Is the $S$-matrix a scalar operator?

Physics Asked on January 2, 2022

The scattering $$S$$ operator which is defined to be the operator corresponding to $$S$$ matrix should be rotational invariance, does this imply $$S$$ operator is a scalar operator?

In a typical collider experiment, two particles, generally in approximate momentum eigenstates at $$t = -infty$$, are collided with each other and it is measured the probability of finding particular outgoing momentum eigenstates at $$t = +infty$$.

In the Heisenberg picture the probability amplitude for the initial states $$vert i rangle$$ to evolve to the final states $$vert f rangle$$ is defined as $$langle f vert S vert i rangle$$ where the time evolution is put in the scattering or S-matrix.

The S-matrix element $$langle f vert S vert i rangle$$ for $$n$$ asymptotic momentum eigenstates is given by the LSZ (Lehmann-Symanzik-Zimmermann) reduction formula which for scalar quantum fields $$phi (x)$$ states
$$langle f|S| i rangle = [i int d^4 x_1 exp (-i p_1 x_1) (Box_1 + m^2)] cdot cdot cdot [i int d^4 x_n exp (i p_n x_n) (Box_n + m^2)] langle Omega | T { phi (x_1) cdot cdot cdot phi (x_n)} | Omega rangle$$
where $$−i$$ in the exponent applies to initial states and $$+i$$ to final states.

The LSZ formula is constructed from Lorentz covariant fields, hence the S-matrix element is an invariant.

Answered by Michele Grosso on January 2, 2022