💼 Case: consider the scattering of a particle from a static potential.
If we assume a static potential (no time dependence) we can write the scattering Amplitude as
$$ \mathcal M_{fi}=\int \psi^*_f(r) V(r) \psi_i(r)\,\text d^3 \vec r $$
🧽 Assume: This is the Born approximation: valid in nonrelativistic with a static potential
We can use a plane wave solution
$$ \psi(r)=e^{i\vec p \cdot \vec r} $$
Plugging this in we get
$$ \mathcal M_{fi}=\int \text d^3 \vec r \,e^{i\vec q\cdot \vec r} \, V(r) $$
where $\vec q=\vec p _f - \vec p_i$ is the momentum transfer
🗒️ Note: this is the Fourier transform of the potential
💼 Case: lets solve it for the Yukawa potential
For the Yukawa potential, 🧠 Remember: $V(r)=-\frac{g^2}{4\pi} \frac{\exp(-M_X r)}{r}$
$$ \begin{aligned} \mathcal M_{fi} &= \int \mathrm{d}^3\vec{r}\, e^{i\vec{q}\cdot\vec{r}}\,V(r) =\; -\frac{g^2}{4\pi}\int \mathrm{d}^3\vec{r}\, e^{i\vec{q}\cdot\vec{r}}\frac{e^{-m_X r}}{r} \\[1mm] &=\; -\frac{g^2}{4\pi}\int_0^\infty dr\,r^2\frac{e^{-m_X r}}{r}\int_0^\pi d\theta\,\sin\theta\, e^{i|\vec{q}|r\cos\theta}\int_0^{2\pi}d\phi \\[1mm] &=\; -\frac{g^2}{4\pi}(2\pi)\int_0^\infty dr\,r\, e^{-m_X r}\int_{-1}^{1} du\, e^{i|\vec{q}|r u} \quad (u=\cos\theta)\\[1mm] &=\; -\frac{g^2}{4\pi}(2\pi)\int_0^\infty dr\,r\, e^{-m_X r}\left(\frac{e^{i|\vec{q}|r}-e^{-i|\vec{q}|r}}{i|\vec{q}|r}\right)\\[1mm] &=\; -\frac{g^2}{4\pi}\frac{4\pi}{|\vec{q}|}\int_0^\infty dr\, e^{-m_X r}\sin(|\vec{q}|r)\\[1mm] &=\; -\frac{g^2}{|\vec{q}|}\int_0^\infty dr\, e^{-m_X r}\sin(|\vec{q}|r) \\ &=-\frac{g^2}{|\vec{q}|}\cdot \frac{|\vec{q}|}{m_X^2+|\vec{q}|^2} \\ &= -\frac{g^2}{|\vec{q}|^2+m_X^2}
\end{aligned}
$$
Where we used the standard integral $\int_0^\infty e^{-ar}\sin(br)\,dr = \frac{b}{a^2+b^2}$ for $(a>0)$
Since we are using the Born approximation then the momentum $|q|^2 \ll m_X^2$ thus we get
$$ \mathcal M_{fi}=-\frac{g^2}{m_X^2}\equiv \text{constant} $$
🗒️ Note: If we look at the relativistic case ie not the Born approximation then we have
$$ \mathcal M_{fi}=-\frac{g^2}{|q^\mu_X|^2-m_X^2} $$
where $q^\mu_X$ is the momentum 4 vector, this is maximised when the momentum component is minimised which leads to a peak, explaining the spectral lines.
The denominator $|q^\mu_X|^2-M_X^2$ is called the propagator of $X$ and for all interactions the scattering amplitude $M_{fi}$ will have this format
<aside> <img src="attachment:21da6a56-7310-48b9-bb8c-45fe3886d171:the_cross_section.png" alt="attachment:21da6a56-7310-48b9-bb8c-45fe3886d171:the_cross_section.png" width="40px" />
The cross section $\sigma$ is proportional to the scattering amplitude squared#
$$ \sigma \propto |\mathcal M_{fi}|^2 $$
It is a measure of the probability that a given interaction can happen
</aside>
We care about the cross section because it allows us to know the rate of events seen in a particle experiment using
$$ N=\sigma L $$
where $N$ is the number of events and $L$ is the integrated luminosity
💼 Case: this is data from atlas.
we use our theoretical model, ie our Feynman diagrams to find the scattering amplitude $M_{fi}$ from which we can find the cross section using $\sigma \propto |M_{fi}|^2$ (continuous grey line)
Now on the experimental side we measure events and luminosity to find the cross section using $N=\sigma L$ (data points)
💎 Conclusion: Here the theory matches very well the data
