Problem with RG beta function in toy model In the article "A Hint of Renormalization" (https://arxiv.org/abs/hep-th/0212049), the author considers a toy model in which a physical quantity (say, a scattering amplitude) $F$, depending on some physical parameter $x$, is perturbatively calculated in a coupling constant $g_0$, giving:
$$F(x)=g_0+g_0^2F_1(x)+g_0^3F_2(x)+\cdots.$$
The author then for concreteness chooses:
$$F_1(x)=\alpha\int_0^\infty \frac{dt}{t+x},$$
and describes how if we have the physical measurement $F(\mu)=g_R$ taken at $x=\mu$, then by introducing a cutoff and eliminating $g_0$ in favor of $g_R$ we may obtain a finite expansion for $F(x)$.
Here I stopped reading and I tried to compute what I think is the beta function, following exercise III.1.3 of Zee. If I understood the discussion in Zee (page 149) correctly, when we introduce a cutoff the couplings are taken to depend on the cutoff:
$$F_\Lambda(x)=g(\Lambda)+g(\Lambda)^2F_{1,\Lambda}(x)+\cdots,$$
where the $g(\Lambda)$ dependence is chosen so that $F_\Lambda$ doesn't depend on $\Lambda$. For instance, in Zee, we have:
$$\mathcal{M}=-i\lambda(\Lambda)+iC\lambda(\Lambda)^2\left[\log\left(\frac{\Lambda^2}{s}\right)+\log\left(\frac{\Lambda^2}{t}\right)+\log\left(\frac{\Lambda^2}{u}\right)\right]+\mathcal{O}(\lambda^3),$$
so to solve exercise III.1.3 we simply demand that $M$ doesn't depend on $\Lambda$:
$$\frac{d\mathcal{M}}{d\log\Lambda}=0,$$
and this seems to work, giving:
$$\frac{d\lambda}{d\log\Lambda}=6C\lambda(\Lambda)^2+\mathcal{O}(\lambda^3).$$
As expected, $\Lambda$ only enters this expression as an argument of $\lambda$. If we do something similar for the "Hint of Renormalization" paper, I obtain instead:
$$\frac{dg(\Lambda)}{d\log\Lambda}=-\frac{\alpha\Lambda g^2}{x+\Lambda}+\mathcal{O}(g^3),$$
which is clearly not good (not only does it depend on $\Lambda$ explicitly; it depends on $x$!)
My first question would be what is going on here? What is the right way to get a beta-function-type equation for the flow of $g$?
The more general question is: Within the framework of perturbative renormalization like in the "Hint" paper and in Zee, what is the right way to think of running couplings? If I understand Zee correctly, there the idea is "Introduce a cutoff to the divergent integrals; then the coupling constants depend on the cutoff such that scattering amplitudes don't change," while briefly skimming the later part of "Hint" seems to suggest that there the idea is "replace $g_0$ in favor of $g_R$; then the coupling constants depend on $\mu$ [which is analogous to $s$,$t$,$u$ in Zee] such that the scattering amplitudes don't change." It is unclear to me whether or not these are the same idea.
I apologize if the question is muddled but I don't think I know enough yet to ask a well-posed question.
 A: The source of confusion that prompted this question was that people seem to mean lots of things when they talk about things "flowing."


*

*Amplitudes for various processes depending on the momenta of the involved particles

*Coupling constants for a theory changing when we change the scale we probe the system at

*Coupling constants varying with the cutoff of a cutoff-dependent theory so that predictions of the theory are cutoff-independent


Obviously these are all related but the details were unclear to me. I am going to attempt to answer my question by providing a broad overview of my understanding of how the various aspects of RG theory fit together, and narrowing in on the issues raised in the question at the very end (so it is very plausible to just skip to the section titled "Answers".)
Renormalization
Physical theories make predictions. Suppose we have a model with parameters $\{h_0\}$, and we query the model for predictions $\text{Pred}(\{h_0\},\vec{x})$, where here $\vec{x}$ are some length-dimesion dynamical variables that describe the details of the setup. Examples:


*

*Spin lattices: $h$'s characterize interactions between spins, $\vec{x}$ describes positions on a lattice of spins, and $\text{Pred}$ is a correlation function of spins living on those sites.

