= Solution
For the stable branch found above, $t=x+\beta=1$ and $s=1$. Hence the <vacuum expectation value> is
$$
\boxed{\langle z\rangle=\sqrt3-1,\qquad C=3,\quad D=-1,}
$$
in 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,
$$
H_{xx}=4s\sqrt3,\qquad H_{yy}=8-4s\sqrt3,\qquad H_{xy}=0.
$$
Because $H$ and its first derivatives vanish there, the <Hessian matrix> of $V$ is just $|m|^4e^{x^2}$ 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 $\beta=2+\sqrt3$, $\langle z\rangle=-\sqrt3-1$ is a <saddle point> and is excluded from the <stable zero-energy Polonyi vacuum>.
There is also a useful global check. Set $u=x-(\sqrt3-1)$ on the stable branch. Directly completing squares gives
$$
H=\left(u^2+y^2+\sqrt3\,u\right)^2+(2\sqrt3-3)u^2+(4-2\sqrt3)y^2.
$$
Both remaining coefficients are positive. Thus $H\geq0$ 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 $D_zW=\sqrt3\,m^2$. Its <supergravity auxiliary field> has
$$
\boxed{|F^z|=\sqrt3\,|m|^2e^{(\sqrt3-1)^2/2}>0\qquad(m\ne0).}
$$
Thus \b[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.
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