# Challenge: Non-Gaussian quartic integral and path integral in Quantum field theory

(1) It is well-known that we can get a Gaussian integral of this type, where $$x$$ is in $$\mathbb{R}$$: $$\int_{-\infty}^{\infty} dx e^{-ax^2}=\sqrt{(2\pi)/a}. \tag{i}$$

We can generalize this integral from a number $$a$$ to a matrix value $$A$$, from a number $$x$$ to a vector $$X=(x_1,x_2,...,x_N)$$.

So we generalize $$\int_{-\infty}^{\infty} Dx \; e^{\frac{i}{2}X^T \cdot A \cdot X+ i JX } = \int_{-\infty}^{\infty} dx_1 \int_{-\infty}^{\infty} dx_2 \dots \int_{-\infty}^{\infty} dx_N e^{\frac{i}{2}X^T \cdot A \cdot X+ i JX }=\left(\frac{(2\pi i)^N}{\det(A)}\right)^{1/2} e^{-\frac{i}{2} J^T \cdot A^{-1} \cdot J}. \tag{ii}$$

(2) We can generalize again the integral to a functional integral or the so-called path integral in Quantum Field Theory.

So for a real scalar field $$\phi$$, can be regarded as the earlier Eq(i) real valued $$x$$ generalized to the field $$x \to \varphi,$$ and the matrix value $$A$$ is generalized to the matrix operator $$M$$, so we can evaluate the path integral in the $$N$$-dimensional spacetime manifold $$\Sigma_N$$: $$\int_{-\infty}^{\infty} D \varphi e^{i \int_{\Sigma_N} (d^N x)\big( -\frac{1}{2}\varphi \cdot M \cdot \varphi + J \cdot \varphi \big)} . \tag{iii}$$ where the Green's function $$G(x_i-x_f)$$ is a solution of the delta function $$\delta^N(x_i-x_f)$$ on right hand side (r.h.s): $$-M G(x_i-x_f)=\delta^N(x_i-x_f).$$

Therefore, the $$A$$ matrix in Eq.(ii) is related to $$M$$, while also related to $$G$$ via the inverse generalization $$A \to M,$$ $$A^{-1}_{if}=G(x_i-x_f).$$ Thus we obtain:

$$\int_{-\infty}^{\infty} D \varphi e^{i \int_{\Sigma_N} (d^N x)\big( -\frac{1}{2}\varphi \cdot M \cdot \varphi + J \cdot \varphi \big)}=Constant\cdot e^{\frac{-i}{2} \int d^N x_i \int d^N x_f J(x_i) G(x_i-x_f) J(x_f) }. \tag{iii}$$ involving the Green's function $$G$$. up to a not crucial constant $$Constant$$ which I do not care.

(3) Question:

Suppose I have a quartic integral from (i) to (i')

$$\int_{-\infty}^{\infty} dx \; e^{h x - a x^2-b x^4}=?. \tag{i'}$$ I suppose that we can still easily get an answer.

How about the matrix value integral from (ii) to (ii'):

$$\int_{-\infty}^{\infty} Dx \; e^{\frac{i}{2}X^T \cdot A \cdot X+ i J \cdot X +(X^T \cdot X)^2 } =?. \tag{ii'}$$ Do we also have a QFT path integral analogy: $$\int_{-\infty}^{\infty} D \varphi e^{i \int_{\Sigma_N} (d^N x)\big( -\frac{1}{2}\varphi \cdot M \cdot \varphi + J \cdot \varphi + \varphi^4 \big)} =?. \tag{iii'}$$

My question concerns the neat evaluation of (i'), (ii') and (iii'). Thank you!

Here we can focus on the real valued function field $$\varphi(x) \in \mathbb{R}$$

• The $\phi^4$ theory in four dimensions is generally believed to have a Landau pole, which means that the final integral does not exist unless you impose a cutoff. en.m.wikipedia.org/wiki/Quartic_interaction Sep 8, 2019 at 2:52
• Well ... either that, or we let the coupling run to zero, i.e., the theory becomes free. Sep 8, 2019 at 3:15
• The terms "neat" and "easily" aren't exactly ones that immediately occur to me as far as even (i') is concerned. Sep 8, 2019 at 4:32
• Can we focus on 4-dimensions ($N=4$ in the above notation)? Sep 8, 2019 at 16:09
• If you can get a nontrivial theory out of (iii’), you should definitely publish. Sep 9, 2019 at 0:52

I keep seeing this question percolate up. I think it deserves at least one more or less correct answer.

