Scalar potential from a field-dependent momentum cutoff ($V(\phi )=\frac{1}{4{\pi }^2}{\int }_0^{|\phi |}{p}^2\sqrt{p^2+m^2}dp$)

If a free real scalar field's three-momentum ultraviolet cutoff is identified with the absolute vacuum expectation value of another scalar field, the displayed scalar potential is even and has its unique global minimum at zero. It grows as ${\phi }^4/(16{\pi }^2)$ for $|\phi |\gg m$, but as $m|\phi {|}^3/(12{\pi }^2)$ for $|\phi |\ll m$ with $m>0$. The positive derivative with respect to $|\phi |$ proves that this contribution alone does not produce spontaneous symmetry breaking. This cutoff identification is a model assumption, distinct from deriving a complete effective potential for a dynamical field.