*Scattering: $h$'s are coupling constants, $\vec{x}$ is the wavelengths of the incident particles, and $\text{Pred}$ is a scattering amplitude.


We are interested in low-energy physics; i.e., $\text{Pred}(\{h_0\},\lambda\vec{x})$ as $\lambda\to\infty$. The clever idea of renormalization group is to shift the $\lambda$-dependence onto the $\{h\}$'s. This is encapsulated in Zinn-Justin eqn. (9.9), which states:
$$\text{Pred}(\{h(\lambda)\},\vec{x})\sim\frac{1}{Z(\lambda)}\text{Pred}(\{h_0\},\lambda\vec{x}),\tag{*}$$
where $\sim$ means "up to things which vanish really fast as $\lambda\to\infty$," which I am going to treat as equality from here out. It seems like it should be possible to set something like this up without the $Z(\lambda)$, but it will prove convenient for us to include it.
The basic strategy is to find flows $\{h(\lambda)\}$ in the "theoryspace" of all possibilities for $\{h\}$ such that (*) is true for any prediction we might care to extract. If we can find such a flow, analyzing the fixed points gives insight into the way the predictions of the $h_0$ theory behave in the $\lambda\to\infty$ limit. Let's try to find an example.
Engineering Dimension
For concreteness, let's say Pred is a correlation function:
$$\text{Pred}(h_0,\vec{x})=\langle\phi(x_1)\phi(x_2)\dots\phi(x_n)\rangle=\langle \mathcal{O}[\phi]\rangle_{h_0}.$$
Because I am scared of active vs. passive and primed vs. unprimed confusion, let's define the squeezing operator:
$$\left(D_\lambda\left[\phi\right]\right)(x):=\frac{1}{\sqrt{Z(\lambda)}}\phi(\lambda x),$$
which allows us to write (*) as:
$$\langle \mathcal{O}[\phi]\rangle_{h(\lambda)}=\langle\mathcal{O}[D_\lambda[\phi]]\rangle_{h_0}.$$

