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Assume $m\ne 0$, 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 $|m{|}^4{e}^{x^2+y^2}$ is positive. The derivatives areThese conditions also show that the zero-energy stationary point must be real. If $y\ne 0$, 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 $x^2+y^2=2$. Butwhich 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 $F=s\sqrt{3}t$, $s\in \{1,-1\}$. Stationarity givesCombining these equations gives $1=(s\sqrt{3}-x)t=t^2$. Writing $t=\epsilon \in \{1,-1\}$ producesThe condition $\beta >0$ leaves exactly $(s,\epsilon )=(1,1)$ 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 selectsZero energy alone, without stationarity and stability, would not imply this parameter value. Even zero energy plus stationarity also admits $\beta =2+\sqrt{3}$ on the unstable branch.

For the stable branch found above, $t=x+\beta =1$ and $s=1$. Hence the vacuum expectation value isin the stated Planck units. To verify that it is a vacuum rather than merely a zero-energy stationary point, evaluate the Hessian matrix. At either zero-energy stationary branch,Because $H$ and its first derivatives vanish there, the Hessian matrix of $V$ is just $|m{|}^4{e}^{x^2}$ times this Hessian matrix. For $s=1$, both eigenvalues are positive. This proves a strict local minimum in both real scalar directions. For $s=-1$, $H_{xx}<0$, so the alternative $\beta =2+\sqrt{3}$, $⟨z⟩=-\sqrt{3}-1$ is a saddle point and is excluded from the stable zero-energy Polonyi vacuum.

There is also a useful global check. Set $u=x-(\sqrt{3}-1)$ on the stable branch. Directly completing squares givesBoth remaining coefficients are positive. Thus $H\ge 0$ everywhere, with equality only at $u=y=0$. The positive exponential prefactor proves that this is the unique global minimum, not just a metastable vacuum.

Finally, at the stable vacuum $z+\beta =1$ and $D_{z}W=\sqrt{3}{m}^2$. Its supergravity auxiliary field hasThus the Minkowski vacuum breaks supersymmetry, even though its cosmological constant vanishes. If $m=0$, the potential is flat and the displayed tuned parameter and scalar value are not selected.