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Peskin & Schroeder eq (3.78) states that

$$(\bar{u}_{1R}\sigma^\mu u_{2R})(\bar{u}_{3R}\sigma_\mu u_{4R})=\cdots$$

But I don't understand what the $u_{1R}$ means. Since 4-component Dirac spinor consists of left-handed and right-handed 2-component spinors, we can say

$$\psi(x)=\begin{pmatrix}u_1(p)\\u_2(p)\\u_3(p)\\u_4(p) \end{pmatrix}e^{-ipx}=\begin{pmatrix}\psi_L\\ \psi_R\end{pmatrix} $$ for the positive frequency solutions. That is, $u_1,u_2$ correspond to the left-handed part and $u_3,u_4$ correspond to the right-handed part. But then, the above notations like $u_{1R},u_{2R}$ do not make any sense to me.

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It is stated above the quoted formula: "By sandwiching identity (3.77) between the right-handed portions (i.e., lower half) of Dirac spinors $u_1$, $u_2$, $u_3$, $u_4$, we find the identity" (3.78).

Thus $u_{1R}$ simply means the right-handed component of the spinor $u_1$; the "$1$" does not indicate a component here. Analogously for the other spinors.

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