Our goal is to determine exactly how $h$ must depend on $\lambda$ in order to make this be true for whatever $\mathcal{O}$ we might want to study.
Using the path integral formulation, we can write:
$$\begin{split}
\langle\mathcal{O}[\phi]\rangle_{h(\lambda)}&=\frac{\int \mathcal{D}\phi\, \exp\left(iS[h(\lambda);\phi]\right) \mathcal{O}[\phi]}{\int \mathcal{D}\phi\, \exp\left(iS[h(\lambda);\phi]\right) }\\
\langle \mathcal{O}[D_\lambda[\phi]]\rangle_{h_0}&=\frac{\int \mathcal{D}\phi \,\exp\left(iS[h_0; \phi]\right) \mathcal{O}[D_\lambda[\phi]]}{\int \mathcal{D}\phi \,\exp\left(iS[h_0; \phi]\right)}
\end{split}$$
How can we make these look similar to each other? Well, note that integrating $\mathcal{D}\phi$ is the same as integrating $\mathcal{D}\left\{D_\lambda[\phi]\right\}$, up to, perhaps, an overall multiplicative constant that will cancel between the numerator and the denominator. This is because $D_\lambda$ is a bijection of field configurations. If we do that replacement in the lower correlation function, we obtain:
$$\langle \mathcal{O}[D_\lambda[\phi]]\rangle_{h_0}=\frac{\int \mathcal{D}\left\{D_\lambda[\phi]\right\} \,\exp\left(iS[h_0; \phi]\right) \mathcal{O}[D_\lambda[\phi]]}{\int \mathcal{D}\left\{D_\lambda[\phi]\right\} \,\exp\left(iS[h_0; \phi]\right)}$$
We can rewrite the expression for the upper correlation function by renaming the integration variable: replace $\phi$ with $D_\lambda[\phi]$ everywhere.
$$\langle\mathcal{O}[\phi]\rangle_{h(\lambda)}=\frac{\int\mathcal{D}\left\{D_\lambda[\phi]\right\}\,\exp\left(iS[{h(\lambda)};D_\lambda[\phi]]\right)\,\mathcal{O}[D_\lambda[\phi]]}{\int\mathcal{D}\left\{D_\lambda[\phi]\right\}\,\exp\left(iS[{h(\lambda)};D_\lambda[\phi]]\right)}$$
These expessions would be equal for any $\mathcal{O}$ if we chose $h$ to depend on $\lambda$ so that:
$$S[\{h_0\};\phi]=S[\{h(\lambda)\};D_\lambda[\phi]]\tag{**}$$
holds. As an example, consider $\phi^4$ theory in $D$ dimensions:
$$S[\phi]=\int d^D x\,\left[\frac{1}{2}((\nabla\phi)(x))^2+\frac{1}{2}m^2\phi(x)^2+\frac{u}{4!}\phi(x)^4\right].$$
We must have $Z(\lambda)=\lambda^{2-D}$, $m(\lambda)=m_0\lambda$, and $u(\lambda)=u_0\lambda^{4-D}$ in order to make (**) be true. Generically, the couplings will be multiplied by $\lambda$ to some power, which is the engineering scaling dimension of the coupling.
Cutoff Dependence
Sometimes, because of UV divergences in the mathematical expressions for $\text{Pred}$, a theory doesn't make sense without a cutoff. Let $\Lambda$ be a momentum-dimension cutoff, and suppose we only consider field configurations with modes less than $\Lambda$. If we try to repeat the engineering-dimension argument, we'll run into a problem because $D$ is no longer injective.

Consider two field configurations $\phi_1$ and $\phi_2$ that differ only in very high-frequency behavior, in modes just below the cutoff. Then $D_\lambda[\phi_1]$ and $D_\lambda[\phi_2]$ differ only in even-higher-frequency behavior, above the cutoff, and are hence indistinguishable as far as our theory is concerned. For lattice models this situation is even more obvious: $D$ corresponds to a Kadanoff-esque blocking of lattice sites, which clearly destroys information. Thus, the only way the previous argument will go through is if we combine our change $\mathcal{D}\phi\to\mathcal{D}\{D_\lambda[\phi]\}$ with raising the cutoff, so that $D$ becomes injective again.
Let's go through the argument again, keeping track of the momentum cutoff $\Lambda$. 
$$\begin{split}
\langle \mathcal{O}[D_\lambda[\phi]]\rangle_{h_0}&\sim \int^\Lambda \mathcal{D}\phi\,\exp\left(iS[h_0;\phi]\right)\,\mathcal{O}[D_\lambda[\phi]]\\
&=\int^{\lambda\Lambda} D\left\{D_\lambda[\phi]\right\}\exp\left(iS[D_\lambda[\phi],\{h(\lambda)\}]\right)\,\mathcal{O}[D_\lambda[\phi]]\\
&=\int^{\lambda\Lambda} D\phi\,\exp\left(iS[\phi,\{h(\lambda)\}]\right)\,\mathcal{O}[\phi]
\end{split}\tag{***}$$
where here, as before, the flow of $h(\lambda)$ is the engineering-dimension flow determined by (**). This is almost $\langle O[\phi]\rangle_{h(\lambda)}$, except the cutoff is wrong. What to do?
Anomalous Dimension
There is a deeper problem with introducing a cutoff to the theory. The cutoff is really just another parameter the theory depends on, so we should add it to the list of $\{h\}$'s. But this list of parameters (with $\Lambda$ appended) no longer uniquely specifies theories:

