Why coupling constants with negative mass dimensions lead to non-renormalizable theories?

Question

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

Answer 1

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:

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).

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

4. Why do all Feynman diagrams with same number of external legs have the same mass dimension?

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$^1$ 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 $d\geq 2$.