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I cannot reproduce the one-loop Higgs mass expression with a cutoff regularization. The standard result found in literature is (which leads to the so-called Veltman condition): \begin{equation}\delta m_h^2 = \frac{3\Lambda^2}{16\pi^2 v^2}(m_h^2+m_Z^2+2m_W^2-4m_t^2)\end{equation} (ignoring contribution from lighter fermions), where $v=246$ GeV.

I manage to recover the expression for the gauge bosons and fermions contributions but get an additional factor $1/2$ for the Higgs contribution.

Defining $V_{Higgs}= -\mu^2|H|^2 + \lambda|H|^4 \subset -\frac{\lambda}{4} h^4$, where $H=(0, \frac{v+h}{\sqrt 2})^T$, the tree-level Higgs mass is $m_h^2=2\mu^2=2\lambda v^2$. The relevant self-energy function is (neglecting the mass in the propagator): \begin{equation} -i \Pi_h = \frac{1}{2}\left(\frac{-i4!\lambda}{4}\right)\int \frac{d^4q}{16\pi^4} \frac{i}{q^2}=3\lambda (-i)\frac{\pi^2}{16\pi^4}\int_0^{\Lambda^2}dq^2_E = (-i) \frac{3\lambda}{16\pi^2}\Lambda^2=(-i) \frac{3\Lambda^2}{16\pi^2 v^2}\frac{m_h^2}{2} \end{equation} where the first $1/2$ factor is a symmetry factor (which can easily be recovered from Wick's theorem). Therefore my result differs by a factor 1/2 as compared to the literature.

For the top contribution for instance, I have no problem:

\begin{equation} -i\Pi_t = -3\left(\frac{iy_t}{\sqrt 2}\right)^2\int \frac{d^4q}{16\pi^4}\text{Tr} \frac{i q_\mu\gamma^\mu}{q^2}\frac{i q_\nu\gamma^\nu}{q^2} = 3\frac{y_t^2}{2}(-4)\int \frac{d^4q}{16\pi^4}\frac{1}{q^2}=- (-i)\frac{3\Lambda^2}{16\pi^2}\frac{4m_t^2}{v^2} \end{equation} which indeed leads to the right result.

Do you see where the problem is ?

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One should actually (obviously..) add the Goldstone contributions. The Feynman rules are (consistent with my previous conventions): \begin{equation} hh\phi_+\phi_-: -2i\lambda , \ \ hh\phi_0 \phi_0: -2i\lambda \end{equation} Since the diagram involving the neutral Goldstone has a symmetry factor of 1/2, the total contribution to the Higgs self-energy from the Goldstones is (Landau gauge): \begin{equation} -i\Pi_\phi=-i\lambda(2+\frac{2}{2})\int \frac{d^4q}{16\pi^4} \frac{i}{q^2} = (-i)\frac{3\lambda}{16\pi^2}\Lambda^2 \end{equation} which is the same as the Higgs contribution, hence the problem is solved.

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