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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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?
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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 $d\geq 2$.
answered May 20, 2019 at 18:09
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$\begingroup$ Correction to answer (v7): Replace the word Feynman diagram with the word Feynman loop diagram. Notes for later: 1. Q: Can we generalize to not putting $\hbar=1$? A: All quantities $Q$ are of the form $[Q]=(\text{reciprocal length})^{\#}(\text{angular momentum})^{\#\#}=L^{-\#}J^{\#\#}$. The corresponding $\hbar$-reduced quantity $\overline{Q}:=\frac{Q}{\hbar^{\#\#}}$ with powers of $\hbar$, so that $[\overline{Q}]=L^{-\#}$ only. Now let $[\cdot]$ denotes inverse length dimension. Momentum $p=\hbar k$ should be replaced with wavevector $k=\overline{p}$. Action $S=\hbar\overline{S}$. $\endgroup$– Qmechanic ♦Commented Mar 27, 2020 at 9:29
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$\begingroup$ A field $\overline{\phi}_f=\frac{\phi_f}{\sqrt{\hbar}}$. A source $\overline{J}_f=\frac{J_f}{\sqrt{\hbar}}$. A coupling constant $\lambda_i$ should be replaced with $\overline{\lambda}_i=\hbar^{n_i/2-1}\lambda_i$, where $n_i=\sum_f n_{if}$ is the number of legs on a vertex of $i$th type. 2. Q: How about _not_ putting $c=1$ for non-relativistic physics? 3. Q: Could vertex numerator momentum factors conspire to become projection operators with an effective lower UV momentum dependence than their power suggests? A: No. $\endgroup$– Qmechanic ♦Commented Feb 7, 2022 at 12:15
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$\begingroup$ NB: Note that SDOD may underestimated the dependence of the UV cut-off: Toyexample in $d=1$: $\quad\int^{\Lambda_H}_{\Lambda_L}\!\frac{mdp}{p^2+m^2}$ $=\int^{\Lambda_H}_{\Lambda_L}\!\frac{mdp}{2i}\left(\frac{1}{p-im}-\frac{1}{p+im}\right)$ $=[\arctan\frac{p}{m}]^{\Lambda_H}_{\Lambda_L}$ $=[{\rm sgn}(p)\frac{\pi}{2}-\frac{m}{p}+{\cal O}((\frac{m}{p})^2)]^{\Lambda_H}_{\Lambda_L}$. Naively $D=-1$ but effectively $D=0$! Notes for later: userswww.pd.infn.it/%7Eferuglio/Falkowski.pdf p.6: eq. (1.7) blocking/dimless coupling constants. p. 6-7: $\hbar$. $\endgroup$– Qmechanic ♦Commented May 22, 2022 at 20:19
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$\begingroup$ Notes for later: 1. For a disconnected diagram with $\pi_0$ connected components, the first term $d\to \pi_0d$ is replaced in eq. (1). 2. In QM $d=1$ the scalar field $[\phi]=\frac{d-2}{2}=-\frac{1}{2}$, so $D$ increased with number of external legs. $\endgroup$– Qmechanic ♦Commented Jul 6, 2022 at 11:31
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$\begingroup$ Notes for later: $\quad \sum_i V_i ( n_i -2) =2(L-1) +\sum_f E_f $. So at a certain $\hbar$/loop-order, the number of vertices $V_i$ with $n_i\geq 3$ legs (in the amputated correlation function) is finite. 2-vertices (apart from the kinetic terms) contain derivatives. If we assume Lorentz symmetry such 2-vertices are irrelevant $[\lambda_i]<0$. If the kinetic term is unique (e.g. by homogeneity & isotropy), then all other 2-vertices has $[\lambda_i]\neq 0$. It is natural to assume that all relevant or marginal 2-vertices belong to the free (rather than the interaction) part of the action. $\endgroup$– Qmechanic ♦Commented Jun 15, 2023 at 7:33