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I’ve been reading Student Friendly Quantum Field Theory and I’m discouraged by the fact that I’ve yet to encounter an actual numerical calculation in the book. I feel like I would be able to understand the theory better if I could just find an explanation of a calculation for a specific scenario, such as a transition amplitude for Bhabha scattering. Is there a textbook or website that has a section explaining the steps for a calculation? I feel like it would help with my motivation for understanding it and I feel like working backwards from it would be an easier way to learn.

To clarify, I am looking for a calculation that is a concrete example - there are no variables. The output of the calculation must be a number.

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  • $\begingroup$ Part of the problem is that when we actually want to calculate a correct answer, it would first require renormalisation, which is horribly complicated and dependent upon how many loops you are computing to. One of the "easiest" computation is to reproduce the Coulomb potential, and that itself requires infinitely many diagrams of the ladder type. Life is difficult in QFT. $\endgroup$
    – naturallyInconsistent
    Commented Jul 6, 2023 at 8:42
  • $\begingroup$ Hi Ryan Parikh. By a number do you mean like a scattering cross section or a decay rate? $\endgroup$
    – Qmechanic ♦
    Commented Jul 6, 2023 at 8:48
  • $\begingroup$ @naturallyInconsistent There are many scattering and decay processes in QED that can be calculated at tree level, and therefore don't require renormalization to get an answer which is correct to leading order in $\alpha$. The coulomb potential (there's a short section in Peskin+Schroeder about it) can be derived from tree-level QED - there are not infinitely many diagrams, unless you mean that there's an integral involved. $\endgroup$
    – AXensen
    Commented Jul 6, 2023 at 10:47
  • $\begingroup$ @AXensen It should be a ladder, especially if you are considering the bound state, where two particles are exchanging a lot of single photons between them. I suppose there could be simpler versions if all we want is to get a repulsion or something like that. $\endgroup$
    – naturallyInconsistent
    Commented Jul 6, 2023 at 11:22
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    $\begingroup$ Most derivations of tree level scattering bring the final result to a very small expression involving fundamental constants and things like the mass of the particle, what additional insight do you think will be gained by replacing this with a number? In fact usually this result, since it depends on things like center of mass energy are better viewed as functions to be plotted over a range to see trends in the results. The understanding of the derivation is in the rest of the steps where numbers really won't help you. $\endgroup$
    – Triatticus
    Commented Jul 6, 2023 at 14:52

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Bhabha scattering data from DESY PETRA experiment

See www.hepdata.net/record/ins191231, Table 3, 14.0 GeV.

\begin{equation*} \begin{matrix} x & y\\ -0.7300 & 0.10115\\ -0.6495 & 0.12235\\ -0.5495 & 0.11258\\ -0.4494 & 0.09968\\ -0.3493 & 0.14749\\ -0.2491 & 0.14017\\ -0.1490 & 0.18190\\ -0.0488 & 0.22964\\ \phantom{+}0.0514 & 0.25312\\ \phantom{+}0.1516 & 0.30998\\ \phantom{+}0.2520 & 0.40898\\ \phantom{+}0.3524 & 0.62695\\ \phantom{+}0.4529 & 0.91803\\ \phantom{+}0.5537 & 1.51743\\ \phantom{+}0.6548 & 2.56714\\ \phantom{+}0.7323 & 4.30279\\ \end{matrix} \end{equation*}

Data $x$ and $y$ have the following relationship with the cross section formula. \begin{equation*} x=\cos\theta, \quad y=\frac{d\sigma}{d\Omega}\text{ in units of nanobarns} \end{equation*}

The Bhabha scattering cross section formula is \begin{equation*} \frac{d\sigma}{d\Omega} =\frac{\alpha^2}{4s} \left(\frac{\cos^2\theta+3}{\cos\theta-1}\right)^2\times(\hbar c)^2 \end{equation*}

To compute predicted values $\hat{y}$, multiply by $10^{37}$ to convert square meters to nanobarns. \begin{equation*} \hat{y} =\frac{\alpha^2}{4s} \left(\frac{x^2+3}{x-1}\right)^2 \times(\hbar c)^2 \times10^{37} \end{equation*}

The following table shows predicted values $\hat{y}$ for $s=(14.0\,\text{GeV})^2$.

\begin{equation*} \begin{matrix} x & y & \hat{y}\\ -0.7300 & 0.10115 & 0.110296\\ -0.6495 & 0.12235 & 0.113816\\ -0.5495 & 0.11258 & 0.120101\\ -0.4494 & 0.09968 & 0.129075\\ -0.3493 & 0.14749 & 0.141592\\ -0.2491 & 0.14017 & 0.158934\\ -0.1490 & 0.18190 & 0.182976\\ -0.0488 & 0.22964 & 0.216737\\ \phantom{+}0.0514 & 0.25312 & 0.264989\\ \phantom{+}0.1516 & 0.30998 & 0.335782\\ \phantom{+}0.2520 & 0.40898 & 0.443630\\ \phantom{+}0.3524 & 0.62695 & 0.615528\\ \phantom{+}0.4529 & 0.91803 & 0.907700\\ \phantom{+}0.5537 & 1.51743 & 1.451750\\ \phantom{+}0.6548 & 2.56714 & 2.609280\\ \phantom{+}0.7323 & 4.30279 & 4.615090\\ \end{matrix} \end{equation*}

The coefficient of determination $R^2$ measures how well predicted values fit the data. \begin{equation*} R^2=1-\frac{\sum(y-\hat{y})^2}{\sum(y-\bar{y})^2}=0.995 \end{equation*}

The result indicates that the model $d\sigma$ explains 99.5% of the variance in the data and $d\sigma$ follows directly from QFT.

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