Often, in many books such as Peskin and Schroeder, a Feynman diagram or the effective potential is expanded as a function of the external momenta or the classical fields respectively. Consider the case of the Feynman diagram. The authors say that, upon expanding the Feynman diagram as a function of the external momenta in a Taylor series, all the Taylor coefficients can't be divergent because the mass dimension of the entire expansion is fixed. Therefore any increment in the mass dimension due to the power of the external momenta must be balanced by the decrease in the mass dimension of its coefficient. Therefore, as a function of the momentum cutoff, $\Lambda$, the coefficients should become finite after some point in the Taylor series. I am confused because I don't understand why can't the coefficient be something like $\frac{\Lambda^2}{m^4}$, where $m$ is the mass parameter? This quantity has a negative mass dimension, therefore it can be used to balance the increment in the mass dimension due to power of the external momentum. However, at all orders it's divergent. Something similar happens during the computation of the effective potential, the authors say that since the mass dimension of the classical field is increasing, the mass dimensions of the coefficients should reduce in order to make sure that the effective potential has a mass dimension of four. A similar argument can also be applied to this situation. Why is this incorrect?

EDIT: P&S gives the power counting argument regarding Feynman diagrams in the last two paragraphs of page 319.

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    $\begingroup$ It's because in the relevant diagrams they're talking about, you only get positive powers of $m$. $\endgroup$ – knzhou Dec 18 '19 at 18:54
  • $\begingroup$ I see, what about the effective potential? $\endgroup$ – Sounak Sinha Dec 18 '19 at 18:56
  • $\begingroup$ Their effective potential is computed by summing rather similar diagrams, in which you again only get positive powers of $m$. $\endgroup$ – knzhou Dec 18 '19 at 18:57
  • $\begingroup$ I'm sorry, but could you please elaborate your comment in an answer? I really don't understand which are the relevant diagrams and why do they have only positive powers of $m$. $\endgroup$ – Sounak Sinha Dec 18 '19 at 19:03

The $m$-dependence enters on the same footing as the momenta $p$ through propagators. One may show that the divergent part of a Feynman diagram with superficial degree of divergence $D$ is at most a $D$-order polynomial in external momenta $p$ and masses $m$, cf. e.g. Ref. 1 or my related Phys.SE here. Hence OP's suggested divergent term with negative powers of $m$ is not possible.


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