2
$\begingroup$

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...

$\endgroup$
2
  • $\begingroup$ Read where? Which page? $\endgroup$
    – Qmechanic
    Jan 27 at 8:36
  • $\begingroup$ @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. $\endgroup$
    – Albert
    Jan 27 at 12:57

1 Answer 1

1
$\begingroup$

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 $1$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.