The idea is that shifting $\Lambda$ may be compensated by shifting the $h$'s so that the theory gives the same predictions. The theories on each vertical curve make exactly the same predictions, and so they are all really the same theory. To label them without redundancy, we fix some $\Lambda_0$, and characterize each theory-curve by where it intersects the $\Lambda=\Lambda_0$ plane, calling those values $h_0$. Now, $h$ may be thought of as varying with both $\lambda$ and $\Lambda$, with the flow along the theory-curves determined by demanding the predictions don't change:
$$\text{Pred}(h_0;\Lambda_0,\vec{x})=Z(\lambda)\text{Pred}(h(\lambda,\Lambda),\Lambda;\vec{x}/\lambda)$$
The idea now is to interpret $\text{(***)}$ as describing the flow of $h$ along a path through $(\lambda,\Lambda)$-space. We can express the results of $\text{(***)}$ by saying:
$$\text{Pred}(h_0,\Lambda_0;\vec{x})=Z(t)\text{Pred}(h(t),t\Lambda_0;\vec{x}/t),$$
where $h(t)$ is engineering-dimension flow. Engineering-dimension flow therefore corresponds to moving along the path $(\lambda(t),\Lambda(t))=(t,\Lambda_0 t)$.

Alright, so let's put everything together.
$$\begin{split}
\frac{d}{d\log t} h(t,t\Lambda_0)&=\text{[engineering dimension of h] } h(t,t\Lambda_0)\\
\frac{\partial}{\partial\log\Lambda} h(\lambda,\Lambda)&=-\text{[anomalous dimension of h] } h(\lambda,\Lambda)\\
\frac{d}{d \log\lambda}h(\lambda,\Lambda_0)&=\left(\text{[engineering dimension of h]} +\text{[anomalous dimension of h]} \right) h(\lambda,\Lambda_0)
\end{split}$$
Field-theoretic renormalization
Rather than characterize the theory-curves by $h_0$, we could, instead, choose for each $h_i$ a particular prediction/Green's function $\text{Pred}_i$ (which would also be invariant along the theory-curves). These are the "renormalized coupling constants" and their choice is the "renormalization prescription." These will of course also flow with $\lambda$. In principle, in our expressions for various predictions, we can replace the $h_i$'s with expressions involving the $\text{Pred}_i$'s, though whether or not a finite number of $\text{Pred}_i$'s suffices is the issue of renormalizability. 
It is still unobvious to me why we expect the $\lambda$-flow of the $\text{Pred}_i$'s to be Markovian, or to resemble the $\lambda$-flow of the $h_i$'s at all. In fact, I don't really know why the $h_i$ flow should be Markovian.
Answers
Thus, the answer to my first question is: The right way to get a beta function in the context of momentum-shell renormalization is to determine the flow of the couplings with rescaling $\Lambda$ and to add that dependence to the flow induced by the engineering scaling dimensions of the couplings. There is no a-priori reason to expect that for an arbitrary toy model the flow will be Markovian. As for the Zee exercise, because the engineering dimension of the coupling in 4-dimensional $\phi^4$ theory is zero, the flow of that coupling with increasing $\log\lambda$ is precisely the same as the flow of the coupling with decreasing $\log\Lambda$, which is what the exercise wants us to compute.
The answer to my second question is: Those two procedures are not really the same idea. The first almost describes momentum-shell renormalization, neglecting the tree-level contribution. The second describes field-theoretic renormalization, which isn't guaranteed to produce the same beta-functions because several different renormalization prescriptions are available.
Bibliography


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*Phase Transitions and Renormalization Group by Jean Zinn-Justin [chapter 9]

*Quantum Phase Transitions by Subir Sachdev [chapter 4 for momentum-shell RG, section 12.3.1 for field-theoretic RG]

*The Theory of Critical Phenomena by J. J. Binney et al [chapter 9]

*Quantum Field Theory in a Nutshell by A. Zee [section III.1]

