# 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.

• Do you want the QFT calculation or the pre-QFT one? The former can be found in any QFT book, and the latter can be found in the original paper (which is 16 pages long; the calculation is very cumbersome and too long to reproduce here). Commented Jul 11, 2018 at 15:28
• 16 pages of calculations ? Geez, I was expecting something relatively simple for a pre-QFT calculation at the lowest order. I'm surprised by this. How can I find the original paper in PDF format? I don't have access to journals.
– Cham
Commented Jul 11, 2018 at 15:46
• @accidentalfouriertransform Could you provide a reference? Commented Jul 11, 2018 at 20:38
• @my2cts the wikipedia page contains the exact two references I would cite here. Commented Jul 11, 2018 at 21:14
• Eurk! The original paper is in german. Can't read this. :-( I'm pretty convinced the Klein-Nishina formula could be derived relatively easily. Or I would be interested to see the main steps to it.
– Cham
Commented Jul 12, 2018 at 0:09

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

$$$$\langle|\mathcal{M}|^2\rangle = \frac{e^4}{4} \left( \frac{f_{11}}{(s-m^2)^2} +\frac{f_{12}}{(s-m^2)(u-m^2)} +\frac{f_{12}^*}{(s-m^2)(u-m^2)} +\frac{f_{22}}{(u-m^2)^2} \right)$$$$

where

\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}

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{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}

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{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}

It follows that \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}

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 $$\begin{equation*} \frac{d\sigma}{d\Omega}\propto \left(\frac{\omega_L'}{\omega_L}\right)^2\langle|\mathcal{M}|^2\rangle \end{equation*}$$

• For a complete derivation see eigenmath.org/compton-scattering.pdf
– user250986
Commented Jan 6, 2020 at 19:27
• @user250986 The pdf file is not available now. Commented Mar 2, 2023 at 1:32