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My notes introduce the truncated expectation in the following way: given $S_0$ a quadratic form, consider a generating function $$e^{W(J)} = \int \prod_x d\varphi_x e^{-S_0+ JO} = \int P(d\varphi) e^{JO}. $$ Let define the truncated expectation of $O$ as $$ E^T(O;n) = \frac{\partial^n}{\partial J^n} W(J) \bigg|_{J=0}= \frac{\partial^n}{\partial J^n} \log \int P(d\varphi) e^{JO} \bigg|_{J=0} $$ And so for example $E^T(O;1)=\frac{\partial}{\partial J} \log \int P(d\varphi) e^{JO} \bigg|_{J=0} =\frac{\int P(d\varphi) O}{\int P(d\varphi) }=\langle O\rangle $.

Now, using the Wick rule, I calculate $$E^T(\varphi_1 \varphi_2 \varphi_3 \varphi_4;1)=\langle \varphi_1 \varphi_2 \varphi_3 \varphi_4\rangle = \langle \varphi_1 \varphi_2\rangle \langle \varphi_3 \varphi_4\rangle +\langle \varphi_1 \varphi_4\rangle \langle \varphi_2 \varphi_3\rangle +\langle \varphi_1 \varphi_3\rangle \langle \varphi_2 \varphi_4\rangle. $$

The rules for Feynman diagrams are: construct a Feynman diagram drawing a segment to each field and then join the different segments if there is a contraction between the field, that is if I have the expectation between the field. Then, following this rule I have that $\langle \varphi_1 \varphi_2 \varphi_3 \varphi_4\rangle $ is composed by disconnected diagram, because, for example for the first term that comes from the Wick rule, $\langle \varphi_1 \varphi_2\rangle \langle \varphi_3 \varphi_4\rangle $ I have one segment that connects the vertices $1-2$ and one segment for $3-4$.

Where am I wrong?

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  • $\begingroup$ I don't think you're wrong. The $O$ vertex should be treated as a connected object and the three terms you wrote are connected vacuum graphs made of a single quartic vertex attached to two tadpoles. $\endgroup$ Jan 20, 2021 at 15:44
  • $\begingroup$ Ok thank you, even if I don't really understand what you mean for quartic vertex and tadpoles... I am mathematician and maybe I am not in feeling with these argument... $\endgroup$ Jan 20, 2021 at 16:53
  • $\begingroup$ See my answer below. $\endgroup$ Jan 20, 2021 at 20:50
  • $\begingroup$ Are your notes available? $\endgroup$
    – Qmechanic
    Jan 20, 2021 at 21:48

2 Answers 2

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If your field don't interact, the only this they do is propagating, so there cannot be connected Feynman diagrams with more than two legs. You may have to add an interaction term somewhere in order to get a connected diagram with four legs; although I don't know exactly how this works in the case you describe.

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  • $\begingroup$ Thanks for your answer @Emmy. Could you be more precise? $\endgroup$ Jan 19, 2021 at 21:46
  • $\begingroup$ I'm sorry but I can't, I'm not fluent enough with these things :-/ $\endgroup$ Jan 27, 2021 at 7:58
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$$ Z=\int P(d\varphi)e^{JO} $$ is the partition function of $\varphi^4$ type model. You can rigorously prove in the sense of power series in $\mathbb{C}[[J]]$ that $Z$ is a sum over all graphs with degree 4 vertices (this is math i.e. graph theory terminology) corresponding to the $O$'s, connected or not. When you take the log $W$ you get a similar sum where you only keep connected graphs. In the example above, graphs look like the number 8. You have one central vertex with 4 half edges emanating from it, with two loops (math terminology), i.e., tadpoles (physics terminology).

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