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I have some questions about the $i\epsilon$ factor in Feynman diagrams. First, what is the physical meaning of $i\epsilon$ in loop amplitudes. Second, how does it ensures unitarity? And third, Dyson series assume that incoming and outgoing particles are free, this can be implemented by assuming that the interaction Hamiltonian switches off adiabatically, $e^{-\eta\,t}H_{I}(t)$. Is this $\eta$ related with the $i\epsilon$?

Qmechanic
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Nausea
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  • Related: http://physics.stackexchange.com/q/110046/ – Mark Mitchison May 26 '14 at 11:47
  • As for the last part of your question, even though I haven't checked it carefully enough myself, I think that indeed the adiabatic switch off of the interactions gives rise to the $i\epsilon$ prescription. You can indeed check it out in section 9.2 of the first volume of QFT book by Weinberg (around eq. 9.2.14) – TwoBs May 26 '14 at 12:07
  • This will likely help: http://physics.stackexchange.com/q/138217/ – DanielSank Nov 17 '14 at 18:29
  • In general, whenever you find yourself enumerating multiple questions as "first", "second" etc, that's a pretty clear sign that you should post those questions separately. The first question you're asking is already explained here. The second question should definitely be posted separately as it is a well defined and self contained question. The third question is also clearly separate and should be posted separately. – DanielSank Jan 22 '15 at 21:11

2 Answers2

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$\rm{i}\varepsilon$ is not a factor, but an addendum. It allows to integrate straight along the axis. It removes poles from axes in a "physical way" - by obtaining the right propagator. The right propagator does not need any adiabatic tricks.

Incoming and outgoing particles are not always free. In particular, the meaningful electron line has many soft photon lines emitted. It means the electron is coupled to the electromagnetic field even in the asymptotic regions. Only perturbatively (which is rather inexact) the lines may be "free".

Vladimir Kalitvianski
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At first, the $i \epsilon$ prescription: In Feynman diagrams we have a lot of Green's functions connected by some rules, the Feynman rules. Actually, the $i \epsilon$ prescription is more related to the Green's function. The prescription is responsible for a choice in the boundary conditions (asymptotic behaviour). We may put this $i \epsilon$ inside the expression $(\omega + i \epsilon)^2$ or $(\omega - i \epsilon)^2$. The first is related to retarded Green's function, the second is related to advanced Green's function. This choices of boundary condition are related to some adapted Feynman's rules. Calculating things in this prescriptions is not manifest relativistically and not so compact. The Feynman Green's function is the best for doing calculations and very natural.

Now, let's be more general. This prescriptions are aroused when we have operators like this:

$$ (H-E)G(t)=\delta(t) $$

We want to find an expression for $G(t)$. Applying Fourier transform, we can find that

$$ G(t) = \frac{1}{(2 \pi)} \int d\omega \, \frac{e^{-ip(t)}}{\omega - E\pm i\varepsilon} $$

If you see this function in complex plane, we can se that we have a cut in real axis made by eigenvalues of $H$. The prescription $\epsilon$ determine how far we pass about the singularity $\omega=E$. The choice of signal tell in what side.

This precription is simply procedure to getting inverse of operators with real eigenvalues, satisfying some asymptotic behavior.

We can interpret this $\epsilon$ as a inverse of timelife of the signal describe by the repective Green's funciton. This is because we can identify a relativistic Breit–Wigner distribution

The untary is hold because we get $\epsilon \rightarrow 0$, so the signal persist at infinite time.

The adiabatic turn on and turn off interaction is another mathematical departure: adiabatic theorem and Gell-mann and Low theorem, but physically is very close to the $i\epsilon$. Is a inverse of a caracteristic time that is send to infinity.

Nogueira
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