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In Peskin&Schroeder they explain in a graphical way why the Schwinger functional generates only connected diagrams. However I don‘t understand why they get 2 diagrams since the first diagram is just a symbol for the propagator $\langle\phi(x)\phi(y)\rangle$.

enter image description here

Qmechanic
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Silas
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3 Answers3

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The propagator has infinitely many diagrams in it, not just two. This page says that they can be categorized into ones which are connected and ones which are not. Are you getting confused with just the free propagator, which has a closed form solution?

doublefelix
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  • I know that the diagrams represents infinitely many diagrams. What confuses me is the second diagram because I don‘t know what it represents physically. – Silas Apr 18 '23 at 06:25
  • Well, in the end there is no proven rule that diagrams must represent something physical. It is if anything more of a pleasant surprise that they correspond to some reasonable-seeming interpretation of how particles interact, given that they are originally just a way to keep track of combinatorics. The disconnected diagrams seem to depict two particles which are capable of interaction but don't significantly influence each other. Notably, all of these disconnected diagrams cancel out and thus do not contribute to scattering cross sections. – doublefelix Apr 18 '23 at 07:51
  • Note also that contrary to what you stated in your question, the first diagram in the screenshot does not include all diagrams that contribute to the propagator. It only includes the connected ones. – doublefelix Apr 18 '23 at 07:54
  • But I thought always that the first diagram represents the full propagator and the disconnected diagrams in the interaction picture are cancelled by the Gell-Mann Low Theorem. – Silas Apr 18 '23 at 08:32
  • Good question. There is an important distinction here that the diagram you've asked about is not a vacuum diagram. That is why it did not cancel - it is the disconnected vacuum diagrams which do. Check out this screenshot from my lecture notes from a few years back (sorry the equation numbering is weird): https://imgur.com/a/DMqfVoG – doublefelix Apr 18 '23 at 09:11
  • So the propagator is actually $\langle\phi(x)\phi(y)\rangle_c$ but in the operator formalism we have no source and therefore the propagator is $\langle\phi(x)\phi(y)\rangle$ due to $\langle\phi(x)\rangle=0$? – Silas Apr 18 '23 at 17:26
  • The propagator is just $\langle \Omega | T \phi(x) \phi(y)|\Omega \rangle$ which P&S writes as $\langle \phi(x) \phi(y) \rangle$, and whose perturbation expansion contains no disconnected vacuum diagrams (which have no external lines and are disconnected), but it DOES contain disconnected diagrams that are not vacuum diagrams (in these, all points are connected to an external line, but not all points are connected to all others, as in your example). So the propagator is not $\langle \phi(x)\phi(y)\rangle_C$ but just $\langle \phi(x)\phi(y)\rangle$. In phi^3 you get these – doublefelix Apr 18 '23 at 17:51
  • But isn’t the propagator generated from the Schwinger functional and hence must be connected? – Silas Apr 19 '23 at 10:10
  • Hi Silas, I learned QFT mostly from a physicist perspective; I do have some minor background in mathematical QFT as well but not with Schwinger functionals. They don't seem to appear in Peskin & Schröder. Are you talking about the generating functional which is used in expressions like $\langle \phi(x)\phi(y)\rangle=\frac{1}{i^2}\frac{\delta}{\delta J(x_1)} \frac{\delta}{\delta J(x_2)} Z[J]$? – doublefelix Apr 19 '23 at 16:20
  • What P&S call E[J] is the Schwinger functional. I have looked in Schwartz‘s book on p. 740 and I‘m now pretty sure that the propagator is only the connected part. – Silas Apr 19 '23 at 21:01
  • Ahh I understand the source of confusion. It is not true that the green's function is generated by $E[J]$ (which is called $W[J]$ in Schwartz). $E[J]$ only generates the connected part of the Green's function. In contrast, $Z[J]$ generates the whole Green's function. Schwartz mentions himself on p.738 bottom, that the full green's function contains both connected and disconnected, and that $W$ only generates the connected part of the function. – doublefelix Apr 19 '23 at 22:25
  • Yes that is clear but my problem is that the propagator should be connected since it is the inverse of $\Gamma^{(2)}$ and the definition of $\langle\phi(x)\phi(y)\rangle$ is only true for fields with zero vacuum expectation value. – Silas Apr 20 '23 at 19:14
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  1. Eq. (11.83) in P&S expresses the fact that the 2-point correlation function $$ \langle\phi(x)\phi(y)\rangle~=~\langle\phi(x)\phi(y)\rangle^c +\langle\phi(x)\rangle^c\langle\phi(y)\rangle^c,\tag{11.83}$$ decomposes into connected correlation functions. The similar decomposition for the 1-point function is $$ \langle\phi(x)\rangle~=~ \langle\phi(x)\rangle^c.$$ A proof uses the linked cluster theorem, cf. e.g. eqs. (4) and (5) in my Phys.SE answer here.

  2. The connected 2-point function $\langle\phi(x)\phi(y)\rangle^c$ can in turn be decomposed in terms of free propagators and self-energy, cf. e.g. this Phys.SE post.

Qmechanic
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The first diagram is not a symbol for the propagator $\langle\phi(x)\phi(y)\rangle$. It is a symbol representing a sum of connected diagrams, which is indicated in the first line of Eq. (11.83). The full propagator $\langle\phi(x)\phi(y)\rangle$ is a sum of all possible diagrams (not only connected) constructed with the vertices which are generated with the Lagrangian. As Peskin&Schroder consider a theory with a source current $J$ in this section, one of the possible vertices is

enter image description here

Because of the presence of such a vertex, there exist disconnected diagrams which contribute into $\langle\phi(x)\phi(y)\rangle$. The simplest diagram of such kind is enter image description here Without a source current, such diagrams are usually absent.

E. Anikin
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