I understand that the renormalization condition $\langle0|\phi|0\rangle=0$ permits one to disregard tadpole diagrams in most calculations due to the leg of the tadpole carrying zero momentum. If an operator is being inserted into a diagram containing a tadpole, can one still make such a claim in general? This affects momentum conservation in such a way, depending on whether the operator is inserted at a vertex or propagator, that tadpole diagrams seemingly have a legitimate contribution to the matrix element. An example would be a 2-point insertion in $\phi^3$ theory:

$$\int d^4z_1 d^4z_2\langle p'|\phi(x)^2\phi(z_1)^3\phi(z_2)^3|p\rangle$$

where an example contraction is the following diagram

enter image description here

where the black blob denotes the operator insertion. In this case, the tadpole leg does not necessarily carry zero momentum which makes me very skeptical on discarding such a diagram.

  • $\begingroup$ Which operator is being inserted in your 2-point example? $\phi(x)^2$? $\endgroup$
    – Qmechanic
    Commented Jul 16, 2021 at 21:43
  • $\begingroup$ Correct. Also derivatives $\partial^\mu_x\partial^\nu_y\phi(x)\phi(y)$ where we take the local limit $x=y$ after the derivatives have been taken. The $\phi(x)^2$ multiplies $g^{\mu\nu}$ in case you're worried about the tensor structure being inconsistent. $\endgroup$
    – Adots005
    Commented Jul 16, 2021 at 21:46

1 Answer 1


No, OP's connected diagram with the $\phi(x)^2$ operator insertion is no longer a tadpole diagram, because no matter how we cut the diagram in two by cutting a single line, each of the two connected components will still contain a source term (i.e. an operator insertion or an external line), cf. e.g. the definition given in Ref. 1.


  1. M. Srednicki, QFT, 2007; Chapter 9 see text above eq. (9.24). A prepublication draft PDF file is available here.

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