# How is dimensionality of $S$ preserved term by term in a perturbative expansion?

In a schematic notation, the scattering matrix element $$\langle p_{out}|S|p_{in}\rangle := 1 + i (2 \pi)^4 \delta^4(p_{in} -p_{out}) M$$ between an incoming state with momentum $$|p_{in}\rangle$$ and an outgoing one with momentum $$\langle p_{out}|$$ is by construction a dimensionless quantity, which implies that $$M$$ has mass dimension $$4$$ due to the Dirac delta factor. Now, how is this dimensionality ensured term by term in a perturbative (diagramatic) expansion, while different diagrams will contribute additive terms of different dimensionality?

As an example, in a real scalar theory with $$\mathcal{L}_{int} = - \frac{\lambda_3}{3!}\phi^3 - \frac{\lambda_4}{4!}\phi^4$$, the tree level contribution to $$M$$ for a $$\phi\phi\to\phi\phi$$ scattering is $$-\lambda_4$$, which is dimensionless, but the tree level contribution for the process $$\phi\phi\to\phi$$ is $$-\lambda_3$$, which has dimension $$1$$, and yet the dimensionality of $$M$$ should be the same for any process.

As a side note, it is worth mentioning that the question arose thinking of effective field theories, because I read a reasoning justifying that the power counting parameter should be chosen to be $$p/\Lambda$$, for $$p$$ the typical energy of the problem addressed by the EFT and $$\Lambda$$ the UV threshold, based on the fact that by dimensional analysis, since the probability amplitude is dimensionless and each non-renormalizable operator of dimension $$d$$ will contribute a factor $$1/\Lambda^{d-4}$$, then some kinematic factor must contribute a $$p^{d-4}$$ to ensure the amplitude remains dimensionless; but of course some diagrams have no kinematic contribution at all...

• Read where? Which page? Jan 27 at 8:36
• @Qmechanic, I read it on some closed access lecture notes. In any case, as I mentioned this was only a comment aside, and I believe that talking about how is the dimensionality preserved term by term in the perturbative expansion of the $S$ matrix element in EFTs is worth a separate question, so I won't delve further into it here. Jan 27 at 12:57

It is the $$S$$ operator that is equal to the dimensionless (schematic) expression $$1+iT$$, not the matrix element calculated between the states $$\langle p_{out}|S|p_{in}\rangle$$, which is like $$\langle p_{out}|S|p_{in}\rangle= \langle p_{out}|p_{in}\rangle +i(2\pi)^4\delta^{(4)}(p_{out}-p_{in})M$$Those states are not dimensionless in the standard relativistic normalization in which you have the matrix element $$M\propto \lambda_4$$ or $$\lambda_3$$. For scattering of two particles with initial energies and spatial momentum $$(E_1, k_1), (E_2, k_2)$$, the identity term is $$\langle p_{out}|p_{in}\rangle \propto E_1E_2 \delta^{(3)}(k_{1,out}-k_1)\delta^{(3)}(k_{2,out}-k_2)$$ Since $$E$$ has energy dimension 1, and a n-dimensional energy or momentum delta function $$\delta^{(n)}$$ has dimension $$-n$$, you can see this is consistent with a dimensionless $$M$$.
You obviously can't compare with the identity term for 2->1 scattering, but you can consider 2->2 scattering with two cubic vertices and an internal propagator, and you have matrix element $$M\propto\frac{\lambda_3^2}{p^2+m^2}$$ which is dimensionless if $$\lambda_3$$ has energy dimension $$1$$.