I'm doing a scattering exercise with fermions and I'm working with the expression of Dirac's spinor that arises from the field solutions of the Euler-Lagrange equation:
$$\psi(x) = \frac{1}{\sqrt{(2\pi)^3}}\int\frac{d^3p}{\sqrt{2E_p}}\sum_s [a_s(p)u(p)e^{-ip\cdot r} + b^{\dagger}_s(-p)v(-p)e^{ip\cdot r} ]$$ Where:
$$u_s(p)=\frac{{p\!\!\!/}+m}{\sqrt{2(E_p+m)}}\begin{bmatrix}\xi_s \\ \xi_s\end{bmatrix}$$
$$v_s(p)=-\frac{{p\!\!\!/}-m}{\sqrt{2(E_p+m)}}\begin{bmatrix}\eta_s \\ -\eta_s\end{bmatrix}$$
To evaluate the matrix element $T_{fi}$ I need to evaluate the following quantity:
$|\overline{u}_s(p)v_{s'}(-p)|^2$
where the overbar comes from the lagrangian interaction of two fermions with a scalar boson: $\mathcal{L}=-g\phi \overline{\psi}\psi$.
I know that the overbar is introduced to denote a change in sign of the Pauli matrices
$\overline{\sigma}_i = -\sigma_i$
But if this is the only change I need to apply to calculate overbar quantities, the quantity
$|\overline{u}_s(p)v_{s'}(-p)|^2$
does not make any sense to me, because the two column vectors containing the Weyl's spinors $\xi_s$ and $\eta_s$ can not be multiplied together.
However, by supposing that in $\overline{u}_s$ I also need to apply a hermitian conjugation I get the correct result for the previous quantity:
$ |\overline{u}_s(p)v_{s'}(-p)|^2 = |[\xi_s^\dagger,\xi_s^\dagger] \begin{bmatrix}0 & p\sigma_\mu \\ p \overline{\sigma}_\mu & 0\end{bmatrix} \begin{bmatrix}0 & p\overline{\sigma}_\mu \\ p \sigma_\mu & 0\end{bmatrix} \begin{bmatrix}\eta_{s'}\\ -\eta_{s'}\end{bmatrix}|^2 = 16 E_p |\xi_s^\dagger p_i \sigma_i \eta_{s'}|^2 = 16E_p |p|^2 \delta_{s,s'}$
So, is it right to apply a hermitian conjugation when calculating overbar quantities?
Also, in the last equality, I used $\xi_s^\dagger\eta_{s'}=\delta_{s,s'}$ which gives me the correct result but is not obvious to me, since Weyl's spinors are orthonormal but $\eta_s$ and $\xi_s$ could be different in general(?).
