I've read somewhere that disconnected diagrams do not contribute to the S-matrix. I don't see why this is the case. I do know why vacuum bubbles do not contribute: given a generating functional for a scalar field the n-point correlation function follows from: $$ G(x_1, \dots,x_n) = \frac{1}{i^n}\frac{\delta}{\delta J(x_1)}\frac{\delta}{\delta J(x_n)}Z_0[J]|_{J=0} $$ If you treat the generating functional in perturbation theory, expaning out the denominator of $$ Z[J] = \frac{\int\mathcal{D}\phi\exp(iS + i\int J\phi dx)}{\int\mathcal{D}\phi\exp(iS)}$$ will cancel all vacuum bubbles from the numerator. Does something similar hold for the disconnected diagrams?

  • $\begingroup$ Have a look at Peskin and Schroeder section 4.6. $\endgroup$
    – user7757
    Jan 24, 2013 at 5:52

1 Answer 1


I think they do contribute to the S-matrix.The amplitude of disconnected diagrams is the product of the amplitudes of all disconnected pieces. For example, putting two connected 2-particle scattering diagrams will give you a 4-particle scattering process, but it's not that physically interesting because this process is not a "genuine" 4-particle process in the sense that it's really just two 2-particle scattering processes happening independently(and the jargon is "cluster-decomposed" process), so that it's best to study the connected pieces separately(these pieces correspond to the connected part of S-matrix). A good reference on this is Weinberg's QFT Vol1 chapter 4.

So in a word, disconnected diagrams do contribute to the S-matrix, but not the connected part of S-matrix(the second half of the sentence is a tautology if one uses connected diagram as the definition of "connected part of S-matrix", but it won't be a tautology if one uses Weinberg's recursive definition)

  • $\begingroup$ Is the point you are trying to make the fact that the disconnected diagrams already contribute to the S-matrix via the connected diagrams? So including them would be some sort of double counting? $\endgroup$
    – Funzies
    Jan 27, 2013 at 13:54
  • $\begingroup$ No, what I mean is they do contribute to S-matrix, and they are not double counting, it's just that they don't contribute to the "interesting" part of the S-matrix. $\endgroup$
    – Jia Yiyang
    Jan 27, 2013 at 14:28

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