# Considering the mass term of Lagrangian as a perturbation

Suppose we have we have a Lagrangian $$\mathcal{L} = \frac{1}{2}\partial_\mu\phi\partial^\mu\phi - \frac{m^2}{2}\phi^2$$ for a scalar field $$\phi.$$ We now treat the first term as the free Lagrangian $$\mathcal{L}_0$$ with propagator $$\frac{i}{p^2+i\epsilon}$$. The second term will be our interaction term $$\mathcal{L}_I$$. How can we find the 'real' propagator again, i.e. $$\frac{i}{p^2-m^2+i\epsilon}$$?

Geometric series: $$\frac{i}{p^2+i \epsilon}+ \frac{i}{p^2+i \epsilon} (-i m^2) \frac{i}{p^2+ i \epsilon}+ \frac{i}{p^2 + i \epsilon} (-i m^2) \frac{i}{p^2 +i \epsilon} (-i m^2) \frac{i}{p^2 + i \epsilon}+ \ldots \\= \frac{i}{p^2 + i \epsilon}\left(1 + \frac{m^2}{p^2 +i \epsilon} + \left(\frac{m^2}{p^2+i \epsilon} \right)^2+\ldots \right)= \frac{i}{p^2 +i \epsilon} \cdot \frac{1}{1-\frac{m^2}{p^2 + i \epsilon}}=\frac{i}{p^2 -m^2+i \epsilon}.$$

• 2 questions: why can we consider our external lines in the Feynman diagrams to also be propagators? Also we have a condition that $m^2 < p^2$ for the geometric series, is this always true? Commented Mar 27, 2023 at 20:10
• @Geigercounter 1. Just compute the two-point function with the interaction term $-m^2 \phi^2/2$ (use the Wick theorem) and convince yourself that you obtain the series given above. 2. Compute the series for $|m^2/p^2| \lt 1$ and continue the result analytically. Commented Mar 27, 2023 at 20:23
• so it also holds for $m^2 > p^2?$ Commented Mar 27, 2023 at 20:27
• As you can compute the propagator $\frac{i}{p^2-m^2 +i \epsilon}$ directly (without treating the mass term as a perturbation), this is obviously the case. Commented Mar 27, 2023 at 20:33
• You are using perturbation theory, so the coupling ($m$) is small. Commented Mar 28, 2023 at 9:35

The interaction term can be dealt with by comparison with the usual interaction term $$\frac{\lambda}{n!} \phi^n$$ of the $$\phi^n$$ theory, which has the vertex $$-i\lambda$$.

So we basically have the case $$n=2$$ with the replacement $$\lambda \to m^2$$, thus one obtains the Feynman rule $$-i m^2$$ for the two-point vertex. The original propagator is then recovered by summing up all contributions to the propagator to all orders in $$m^2$$, i.e. the massless propagator with $$0,1,2,\dots$$ vertices attached.

Edit: I actually had written out the exact same geometric series as in Hyperon's answer, which was posted while I was typing...

Of course this already assumes that the geometric series converges. The QFT script that I found this in states that one can make use of analytic continuation in case it doesn't