I am studying the terms in the dual Taylor expansion of [itex]Z_{1}(J)[/itex] in [itex]\phi^{3}[/itex] theory, and being introduced to Feynman diagrams in the process. I thought I would try to simplify one of the terms in the expansion so that, after taking derivatives of all the sources, I ended up with integrals that clearly corresponded to each part of the diagram. As an example, I decided to try this for the tadpole with one source, two propagators, and one vertex. This term is complicated enough to have functional derivatives, which I wanted to work with, without appearing too horrendous.(adsbygoogle = window.adsbygoogle || []).push({});

But when I start with the Taylor series term,

[tex]\frac{iZg}{6} \int d^{4}x\left(\frac{1}{i}\frac{\delta}{\delta J(x)}\right)^{3}\frac{i^{2}}{2\cdot4}\int d^{4}yd^{4}zd^{4}y'd^{4}z' J(y)\Delta(y-z)J(z)J(y')\Delta(y'-z')J(z'),[/tex]

and begin taking derivatives, I will get terms where I act a functional derivative on the [itex]J(y)[/itex] and then the [itex]J(z)[/itex], for example. Upon doing the y and z integrals over the resulting dirac deltas, this gives me a propagator of the form [itex]\Delta(x-x) = \Delta(0).[/itex] In fact, all of the terms in my result pick up a factor of [itex]\Delta(0).[/itex] As a consequence, I don't quite get the tadpole diagram I am after, with two propagators, one source, and one vertex, as I "lose" one of the propagators in this manner.

I am sure that I am making some dumb mistake here, but I can't figure out what's happening.

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# A Question about QFT Diagrams and their Integrals

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