Fermion annihilation amplitude Is there a physical reason why annihilation  of 2 identical fermions with mass m to 2 scalars amplitude for $s=4m^2$ (fermions at rest) is zero? For example we can have 2 scalars annihilating in 2 scalars and the amplitude is non-zero if we have a quadrilinear term in the largangian $\phi_1^2 \phi_2^2$. Also a decay of a scalar to 2 fermions at $s=4m^2$ (m - fermion mass) is zero.
I consider a Yukawa lagrangian $Y\bar{\psi}\psi\phi$, an annihilation in t and u channels at the tree level. I actually am interested in amplitude modulus squared and take an average over all spin states.
 A: For generally valid reasons, the amplitudes analytically depend on the momenta and on $s$ in particular. There can't be any "jumps", the simplest discontinuities. Because the processes you enumerated with the total energy $E_{cm}\lt 2m$ i.e. $s\lt 4m^2$ are strictly prohibited by energy conservation – $2m$ is the minimum energy two fermions may have, as Trimok said – and because of the continuity, the amplitude has to be zero not only for $s\lt 4m^2$ but also for the borderline case $s=4m^2$.
The argument in the previous paragraph was sloppy and neglected the fact that the prohibition of the processes may come from $M=0$ or from the kinematic factors and that's why the cases of bosons and fermions are different. The amplitudes near the threshold of production – in our case, $s=4m^2$ – are increasing as a higher power law i.e. they are more suppressed for the fermions than for the bosons. This ultimately boils down to the normalization of the scalar-like or spinor-like solutions of the free equations of motion. 
To fully understand how it works, just do the careful calculation of your simple two cases and carefully watch the powers of $(E_{fermion}-m)$ that appear in the kinematic factors and the dynamic amplitudes both in the bosonic and fermionic case.
