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I would like to understand some shortcuts people are using to calculate $\beta$ functions using dim. reg. with mass scale $\mu$ and/or the hard cutoff $\Lambda$. My end goal is to use equation 12.53 in Peskin & Schroeder:

\begin{equation} \beta(g)=M\frac{\partial}{\partial M}\left(-\delta_g+\frac{1}{2}g\sum_i \delta_{Zi}\right)\tag{12.53}, \end{equation}

where $M$ is the renormalization scale, i.e. the scale at which you impose your renormalization conditions (not the mass scale $\mu$ introduced to make up the dimensions in dim. reg.). Ultimately, I would like to understand the existing shortcuts to extend them to 2d.

On the one hand, we have the brute force method. Consider 4d $\lambda\phi^4$ theory. You can use dim. reg. with $d=4-\epsilon$, and impose your renormalization conditions at scale $M$ (i.e. at $s=t=u=-M^2$ for the $\lambda$ vertex, and at $p^2=-M^2$ for the propagator). You end up with this:

\begin{equation} \delta_{\lambda}=\frac{3\lambda^2}{32\pi^2}\left(\frac{2}{\epsilon}+\log\frac{\mu^2}{M^2} +\rm{finite\ stuff\ independent\ of\ M}\right). \end{equation}

So far so good. Then, without justification, I have seen people make the general identification that these terms should always (?) come up as a pair:

\begin{equation} \frac{2}{\epsilon}+\log\frac{\mu^2}{M^2}. \end{equation}

If you can argue that these terms always show up as a pair, you are free to make whatever renormalization conditions you like to derive just the divergent part of your $\delta$. Then, you take your result and replace $\frac{2}{\epsilon}\to\frac{2}{\epsilon}+\log\frac{\mu^2}{M^2}$. This means that you can choose potentially more convenient renormalization conditions and toss out any complicated finite stuff that you otherwise would have had to deal with to derive the $M$ dependence explicitly.

This is the first shortcut I have seen. For example here in 12.1 and 12.2 it works for Yukawa and Gross-Neveu (so apparently it works in 2d too!). My questions:

(1) Do these terms $\frac{2}{\epsilon}+\log\frac{\mu^2}{M^2}$ always come up as a pair? If so, how can you show that in general? If not, in what conditions (type of theory, dimension, etc.) do they appear as a pair?

Now, there is a second shortcut I have seen in 4d pseudoscalar Yukawa theory. You first - using dim. reg. - calculate the divergent part of your $\delta$ using any renormalization conditions you like. Then, you make an equivalence between the divergent terms in dim. reg. and the divergent terms in the hard cutoff method to rewrite your counterterms with $\Lambda$ dependence instead of $\epsilon$ dependence. So, in the example just cited, it goes like this:

Using dim. reg. and without any reference to a scale $M$, you find e.g.:

\begin{equation} \delta_{\lambda}=\left(\frac{3\lambda^2}{32\pi^2}-\frac{3g^4}{2\pi^2}\right)\frac{2}{\epsilon}. \end{equation}

Then, you match the divergences and rewrite in terms of hard cutoff variables:

\begin{equation} \left(\frac{3\lambda^2}{32\pi^2}-\frac{3g^4}{2\pi^2}\right)\frac{2}{\epsilon}\to \left(\frac{3\lambda^2}{32\pi^2}-\frac{3g^4}{2\pi^2}\right)\log\frac{\Lambda^2}{M^2}. \end{equation}

This is in agreement with the relationship in (for example) equation 4 here (note their $\epsilon$ is defined differently). So, for this second shortcut, I would like to know:

(2) In what conditions (type of theory, dimension, etc.) is the matching $\frac{2}{\epsilon}\to\log\frac{\Lambda^2}{M^2}$ legitimate? What is the generalization in 2d?

