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Adding $-({\varphi }^2+{\varphi }^{\ast 2})/2$ to the exponential complex-scalar potential gives $V=-2x^2+e^{\lambda (x^2+y^2)}$, where $\varphi =x+iy$. Phase rotations retain only $\varphi \mapsto -\varphi$; the full real-field internal symmetry also permits $\varphi \mapsto {\varphi }^{\ast }$, giving two independent sign reflections. For $0<\lambda <2$ the minima are $\varphi =\pm v$, $v^2=\ln (2/\lambda )/\lambda$. Around either minimum, $\varphi =v+(h+i\pi )/\sqrt{2}$ gives squared masses $4\ln (2/\lambda )$ and $2$. For $\lambda \ge 2$ the unique minimum is zero. Explicit anisotropy removes the continuous vacuum circle, so discrete symmetry breaking supplies no mandatory Goldstone boson.

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.