How to derive the Klein-Nishina formula from the Dirac equation? I'm looking for the simplest demonstration of the Klein-Nishina formula, from the Dirac equation without the field described as a quantum operator:
https://en.wikipedia.org/wiki/Klein%E2%80%93Nishina_formula
Consider $\psi$ as a "classical" spinor field (not a quantum operator), satisfying the Dirac equation :
$$\tag{1}
\gamma^a \partial_a\psi + i m \psi = 0.
$$
How can we deduce the following Klein-Nishina formula?
$$\tag{2}
\frac{d\sigma}{d\Omega} = \frac{r_{\mathrm{c}}^2}{2} \Big( P(E, \vartheta) + \frac{1}{P(E, \vartheta)} - \sin^2 \vartheta  \Big)  P^2(E, \vartheta),
$$
where $r_{\mathrm{c}}$ is the classical electron radius and
$$\tag{3}
P(E, \vartheta) = \frac{1}{1 + \frac{E}{m c^2}(1 - \cos{\vartheta})}.
$$
The formula (2) was derived in 1928 to the lowest non-trivial order, after Dirac published his equation and before QFT was formulated (i.e. QED), so I'm expecting that the derivation isn't very complicated.
 A: In the center of mass frame, let $p_1$ be the inbound photon, $p_2$ the inbound electron, $p_3$ the scattered photon, $p_4$ the scattered electron.
\begin{equation*}
p_1=\begin{pmatrix}\omega\\0\\0\\ \omega\end{pmatrix}
\qquad
p_2=\begin{pmatrix}E\\0\\0\\-\omega\end{pmatrix}
\qquad
p_3=\begin{pmatrix}
\omega\\
\omega\sin\theta\cos\phi\\
\omega\sin\theta\sin\phi\\
\omega\cos\theta
\end{pmatrix}
\qquad
p_4=\begin{pmatrix}
E\\
-\omega\sin\theta\cos\phi\\
-\omega\sin\theta\sin\phi\\
-\omega\cos\theta
\end{pmatrix}
\end{equation*}
where $E=\sqrt{\omega^2+m^2}$.
It is easy to show that
\begin{equation}
\langle|\mathcal{M}|^2\rangle
=
\frac{e^4}{4}
\left(
\frac{f_{11}}{(s-m^2)^2}
+\frac{f_{12}}{(s-m^2)(u-m^2)}
+\frac{f_{12}^*}{(s-m^2)(u-m^2)}
+\frac{f_{22}}{(u-m^2)^2}
\right)
\end{equation}
where
\begin{equation}
\begin{aligned}
f_{11}&=-8 s u + 24 s m^2 + 8 u m^2 + 8 m^4
\\
f_{12}&=8 s m^2 + 8 u m^2 + 16 m^4
\\
f_{22}&=-8 s u + 8 s m^2 + 24 u m^2 + 8 m^4
\end{aligned}
\end{equation}
for the Mandelstam variables $s=(p_1+p_2)^2$, $t=(p_1-p_3)^2$, $u=(p_1-p_4)^2$.
Next, apply a Lorentz boost to go from the center of mass frame to the lab frame in which the electron is at rest.
\begin{equation*}
\Lambda=
\begin{pmatrix}
E/m & 0 & 0 & \omega/m\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
\omega/m & 0 & 0 & E/m
\end{pmatrix},
\qquad
\Lambda p_2=\begin{pmatrix}m \\ 0 \\ 0 \\ 0\end{pmatrix}
\end{equation*}
The Mandelstam variables are invariant under a boost.
\begin{equation}
\begin{aligned}
s&=(p_1+p_2)^2=(\Lambda p_1+\Lambda p_2)^2
\\
t&=(p_1-p_3)^2=(\Lambda p_1-\Lambda p_3)^2
\\
u&=(p_1-p_4)^2=(\Lambda p_1-\Lambda p_4)^2
\end{aligned}
\end{equation}
In the lab frame, let $\omega_L$ be the angular frequency of the incident photon
and let $\omega_L'$ be the angular frequency of the scattered photon.
\begin{equation}
\begin{aligned}
\omega_L&=\Lambda p_1\cdot(1,0,0,0)=\frac{\omega^2}{m}+\frac{\omega E}{m}
\\
\omega_L'&=\Lambda p_3\cdot(1,0,0,0)=\frac{\omega^2\cos\theta}{m}+\frac{\omega E}{m}
\end{aligned}
\end{equation}
It follows that
\begin{equation}
\begin{aligned}
s&=(p_1+p_2)^2=2m\omega_L+m^2
\\
t&=(p_1-p_3)^2=2m(\omega_L' - \omega_L)
\\
u&=(p_1-p_4)^2=-2 m \omega_L' + m^2
\end{aligned}
\end{equation}
Compute $\langle|\mathcal{M}|^2\rangle$ from $s$, $t$, and $u$ that involve $\omega_L$ and $\omega_L'$.
\begin{equation*}
\langle|\mathcal{M}|^2\rangle=
2e^4\left(
\frac{\omega_L}{\omega_L'}+\frac{\omega_L'}{\omega_L}
+\left(\frac{m}{\omega_L}-\frac{m}{\omega_L'}+1\right)^2-1
\right)
\end{equation*}
From the Compton formula
\begin{equation*}
\frac{1}{\omega_L'}-\frac{1}{\omega_L}=\frac{1-\cos\theta_L}{m}
\end{equation*}
we have
\begin{equation*}
\cos\theta_L=\frac{m}{\omega_L}-\frac{m}{\omega_L'}+1
\end{equation*}
Hence
\begin{equation*}
\langle|\mathcal{M}|^2\rangle=
2e^4\left(
\frac{\omega_L}{\omega_L'}+\frac{\omega_L'}{\omega_L}+\cos^2\theta_L-1
\right)
\end{equation*}
The differential cross section for Compton scattering is
\begin{equation*}
\frac{d\sigma}{d\Omega}\propto
\left(\frac{\omega_L'}{\omega_L}\right)^2\langle|\mathcal{M}|^2\rangle
\end{equation*}
