# Deriving a Massive Propagator from a Massless Propagator

I'm trying compute something I already know the answer to in order to test myself and gain confidence in my QFT computational skills, but I'm not getting the right factors. The text I'm following and using conventions from is Quantum Field Theory by Mark Srednicki.

For a massless propagator, we have:

$$\Delta(x-y)=\int\frac{d^4k}{(2\pi)^4}\frac{e^{ik(x-y)}}{k^2}$$

Now my reasoning is that I can introduce a mass term as an interaction, and after adding up all the perturbations I should retain exactly the massive propagator. What I need to compute is this:

$$\Delta_\text{massive}(x-y)=$$

$$\langle\phi(x)\phi(y)\exp\left(-\frac{im^2}{2}\int d^4z \thinspace \phi^2(z)\right)\exp \left(\frac{i}{2}\int d^4v\int d^4u\thinspace J(u)\Delta(u-v)J(v) \right)\rangle=$$

$$\frac{\delta}{i\delta J(x)} \frac{\delta}{i\delta J(y)}\exp\left(-\frac{im^2}{2}\int d^4z \thinspace \frac{\delta^2}{i^2\delta J^2(z)}\right)\exp \left(\frac{i}{2}\int d^4v\int d^4u\thinspace J(u)\Delta(u-v)J(v) \right)$$

Where we take the source term $J$ to zero after the expansion. Since I have two derivatives of $J$ acting on the whole expression, the only surving n-th order term of the first exponential expansion should act on the (n+1)-th order term of the second.

Letting $\frac{\delta}{i \delta J(x)}=-i\delta_x$, the n-th order correction in $m$ of the propagator should be:

$$\Delta^{(n)}(x-y)=$$

$$-\delta_x \delta_y \frac{1}{n!}\frac{1}{(n+1)!}\left(\frac{im^2}{2}\int d^4z \thinspace \delta_z^2\right)^n \left(\frac{i}{2}\int d^4v\int d^4u\thinspace J(u)\Delta(u-v)J(v) \right)^{n+1}=$$

$$- i \frac{1}{n!}\frac{1}{(n+1)!}\left(-\frac{m^2}{4}\right)^n \delta_x \delta_y \int^n \delta^{2n} \int^{2n+2} J^{2n+2}\Delta^{n+1} =$$

Where I got a little lazy with notation and am just trying to represent the number of integrals, factors of $J$, and $\delta$s. Now here is maybe where I went wrong, but I though about if carefully and tried it on smaller computations, and it worked out. I have $2n+2$ derivatives acting on $2n+2$ sources, making $2n+2$ delta functions which will kill $2n+2$ integrals and give my a factor of $(2n+2)!$ Giving me:

$$\Delta^{(n)}(x-y)=- i \frac{1}{n!}\frac{(2n+2)!}{(n+1)!}\left(-\frac{m^2}{4}\right)^n \int^n \Delta^{n+1}$$

As a sanity check I convinced myself this makes sense since I will have 2-free coordinates of my $n+1$ propagators in $n$ integrals, and I'm basically integral over the location of $n$ 'mass' interactions in-between my starting and ending point. Applying a Fourier transform to both sides and plugging in the original massless propagator yields:

$$\tilde{\Delta}^{(n)}(p)= - i \frac{1}{n! }\frac{(2n+2)!}{(n+1)!} \frac{1}{4^n}\left(-\frac{m^2}{p^2}\right)^n \frac{1}{p^2}$$

Which is dauntingly close if only that big mess of an expression out front was a constant. I plotted it and it diverges as $O(\sqrt n)$. If I can get rid of that then the sum would be a geometric series and I'd get:

$$\Delta_\text{massive} \propto \frac{1}{p^2+m^2}$$

Let me know where I went wrong, or if the reasoning is sound. If it isn't, I'd very much like to know why.

• I think extra care should be taken with delta-functions canceling the integrals. In particular, there’s a term corresponding to each of the possible cancelation patterns, and those terms should cancel out your factors depending on $n$. Alternatively, you can show that only a single Fourier mode contributes in all propagators and just work with the momentum representation, in which case it is straightforward to obtain the correct result. – Prof. Legolasov Mar 5 '18 at 12:42
• I'm a little uneasy working with the momentum representation from the start. I'm at the point in Srednicki's book where he's worked examples in the position representation, and the momentum representation has these annoying minus signs (for examples, I think the argument of the exponential of the source term in momentum is $i\int \tilde{J}(k) \tilde{J}(-k)/k^2$). Plus, the units when working with functional derivatives are a little screwy as it is... If you could show me how to set up a momentum perturbation expansion for this example it'd be much appreciated. – Connor Dolan Mar 5 '18 at 16:06
• Also note that theres a $Z[0]^{-1}$ factor missing from your calculations. Do you also sum over bubble graphs? – Prof. Legolasov Mar 6 '18 at 5:23
• I thought I had the right to set Z[0] to 1. It doesn't have a $J$ so it would just end up as a constant in expansion. And no, I didn't sum over bubble graphs. Do I have to? Can't I absorb all vacuum fluctuations into $Z[o]$ so I only have to deal with the propagator? – Connor Dolan Mar 6 '18 at 13:21
• Well usually that $Z[0]$ factor isn’t accounted for because only connected diagrams are considered (which is equivalent to dividing by $Z[0]$). – Prof. Legolasov Mar 7 '18 at 1:31