 Solution

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