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:


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


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(?).


1 Answer 1


So, is it right to apply an hermitian conjugation when calculating overbar quantities?

Yes, that is correct.

$$\bar{\psi} = \psi^\dagger \gamma_0$$ $$\bar{u} = u^\dagger \gamma_0$$

and as always, for finite dimensional vectors the hermitian conjugate is

$$\psi^\dagger = (\psi^*)^T$$

As for $\xi_s^\dagger\eta_{s'}=\delta_{s,s'}$ I don't think that is true. I know that one example of a basis chosen is

$$\xi_{1/2} = \begin{bmatrix} 1 \\ 0\end{bmatrix} \xi_{-1/2} = \begin{bmatrix} 0 \\ 1\end{bmatrix}$$

$$\eta_{1/2} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \eta_{-1/2} = \begin{bmatrix} -1 \\ 0 \end{bmatrix}$$

Which, instead of your relation, gives

$$\xi^\dagger_{1/2} \eta_{1/2} = \xi^\dagger_{-1/2} \eta_{-1/2} =0$$ $$\xi^\dagger_{1/2} \eta_{-1/2} = -1$$ $$\xi^\dagger_{-1/2} \eta_{1/2} = 1$$

Usually when you are calculating the scattering matrix, you don't want to expand out the form of $u,v$ explicitly. Rather, you use the following properties to just get the result in terms of gamma matrices:

$$\sum_{s = \pm 1/2} u_s(p) \bar{u}_s(p) = \not p + m$$

$$\sum_{s = \pm 1/2} v_s(p) \bar{v}_s(p) = \not p - m$$

The sums above come from spin-averaging... you should already have them in your expression. Remember that something like $\bar{u}_s v_{s'}$ is a scalar, and so it commutes with its complex conjugate $\bar{v}_s' u_{s}$. They both come from the norm $|\bar{u}_s v_{s'}|^2$.

The next step is to expand the Feynman slash into gamma matrices - generally you will be able to rewrite the sum in terms of a trace of gamma matrices - and then use the properties of gamma matrices to simplify the result.


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