# Do "tadpole diagram with two external legs" always vanish in massless theories?

What I think the physics community means when talking about tadpole diagrams is something like this: and I don't understand why these diagrams always cancel (or why they don't for the Higgs boson) but this isn't the focus of this question. What my professor insists on calling tadpole diagrams are, instead, these: and any diagram where the internal momentum isn't linked to the external one and can therefore assume any value. They claim that these vanish for every massless theory in dimensional regularization because they have a pole in $$d=2$$ but they're "just a number" and can be thrown away. I can't find any similar proof anywhere, but it seems like there is always a reason why these diagrams cancel in massless theories: for $$\lambda \phi^4$$ it is because the mass counterterm can be chosen to cancel it (Peskin & Schoeder, p. 412), for non-Abelian gauge theories it is because the two other 1-loop diagrams cancel it (Peskin & Schoeder, p. 524)... so my question is: can I safely say that in massless theories any diagrams with internal momentum not linked to the external ones is zero?

As I see you understand that these kind of diagrams, in massless theories, lead to so-called scaleless integrals, that is they do not depend on any Lorentz-invariant scale. These integrals always "vanish" in dimensional regularization. Indeed, these integrals can always be reduced to factors of $$I(d) = \int d^dk~(k^2)^{\alpha},$$ where $$\alpha$$ is an arbitrary number. The properties of dimensional regularization include scaling of the measure, so we can conclude for arbitary $$s$$ $$I(d) = \int d^d(s k)~(s^2 k^2)^{\alpha} = s^{d+2\alpha} I(d).$$ Hence $$I(d) = 0$$, unless $$d + 2\alpha = 0$$, but since we want $$I$$ to be continuous $$I = 0$$. Another, surely more satisfying explanation is found in Analytical continuation in QFT.
But this is not really the whole story: Although scaleless integrals never produce finite contributions, the can produce UV and IR poles. Let us consider the integral ($$d = 4-2\epsilon$$) $$\int d^d k (k^2)^{-2} = - \int d^dk \int_0^{\infty} d\lambda \frac{1}{(k^2 - \lambda)^3} \propto \int_0^{\infty} \frac{d\lambda}{\lambda^{1+\epsilon}}.$$ Note that $$\int_0^{\infty} \frac{d\lambda}{\lambda^{1+\epsilon}}$$ does not converge for any $$\epsilon$$. As (more properly than here) explained in Analytical continuation in QFT, we can split up the integral $$\int_0^{\infty} \frac{d\lambda}{\lambda^{1+\epsilon}} = \int_0^1 \frac{d\lambda}{\lambda^{1+\epsilon}} + \int_1^{\infty} \frac{d\lambda}{\lambda^{1+\epsilon}} = \frac{1}{\epsilon_{\text{UV}}} - \frac{1}{\epsilon_{\text{IR}}},$$ where we had to take $$\epsilon = \epsilon_{\text{UV}} > 0$$ in one integral and $$\epsilon = \epsilon_{\text{IR}} < 0$$ in the other. Dim.reg. does not distinguish between $$\epsilon_{\text{UV}}$$ and $$\epsilon_{\text{IR}}$$ so it gives zero. However these poles can still contribute to renormalization, in the case of $$\epsilon_{\text{UV}}$$, or infrared structure, in the case of $$\epsilon_{\text{IR}}$$. So they are indeed important, so it is kind of misleading to say that scaleless integrals are zero. More precisely one should say that the produce only pure UV and IR poles (that precisely cancel each other). However any QFT observable is defined with respect to some subtractive procedure to get rid of these divergences, and after this the contributions from these diagrams are truly zero.