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

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?

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.

• Thank you, and sorry for my stupid questions..., but why does it mean that it is left-/right-handed if $\gamma^5u_\pm=\pm u_\pm$ holds? Maybe my confusion is rather, why do we need ultrarelativistic limit to project these spinors to the left- or righthanded Weyl fields? If we use the projection operator, e.g. $1/2(1+\gamma_5)$, it will only give the right-handed Weyl field, no matter what we project. However, clearly the 1 on the top row in $u_+$ would be projected to the left-handed Weyl field instead, if we use the corresponding operator, so isn't $u_+$ rather a mix of left/right?
– a20
May 25 '20 at 20:13
• Don't be sorry, but I can't help you too much since what you ask is a bit too long to explain for a comment. Nevertheless, for the first question, recalling the basics of quantum mechanics can help to interpret the property. I recommend you do a little research on the difference of chirality/helicity and why for massless particles they are the same. May 25 '20 at 21:22