Finding the Lagrangian term from a scattering amplitude

Question

I am reading the book Scattering Amplitudes in Gauge Theory and Gravity from Elvang and Huang. In section 2.6 they seem to suggest that the mass-dimension of the kinematic part of the amplitude is in 1-to-1 correspondence to the number of derivatives in the interaction term in the Lagrangian. For example, for the 3-gluon amplitude $$A_3(g_1^- g_2^- g_3^+) = g \frac{\langle 12\rangle^3}{\langle13\rangle\langle 23\rangle} $$ has a kinetic part with mass-dimension 1, which means it is compatible with the $AA\partial A$ interaction term in $\mathrm{Tr}F_{\mu\nu}F^{\mu\nu}$. Similarly they argue that the amplitude $$A_3(g_1^-g_2^-g_3^+) = g' \frac{[13][23]}{[12]^3}$$ has mass-dimension -1, and thus must come from $g' AA \frac{\partial}{\Box} A$ (and is thus nonlocal and unphysical). Later on a similar claim is made about the amplitude $$A_3(g_1^-g_2^-g_3^-) = a \langle12\rangle\langle13\rangle\langle23\rangle$$ coming from an interaction term with three derivatives

This seems intuitive, but it also seems to contradict an earlier result from QED with massless fermions discussed in section 2.4. In that section we derive the 3-particle amplitude $$A_3(f^- \bar{f}^+ \gamma^-) = \tilde{e} \frac{\langle 13\rangle^2}{\langle12\rangle}.$$ By the same logic you would expect this to come from a Lagrangian interaction term with one derivative, but in fact it seems to arise from the $\gamma^\mu\bar{\Psi} A_\mu \Psi$ interaction term, which does not have any derivatives.

Can someone clarify this for me?

Answer 1

The logic goes as follows, as the another answer says : you start with the knowledge, that the Lagrangian density has mass dimension $[\mathcal{L}] = 4 $, then subtract the dimension of all fields, involed in the interaction, for example (assuming the coupling is dimensionless): $$ A_3 (f^{-} \bar{f}^{+} \gamma^{-}) = 4 - 2 \cdot(3/2)-1=0 \quad A_3 (g_1^{-} g_2^{-} g_3^{+}) = 4 - 3= 1 \quad $$ Therefore, in the former case you do not need a derivative, whereas, in the latter, you need 1 derivative. To get the second and third example from your post, you need a dimensionful coupling and nonlocal coupling.

Answer 2

In natural units, that is $\hbar = c = 1$ (dimensionless), the dimensional analysis provides
$[E] = [m] = [p] = 1$
$[x^\mu] = -1$
$\partial_\mu = 1$

As the action $S = \int d^4x \mathcal L$ is dimensionless, the Lagrangian density $\mathcal L$ has dimension $[\mathcal L] = 4$.

According to the Lagrangian density terms, the electromagnetic four-vector $A^\mu$ has dimension $[A^\mu] = 1$, a scalar field $\phi$ has dimension $[\phi] = 1$ and a Dirac spinor field $\psi$ has dimension $[\psi] = \frac{3}{2}$.

Both in scalar and spinor QED (quantum electrodynamics) the coupling $e$ is dimensionless, i.e. $[e] = 0$.