= Solution
The <scalar potential> is $V=-m^2|\varphi|^2+\lambda|\varphi|^4/4$. Completing the square gives
$$
V=\frac\lambda4\left(|\varphi|^2-\frac{2m^2}{\lambda}\right)^2-\frac{m^4}{\lambda}.
$$
The <Hamiltonian density> also contains the nonnegative terms $|\dot\varphi|^2+|\nabla\varphi|^2$. Its lowest-energy configurations therefore have no time or spatial variation and minimize the displayed square. Thus
$$
\boxed{|\varphi|^2=v^2=\frac{2m^2}{\lambda},\qquad v=\frac{\sqrt2\,m}{\sqrt\lambda}>0.}
$$
The minima form a circle of phases. Selecting the real positive representative $\varphi=v$ chooses one <vacuum> and spontaneously breaks the continuous <global symmetry in field theory>. The stationarity at $\varphi=0$ is a local maximum of the radial potential, not the <ground state>.
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