# Spinor outer products

I am confused about the use of the formulas for the spinor outer products $$\sum_{s=1}^2 u_s(p)\bar u_s(p) = \gamma^\mu p_\mu + m$$ and $$\sum_{s=1}^2 v_s(p)\bar v_s(p)= \gamma^\mu p_\mu - m,$$ where $$u_s(p)$$ and $$v_s(p)$$ are a basis of spinor solutions to the Dirac equation $$(i\gamma^\mu p_\mu - m)\psi = 0$$, i.e. $$\psi = \sum_{s=1}^2\int\frac{d^3p}{(2\pi)^3}(u_s(p)e^{-ip\cdot x}+v_s(p)e^{ip\cdot x}).$$ It seems I can get different results by using either formula. For example, consider the quantity $$(\gamma^\mu p_\mu+m)\gamma^5 u_1(p).$$ Depending on which formula I use, this can either be $$\sum_{s=1}^2 u_s(p)\bar u_s(p)\gamma^5 u_1(p)$$ or $$-\sum_{s=1}^2 v_s(-p)\bar v_s(-p)\gamma^5 u_1(p)$$ Now, explicit forms for the spinors in the Weyl representation are (if $$p^\mu = (E,0,0,p_z)$$), $$u_1(p)= \begin{pmatrix}\sqrt{E-p_z}\\0\\\sqrt{E+p_z} \\0\end{pmatrix}$$ $$u_2(p)=\begin{pmatrix}0\\\sqrt{E+p_z}\\0\\\sqrt{E-p_z}\end{pmatrix}$$ $$v_1(-p)=\begin{pmatrix}\sqrt{E+p_z}\\0\\-\sqrt{E-p_z} \\0\end{pmatrix}$$ $$v_2(-p)=\begin{pmatrix}0\\\sqrt{E-p_z}\\0\\-\sqrt{E+p_z}\end{pmatrix}$$ as well as $$\gamma^5 = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}.$$ One can then see that since $$\gamma^5 u_1(p) = \begin{pmatrix} -\sqrt{E-p_z} \\ 0\\ \sqrt{E-p_z} \\ 0\end{pmatrix}$$, all the $$\bar u_s(p)\gamma^5 u_1(p)$$ vanish while $$\bar v_1 (-p) \gamma^5 u_1(p)$$ does not ($$\bar v_2 (-p) \gamma^5 u_1(p)$$ vanishes also.) So one formula tells me the expression is zero, and the other does not. Am I doing something wrong?

• Not 100% sure, but I think $\sum v_s(-p)\bar{v}_s(-p)\neq -\gamma^\mu p_\mu-m$ because $p_\mu=(E,-p_1,-p_2,-p_3)$ Apr 7, 2023 at 8:02