I would like to understand how to extract renormalization constants of vacuum polarization diagram in pseudoscalar Yukawa theory with interaction $ig\bar{\psi}\gamma^5\psi$. This diagram is represented by the following integral: $$i\Pi(k^2)=-4g^2\int_p\frac{p^2+p\cdot k-m^2}{(p^2-m^2)((p+k)^2-m^2)},$$ where $\int_p=\int d^4p/(2\pi)^4$ and should give two renormalization constants: $$\delta Z_{\phi}=\frac{g^2}{4\pi^2}\frac{1}{\epsilon};\quad \delta Z_m=-\frac{g^2}{2\pi^2}\frac{1}{\epsilon},$$ which comes from singular part of polarization: $$\Pi_{\text{sing}}=\frac{g^2}{4\pi^2}\left[\frac{1}{\epsilon}(k^2-2m^2)\right]$$ I know the following statement:

It is enough to capture out divergence to find renormalization constants

I would like to avoid full calculation of the integral and only calculate these constants. Thus, I consider big momenta $p$ and expand $$\frac{1}{(p+k)^2-m^2}=\frac{1}{p^2}-\frac{2(p\cdot k)}{p^4}+\mathcal{O}(p^{-4}).$$ Substituting this expansion, I try to define singularity: $$i\Pi_{\text{sing}}=-4g^2\int_p\frac{p^2}{(p^2-m^2)p^2},$$ which can be calculated with dimension regularization and I find $$\Pi_{\text{sing}}=-\frac{g^2m^2}{2\pi^2}\rightarrow \delta Z_m=-\frac{g^2}{2\pi^2}\frac{1}{\epsilon}.$$ Then I have cast doubts about my derivation and do not understand how to extract the singular term which is $\propto k^2$.


The scalar boson self-energy integral you are looking at has a naive quadratic divergence. This means that if you shift the integration variable to $\ell=p-xk$, the surface term you get is still divergent. Since the integrand of the surface term has to be proportional to $(xk)^{2}$, the surface term has a logarithmic divergence (two powers of $p$ in the numerator replaced with two powers of the fixed external momentum $k$). This term, with its proportionality to $k^{2}$, gives the field strength renormalization constant that you missed.

In an initially logarithmically divergent expression, you could just drop all other masses and momenta in comparison with the integration momentum $p$ and get the infinite residue that way. However, with a quadratic divergence, there are two divergences. In dimensional regularization, the residue of the divergence $\Gamma(1-d/2)$ (as $d\rightarrow 4$) is, for dimensional reasons, independent of $k$ and gives the mass renormalization. After you introduce a counterterm to cancel that divergence, there is another divergence (proportional to $k^{2}$) coming from $\Gamma(2-d/2)$, which is canceled by a field strength renormalization counterterm.

  • $\begingroup$ Can You please clarify what is "surface term"? If I shift variables, I obtain for numerator $q^2+(q\cdot k)(1-2x)-(k^2x(1-x)+m^2)$ and for denominator $q^2-(m^2-k^2x(1-x))$. Also, the second part of Yours answer seems unclear for me. You mean that afterneglecting of all masses and momenta the value of the integral is defined by the residues at denominator poles? $\endgroup$ – Artem Alexandrov Jun 17 '19 at 14:25

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