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In evaluating contributions to the two-point function in say $\phi^3$ theory to:

$$\langle 0|\phi(x)\phi(y)e^{-i\int d^4z\frac{\lambda}{3!}\phi^3(z)}|0\rangle,$$

at $\mathcal{O}(\lambda^2)$, one of the possible contractions is the usual tadpole diagram. However, the literature often says that diagrams which can be rendered disconnected with a single cut do not contribute to this matrix element (Collins Renormalization pg. 41, Peskin & Schroeder pg. 219).

My question: Does this mean that tadpoles don't contribute to the self energy since one can separate the bubble from the source with a cut? That doesn't sound right to me, but maybe I could keep them but also include a counterterm

$$\langle 0|\phi(x)\phi(y)e^{i\int d^4z\left(\frac{\lambda}{3!}\phi^3(z)+c\phi\right)}|0\rangle.$$

One could generate an $\mathcal{O}(\lambda^2)$ contribution via the mixed term $$-\int d^4z_1 d^4z_2\frac{\lambda^2 c}{3!}\phi^3(z_1)\phi(z_2)$$ which would generate something like a tadpole counterterm contribution to the two-point function. I suppose my confusion can be summed up as follows:

  1. If I don't disregard tadpoles in the two-point function, must I include the counterterm $c\phi$ in the interaction Lagrangian?

  2. If so, does the counterterm remove the entire diagram anyway, or just the divergent part (assuming the contribution is finite + divergent)?

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  • Yes, in general the self-energy $\Sigma=G_0^{-1}-G_c^{-1}$ may contain tadpoles$^1$ even though they are not 1PI.

  • However, if one imposes the renormalization condition $\langle \phi \rangle_{J=0}=0$, then one may show that the self-energy only contains 1PI diagrams, and hence no tadpoles, cf. my Phys.SE answer here.

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$^1$ NB: Note that a self-loop diagram is not necessarily a tadpole diagram, cf. Wikipedia.

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  • $\begingroup$ So I imagine this means whether I solve for $c$ by requiring $\langle 0|\phi(x)\phi(y)\exp(i\int\left(-\frac{\lambda}{3!}\phi^3+c\phi\right)|0\rangle$ not have contributions from tadpoles or by requiring $\langle 0|\phi(x)|0\rangle=0$ both give me the same value for $c$. Is that correct? $\endgroup$
    – Adots005
    Nov 3, 2021 at 20:30
  • $\begingroup$ ^Update: I just checked, it does. Thank you! $\endgroup$
    – Adots005
    Nov 3, 2021 at 22:07

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