How can an asymptotic expansion give an extremely accurate prediction, as in QED?

Question

What is the meaning of "twenty digits accuracy" of certain QED calculations? If I take too little loops, or too many of them, the result won't be as accurate, so do people stop adding loops when the result of their calculation best agrees with experiment? I must be missing something.

Answer 1

Suppose you're interested in computing some quantity $F(\alpha)$, like the excess magnetic moment $g-2$, which depends on the fine structure constant $\alpha \simeq .007$.

Perturbation theory gives a recipe for the coefficients $F_i$ of an infinite series $\sum_{i \geq 0} F_i \alpha^i$, which is expected to be asymptotic to $F$. This means that, if you add up the first few terms, you should get a decent approximation to $F$.

$F \simeq F_0 + F_1 \alpha + F_2 \alpha^2 + F_3 \alpha^3 + ... + F_N \alpha^N$

If you keep adding higher order corrections, making $N$ bigger, the approximation will get better for a while, and then, eventually, it will start to get worse and diverge in the limit $N \to \infty$. In QED, we don't know exactly where the approximation gets worse. However, one expects on general grounds that the series approximation will start to get worse when $N \simeq 1/\alpha \simeq 137$. (And this is indeed what we see in toy models, where we can make everything completely rigorous.)

In principle, then, we should add up all the Feynman diagrams of order $\leq 137$. In practice, we don't have the ability to do this. Computing all those diagrams is very time-consuming. Even with computer assistance, we have difficulty going beyond 4 or 5 loops.