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From this article Fine Tuning Problem (Perez, Weizmann Institute of Science, Lecture Note) in section C, the hierarchy problem arises from the fact that there are quadratically divergent loop contribu18tions to the Higgs mass. The most significant of these divergences come from three sources. They are the top quark, the electroweak gauge bosons, and the Higgs loop. Let us focus on the gauge (V) loop contribution in equation (18)

$$\Pi^{VV}_{hh} =\frac{1}{16\pi^2}g^2\Lambda^2$$

How do we calculate this gauge loop contribution?

In Equation (2.147) Cheng and Li Gauge Theory of Elementary Particle Physics book, the vector propagator in momentum space is $$D_{\mu\nu}(k)=\frac{-i(g_{\mu\nu}-k_\mu k_\nu/M_V^2)}{k^2-M_V^2+i\epsilon}$$

Do we use this propagator? Or we can simplify this propagator to this $$D_{\mu\nu}(k)=\frac{-i}{k^2-M_V^2}$$

Thank you.

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1 Answer 1

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Firstly, the final term that you have written does not have Lorentz indices so the expression is incorrect.

Assuming that what you meant is $$ D_{\mu \nu}(k) = \frac{-i g_{\mu \nu}}{k^2 - M^2_V},$$

this is still incorrect. An easy way of seeing this is, recognizing that computing a loop with such as

$$ \int d^4k \frac{(-i)^2 g_{\mu \nu} g^{\mu \nu}}{(k^2 -M^2_V)((p-k)^2 -M^2_V)},$$

would produce logarithmic UV divergence at most, since at high momenta we have

$$ \int d^4k \frac{(-i)^2 g_{\mu \nu} g^{\mu \nu}}{(k^2 -M^2_V)((p-k)^2 -M^2_V)} \sim \int_0^{\Lambda} \frac{k^3}{k^4} dk \sim \ln \Lambda.$$

More importantly, the symmetry breaking mechanism generates a mass for the $W$ and $Z$ bosons explaining the short range of the weak interactions, so the mass contribution to the propagator cannot be ignored.

In short, you should use the first form of the propagator, which you state.

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  • $\begingroup$ But there is a quartic vertex $\sim H^2 A^2$ in the Standard Model, so there is also a contribution at 1 loop where the integrand has only have one propagator factor (1 quartic vertex + 1 gauge boson propagator in the loop), which is a quadratic divergence. Furthermore the cubic vertex $\sim \partial H A H$ has a derivative, so the correct one-loop integrand corresponding to the diagram in your answer should also include a factor $\sim k^2$ in the numerator coming from the two vertex factors (2 cubic vertices + 1 gauge boson propagator in the loop), again leading to a quadratic divergence. $\endgroup$
    – Andrew
    Commented Nov 27, 2022 at 3:12

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