I'm looking for the simplest demonstration of the Klein-Nishina formula, from the Dirac equation without the field described as a quantum operator:


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.

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    $\begingroup$ 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). $\endgroup$ Jul 11, 2018 at 15:28
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    $\begingroup$ 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. $\endgroup$
    – Cham
    Jul 11, 2018 at 15:46
  • $\begingroup$ @accidentalfouriertransform Could you provide a reference? $\endgroup$
    – my2cts
    Jul 11, 2018 at 20:38
  • $\begingroup$ @my2cts the wikipedia page contains the exact two references I would cite here. $\endgroup$ Jul 11, 2018 at 21:14
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    $\begingroup$ 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. $\endgroup$
    – Cham
    Jul 12, 2018 at 0:09

1 Answer 1


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}


\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*}

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

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