What is the Asymptotic Behavior of the Propagator in QFT?

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Please demonstrate for me that:
In any theory,the propagator \Delta_{f}(k) of a field of type f has asymptotic behavior:

\Delta_{f}(k)~k^{-2+2sf}

where sf is ''spin'' of the field.For massive fields of Lorentz type (A,B) then sf=A+B.
(However,dropping terms that because of gauge invariance have no effect,eg. photon has sf=0)
(QFT of Weinberg Vol 1,&12.1 Degrees of Divergence)
 
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I know this is true for photon,W-Z boson,gluon,graviton,spin 1/2 fermion.But what about the other particles?
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!

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