4

I am currently studying functional integration in the context of classical and quantum equilibrium thermodynamics. However one thing puzzles me:

In the book "Phase Transitions and Renormalization Group" by Zinn-Justin he explains several times, for example in 8.9 that corrections to a Gaussian model described by the partition function $$ \mathcal{Z}=\int \Pi_id\sigma_i\;\exp\left(-H_0(\sigma)-b_4 \sum_i\sigma_i^4+\sum_iH_i\sigma_i \right)\tag{p.203} $$ can be either derived by expanding the quartic term like this: $$ \mathcal{Z}=\int \Pi_id\sigma_i\;(\sum \frac{(-1)^l}{l!}(b_4 \sum_i\sigma_i^4)^l)\;\exp\left(-H_0(\sigma)+\sum_iH_i\sigma_i \right)\tag{p.203} $$ which allows you to compute the vertex function at vanishing magnetic field $\Gamma^{(2)}(M=0)$, in other words the heat capacity perturbatively (see Eq.8.75). You can do this for example until $l=1$ which is first order perturbation theory. Alternatively, he shows that the same result can be obtained by the steepest descent method which gives you directly the generating functional $$ \Gamma(M)=\Gamma(M)_{\text{Gaussian}}+\frac{1}{2}\text{tr}\log(\frac{\partial^2H(M)}{\partial M\partial M}).\tag{6.31} $$ The second derivative with respect to $M$ then gives the same result for the heat capacity. I take the message that steepest descent gives the same result as first order perturbation theory.

My question is: Is this always the case? In general, I thought the saddle point approximation is accurate when the pre-factor of the action in the exponent is large, and perturbation theory is the way to go when the parameter in front of the quartic term is small. In this case here, I understand that when the parameter $b_4$ is small, the distribution is close to Gaussian so I see why the saddle point approximation gives the same result as simple perturbation theory. So, here this is the same limit. But this is not a general result right? When are these two approaches a different limit?

Qmechanic
  • 201,751

1 Answers1

1

TL;DR: The concept of a perturbation theory depends by definition on a small$^1$ parameter. More broadly, OP seems to be asking if something universal can be said if a physical model has 2 or more small parameters? The answer is in general No: It is often possible to obtain different perturbative expansions by imposing different hierarchies among the small parameters.

Concerning Ref. 1 (that OP mentions) it is careful to mention the assumptions behind the steepest descent/saddle point method, which requires a large positive parameter. In sections 2.6-2.7 the large parameter is called $1/\lambda$, i.e. $\lambda$ is a small positive parameter. E.g. it is adapted to the large $n$ limit in section 3.1, and the thermodynamic limit $\Omega\to \infty$ in section 6.1. In OP case on p. 203 it is mentioned that $\lambda=b_4\to 0$ yields the Gaussian model. This can be viewed as a steepest descent limit by rescaling the field $\sigma\to \frac{\sigma}{\sqrt{b_4}}$.

Example. In QFT the small parameter in the steepest descent method is often played by Planck's constant $\hbar$. Let's also imagine that the QFT has a coupling constant $g$ that we want to treat perturbatively. The 1PI effective/proper action $\Gamma[\phi_{\rm cl}]$ could contain terms beyond 1-loop, cf. e.g. my Phys.SE answer here.

References:

  1. J. Zinn-Justin, Phase Transitions and Renormalization Group, 2007; p. 203.

--

$^1$ If the parameter is dimensionful, then the words large/small are defined relative to some physical scale.

Qmechanic
  • 201,751
  • Thank you very much for your insightful comment. Just to be clear, in this particular example from Zinn-Justin, the perturbative expansion in the small parameter $b_4$ is in this case the "same" as the saddle point approximation because a rescaling $\sigma \to \sigma /\sqrt{b_4}$ shows that the action becomes $\frac{1}{b_4} S[\sigma]$ which is indeed a large pre-factor for small $b_4$. – user203417 Aug 21 '23 at 07:40
  • However, this argument holds only if the action $S$ becomes independent of the small parameter. So if the action would have the form $S=S_0+\int b_4\cos(\sigma)$ for example, the perturbative expansion and the saddle point approximation generally don't give the same result. – user203417 Aug 21 '23 at 07:41
  • $\uparrow$ I agree. – Qmechanic Aug 21 '23 at 07:50