The differential cross-section of relativistic point electron off a static Coulomb potential provided by a point charge with density $\rho(\vec x)=e\delta^{(3)}(\vec x)$ is given by the Mott formula (see Mandl and Shaw, Quantum Field Theory) $$\left(\frac{d\sigma}{d\Omega}\right)_{\rm Mott}=\frac{\alpha^2}{4E^2 v^4\sin^4(\theta/2)}\left[1-v^2\sin^2(\theta/2)\right]\approx\frac{\alpha^2}{4E^2\sin^4(\theta/2)}\cos^2(\theta/2)$$ where in the last step we assumed $v\approx 1$ and $\alpha=e^2/4\pi$ is the fine structure constant. This formula does not take the target recoil into account.

However, if the target recoil is taken into consideration, the formula becomes (see Burcham and Jobes, Nuclear and Particle Physics) $$\left(\frac{d\sigma}{d\Omega}\right)_{\text{Mott+Recoil}}\approx \frac{\alpha^2}{4E^2\sin^4(\theta/2)}\frac{\cos^2(\theta/2)}{1+(2E/M)\sin^2\theta/2}$$ where $M$ is the mass of the target.

Can someone explain how to derive the second formula using quantum field theory?


1 Answer 1


Of course you can do.

For example, let us consider the elastic process such that one heavy particle with mass $M$ and momentum $p_{i,f}$ is scattered by some light particle with mass $m$ and momentum $k_{i,f}$; i.e. $m\ll M.$ For simplicity we take the rest frame of the heavy particle, i.e. $p_i=(M,0)$.

Formally, the differential scattering amplitude for this process is $$d\sigma=\int\frac{d^3k_f}{(2\pi)^3} \frac{d^3p_f}{(2\pi)^3} \frac{(2\pi)^4\delta^4(p_i+k_i-p_f-k_f)}{4E’_{p_i}E’_{p_f}4E_{k_i}E_{k_f}v } |\mathcal{M}|^2,$$ where $v$ is a relative velocity and $E_p $ is a energy of light particles and $E’_p$ is of heavy particles. Here all the information about interaction is included in the term $|\mathcal{M}|^2$. We are interested in the Mott scattering, so we should use the QED calculation, but this is not important to understand the recoil effect because it does not depend on the detailed form of interactions but kinetic factor such as the term $\frac{(2\pi)^4\delta^4(p_i+k_i-p_f-k_f)} {4E’_{p_i}E’_{p_f}4E_{k_i}E_{k_f}v } $.

Let’s check this explicitly. Firstly, we will do the integration over $p_f$, then we will obtain $$d\sigma=d\Omega\int\frac{k_f^2d k_f}{(2\pi)^2} \frac{(2\pi)^4\delta^4(E_{k_i}+M-E’_{q}-E_{k_f})}{4ME’_{q}4E_{k_i}E_{k_f}v } |\mathcal{M}|^2,$$ where we use $d^3 k_f=k_f^2 dk_f d\Omega$ and define $q=k_i-k_f$. As usual, we assume $q^2\ll M^2.$ Also, here after we assume $\theta$ involved in $\Omega$ to be constant, because we are interested in differential scattering cross section.

The integration over $k_f$ is not difficult thanks to the remaining delta function. Actually, if we notice the formula $\delta(f(x))=\frac{1}{|f(’x_0)|}\delta(x-x_0)$ and the following kinetic formulae $E_{k_f}E’_{q}\frac{d(E_q’+E_{k_f})}{dk_f}\overset{m\to 0}{\simeq}Mk_f(1-\frac{q^2}{2ME_{k_f}})\ $ and $\ v\simeq \frac{k_i}{E_{k_i}}$, then under these approximations we will obtain

$$d\sigma=d\Omega\frac{k_f}{k_i} \frac{1}{(2\pi)^2}\frac{|\mathcal{M}|^2}{16M^2}\overset{m\to 0}{\simeq} d\Omega\frac{E_{k_f}}{E_{k_i}} \frac{1}{(2\pi)^2}\frac{|\mathcal{M}|^2}{16M^2 (1-\frac{q^2}{2ME_{k_f}})}. $$

Here we substitute the result from QED: $|\mathcal{M}|_{\mathrm{SpinAveraged}}^2 \simeq \frac{4M^2e^4}{q^2}\cot(\frac{\theta}{2})^2$. Then we will find that the last factor $\frac{1}{(2\pi)^2}\frac{E_{k_f}}{E_{k_i}} \frac{|\mathcal{M}|^2}{16M^2 }$ gives the Mott amplitude.

The remaining factor $(1-\frac{q^2}{2ME_{k_f}})^{-1}$ is recoil factor, which gives $$\frac{1}{1+\frac{2E_{k_i}}{M}\sin^2(\frac{\theta}{2})},$$ where we use $q^2\simeq 4E_{k_f}E_{k_i}\sin^2(\frac{\theta}{2})$.

  • $\begingroup$ In this derivation, did you assume both $m$ and $M$ to be spin-1/2 particles? $\endgroup$ May 7, 2023 at 17:38
  • $\begingroup$ Yes. As a $|\mathcal{M}|$, I took the value of the scattering of two spin-1/2 particles, and assumed that heavy particles are well-described by Dirac spinor. Of course, heavy particles (e.g. proton) is composite particle, this is a rough approximation. $\endgroup$
    – Siam
    May 8, 2023 at 2:55
  • $\begingroup$ I needed an exact calculation where the target is pointlike and the spin of the target is not taken into account (as in Mott scattering) but target recoil is included. $\endgroup$ May 8, 2023 at 4:34
  • $\begingroup$ It is not important. In the QED framework, nuclei are not elementary particles, so $|\mathcal{M}|$ cannot be calculated directly. However, since the details of such interactions are contained only in $|\mathcal{M}|$, one can, for example, adopt the invariant amplitude of relativistically extended Rutherford scattering as $|\mathcal{M}|$. This is what you want to know. The important point is that microscopic information about what particles interact is contained only in $|\mathcal{M}|$, and the recoil effect is independent of it. $\endgroup$
    – Siam
    May 8, 2023 at 6:08

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