This question is probably going to be somewhat trivial, but I am a little confused about the dimension of the matrix element that appears in the formula for the cross section of a scattering process. I had always assumed that this kind of matrix elements should be dimensionless since they represent a probability amplitude. Anyway, this seems not to be the case looking at Schwartz's formula [5.22]: $$ d \sigma=\frac{1}{\left(2 E_{1}\right)\left(2 E_{2}\right)\left|\vec{v}_{1}-\vec{v}_{2}\right|}|\mathcal{M}|^{2} \prod_{\text {final states } j} \frac{d^{3} p_{j}}{(2 \pi)^{3}} \frac{1}{2 E_{p_{j}}}(2 \pi)^{4} \delta^{4}(\Sigma p).\tag{5.22} $$ On the LHS the dimension is $E^{-2}$, while on the RHS it is $E^{-2}E^{2n_f-4}[\mathcal{M}]^2$, where $n_f$ is the number of final particles. So for 2 particles in the final state the matrix element looks dimensionless, but for $n_f \geq 3$ it should have a negative mass dimension, e.g. for 3 particles $E^{-2}$. Is this correct?
2 Answers
We wondered the same thing with a group of friends and came to the conclusion that you are right.
As a confirmation, if you have a copy of Peskin and Schroeder, you can deduce from equation (4.74) that $|\mathbf{p_1} \dots \mathbf{p_n}\rangle$ has mass dimension $-n$, and going back to the definition of the matrix elements $\mathcal{M}$ in equation (4.73), with the fact that the momentum delta function has mass dimension 4, you can deduce that the mass dimension of the matrix elements is, in general $$[\mathcal{M}] = 4 - N$$ where $N$ is the total number of particles involved in the process. In the $3 \to 2$ case you mentioned, this is consistent with your conclusion that $[\mathcal{M}^2] = -2$.
Schwartz answers this on page 388, giving $$\dim(\mathcal{M}) = 4 - \frac{3}{2}f - b.$$
