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I've been playing around with scalar theories as a personal exercise in getting beta functions by calculating the effective potential. Hollowood's Renormalization Group and Fixed Points in Quantum Field Theory does this exact sort of calculation for the $\phi^4$ theory in $d=4$ (Sec. 3.1).

I then decided to try to do the same calculation in $d=3$. The effective potential is given by $$V_{\text{eff}}(\phi) = V(\phi) + \frac{\mu^{3-d}}{2 (2\pi)^d} \int \log(p^2 + V''(\phi)) \mathrm{d}^d{p},$$ where $\mu$ is some renormalization scale we are using to keep units consistent. $V(\phi)$ is the classical potential. This is the usual expression for the effective potential at one-loop in Euclidean signature.

Computing the integral in $d$ dimensions, we find $$\frac{1}{(2\pi)^d} \int \log(p^2 + V''(\phi)) \mathrm{d}^d{p} = - \frac{V''(\phi)^{\frac{d}{2}}}{(4\pi)^{\frac{d}{2}}} \Gamma\left(-\frac{d}{2}\right).$$ Hence, $$V_{\text{eff}}(\phi) = V(\phi) - \frac{\mu^{3-d} V''(\phi)^{\frac{d}{2}}}{2 (4\pi)^{\frac{d}{2}}} \Gamma\left(-\frac{d}{2}\right).$$

Here comes the interesting bit. If we were to take the limit $d \to 4$ (and changed $\mu^{3-d}$ to $\mu^{4-d}$ to keep units right), this expression would diverge and we would need to regularize it by adding counterterms. Nevertheless, it converges if we take $d \to 3$, since $\Gamma\left(-\frac{3}{2}\right) = \frac{4 \sqrt{\pi}}{3} < \infty$. Hence, it appears the one-loop effective potential is finite in $d=3$.

I find this particularly counterintuitive. I never specified the details of the potential $V(\phi)$, and hence I could now pick terms $\phi^{2n}$ with $n$ as large as I like, and still, it seems like they won't yield any divergences. I would think their divergences could be simply power-like and not appear on dimensional regularization, but $\phi^6$ has a dimensionless coupling, and hence I would expect it to lead to a logarithmic divergence, meaning it should make the one-loop effective potential be divergent in some way.

Why is this happening? In a more straightforward manner, why is the one-loop effective potential for a scalar theory in $d=3$ finite in dimensional regularization? Furthermore, how can there be a renormalization group flow in the absence of divergences?


Integral Calculation

I'll sketch here the computation I'm doing for the integral. It can also be found in Peskin & Schroeder (p. 374, Eq. (11.72)), but they are working in Lorentzian signature, while I'm already in Euclidean.

First we integrate over the angular variables to get to $$\frac{1}{(2\pi)^d} \int \log(p^2 + V''(\phi)) \mathrm{d}^d{p} = \frac{2}{(4\pi)^{\frac{d}{2}} \Gamma\left(\frac{d}{2}\right)} \int_0^{+\infty} \log(p^2 + V''(\phi)) p^{d-1} \mathrm{d}{p}.$$ Now we use the expression $$\log(p^2 + V''(\phi)) = - \left.\frac{\partial}{\partial \alpha} \frac{1}{(p^2 + V''(\phi))^\alpha}\right|_{\alpha=0}.$$ This leads us to \begin{align*} \frac{1}{(2\pi)^d} \int \log(p^2 + V''(\phi)) \mathrm{d}^d{p} &= - \frac{2}{(4\pi)^{\frac{d}{2}} \Gamma\left(\frac{d}{2}\right)} \left.\frac{\partial}{\partial \alpha}\int_0^{+\infty} \frac{p^{d-1}}{(p^2 + V''(\phi))^\alpha} \mathrm{d}{p}\right|_{\alpha=0}, \\ &= - \frac{2}{(4\pi)^{\frac{d}{2}} \Gamma\left(\frac{d}{2}\right)} \frac{\partial}{\partial \alpha}\left[\frac{V''(\phi)^{\frac{d}{2} - \alpha} \Gamma\left(\frac{d}{2}\right) \Gamma\left(-\frac{d}{2} + \alpha\right)}{2 \Gamma\left(\alpha\right)}\right]_{\alpha=0}, \\ &= - \frac{V''(\phi)^{\frac{d}{2}}}{(4\pi)^{\frac{d}{2}}} \Gamma\left(-\frac{d}{2}\right). \end{align*}

