The momentum space propagator in statistical field theory can be defined (see for example Abrikosov's book) as follows: \begin{equation} G(\omega,\vec{p})=\frac{E^{2}(\vec{p})}{\omega^{2}-E^{2}(\vec{p})+i\epsilon} \end{equation} where E is the non-relativistic expression for the energy of the (quasi) particle. When considering loop integrals, we want to evaluate this integral over 4 momentum integral measure. My question is how to evaluate this integral.

My attempt:

1- Do first the energy integral: the poles are $\omega= \pm \sqrt{E^{2}-i\epsilon}\approx \pm(E-i\epsilon)$. This can be done easily using Cauchy's residue theorem

2- Do the 3-momentum integral: first use polar coordinates in 3D $d^{3}p=4\pi p^{2}dp$ and then integrate.

However, this 3 momentum integral gets very complicated. first my $E(p)$ expression contains terms like $p,p^{2}, \sqrt{p^2 -...}$ and the for nonzero $\epsilon$ I get something like: \begin{equation} \int dp \frac{-\epsilon E^2 p^2 +i E^3 p^2}{E^2+\epsilon^2} \end{equation}

Compared to the relativistic case where we have in the denominator the nice mass shell expression $p^2-m^2$. There we integrate over 4 momentum right away (after applying Wick rotation to get 4D polar coordinates). Is it possible to treat the non-relatitivsitc integral relativistically (thus using mass shell expression) and take the non-relativistic limit at the end?

Edit: The expression for $E(p)$ is: \begin{equation} E(p)= Ap^2 + Bp +\sqrt{(Ap^2 - Bp)^2 + C} \end{equation}

  • $\begingroup$ What is the expression of $E(p)$ ? To take the non-relativistic limit as you say, one necessary condition is that you get $E(p)$ when getting the non-relativistic limit of $E^2=p^2+m^2$ (for a free particle). $\endgroup$
    – Frotaur
    Commented Oct 20, 2022 at 11:07
  • $\begingroup$ isn't that the case when looking at the denominator? for example $p^{2}-m^{2}=\omega^{2}-\vec{p}^2 - (E^2-\vec{p}^2)=\omega^2 - E^2$.So regardless of the explicit form of $E$, it seems to match the relativistic expression $\endgroup$
    – M91
    Commented Oct 20, 2022 at 11:12
  • $\begingroup$ Sorry but in the above comment what is $\omega$ ? You write $p^2=\omega^2-\vec{p}^2$ and also $m^2=E^2-\vec{p}^2$ but since $p^2=m^2$, it seems you assumed $\omega = E$ (which is certainly true in the relativistic case) $\endgroup$
    – Frotaur
    Commented Oct 21, 2022 at 11:31


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