My question is related with this question. Several years ago, I posted an answer to the question, and the author of the reference removed the link permanently, now I have no clue what's going on.

In the so-called zero-dimensional QFT, one can compute the path-integral $$\mathcal{Z}=\int_{-\infty}^{+\infty}dx e^{-x^{2}-\lambda x^{4}}$$

by mathematica, and the result is $$\mathcal{Z}=\frac{e^{\frac{1}{8\lambda}}K_{\frac{1}{4}}(\frac{1}{8\lambda})}{2\sqrt{\lambda}}$$

for $\mathrm{Re}(\lambda)>0$, and $K_{n}(x)$ is is the modified Bessel function of the second kind. The question is how to show it by hand.

The answer appears in one of the lecture notes which I shared there, but has been permanently removed by the author. The idea was as follows:

The partition function is given by

$$\mathcal{Z}(\hbar)=\int dx e^{-\frac{1}{\hbar}S(x)}\equiv\int_{C}\frac{dz}{2\pi i}e^{-z/\hbar}B(z).$$ where $B(z)$ is the modified Borel transform, given by

$$B(z)=-\int\prod_{i=1}^{N}dx_{i}\frac{\Gamma(\frac{N}{2}+1)}{(S(x)-z)^{\frac{N}{2}+1}}. \tag{$\star$}$$

The coutour $C$ encloses the range of $S(x)$.

This is the only thing I could remember from the lecture notes, and I have no clue about what the modified Borel transform is, and have no idea how to compute that $N$ dimensional integral.

Does anyone know what's going on with equation $(\star)$? Please help me figure out the rest of the steps to obtain the result.


1 Answer 1


I'll write what I understand from XiYin's notes, of which there is a copy on the wayback machine. Probably I miss some details. We would like to evaluate $$Z = \int dx e^{- S[x]} = \int dx e^{-(\mu x^2 + g x^4)}, \quad S[x] := \mu x^2 + g x^4$$ In the regime $\mu > 0$, you have a single minima of the potential at $x = 0$ that you can expand around, and for $\mu < 0$ you have a choice between two minima (as well as kink instanton solutions, that will be important for later). If you try to do perturbation theory around $g = 0$: $$\mu > 0 : \quad Z = \sqrt{\frac{\pi}{\mu}}\left( 1 - \frac{3g}{4 \mu^2} + \frac{105g^2}{32\mu^4} - \cdots (-1)^n\frac{g^n(4n-1)!!}{\mu^{2n}2^{2n}n!}\cdots\right)$$ You notice that the radius of convergence, $r \sim \left(\frac{(4n-1)!!}{\mu^{2n}2^{2n}n!}\right) \big/ \left(\frac{(4n+3)!!}{\mu^{2n+2}2^{2n+2}(n+1)!}\right) = \frac{4\mu^2(n+1)}{(4n+3)(4n+1)} \to 0$ is zero, so the series doesn't converge. This is generic behaviour, (along the lines of an argument Dyson gave), $g < 0$ behaviour is very different to $g > 0$, so you don't really expect it to converge. A general approach to resumming such divergent series is the Borel Summation. I'll give a rough overview of what I understand of it, (which I learned roughly from Iain Stewarts EFT notes). Take a not-convergent asymptotic series $f$, and do the Borel transform (be careful of the strange indexing here, I'm just following convention) $$f(\alpha) = \sum_{n=-1}^\infty f_n \alpha^{n+1}, \quad F(b) := f_{-1} \delta(b) + \sum_{n = 0}^\infty \frac{1}{n!} f_n b^n$$ Notice that, being a bit lax, we see that we can recover $f$ by doing inverse borel transform: $$\int_0^\infty db e^{-b/\alpha} F(b) = \int_0^\infty db e^{-b/\alpha} \left( f_{-1} \delta(b) + \sum_{n = 0}^\infty \frac{1}{n!} f_n b^n \right) = \sum_{n=-1}^\infty f_n \alpha^{n+1} = f(\alpha)$$ The idea is that we improve the convergence properties of our series in our borel space since we get this $n!$ suppression of the coefficients, sum it up in borel space, and then transform back to regular space. Applying this idea to our series, we get that: $$F(b) = \sqrt{\frac{\pi}{\mu}}\left( \delta(b) + \sum_{n=0}^\infty (-1)^{n+1}\frac{b^n(4n+3)!!}{\mu^{2n+2}2^{2n+2}(n+1)!n!}\right) = \sqrt{\frac{\pi}{\mu}} \left( \delta(b) -\frac{3}{4 \mu^2} \ {}_2 F_1\left(\frac{5}{4},\frac{7}{4},2,-\frac{4b}{\mu^2}\right)\right)$$ At this point, notice that the extra $n!$ in the denominator has given us a nonzero radius of convergence so we have some hope, you can stare at the series definition of the hypergeometric function and notice how to write it as a hypergeometric function. The final step is to do the reverse borel transform - I am not very good at integrals - but you should recover: $$f(\alpha) = \sqrt{\frac{\mu}{4g}} e^{\frac{\mu^2}{8 \alpha}} K_{\frac{1}{4}}\left(\frac{\mu^2}{8 g}\right)$$ The business about using modified Borel Summations comes from a paper by Crutchfield, that I have not yet read through and understood - the idea though is that for $\mu < 0$ when you have instanton solutions, you expect to see a pole on the real positive $b$ axis in your borel transformed function, which is signifying nonperturbative effects that you cannot resum. Crutchfields modified Borel transformation should be a way to deal with this. At the current point, I am confused because I don't see the borel pole right now - as I wrote it above $F(b)$ (aside from an overall normalization factor) is a function of $\mu^2$, so I don't know how the pole arises when $\mu < 0$, but if I figure it out I will edit this answer (also anyone please help me)

  • 1
    $\begingroup$ Thank you so much Joshua. Thank you for writing down the answer in such great detail! $\endgroup$
    – Valac
    Jun 6, 2022 at 6:09
  • $\begingroup$ The pole in the instanton case come from the fact that the alternating factor $(-1)^{n}$ in the asymptotic expansion is absent. This corresponds to a pole of the integrand of the Laplace integral (aka the inverse Borel transform) on the positive real axis. I've been reading Crutchfield's article but it's so obscure. $\endgroup$
    – Valac
    Jan 24, 2023 at 8:07
  • $\begingroup$ Hi. Do you mind sending me a link to the copy on the wayback machine? I couldn't find it there. $\endgroup$
    – Valac
    Jan 27, 2023 at 11:06
  • $\begingroup$ Perhaps: web.archive.org/web/20210506222821/http://… I haven't looked at this in a while $\endgroup$ Jan 27, 2023 at 15:50
  • $\begingroup$ OMG Thank you so much! $\endgroup$
    – Valac
    Jan 27, 2023 at 16:06

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