What is the Asymptotic Behavior of the Propagator in QFT?

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SUMMARY

The asymptotic behavior of the propagator \(\Delta_{f}(k)\) in Quantum Field Theory (QFT) is defined as \(\Delta_{f}(k) \sim k^{-2 + 2sf}\), where \(sf\) represents the spin of the field. For massive fields of Lorentz type, the spin is calculated as \(sf = A + B\). This behavior is confirmed for particles such as the photon, W-Z boson, gluon, graviton, and spin 1/2 fermions. The discussion emphasizes the importance of gauge invariance in determining the effective spin for certain particles, notably indicating that the photon has an effective spin of zero.

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ndung200790
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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?
 

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