# A Question about QFT Diagrams and their Integrals

1. Nov 13, 2017

### Opus_723

I am studying the terms in the dual Taylor expansion of $Z_{1}(J)$ in $\phi^{3}$ 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.

But when I start with the Taylor series term,

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

and begin taking derivatives, I will get terms where I act a functional derivative on the $J(y)$ and then the $J(z)$, for example. Upon doing the y and z integrals over the resulting dirac deltas, this gives me a propagator of the form $\Delta(x-x) = \Delta(0).$ In fact, all of the terms in my result pick up a factor of $\Delta(0).$ 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.

2. Nov 13, 2017

### Opus_723

And naturally, after puzzling over what was happening all day, I think it finally clicked about 10 seconds after posting this question, and I feel really silly.

If I'm right, the appearance of that degenerate propagator is completely fine, since it corresponds to the loop in the tadpole? This is basically the same infinity that we are attempting to eliminate when we introduce counterterms, I think? I think that didn't click earlier because of some different choices of variables in my book vs. what I'm doing, but I think maybe that's all that has happened.

3. Nov 14, 2017

### vanhees71

Yes, indeed. In the time-position domain you get the singular expression $\propto \Delta(0)$. This is in terms of Feynman diagrams a closed loop connected only one vertex point with a propagator line. It's called tadpole diagram for semi-obvious reasons (the naming "spermium diagram" was rejected by Phys. Rev. as rumor has it).

In momentum space the diagram of course has the meaning of the quadratically divergent integral
$$\int \mathrm{d}^4 p \frac{1}{m^2-p^2-\mathrm{i} 0^+}.$$
If you argue in the operator formalism including normal ordering of the Hamiltonian density, you get 0. In vacuum QFT all tadpoles can be set to 0 within some class of renormalization schemes. In gauge theories, one should not normal order since the tadpole terms are important for off-shell vertex functions to fulfill the Ward-Takahashi (or Slavnov-Taylor identities).