# 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?
• The expansion is the usual Taylor expansion of a funcion $f(k^2)$ near $k^2=0$. You can always do that (expect $k^2=0$ is a singularity; but this is not the case). The coefficients are defined as the $n-$th derivative evaluated at $k^2=0$. Dec 19 '17 at 22:03

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.

• How do I know or prove that $G^{(2)}_{amp}(p)=k^2-m^2+a+b+k^2$ in this case? Dec 20 '17 at 8:35
• @Turbotanten I edited my answer. Let me know if it is more clear now. Dec 20 '17 at 11:04
1. Because $I$ is a Lorentz-invariant scalar, and it depends only on the 4-vector $k$. There are just no other scalars than $k^2$ available! Sorry, I got your question wrong. Looks like it was answered in the comments. Anyway, just to re-state it here - it is a regular Taylor series formula that can be applied to $I$ and allows expansion in powers of $k^2$.
2. It looks like the author omits some details for the sake of simplicity. In fact what happens here is that you calculated a 1PI sub-diagram (One Particle Irreducible). Usually afterwards the students are presented with this formula:

$$G_\text{full} = \frac{1}{k^2 - m^2} + \frac{1}{k^2 - m^2} I \frac{1}{k^2 - m^2} + \frac{1}{k^2 - m^2} I \frac{1}{k^2 - m^2} I \frac{1}{k^2 - m^2} + \dots$$

The idea is that the full propagator is not just a sum of bare propagator and 1PI diagram, but an infinite sum of combinations of them (1PI is "amputated" and has no incoming/outgoing propagators in it). Through the aid of geometric progression formula, this turns into:

$$G_\text{full} = \frac{1}{k^2 - m^2 + I}$$

Which shifts the pole to the new location.

• I fully understand the first part now but I can't still wrap my head around the second part. Luckily I have Peskin and Schroeder book at hand which is more rich in details. I've read some about the 1PI sub-diagrams now but I don't know how to translate it to my case to get the pole shift $m+\delta m =(m^2-a)(1+b)^{-1}$ Dec 20 '17 at 8:33
• You've to be looking for the poles of the full propagator $G_\text{full}$. The 1PI part that you computed automatically gets into the denominator of it as I briefly derived. From there: $k^2 - m^2 + D + E k^2 = (1 + E) k^2 - (m^2 - D) = (1 + E) (k^2 - \frac{m^2 - D}{1 + E})$. Dec 20 '17 at 9:31