Why coupling constants with negative mass dimensions lead to non-renormalizable theories? can somebody explain or point to the relating mathematics showing Why coupling constants with negative mass dimensions lead to non-renormalizable theories?  
 A: The standard argument goes as follows. For a connected Feynman diagram$^1$ the superficial degree of (UV) divergence $D$ is equal to$^2$
$$\begin{align} D~:=~& \#\{\text{$\mathrm{d}p$ in int. measure}\} ~+~ \#\{\text{$p$ in numerator}\}\cr
&~-~ \#\{\text{$p$ in denominator}\}\cr\cr
~=~&  Ld +\sum_i V_i d_i + \sum_f  [\widetilde{G}_{0f}]I_f\cr
~\stackrel{\text{Ref. }3}{=}& \left(\sum_f I_f -(\sum_i V_i -1)\right)d  \cr
& +\sum_i V_i d_i + \sum_f(2[\phi_f]-d) I_f  \cr
~=~&d- \sum_i(d-d_i) V_i + \sum_f[\phi_f] ~2I_f       \cr
~=~&d- \sum_i(d-d_i) V_i  +\sum_f[\phi_f] \left(\sum_i V_i n_{if}-E_f\right)    \cr
~=~& d -  \sum_i \left(d - d_i - \sum_f [\phi_f] n_{if}\right) V_i - \sum_f [\phi_f] E_f\cr
~=~& d  - \sum_f [\phi_f] E_f   - \sum_i [\lambda_i] V_i \cr
~\stackrel{\text{Ref. }4}{=}&  [\text{amputated diagram}]  - \sum_i [\lambda_i] V_i,  \end{align}\tag{1} $$
where

*

*$d$ is the number of spacetime dimensions;


*$[\cdot]$ denotes the mass dimension in units where $\hbar=1=c$;


*$L$ is the number of independent loops;


*$I_f$ is the number of internal lines with a free propagator $\widetilde{G}_{0f}$ in the Fourier momentum space of a field $\phi_f$ of type $f$;


*$V_i$ is the number of vertices of $i$'th interaction type with coupling constant $\lambda_i$, $d_i$ number of spacetime derivatives, and $n_{if}$ legs of type $f$;


*$E_f$ is the number of amputated external lines with a field $\phi_f$ of type $f$.
The formula (1) has in principle a simple interpretation in terms of double-entry bookkeeping as follows. Recall that each vertex arises from a dimensionless action term. So instead of debiting the loop momentum variables $p$ [cf. the definition of $D$], we can instead credit [with the opposite sign] the mass dimension of the rest of the Feynman diagram, namely coupling constants and amputated legs [cf. formula (1)].
Let us now return to OP's question. If an interaction vertex, say of type $i_0$, has $[\lambda_{i_0}]<0$, then eq. (1) indicates that we can build infinitely many superficially divergent Feynman diagrams with $D\geq 0$ by using more and more vertices of type $i_0$.  This render the theory non-renormalizable in the old Dyson sense.
References:

*

*S. Weinberg, Quantum Theory of Fields, Vol. 1, 1995; eq. (12.1.8).


*M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; eqs. (10.11) + (10.13).


*Use my example to explain why loop diagram will not occur in classical equation of motion?


*Why do all Feynman diagrams with same number of external legs have the same mass dimension?
--
$^1$ We assume that the sources $J_k$ are either stripped from the Feynman diagram or are delta functions in momentum space so that the external legs carry fixed 4-momenta.
$^2$ It is implicitly assumed that the coefficients in front of the kinetic terms in the action are dimensionless. The quantity $[\phi_f]$ is non-negative for $d\geq 2$.
