# Renormalization group and minimum subtraction

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I have several questions about renormalization group and minimum subtraction scheme in particular.

My first question is:

1) Why is the beta function typically just a function of coupling? In other words, why does it explicitly depend only on the coupling constants but not on the relevant energy scale, such as the cutoff or subtraction point? Of course the beta function implicitly depends on the energy scale, but it seems mysterious that it always appears not to explicitly depend on it.

To make this question more concrete, let us suppose we are doing RG for a massless $\phi^4$ theory with a cutoff $\Lambda$, which is described by the Lagrangian:

\mathcal{L}=\frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{g\mu^D}{4!}\phi^4-\frac{\delta g\mu^D}{4!}\phi^4

I assume in the spacetime dimension considered here, the engineering mass dimension of the $\phi^4$ coupling is $D$, and $\mu$ is an arbitrary energy scale introduced to make $g$ dimensionless. The third term above is the counter term. Notice I have not included the field strength renormalization because at one-loop order it does not show up.

Now to calculate the beta function of $g$, we notice that the bare coupling is independent of scale, so

\mu\frac{d}{d\mu}\left(\mu^D(g+\delta g)\right)=0

then we get

\beta(g)=-D(g+\delta g)-\mu\frac{d\delta g}{d\mu}

According to minimum subtraction scheme, I will choose the counter term to cancel the infinite part of the loop integral, which in 3+1 dimensions has the form $Ng^2\mu^{2D}\ln\Lambda^2$ where $N$ is some numerical factor. Then the counter term must have the form

\delta g=-Ng^2\mu^D\ln\frac{\Lambda^2}{\mu'^2}

Here $\mu'$ is another energy scale introduced so that the argument of the log is dimensionless. My second question is

2) Can I choose $\mu'$ to be $\mu$? In fact I think I have the freedom to make this choice, but it seems I also have the freedom to make other choices, so what will happen if I choose $\mu'$ to be very different from $\mu'$? Should I then have two different beta functions, one for $\mu$ and the other for $\mu'$?

In the following I will choose $\mu'=\mu$, then the beta function is

\begin{split}
\beta(g)
&=-Dg+DNg^2\mu^D\ln\frac{\Lambda^2}{\mu^2}+N2g(-Dg)\mu^D\ln\frac{\Lambda^2}{\mu^2}+DNg^2\mu^D\ln\frac{\Lambda^2}{\mu^2}-2Ng^2\mu^D\\
&=-Dg-2Ng^2\mu^D
\end{split}

Notice the remarkable cancellation of the $\Lambda$ dependence, but the explicit dependence on $\mu$ seems to be still there. But notice to get this log divergence, we must be working in 3+1 dimensions where $D=0$, so we see the explicit $\mu$ dependence also drops out. My third question is

3) It seems very non-trivial that the explicit dependence on energy scale drops out, how general is this?

Next I have several questions about minimum subtraction, where we set the counter terms to be the infinite part of the loop integral.

4) One can say that we have introduced an arbitrary scale $\mu$, so to make the physical prediction independent of $\mu$, we must let the couplings run. But what is the physical meaning of the running of coupling coming out from minimum subtraction? In particular, do the beta functions calculated in this way always coincide the ones calculated from other schemes, like the scheme that fixes some renormalization conditions and requries the counter terms preserve those renormalization conditions? Also, do the fixed points calculated in this way have any meaning of universality, by which I mean the physics at low energies does not depend on the microscopic details.

5) In fact I quite doubt that the beta functions calculated from minimum subtraction will be the same as the ones from other methods. An example could be that if the spacetime dimension is low enough that there is no divergence, according to minimum subtraction there is no counter term, so the beta function is completely determined by the engineering dimensions of the coupling constants, which means the beta function is linear in the couplings. But in the Wilsonian picture, it is hard to imagine this because it seems the beta function should still have higher order terms in the couplings coming from the loops although these loops are convergent.

