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can somebody explain or point to the relating mathematics showing Why coupling constants with negative mass dimensions lead to non-renormalizable theories?

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For an amputated one-particle irreducible (1PI) Feynman diagram the superficial degree of divergence $D$ is equal to$^1$ $$\begin{align} D&~=~\sum_f [\widetilde{G}_{0f}]I_f + Ld +\sum_i V_i d_i \cr &~=~\sum_f(2[\phi_f]-d) I_f + \left(\sum_f I_f -(\sum_i V_i -1)\right)d +\sum_i V_i d_i \cr &~=~d+2\sum_f[\phi_f] I_f - \sum_i(d-d_i) V_i \cr &~=~d+\sum_f[\phi_f] \left(\sum_i V_i n_{if}-E_f\right) - \sum_i(d-d_i) V_i \cr &~=~d - \sum_i \left(d - d_i - \sum_f [\phi_f] n_{if}\right) V_i - \sum_f [\phi_f] E_f\cr &~=~d - \sum_i [\lambda_i] V_i - \sum_f [\phi_f] E_f,\tag{*} \end{align} $$ 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 $f$.

Let us now return to OP's question. If an interaction vertex, say of type $i_0$, has $[\lambda_{i_0}]<0$, then eq. (*) 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:

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

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

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$^1$ 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 $d\geq 2$.

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