First, the 1d integrals $$I(a,b) = \int_{\mathbb{R}} \exp(-\frac{1}{2}x^2 - ax - b x^4) dx$$ certainly exist for $$b \geq 0$$ and are analytic in $$a$$ and $$b$$. But they're not do-able in elementary terms. Noting that $$\partial_bI = \partial_a^4I$$ and integrating by parts gives a 4-th order ODE in $$a$$: $$4b\partial_a^4 I - \partial_a^2 + a\partial_aI + I = 0.$$ You can solve this via Laplace transform. The boundary values along $$a=0$$ are integrals $$\partial_a^kI = C_k\int x^k \exp(-\frac{1}{2}x^2 - b x^4)dx$$, which can be reduced to sums of products of parabolic cyclinder functions and exponential functions. The cylinder functions can be expressed in terms of $$\operatorname{erfc}$$ by recursion, but that's as elementary as they get.

The 2nd kind of integral is (after switching back to Euclidean signature) just a multivariate generalization of the first. It can be done in an eigenbasis of C.

The 3rd integral isn't really an analogue of the first two. (The quadratic case is so degenerate as to be misleading.) That path integral is a shorthand for a limit of finite-dimensional integrals of the second kind. Fix $$\Sigma = \mathbb{R}^n$$. Fix in $$\mathbb{R}^n$$ a finite cubic lattice $$L_{a,N}$$ which has edges of length $$a$$ and $$N^n$$ sites. The path integral $$"\int e^{-\int_\Sigma (q(\phi,\phi) + b\phi^4) dx}\mathcal{D}(\phi)"$$ is meant to be defined to be a certain limit $$a\to 0, N\to\infty$$ of integrals of the form $$\int_{Map(L_{a,N}, \mathbb{R})} e^{-\int_{L_{a,N}} Q(a)(f,f)(l) + B_4(a) f^4(l) dl} df$$ The critical distinction here is that the coefficients $$Q(a)$$ and $$B_4(a)$$ are functions of the lattice spacing. One is supposed to choose these functions so as to fix the lower order moments of the resulting measures to be given functions of $$q$$ and $$b$$. The theory of renormalization tells you how to select these functions: Basically, you can integrate out the field $$f$$ on half of the sites and get a theory on the lattice $$L_{2a,N/2}$$. This theory should still have the same limit $$N\to\infty$$, $$a\to 0$$.

The case $$n=1$$ is basically the theory of Ornstein-Uhlenbeck measures, and its relation to quantum mechanics. No renormalization required here.

In $$n=2$$ and $$n=3$$ some of these path integrals are known to exist -- and known to have moments -- due to the efforts of the constructive QFT school associated with Glimm & Jaffe.

In $$n=4$$, it is generally believed that no such limit exists, except for the degenerate case when $$b=0$$. The evidence is numerical and circumstantial at the moment. I don't believe any general proof has been given.

In $$n \geq 5$$, it is actually proven that no such limits exist, for pure scalar field theories. IIRC, this is due to Aizenman, Brydges, and Spencer. Edit: Per Malek, Aizenman & Froehlich have priority.

• Aizenman and Froehlich. These typically use random walk-like representations. There is the random current representation and also the representation due Brydges, Froehlich and Spencer. Oct 16, 2019 at 14:34
• In $n=4$, a proof of triviality (in a very strong sense) by Aizenman and Duminil-Copin has been announced, but the preprint isn't available yet. Oct 16, 2019 at 15:12
• It is also worth mentioning that what has been shown to converge as $a \to 0$ are the corresponding measures on suitable spaces of distributions, normalised to be probability measures. Furthermore, the divergence of the normalisation constant (the integral written above, also called "partition function") is very well understood. Oct 16, 2019 at 15:23
• @MartinHairer Well, that's the most exciting news I've heard in a while. For anyone who's following from the cheap seats like me, I found the following slides helpful: mathphys.org/Venice19/slides/aizenman.pdf Presumably this is building off the techniques in their paper with Sidoravicius, arxiv.org/abs/1311.1937. Oct 16, 2019 at 15:37
• The paper by Aizenman and Duminil-Copin has been out now, see arxiv.org/abs/1912.07973 Feb 19, 2020 at 15:05

For your integral (i'), at least the $$h=0$$, $$a>0$$, $$b>0$$ case can be done analytically (or using Mathematica or Gradshteyn-Ryzhik): $$\int_{-\infty}^\infty e^{-ax^2-bx^4}{\rm d}x = \frac{1}{2} \sqrt{\frac{a}{b}} e^{\frac{a^2}{8 b}} K_{\frac{1}{4}}\left(\frac{a^2}{8 b}\right).$$ The $$h\not=0$$ case seems not to permit a closed-form solution, as far as I can tell (but of course the integral exists).

For symmetric positive-definite $$A$$, the multidimensional case can be expressed in terms of the one-dimensional integrals in an eigenbasis of $$A$$. I don't think it will have a closed-form solution unless $$J=0$$ and $$A$$ is a multiple of the identity.

The path integral case is very different, because it only makes sense upon renormalization (the parameters of the action will depend on the regulator used to define and evaluate the integral). It is generally believed that the integral exists in a meaningful sense only for $$\lambda=0$$ (i.e. "$$\phi^4$$ theory is trivial").