6) How should I do minimum subtraction if I get power divergence instead of log divergence? When I get log divergence I can make the counter term proportional to $\ln\frac{\Lambda^2}{\mu^2}$, if I have linear divergence, should I make the counter term proportional to $\Lambda-\mu$?

edited Mar 22, 2015

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I haven't mastered the subject well, nevertheless let me attempt at an answer here instead of a comment, since it invites quicker correction and criticism.

1) Why is the beta function typically just a function of coupling? In other words, why does it explicitly depend only on the coupling constants but not on the relevant energy scale, such as the cutoff or substraction point?

Strictly speaking it does depend on energy scale and the cutoff, but most of the times in a negligible way. For many QFT's in general(i.e. the ones with no ultraviolet completion), we define the theory with a cutoff(with the understanding that we are not obliged to push the cutoff to infinity, here's a related post.), but also define it in such a way(by renormalization) that all/most of the observable quantities depend on cutoff negligibly, such as $O(\frac{1}{\Lambda})$, so we are not too wrong if we consider beta function to be independent of $\Lambda$. In some low order calculations, $\Lambda$ gets explicitly eliminated by $\frac{d}{d\mu}$, but there seems to be no reason to expect it to happen at arbitrary high orders. For energy scale dependence, we are often interested to apply RG at a energy scale $E$ much larger than mass scale $m$, and by dimensional analysis, beta function can only depend on energy scale via $\frac{m}{E}$, hence also negligible at high energy. (See Weinberg Vol2 Chap 18).

One word of caution, in many contexts a ultraviolet completion(fix point) is assumed, in such case there's no cutoff hence of course no dependence on cutoff.

2) Can I choose μ′ to be μ? In fact I think I have the freedom to make this choice, but it seems I also have the freedom to make other choices, so what will happen if I choose μ′ to be very different from μ′? Should I then have two diffeerent beta functions, one for μ and the other for μ′?

The point of renormalization group is to find the functional relation between $g_{\mu'}$ and $g_\mu$ in a controlled way(by solving the RG equation), that is, to find the explicit form of $g_{\mu'}=G(g_\mu, \mu'/\mu)$, and beta function by definition would be $\mu \frac{\partial G}{\partial \mu'}|_{\mu'=\mu}$, hence beta function by definition depends on only one energy scale. You are free to choose two scales then call it something else. (EDIT:On a second look I see you are not exactly asking about the above perspective, in that you didn't care about using $\mu'$ scale to probe $g_{\mu'}$,  and you just use it as a different subtraction point(not just a different subtraction point, you make $\mu'$ a variable that gives nontrivial result under differentiation, so it seems like a radically different scheme from MS, but my memory of MS is vague....).　Maybe this is the more proper response: Yes you get two different beta functions, which is no surprise since you are using two different renormalization schemes, see my answer to your question 4) below.)

3) It seems very non-trivial that the explicit dependence on energy scale drops out, how general is this?

As in my earlier argument, I don't think it is very general, you can expect the $\mu$ to drop out only if it's much larger than mass scale.

4)...In particular, do the beta functions calculated in this way always coincide the ones calculated from other schemes, like the scheme that fixes some renormalization conditions and requries the counter terms preserve those renormalization conditions?...

No, beta functions from different renormalization schemes don't necessarily coincide. However, if you are doing a perturbative RG, under mild assumptions the two leading terms of your beta function will be the same for different schemes, so that if you want no more than two terms, you can borrow the result from one scheme to another freely. There is an easy proof of this in Weinberg Vol2 Chap 18, page 138.

4)...But what is the physical meaning of the running of coupling coming out from minimum substraction? ... Also, do the fixed points calculated in this way have any meaning of universality, by which I mean the physics at low energies does not depend on the microscopic details.

Not sure if I understand your question, but to define the same theory you need to implement the physical renormalization condition (e.g. pole mass condition) at some point in your math. MS scheme is very handy to get to finite result, but you still have to link it to observables by some physical renormalization condition. So ultimately you really do have the same underlying theory put in different forms, and if in one form you have "physics at low energies does not depend on the microscopic details.", so will you in the other.

5) Should be resolved by my answer to 4)?

6)I have no idea either.

answered Mar 22, 2015 by (2,605 points)
edited Mar 22, 2015

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