# Do Tadpoles Contribute to Self-energy?

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)?

• 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.
$$^1$$ NB: Note that a self-loop diagram is not necessarily a tadpole diagram, cf. Wikipedia.
• 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? Nov 3, 2021 at 20:30