I know that you can come across non-perturbative effects in QFT, particular when the coupling constant lies outside the radius of convergence of the asympototic perturbation series.

From the perspective of canonical quantization I understand how and why such instantons crop up. Specifically, they give (non-trivial) exact solutions of the equations of motion for the full interacting theory. These may be canonically quantized to produce a quantum instanton.

My question relates to the origins of instantons from the path-integral approach. Weinberg (in Volume II of his QFT text) claims that in perturbation theory

we expand the action $S$ around spacetime independent vacuum values of the fields, keeping the leading quadratic term in the exponential $\exp(iS)$

What does he mean by this? I've definitely never done any expansions around vacuum values to derive perturbation theory for the path integral. Doesn't one usually just split e.g. $\exp(iS_{free})\exp(iS_{int})$ and that immediately gives you the series?

He goes on to say that non-perturbative effects

arise because there are stationary points of the action which are spacetime dependent

I understand this in the context of canonical quantization, but it doesn't make much sense to me with the path-integral. Unless of course one introduces the stationary phase approximation, but I don't know its relation to usual pertubation theory. Hopefully this will be clearer once the previous paragraph makes sense to me!

  • $\begingroup$ Which page in Vol II? $\endgroup$
    – Qmechanic
    Oct 28, 2013 at 20:14
  • $\begingroup$ Page 421 (Chapter 23). $\endgroup$ Oct 28, 2013 at 20:20

1 Answer 1


I) This is discussed around eq. (23.7.1) on p. 462 in Ref. 1. The task is to perform the path integral

$$\tag{1} \int_{BC} [d\phi]e^{\frac{i}{\hbar}S[\phi]} ~=~\sum_{\nu}\int\! du \int_{BC_0} [d\phi_q]e^{\frac{i}{\hbar}S[\phi_{cl}+\phi_{\nu,u}+\phi_q]} $$ over fields $\phi$ with some (possible inhomogeneous) boundary conditions $BC$. This is done by splitting the fields

$$\tag{2} \phi~=~\phi_{cl}+\phi_{\nu,u}+\phi_{q}$$

into the following parts.

  1. A single distinguished classical solution $\phi_{cl}$ (in the trivial instanton sector). The classical solution $\phi_{cl}$ satisfies the Euler-Lagrange equations with the (possible inhomogeneous) boundary conditions $BC$.

  2. A set of instantons $\phi_{\nu,u}$ labelled with discrete topological number $\nu$ and continuous moduli $u$. The instantons $\phi_{\nu,u}$ satisfy the Euler-Lagrange equations with homogeneous boundary conditions $BC_0$. Instantons arise when there isn't a unique solution to the Euler-Lagrange equation with the given boundary conditions $BC$.

  3. And quantum fluctuation $\phi_q$ satisfying the homogeneous boundary conditions $BC_0$.

The action

$$\tag{3} S[\phi_{\rm cl}+\phi_{\nu,u}+\phi_q] ~=~S[\phi_{\rm cl}+\phi_{\nu,u}]+S_{2}[\phi_q]+{\cal O}((\phi_q)^3) ~\approx~S[\phi_{\rm cl}+\phi_{\nu,u}]+S_{2}[\phi_q]$$

is then often expanded to quadratic order (denoted $S_2$) in the quantum fluctuations $\phi_q$ leading to a Gaussian path integral. See also eq. (23.7.2) in Ref. 1. Note that the linear term $S_{1}[\phi_q]=0$ in $\phi_q$ vanishes because of Euler-Lagrange equations.

In ordinary perturbation theory without instantons, there is no summation over instanton sectors and integration over moduli.

II) One may wonder if the summation over instanton sectors in eq. (1) constitutes a kind of over-counting of the field configurations in the path integral? E.g. couldn't one reproduce a non-trivial instanton by including sufficiently many (all?) quantum corrections in the trivial sector, etc.?

From an idealized mathematical point of view, the need to sum over instanton sectors may be seen as the mathematical fact that not all $C^{\infty}$ functions are analytic. (Speaking of analycity, it seems relevant to mention the characteristic hallmark of instantons: The (non-trivial) instanton terms in the partition function have non-analytic dependence of the coupling constants of the theory. In short: One cannot reproduce non-perturbative effects by only applying perturbative methods.)

However in practice the path integral over quantum fluctuations is not well-defined (much) beyond the Gaussian approximation. So in practice one may view the decomposition on the rhs. of eq. (1) as a pragmatic definition of the full path integral on the lhs. of eq. (1).


  1. S. Weinberg, The Quantum Theory of Fields, Vol. 2, p. 421 and p. 462.
  • $\begingroup$ Ah okay - I hadn't read that far. But I'm still confused by the first quotation in my original question. How does that splitting (or the similar one for constant fields) relate to "usual" perturbation theory? By "usual" I mean the splitting of the exponentials as a product of free and interaction terms. $\endgroup$ Oct 28, 2013 at 20:36
  • $\begingroup$ Hmm... but doesn't that now say that solutions without instantons are entirely quantum fluctuations. I think your $\phi'$ isn't quite right somehow. And I'm afraid I still don't understand which tacit assumption I make when I do my "usual" splitting that gives perturbation theory. I must make an assumption, because I don't get instantons. But what is it? I don't understand how it can have anything to do with the classical field equations, because my splitting doesn't relate to them (it's just an algebraic manipulation)! $\endgroup$ Oct 28, 2013 at 22:11
  • $\begingroup$ But wouldn't that mean that instantons just amount to a resumming of the series then? I had thought that they were more fundamental than that. But maybe if I could sum the whole perturbation series from my "usual" perturbation theory I'd be able to see the instanton effects... Or is that logic wrong? If so, why? Thanks again for all your help. Your answer has helped me understand the mechanics of the situation, but not yet the underlying mathematical reasoning why. $\endgroup$ Oct 29, 2013 at 19:36
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    $\begingroup$ Okay. So my question is why does ordinary perturbation theory involve an expansion around the trivial instanton sector? What's wrong with the following argument? The correlation function is given by (a ratio of) quantities of the form $\int exp(iS)$ where $\int$ is a path integral. $S$ is defined to have some form $S_{free}+S_{int}$. So I may rewrite $\exp(iS) = \exp(iS_{free})(1+...)$ and get a perturbation series which tells me exactly the correlation function, if I take all the terms. Note: I assumed nothing about any expansions there. But somewhere I must have done... $\endgroup$ Oct 29, 2013 at 20:59
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    $\begingroup$ It seems now that the heart of your question can be seen as a version of the mathematical fact that not all $C^{\infty}$-functions are analytic. $\endgroup$
    – Qmechanic
    Oct 29, 2013 at 22:39

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