Is the $S$-matrix a scalar operator? 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?
 A: 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.