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  • $\begingroup$ Mathematica does put some convergence conditions on the step where the integral is actually computed (with some beta function identity), but curiously the final result seems finite regardless $\endgroup$ Jan 16, 2023 at 22:01
  • $\begingroup$ all that $\alpha$ chicanery is rather meaningless. You cannot swap $\int$ and $\partial/\partial\alpha$ since the integral doesn't converge in the first place. such formal manipulations are not to be trusted. many such cases! $\endgroup$ Jan 16, 2023 at 22:09
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    $\begingroup$ you can cut the integral off at some high momentum, the integral is meaningful and you get a divergent piece for any $d$, e.g., $\Lambda^3\log\Lambda$ for $d=3$ and $\Lambda^4\log\Lambda$ for $d=4$ (plus other, subleading divergent pieces). $\endgroup$ Jan 16, 2023 at 22:18
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    $\begingroup$ Not exactly the same model, but possibly relevant... in arxiv.org/abs/1004.0982, the authors claim that they only generate a mass term (ie, violate conformal invariance, ie, have non-zero log divergences) at two loops, for their three dimensional susy model. $\endgroup$
    – Andrew
    Jan 16, 2023 at 22:19
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    $\begingroup$ I think it's possible there is an accidental cancellation at one loop, and the leading order RG flow happens in two loops in this theory. There's no reason the RG flow has to start at one loop. $\endgroup$
    – Andrew
    Jan 17, 2023 at 2:16

1 Answer 1

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Any way I look at it yes, apparently the dimreg continuation of that integral only has poles at even dimensions. I'm not sure if the kinetic part of the effective action has divergences or not but that would be a little tiresome to check. If you want to do it anyway, the object to look at is

$$\text{tr}\log(1+(-\partial^2+M^2)^{-1}(V''(\phi)-M^2))=-\sum_{n=1}^\infty\frac{(-1)^n}{n}\int\prod_{\alpha=1}^n\frac{d^dk_\alpha d^dx_\alpha}{(2\pi)^d}\frac{V''(\phi(x_\alpha))-M^2}{k_\alpha^2+M^2}e^{ik_\alpha(x_{\alpha-1}-x_\alpha)},$$

with $x_0=x_n$ for some arbitrary and convenient $M^2$, up to second order partial derivatives of $\phi$. This object is what remains of $\text{tr}\log(-\partial^2+V''(\phi))$ after you take out a $\phi$ independent $\text{tr}\log(-\partial^2+M^2)$. If the kinetic term diverges then due to the renormalization of the field itself you're going to get a flow.

Anyway, the point I want to make is about the culmination of your question: how can non-divergent loop corrections give rise to RG flow? Well, remember that RG depends on not only your theory, but the renormalization scheme. If you're working with MS or $\bar{\text{MS}}$ there's surely no flow without divergence, but if you're doing on-shell or some other thing, there probably is.

Why? Because renormalization is not simply a recipe to do away with divergences. The question it aims to solve is basically "if I systematically ignore some degrees of freedom, is there a change of variables I can do to the remaining ones to essentially get back to the same theory? If so, how exactly does it differ from what I started with?"

It's not intrinsically tied to divergences. The way I particularly like to illustrate this is the following. Take a theory of a field $\phi$ with action $S[\phi;\lambda]$ for some parameters $\lambda$. Separate $\phi=(\phi_\text{care},\phi_\text{don't})$ as independent degrees of freedom. The path integral is then

$$\int D\phi\,e^{-S[\phi]}=\int D\phi_\text{care}D\phi_\text{don't}\,e^{-S[\phi_\text{care},\phi_\text{don't};\lambda]}=\int D\phi_\text{care}e^{-S_\text{eff}[\phi_\text{care};\lambda]},$$

where

$$S_\text{eff}[\phi_\text{care};\lambda]=-\log\int D\phi_\text{don't}\,e^{-S[\phi_\text{care},\phi_\text{don't};\lambda]}.$$

The magic of renormalization then happens if there is a change of variables $\phi_\text{care}\mapsto\tilde\phi$ such that

$$S_\text{eff}[\phi_\text{care};\lambda]=S[\tilde\phi;\tilde\lambda]$$

for some new set of parameters $\tilde\lambda$. You have just mapped your coarse grained theory to your original one, and the RG flow is the map $\lambda\mapsto\tilde\lambda$. This is completely independent of the fact that the parameters involved may or may not be infinite.

Translating to traditional particle physics QFT, the step $\phi=(\phi_\text{care},\phi_\text{don't})$ corresponds to choosing a particular regularization (Wilson cutoffs, dimreg), while $\phi_\text{care}\mapsto\tilde\phi$ corresponds to your renormalization conditions (MS, on shell). Try doing renormalization on you divergence free theory with on-shell or zero-energy conditions, see if you get any flow.

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