 Solution

A spherical three-momentum ultraviolet cutoff givesFor $m>0$ and $\Lambda \gg m$, the asymptotic expansion isIn particular, the leading ultraviolet divergence is quartic. For $m=0$, $\mathcal{E}={\Lambda }^4/(16{\pi }^2)$ exactly.

Write $\phi =⟨\psi ⟩$. Under the proposed identification, the scalar potential from a field-dependent momentum cutoff is $V_{\mathrm{ind}}(\phi )=\mathcal{E}(|\phi |)$. It is even and nonnegative, and its derivative with respect to $|\phi |$ is $|\phi {|}^2\sqrt{{\phi }^2+m^2}/(4{\pi }^2)>0$ away from zero. Consequently its only global minimum is at $\phi =0$: this contribution alone does not cause spontaneous symmetry breaking. At large $|\phi |$ it grows quartically. At small $|\phi |$ with fixed positive $m$, however,Thus the quartic sketch is an asymptotic shape, not the exact small-field behavior for a massive original field. Identifying a regulator with a new field is an additional modeling prescription; an actual effective potential also depends on the new field's dynamics and any other allowed terms.