begin{pmatrix}0 \\ 1 \\ 0 \\ \frac{p_z}{E+m} \end{pmatrix} $$

Now, since $E=\pm \sqrt{p_z^2+m^2}$, the limit where $p_z >> m$ should give:

$$ u_+ = \begin{pmatrix}1 \\ 0 \\ 1 \\ 0 \end{pmatrix} , ~~u_- = \begin{pmatrix}0 \\ 1 \\ 0 \\ 1 \end{pmatrix} $$

but I do not understand how these two solutions are purely left- or right-handed?

In Srednicki, he showed it by using $u_s(\bar{p})\bar{u}_s(\bar{p})\rightarrow \frac{1}{2}(1+s\gamma_5)(-\gamma^\mu p_\mu)$, but should it not be possible to show the same thing by going directly from the two solutions?

Answer 1

You should specify in which representation of the gamma matrices you are working. In your case is the Dirac representation. Also notice that you are missing a minus sign in $u_-$ that comes from $p_z$.

$u_-= \begin{pmatrix} 0 \\ 1 \\ 0\\ -1\\ \end{pmatrix}$

In order to show that they are right or left handed, note that they are eigenvectors of the $\gamma^5$ operator, which in the Dirac basis looks like:

$ \gamma^5=\begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ \end{pmatrix}$

Then you can easily prove $\gamma^5 u_+=u_+$ and $\gamma^5 u_-=-u_-$ which accounts for the two chiralities.