# Disconnected graphs, decomposition principle

1. Aug 29, 2015

### SpookyMulder

Hi, I'm reading Schwartz's book "Quantum field theory and the standard model", section 7.3.2., page 95 (https://books.google.com/books?id=H...ce=gbs_ge_summary_r&cad=0#v=onepage&q&f=false)
where he's talking about disconnected diagrams, the ones that have subsets of external vertices connected to each other without interacting with the other subsets. Schwartz basically states that the disconnected part of the $S$-matrix (or is he talking about the full $S$-matrix here, please correct me if I'm wrong) factorizes into the products of the two $1\to3$ connected processes. Then he says that the connected diagram has only single overall delta-function $\delta^4(\sum p)$ (and cannot produce more delta-functions which is related to the cluster decomposition principle), while the disconnected matrix elements will have extra delta-functions $\mathcal{M}=\delta^4(\sum_{subset} p)(\dots)$.

Finally, he states that disconnected amplitudes are always infinitely larger than the connected ones, so that the interference between connected and disconnected diagrams vanishes. Now, what does it mean that there's no interference? Does it mean that $\left|\mathcal{M}_{disc}+\mathcal{M}_{conn}\right|^2 = \left|\mathcal{M}_{disc}\right|^2+\left|\mathcal{M}_{conn}\right|^2$ or something else? How do we prove it in general? I know I should read Weinberg at some point, but still...

Schwartz also has a problem (7.2) at the end of the chapter related to this discussion (see page 103 https://books.google.com/books?id=H...ce=gbs_ge_summary_r&cad=0#v=onepage&q&f=false)
Attached are the relevant diagrams. Using position-space Feynman rules, I calculated
$$\langle i|S| \rangle_{conn} = i \lambda (2\pi)^4 \delta^{4}(p_1+p_2-p_3-p_4-p_5-p_6)$$
$$\langle i|S| \rangle_{disc} = -g^2 (2\pi)^8 \delta^{4}(p_1-p_3-p_4) \delta^{4}(p_2-p_5-p_6)$$
where $p_1, p_2$ --- incoming, $p_3, p_4,p_5,p_6$ --- outgoing momenta.

How do I show that there's no interference?

Last edited: Aug 29, 2015
2. Aug 31, 2015

### SpookyMulder

One more thing. Is there a book other than Weinberg's that carefully explains these things? Don't get me wrong, I love Weinberg's books, but sometimes you need a less advanced book to understand what Weinberg says and I've never seen anyone except for Schwartz who would talk about cluster decomposition principle.

P.S. Should this thread be moved to homework/coursework forum? Obviously, the question I asked is not related to my hm/coursework, but it seems like you have a rule that requires hm/coursework-TYPE questions to be posted in the hw/coursework forum (sorry I didn't catch this before):

3. Aug 31, 2015

### Avodyne

I don't think Schwartz explains this very well. Here's my understanding of what's going on.

The product of the disconnected diagrams is zero unless both delta-functions are satisfied. This is a much stronger constraint that the single delta-function of the connected diagram. Thus, there are many sets of momenta for which the disconnected diagrams vanish, and then only the connected diagram contributes.

In the case that both delta-functions are satisfied, the product of the disconnected diagrams is infinitely larger than the connected diagram (actually, larger by a factor of VT, where V is the volume of the interaction region and T is the duration of the measurement, in units of the Compton wavelength of the particles, roughly speaking). Thus, in that case, we can ignore the connected diagram.

So we are always in a situation where either only the disconnected diagrams, or only the connected diagram, contributes. I guess that's what Schwartz means by "no interference".

Schwartz also explains why cluster decomposition isn't discussed much: it's automatic in QFT with local interactions. So if you're going to study QFT with local interactions, there's no need to separately worry about cluster decomposition.

4. Sep 1, 2015

### SpookyMulder

Avodyne, thanks for your reply. Now it's more clear. Can you think of any other books where this topic is explained better than in Schwartz's book?
OK, if we go back to the problem that Schwartz assigned, am I supposed to prove that $\left|\mathcal{M}_{disc}+\mathcal{M}_{conn}\right|^2 = \left|\mathcal{M}_{disc}\right|^2+\left|\mathcal{M}_{conn}\right|^2$ (and hence the decay rate is the sum of the decay rates due to the connected and disconnected graphs) by integrating over the phase space?

5. Sep 1, 2015

### Avodyne

Um, not really. I don't think this requires much explanation. In practice, if you want to study a particular connected process experimentally, you simply avoid the regions of phase space where the disconnected processes contribute. I believe Srednicki says a few words about this, but no formulas.
I have no idea what Schwartz wants you to do! I think it's a badly posed problem. As I said, if you are in a region of phase space where the disconnected process is allowed, then it dominates the probability for the total process to occur (assuming it is happening in a large volume over a long time), and if you are not in such a region of phase space, then the disconnected diagrams vanish (again assuming large volume and time), and only the connected process contributes.

6. Sep 4, 2015

### vanhees71

Do you mean the theorem that the (unconnected) $N$-point function the disconnected pieces (i.e., pure vacuum subdiagrams which do not contain an external leg) are just an undefined phase factor and thus irrelevant for observables like cross sections which are $\propto |S_{fi}|^2$? Then you find two proofs in my QFT lecture notes:

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

One, using the operator formalism in Sect. 3.9 and the other, using path-integral methods and generating functions, in Sect. 4.6.

7. Sep 4, 2015

### SpookyMulder

No, we are not talking about pure vacuum subdiagrams here, but disconnected diagrams that have external legs.

I thank Avodyne once again for answering my question.