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  • $\begingroup$ Have you tried to just calculate the dimesionally regulated integrals by hand? You'll see that (1) is simply the result of a series expansion of the term $\left(\frac{4 \pi \mu^2}{M^2}\right)^{\epsilon}$ multiplied by some combination of gamma functions. (Here I'm using the other commonly used convention of $d=4-2\epsilon$ which only really affects the coefficient of the $\epsilon$ pole piece) $\endgroup$
    – Triatticus
    May 20, 2022 at 22:20
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    $\begingroup$ $\mu$ is already the renormalization scale in dimensional regularization. You don't need a separate $M$. In dimensional regularization people usually introduce counterterms which are set to ensure diagrams have no explicit $\epsilon$ dependence (there are various schemes to do this). The inclusion of the counterterms is already defining what your renormalized couplings are. $\endgroup$
    – octonion
    May 20, 2022 at 22:31
  • $\begingroup$ @octonion I am confused about $\mu$ being the renorm scale. I agree that counterterms absorb $\epsilon$ dependence, but don't we need a(nother) scale at which to do this? E.g. eqs. 10.19 & 10.21 in P&S. They write counterterms to absorb $\epsilon$ dependence, but had to choose where to define $\lambda$ to do so. (They don't explicitly include $\mu$'s here.) In this case they picked zero external 3-momenta to impose their renorm conditions, but later they note that this is problematic for massless particles and seem to imply that introducing $M$ is more correct. Can you comment on this? $\endgroup$
    – mkn
    May 26, 2022 at 18:05
  • $\begingroup$ @Triatticus I did manually verify this for $\lambda\phi^4$ theory, and agree this is how the relationship (1) between $\epsilon$ and $M$ comes in in that case, once you include the $\Gamma (\epsilon /2)$. I am hoping that there is some general argument that the $M$ dependence always comes in this way, no matter the dimension or particular diagram you are trying to work out. Perhaps the best approach is to avoid such shortcuts and explicitly work out the $M$ dependence. (Though I am not sure what to make of this approach in light of octonion's comments.) $\endgroup$
    – mkn
    May 26, 2022 at 18:17
  • $\begingroup$ @Maria, What I'm talking about is a different kind of scheme that is introduced later in P&S around (11.77). The $M$ that they have there is what you are calling $\mu$. You could apply this to a Lagrangian like (10.18) if in (10.24) you only shifted $6/\epsilon$ or perhaps $6/\epsilon - 3\gamma +3\log(4\pi)$ into the counterterm instead of the $x$ dependent part. Similarly for the two-point function discussed later. $\endgroup$
    – octonion
    May 26, 2022 at 21:10

1 Answer 1

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Here's a rough argument for why the latter part of your question going from a pole to beta function is valid. (As I mentioned in the comments, the use of an additional scale $M$ in addition to $\mu$ is unnecessary unless you specifically want to relate the dim reg renormalized couplings to some other scheme.)

For simplicity let's consider a theory with one dimensionless coupling in the dimension of interest. When we dimensionally regularize the bare coupling $\lambda_B$ will pick up some dimension parametrized by $\mu$, and it will also involve an $\epsilon$ dependent counterterm factor. $$\lambda_B = \lambda \mu^{\epsilon} Z(\lambda,\epsilon).$$

$Z$ takes the schematic form $$Z(\lambda,\epsilon) = 1 + \sum_{k=1}\frac{C_k(\lambda)}{\epsilon^k}.$$

Now the idea is that $\lambda$ is implicitly a function of $\mu$ so that the original $\lambda_B$ is the same at any scale, so the total $\mu$ derivative of $\lambda_B$ vanishes, $$\mu\frac{d}{d\mu}\lambda_B = \mu \frac{\partial}{\partial\mu}\lambda_B + \hat\beta_\lambda \frac{\partial}{\partial\lambda}\lambda_B=0,$$ where $\hat\beta_\lambda \equiv \mu \frac{d \lambda}{d \mu}.$

As will become clear in a moment, $\hat\beta_\lambda$ will involve a term $\lambda \epsilon$ and it is convenient to define the part of the beta function $\beta_\lambda$ apart from that piece, $$\hat\beta_\lambda \equiv -\epsilon \lambda + \beta_\lambda.$$

Then the equation enforcing RG invariance of $\lambda_B$ may be rewritten as $$ \epsilon \lambda^2 \frac{\partial Z_\lambda}{\partial \lambda}= \beta_\lambda\left(Z_\lambda + \lambda\frac{\partial Z_\lambda}{\partial \lambda}\right)$$ This equation should be true at each order in $\epsilon.$ In particular the zeroth order equation is just, $$\beta_\lambda = \lambda^2 C_1'.$$ So the beta function is completely specified by the lowest $1/\epsilon$ pole. All the other orders of $\epsilon$ lead to various non-trivial relations between the coefficients $C_k$ which nevertheless must hold if the theory is renormalizable. You can read more about this in Anselmi's textbook Renormalization.

The reason why people sometimes just trade $1/\epsilon$ for something like $\log(M/\Lambda)$ is because the beta function is typically very robust at one loop and does not depend on the choice of scheme. At higher loop order you will run into trouble if you try to make naive substitutions like this.

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