S-matrix structure of the lightest particle

[geoduck]

If you have a particle that is the lightest particle, then it cannot decay into other particles. As a consequence of this, its T-matrix amplitude on-shell should be zero, since the S-matrix is S=1+iT, and the amplitude for the particle to be found with the same quantum numbers and momentum is 1 since it can't decay, so T=0. The T-matrix in this case should be the self-energy of the particle. So you should always have $\Pi(p^2=m^2)=0$. But what if you choose a different renormalization scheme so that $\Pi(p^2=m^2) \neq 0$? Then the S-matrix doesn't seem like it can be equal to 1, since this is the lightest particle, so the T-matrix should always be real on-shell, so that S=1+imaginary number, and the amplitude of this is greater than 1.

Unitarity requires:

$$
- \text{Im} \, \Pi (i \leftarrow i)=\sum_n (2 \pi)^4 \delta^4(P_{ni})(\Pi_n \frac{1}{\rho^2})F (n \leftarrow i)^* F (n \leftarrow i)
$$

In this formula, $n$ are all on-shell intermediate states of all possible particles, and $\rho$ are normalization factors of these particles, and $F$ are Feynman amplitudes. Since this is the lightest particle and can't decay, only when $n=i$, $F \rightarrow -i \Pi(p^2=m^2)$ does the RHS contribute. This seems to imply that if the self-energy is not equal to zero on-shell, then the RHS is non-zero, so the imaginary part of the self-energy on the LHS can be non-zero on-shell. Is this correct?

 

[vanhees71]

Of course, you also have to renormalize the wave function. The physical mass of the particle is always defined by the pole of the corresponding single-particle propagator (one-particle Green's function).