= Solution
Assume $m\ne0$, since the trivial $m=0$ theory cannot fix $\beta$ or the <vacuum expectation value>. A zero-energy vacuum must satisfy both $V=0$ and stationarity in both real scalar directions. With the notation from the preceding solution, these become $H=H_x=H_y=0$, because the prefactor $|m|^4e^{x^2+y^2}$ is positive. The derivatives are
$$
H_x=2P(2x+\beta)-6(x+\beta),\qquad
H_y=2y(2P+\beta^2-3).
$$
These conditions also show that the zero-energy <stationary point> must be real. If $y\ne0$, the second equation gives $P=(3-\beta^2)/2$; the first then gives $x=-(\beta^2+3)/(2\beta)$. Substituting into the definition of $P$ gives $x^2+y^2=2$. But
$$
x^2=\frac{(\beta^2+3)^2}{4\beta^2}\geq3,
$$
which is impossible. Thus $y=0$, without assuming a real vacuum in advance.
Set $t=x+\beta$ and $F=1+xt$. Since $t=0$ would give $H=1$, it cannot occur. The zero-energy equation gives $F=s\sqrt3\,t$, $s\in\{1,-1\}$. Stationarity gives
$$
2F(2x+\beta)-6t=0\quad\Longrightarrow\quad 2x+\beta=s\sqrt3.
$$
Combining these equations gives $1=(s\sqrt3-x)t=t^2$. Writing $t=\varepsilon\in\{1,-1\}$ produces
$$
x=s\sqrt3-\varepsilon,\qquad \beta=2\varepsilon-s\sqrt3.
$$
The condition $\beta>0$ leaves exactly $(s,\varepsilon)=(1,1)$ and $(-1,1)$. The second branch is a <saddle point>, as its real-direction curvature is negative; the stability calculation in the next solution verifies this explicitly. The <stable zero-energy Polonyi vacuum> therefore selects
$$
\boxed{\beta=2-\sqrt3,\qquad A=3,\quad B=2.}
$$
Zero energy alone, without stationarity and stability, would not imply this parameter value. Even zero energy plus stationarity also admits $\beta=2+\sqrt3$ on the unstable branch.
Back to article page
