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

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