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:
\begin{equation}
\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
\end{equation}
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
\begin{equation}
\mu\frac{d}{d\mu}\left(\mu^D(g+\delta g)\right)=0
\end{equation}
then we get
\begin{equation}
\beta(g)=-D(g+\delta g)-\mu\frac{d\delta g}{d\mu}
\end{equation}
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
\begin{equation}
\delta g=-Ng^2\mu^D\ln\frac{\Lambda^2}{\mu'^2}
\end{equation}
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{equation}
\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}
\end{equation}
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 $\Lambda-\mu$?
Q&A (4868)
