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]: $$ d \sigma=\frac{1}{\left(2 E_{1}\right)\left(2 E_{2}\right)\left|\vec{v}_{1}-\vec{v}_{2}\right|}|\mathcal{M}|^{2} \prod_{\text {final states } j} \frac{d^{3} p_{j}}{(2 \pi)^{3}} \frac{1}{2 E_{p_{j}}}(2 \pi)^{4} \delta^{4}(\Sigma p) $$ 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?

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?