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As Dirac was the first to realize (Dirac 1933, page 69), the reason the quantum path integral converges to the classical action principle as $h\rightarrow 0$ is that

The only important part in the domain of integration of $q_k$ is thus that for which a comparatively large variation in $q_k$ produces only a very small variation in $F$. This part is the neighbourhood of a point for which $F$ is stationary with respect to small variations in $q_k$.

In QED the paths are replaced by field configurations, but it's still the same idea. So even though the theory says you have to add up all the configurations to get the result, practically you would only have to include those within that "neighborhood" of the configuration of stationary action.

My question is, how big would you have to make that neighborhood in order to get sufficient precision to match the famous electron $g-2$ measurement? Specifically, by how much would you have to allow the action of your configurations to differ from the stationary action? Maybe $\frac12 h$? Or $h$, or $2h$, or $3h$?

I don't think people are actually integrating with respect to action when they calculate Feynman diagram terms, but I'm basically wondering if it's possible to estimate the extent and rate at which the configurations really do cancel each other out as you move away from the stationary one.

Qmechanic
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Adam Herbst
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  • Related: Why does the classical path give the dominant contribution in the path integral? – Qmechanic Sep 11 '22 at 04:55
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    Let's consider a similar but much simpler problem: how big does $X>0$ need to be for $A(X):=\int_{-X}^Xf(x)dx$ to well-approximate $A(\infty)$? That's already a big question, requiring inferences of suitable inequalities with dependence on not only $f$ but also how we compute $A(X)$, which may well be with numerical approximations. So I think you're asking "how long is a piece of string?" until you switch to a very specific example. – J.G. Sep 11 '22 at 06:21
  • @J.G. Point taken, I have updated the question to focus only on the electron g-factor, using only QED (that's all you need there, right?) – Adam Herbst Sep 11 '22 at 15:27
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    1/2 The usual way to compute $g-2$ is to dimensionally regularize a sum of $k$-space integrals, so the obvious "don't do the whole integral" approximation would cut off at a finite energy scale. Your idea is a bit different. You basically want to know how big $n$ needs to be for good approximations of the form$$I_f(n):=\frac{\int_{-\sqrt{2n\hbar/S''0}}^\sqrt{2n\hbar/S''_0}f(x)e^{iS''_0x^2}dx}{\int{-\sqrt{2n\hbar/S''_0}}^\sqrt{2n\hbar/S''_0}e^{iS''_0x^2}dx}\approx I_f(\infty),$$ – J.G. Sep 11 '22 at 17:40
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    2/2 where specifying $f$ hinges on your choice of the electronic $g-2$ problem, $x$ is a deviation from the classical solution, and the integration range is inferred from $S\approx S_0+\frac12S''_0x^2$, so $|S-S_0|\le n\hbar$ is (almost) equivalent to $|x|\le\sqrt{2n\hbar/S''_0}$. – J.G. Sep 11 '22 at 17:40
  • @J.G. That’s helpful, thanks! So it sounds like there might not be a good answer since this isn’t how the integral is actually performed… – Adam Herbst Sep 11 '22 at 19:21
  • In case you're interested, the QED contribution to the electron's $g-2$ is a power series in $\alpha$, whose $\alpha^3$ coefficient is the highest-order one we can compute analytically viz. this paper, with higher coefficients following from similar logic but reducing to integrals we can only do numerically, the highest known being for $\alpha^5$. – J.G. Sep 11 '22 at 19:55

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