Can the $S$-matrix always be decomposed as $S = 1 + iT$?

The LSZ formula for a scalar field $$\phi$$ with $$n$$ out-states and $$r$$ in-states is $$\langle p_1,\dots,p_n\vert S \vert q_1,\dots, q_r \rangle = \left(\mathrm{i}Z^{-1/2}\right)^{n+r}\prod_i (-p_i^2 + m^2)\prod_j (-q_j^2 + m^2)\left\langle \prod_k \phi(p_k)\prod_l \phi(q_l)\right\rangle + \text{disconnected pieces} = \text{LSZ} + \text{disconnected pieces}$$ where the expectation value $$\langle \dots \rangle$$ is the time-ordered vacuum expectation value and the "disconnected pieces" part represents the disconnected part of the $$S$$-matrix where the amplitude factors into the product of two (or more?) smaller amplitudes. (i.e. the 'disconnected pieces' include all Feynman diagrams with one or more 'spectator particles' whose contributions to the first term will be 0)

I would like people to check if my understanding of this is correct:

• The disconnected pieces can all be calculated from lower order S-matrices. So for the above n incoming and r outgoing particles, we could write $$S(n,r) = LSZ + S(n-1,r-1) +S(n-2,r-2) \ + \ ... \ + 1$$. (Where $$S(n-1,r-1)$$ is the S matrix with 1 spectator particle, and $$1$$ is the trivial case where all particles are spectators).

• Of course, we can always define $$T$$ such that $$S = 1 + iT$$, but in this case $$iT$$ will not correspond to the LSZ term but rather the LSZ term + all disconnected pieces except for $$1$$.

• Alternatively we can define $$iT = LSZ$$ (i.e. let $$T$$ only include all interactions with no spectator particles) and then $$S = 1 + iT +$$ (sum of lower order S matrices to account for 1 or more spectator particles)

• Only in certain cases (like 2 to 2 particle scattering where, if one particle is a spectator, so too must the other particle, so you therefore only have the LSZ term + $$1$$) does $$S=1+iT$$ in both definitions.

This makes sense since there is no reason why a Feynman diagram for 9 to 9 particle scattering with just 1 spectator particle and the other 8 particles interacting with each other should not contribute to the scattering amplitude (as it would if we said $$S = 1 + LSZ$$ since its LSZ contribution would be 0)

Of course the $$S=1+iT$$ part is a matter of definition, but I wanted to check the above understanding was correct.

From other questions (Contradictory result for scalar-field propagator from Feynman rules and LSZ formula)

• I believe your resolution may be found here: physics.stackexchange.com/q/82387 Commented Jan 27, 2022 at 21:50
• In short, (a) formally writing $S=I+iT$ is always permissible. Equivalently we can define $T=-i(S-I)$. (b) you seemed to remove the "disconnected" pieces in your LSZ formula, but that's not right. In your example of $9\rightarrow 9$ scattering, the amplitude will definitely also include $8\rightarrow 8$ scattering with 1 spectator possible (with all the different allowed permutations), along with all other disconnected scattering processes you can think of. The only ones that truly don't contribute to $T$ are the "fully disconnected" diagrams, the ones with no vertices whatsoever. Commented Jan 27, 2022 at 21:55
• And am I correct in thinking that the 'fully disconnected diagrams' with no vertices whatsoever are equivalent to $I$, hence the definition? Commented Jan 28, 2022 at 13:51
• @ArturodonJuan Nonetheless, I'm still confused as to how the scattering processes with the spectator particles will have non-zero contributions, when those terms will have (in this case) fewer than 9 $\frac{1}{0}$s (as they have fewer than 9 external on-shell propagators) but there will still be 9 $0$s (from the $(-p^2+m^2)$ prefactors) making the term equal to 0? Commented Jan 28, 2022 at 13:52
• I've edited my question. My theory is that, yes, we can always define $S = I + iT$ but then $T$ will not always be given by the LSZ result in that case Commented Jan 28, 2022 at 14:43