# The Partition Function of $0$-Dimensional $\phi^{4}$ Theory

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.

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)
• 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. Jan 24, 2023 at 8:07