I am a qft noob studying from Quantum Field Theory: An Integrated Approach by Fradkin, and in section 13 it discusses the one loop corrections to the effective potential $$U_1[\Phi] = \sum^\infty_{N=1}\frac{1}{N!}\Phi^N\Gamma^{N}_1(0,…,0)$$
And how the first $D/2$ terms are divergent where $D$ is the dimensionality.
The book then discusses that the solution to this is to define the renormalized mass $\mu^2 = \Gamma^{(2)}(0)$ and renormalized coupling constant $g = \Gamma^{(4)}(0)$, and then expressions relating the bare mass to the renormalized mass, and expressions relating the bare coupling constant to the renormalized constant are found, where the integrals now only run to a UV cutoff $\Lambda$. The effective potential is then written in terms of the renormalized mass and renormalized coupling constant, and the result is magically finite.
Somewhere in this process I am a bit lost. First, I am not really intuitively seeing the motivation for defining the renormalized mass and coupling constant as the two- and four-point vertex function at zero external momenta. What is the motivation behind this? Second I feel I am a bit lost about how the resulting effective potential after all of this becomes finite. I suppose I can see mathematically that the result is finite, but I do not at all understand it. At what point in our scheme does the finiteness really come out of? What is the point in all of this?
Best Answer
Here's the thing: renormalization and divergences have nothing to do with each other. They are conceptually unrelated notions.
Renormalization
Simply put, renormalization is a consequence of non-linearities. Any theory (other than those that are linear) requires renormalization. Even in classical Newtonian mechanics.
Renormalization means: when you change an interaction, the parameters of your theory change. The simplest example is the classical (an)harmonic oscillator. Say you begin with $L=\frac12m\dot q^2-\frac12kq^2$. If you prepare this system in a laboratory, you will observe that $q$ oscillates with frequency $\omega^2=k/m$. Now, say you add the (nonlinear) interaction $\gamma q^4$ to your Lagrangian. The frequency that you will measure in a laboratory is no longer $\omega^2=k/m$, but rather $\omega^2\sim k/m+\gamma$.
This trivial phenomenon also occurs in quantum mechanics, in particular QFT. You typically have a set of measured parameters, such as $\omega$ above, and a set of coefficients in your Lagrangian, such as $k,m$. The latter are not directly observable. Solving your theory gives you a function $\omega=f(k,m,\dots)$ for some $f$. You can use the measured value of $\omega$, and the calculable function $f$, to fix the value of your Lagrangian parameters $m,k,\dots$. If you change your Lagrangian by adding a new term, the function $f$ will change, and therefore the value of your parameters $m,k,\dots$ will change. Of course, $\omega$ stays the same, as this is something you measure in a laboratory (and does not care about which Lagrangian you are using to model the dynamics).
For example, say you have a QFT that describes a particle. The system will have several parameters $c_1,c_2,\dots$, which multiply the different terms in your Lagrangian, say $L=(\partial\phi)^2-c_1\phi^2-c_2\phi^4+\cdots$. This Lagrangian predicts that there is a particle with some mass $m=f(c_1,c_2,\dots)$, where $f$ is a function that you obtain by solving the theory (say, using Feynman diagrams). The value of $m$ is fixed by experiments, you can measure it in a lab. From this measured value, and the known form of the function $f$, you can fix the value of your model-dependent parameters $c_i$. For example, if you begin with the model $L=(\partial\phi)^2-c_1\phi^2$, then the function $f$ takes the form $m=c_1$, and therefore the value of $c_1$ is identical to what you measure $m$ to be. If you take, instead, $L=(\partial\phi)^2-c_1\phi^2-c_2\phi^4$, then you now have $m=c_1+\frac{3}{16\pi}c_2+\mathcal O(c_2^2)$. Using this (and some other measured observable, such as a cross section), you can fix the value of $c_1,c_2$. Note that $c_1$ will not in general take the same value as before, namely $c_1\neq m$. This is what we mean by "interactions renormalize your couplings". We just mean that, by adding interactions to your model, the numerical value of your coupling constants will change.
This is all true even if your theory is finite. Theories without divergences still require renormalization, i.e., the value of the coupling constants are to be determined by comparing to observable predictions, and changing interactions changes the numerical value of your coupling constants. (The exception is, of course, linear theories and some other special cases such as those involving integrability).
Renormalizing a theory typically means: fix the value of your model-dependent parameters $\{c_i\}$ as a certain function of the measurable parameters, such as $m$ and other observable properties of your system. The value of $m$ is fixed by nature, and we have no control over it. The value of $\{c_i\}$ depends on which specific Lagrangian we are using to model the physical phenomenon, and it changes if we change the model.
Divergences
In QFT divergences are the result of mishandling distributions. In $d>1$, quantum fields are not functions of spacetime, but rather distributions. As such, you cannot manipulate them as if they were regular functions, e.g. you cannot in general multiply distributions. When you do, you get divergences.
This is manifested in the fact that the function $f$ from before typically has divergent terms inside. The cute thing is: for a large class of theories, if you write $c_i=\tilde c_i+\delta_i$, with $\delta_i$ a (specifically constructed) divergent contribution, and $\tilde c_i$ a finite part, then the relation $m=f(c_i)=\tilde f(\tilde c_i)$ is rendered finite, i.e., the function $\tilde f$ no longer has divergent terms inside.
But note that renormalization did not actually cure the divergences. The divergences were eliminated by writing $c_i=\tilde c_i+\delta_i$ and carefully compensating the divergences with another divergent part, $\delta_i$. This is not what renormalization is. Renormalization is the statement that, if you were to change the model, the numerical value of the constants $c_i$ (and, by extension, that of $\tilde c_i$), change accordingly.
For a general theory, you need to perform both steps: 1) cancel the divergences by splitting your model-dependent parameters as $c_i=\tilde c_i+\delta_i$, with finite $\tilde c_i$ and a suitable divergent $\delta_i$, and 2) renormalize your theory, i.e., fix the value of your model-dependent parameters $\tilde c_i$ as a function of observable parameters such as $m$.
n-point functions
What measurable parameters can we use beyond $m$? in general there will be multiple parameters $c_i$, so you need as many observables in order to fix the former as a function of the latter. For a completely general theory, it is hard to come up with a concrete list of observable parameters. For specific systems this is easy, e.g. for a thermodynamic material we could use the susceptibility at criticality, while for QCD we could use the pion decay constant. Both of these are measurable in a laboratory, and they can both be predicted as a function of the parameters in the Lagrangian.
But what if we are dealing with a more general QFT, one for which we do not have a clear picture of what it is describing in real life? Which observables can we use then, if we don't even know what the theory is modeling in the first place? In this situation, it is convenient to use correlation functions at specific values of the external momenta as "observables". So for generic theories, instead of using a decay constant as a measureable parameter, we use $\Gamma^{(n)}(0)$, as if this were something we could measure. Often enough, it actually is, but this really depends on which QFT you are working with.