# Why can we simply absorb the positive coefficient of $i\epsilon$ in a propagator?

As far as I know, absorbing of the positive coefficient of $i\epsilon$ in a propagator seems to be a trivial operation without even the need of justification.

In Peskin page 286, he did this: $$k^0\rightarrow k^0(1+i\epsilon)$$ $$(k^2-m^2)\rightarrow (k^2-m^2+i\epsilon)$$

In M. Srednicki's Quantum Field Theory, page 51,

The factor in large parentheses is equal to $E^2-\omega^2+i(E^2+\omega^2)\epsilon$, and we can absorb the positive coefficient in to $\epsilon$ to get $E^2-\omega^2+i\epsilon$.

Why and does this kind of manipulation affect the final result of calculation?
Although $\frac{1}{k^2-m^2+i\epsilon k^2}-\frac{1}{k^2-m^2+i\epsilon}$ is infinitesimal, but the integration of such terms may lead to divergences, and this is my worry.

Also the presence of $k^0$ in the coefficient of $i\epsilon$ could potentially influence the poles of an integrand and consequently influence the validity of Wick Rotation.

• Looks to me like Srednicki's doing $\epsilon'=(E^2+\omega^2)\epsilon$ and then dropping the prime in the latter equation. Apr 28, 2014 at 13:54
• @Kyle Kanos, But I don't think he can simply do that and pretend that ϵ′ doesn't depend on E anymore.
– LYg
Apr 28, 2014 at 13:57
• Why would $\epsilon'$ not depend on $E$ anymore? If $\epsilon$ is a function of $E$ then $\epsilon'$ is a function of $E$ as well. Apr 28, 2014 at 14:01
• @Kyle Kanos, But their subsequent calculations all treat $\epsilon$ as independent of $k$, for example in the formalism of Wick Rotation, such dependence could influence the position of the poles in the $k^0$ plane and therefore influence the validity of Wick Rotation method.
– LYg
Apr 28, 2014 at 14:05

The size of the parameter $\epsilon$ does not matter, as long as it is infinitesimally small. Rescaling it by that function does not change this. Recall that the whole procedure is just a mathematical trick which allows us to perform a contour integral over the real axis of the complex plane. The shift is really arbitrary, as long as it is small. The precise size is should not affect any results we gain from it.

• Correct, only the sign of $\epsilon$ matters, otherwise amplitudes would depend on its value. Nov 26, 2019 at 18:59