 Solution

Use metric signature $(+,-,-,-)$ and the model's rescaled units $c=\hslash =1$. Its exponential implicitly uses dimensionless rescaled field and coupling combinations; the numerical coefficients of the potential are those supplied. Treat $\varphi$ and ${\varphi }^{\ast }$ as independent variables in taking canonical derivatives:The Hamiltonian density isThe kinetic and gradient terms are nonnegative, so a static homogeneous minimum is obtained by minimizing the potential. Put $s=|\varphi {|}^2\ge 0$. For $\lambda =0$, $V=1-s$ is unbounded below. For $\lambda <0$, the exponential decays while $-s$ tends to minus infinity. Thus a stable vacuum requires $\boxed{\lambda >0}$.

For positive $\lambda$,The exponential complex-scalar potential consequently hasFor $0<\lambda <1$, the minima form a circle $\varphi =v{e}^{i\theta }$, with $v^2=-\ln \lambda /\lambda$ and $V_{\min }=(1+\ln \lambda )/\lambda$. For $\lambda \ge 1$ the minimum has $V_{\min }=1$. The symmetry remains global phase rotation in either phase, but a chosen nonzero vacuum breaks it spontaneously. At $\lambda =1$ the origin is a unique minimum with vanishing quadratic curvature and a positive leading quartic term; this critical masslessness should not be confused with a Goldstone mode.