quantum field theory - Understanding renormalizability and bare mass

I'm studying QFT on my own. I'm using the book Quantum Field Theory in a Nutshell by Zee.

I need some help with understanding and clarification with the following. I'm going to quote a paragraph from Zee's book Quantum Field Theory in a Nutshell p.174. which I need help in understanding

Consider the following double integral $$ I\equiv(-i\lambda)^2\int^\Lambda\int^\Lambda\frac{d^4p}{(2\pi)^4}\frac{d^4q}{(2\pi)^4}\frac{i}{p^2-m^2+i\epsilon}\frac{i}{q^2-m^2+i\epsilon}\frac{i}{(p+q+k)^2-m^2+i\epsilon} $$ Counting powers of $p$ and $q$ we see that the integral $\sim \int(d^8P/P^6)$ and so $I$ depends quadratically on the cutoff $\Lambda$. By Lorentz invariance $I$ is a function of $k^2$, which we can expand in a series $$ D+Ek^2+Fk^4+\ldots \tag{$\mathbf{1}$} $$ The quantity $D$ is just $I$ with external momentum $k$ set equal to zero so depends quadratically on the cutoff $\Lambda$. Next we get $E$ from differentiating $I$ with respect to $k$ twice and then set $k$ to zero. This clearly decreases the power of $p$ and $q$ in the integrand by 2 so it depends logarithmically on the cutoff $\Lambda$. Similarly we get $F$ from differentiating with respect to $k$ four times and setting $k$ to zero. This decreases the power of $p$ and $q$ in the integrand by 4 and this F is given by an integral that goes as $\sim \int d^8P/P^{10}$ for large $P$. The integral is convergent and hence cutoff independent. Thus $F$ and all $(\ldots)$ terms are cutoff independent as the cutoff goes to infinity and we don't ahve to worry about them.

Putting it altogether, we have the inverse propagator $k^2-m^2+a+b+k^2$ The propagator is changed to $$ \frac{1}{k^2-m^2}\rightarrow\frac{1}{(1+b)k^2-(m^2-a)} \tag{$\mathbf{2}$} $$ The pole $k^2$ is shifted to $m_\textbf{p}\equiv m+\delta m\equiv (m^2-a)(1+b)^{-1}$, which we identify as the physical mass.

Question

I have marked the parts that I have a hard time understanding with bold text. Those are some parts where I could need some clarification.

1. I'm onboard with that $I$ is a function of $k^2$ but why can we do a series expansion like that in (1)?


2. When he says putting it altogether, what is it that we are actually putting together and how do we obtain that expression?

Answer

As I said in the comment, as long as you don't hit singularities for $k^2=0$, the function $I(k^2)$ is completely regular and you can perform the expansion. As well, you can take the one-dimensional integral (it may be a simplification of a more general case you can find in computing n-point Green functions)

$$ I(k) = \int_0^{+\infty} d q \frac{q^4}{(q+k)^2} $$

where $k$ is the external momenta. This has degree of divergence $D=2$ but you can derive three-times with respect to the external momenta $k$ and you get I'''(k) which is completely convergent

$$ I'''(k) =\int_0^{+\infty} d q -6\frac{6q^2}{(k+q)^4} = -\frac{2}{k} $$

Then, you can integrate back (integrate wrt to external momenta!!)

$$ I''(k) = -2\log(k) + A $$

where A (and in the following all the upper case letters) is a divergent constant. Then,

$$ I'(k) = +2k -2k \log(k) + A k + B\rightarrow I(k) = Bk + \frac{1}{2}\left( 3+A\right)k^2 - k^2\log(k) $$

Regarding your second question, you can express the n-point Green $G^{(n)}(p)$ function in terms of the amputated one $G^{(n)}_{amp}$

$$ G^{(n)}(p_1,...,p_n) = \Pi_{i=1}^n \left[G^{(2)}(p_i)\right] G^{(n)}_{amp}(p_1,...,p_n) $$

The S-matrix is nothing else the amputated Green function in which you add the wave-function polarization and then put everything on shell. In the case of the scalar theory the wave-function polarization is trivial (i.e. it is 1). For $n=2$ you look at the full propagator $G^{(2)}(p)$ $$ G^{(2)}(p) =G^{(2)}(p)G^{(2)}(p) G^{(2)}_{amp}(p) $$

and you see that

$$ G^{(2)}(p) = \frac{1}{G^{(2)}_{amp}(p)} \,,\qquad \qquad (1) $$

(EDITED) In perturbation theory, you can set $G^{(2)}_{amp}(p) = k^2 - m^2 + a +b+k^2$. I guess the term $a +b+k^2$ include some power of the perturbative coupling constant. You can avoid higher order corrections because they are already taken into account in Eq. (1).

Notice that $G^{(2)}_{amp}(p)$ has that value because you are using the Feynman rules of the kinetic term seen as a vertex and the result of the loop-integral. If you want, the loop-integral provide you a contribution into the effective action proportional (roughly) to $(a+b)\phi^2 - (\partial\phi)^2$. Then, the term $a+b+k^2$ is the feynman rule associated to this vertex and it enters in $G^{(2)}_{amp}(p)$.

Notice I am not doing the sum like in the other answer. This is because I should do the same sum but with $1/(k^2-m^2)\rightarrow G^{(2)}(p)$ where $ G^{(2)}(p)$ is the full quantum propagator (the one which include quantum corrections). If you work in perturbation theory, that sum is equivalent to what I did here.