Kallen-Lehmann representation versus perturbation theory According to the Kallen-Lehmann representation of the propagator, the propagator has a branch cut beginning around $p^2\approx4m^2$. This appears to invalidate the use of the usual $\frac{i}{p^2-m^2+i\epsilon}$ for intermediate propagators in computing an amplitude, as the branch cut would dominate it. To cite an example, the tree level graph for two-to-two scattering in $\phi^3$ theory is generally calculated whilst ignoring the branch cut region of the propagator.
 A: Kallen-Lehmann representation originates from the decomposition of $\hat{\phi}(x)|\Omega\rangle$ into the Hamiltonian eigenstates $|\lambda_{\mathbf{p}}\rangle$. If $|\langle\Omega|\hat{\phi}(x)|\lambda_{\mathbf{p}}\rangle=0$ that state doesn't give contribution into the propagator spectral representation.
In the free theory $\hat{\phi}(x)|0\rangle$ is simply a superposition of single particle states. Even though there are other free Hamiltonian eigenstates - many-particle states, they have zero overlap with $\hat{\phi}(x)|0\rangle$ and you get just one pole in the free propagator.
However for interacting theory $\hat{\phi}(x)|\Omega\rangle$ decomposes into all kinds of states including many-particle states that give you branch cut.
The main point of my answer to your question is that Feynman's perturbation theory uses the propagators of the free theory for the lines of the diagrams and they as I said don't have branch cut. The full propagator is given by the sum of all diagrams and you will see that it has branch cuts when you compute loop-level contributions.
