 Solution

For a single chiral superfield, the supergravity F-term potential, with $\kappa =m_p^{-1}$ restored, isThe canonical Kähler potential has $K_{z\bar{z}}=K^{z\bar{z}}=1$, so the Kähler covariant derivative of a superpotential isConsequently the Polonyi model has the scalar potentialThe negative term is essential; retaining only the square of the supergravity auxiliary field would give the global F-term scalar potential rather than the supergravity F-term potential.

Set $x=\kappa z=a+iy$ and $b=\kappa \beta =2-\sqrt{3}$. In this notation $V=|\mu {|}^2{m}_p^2{e}^{|x{|}^2}U$, whereAt $a_0=\sqrt{3}-1$, $y=0$, one has $a_0+b=1$ and $1+a_0(a_0+b)=\sqrt{3}$, so $U=0$. To prove that this is a local minimum in both real directions, rather than checking only the real axis, put $u=a-a_0$. Direct expansion gives the stable zero-energy Polonyi vacuum identityEvery coefficient outside the square is strictly positive, and the exponential factor is positive. Thus $V\ge 0$ everywhere and equality requires $u=y=0$, provided $\mu \ne 0$. This proves the stronger conclusiona unique global minimum of this Polonyi model. If $\mu =0$, the scalar potential vanishes identically and this uniqueness and supersymmetry breaking conclusion do not hold.

For an independent local stability check, the Hessian matrix of $U$ at this global minimum is diagonal:Since $U$ and its first derivatives vanish there, differentiating the exponential introduces no additional Hessian matrix terms at the global minimum. For canonically normalized real fluctuations $z=z_0+(s+it)/\sqrt{2}$, the squared scalar masses are thereforewhich are both positive.

Although the vacuum energy is zero, the supergravity auxiliary field is not. In the convention $F^z=-e^{{\kappa }^{2}K/2}\overline{D_{z}W}$, its vacuum expectation value obeysThe matter fermion transformation contains a term proportional to $F^zϵ$, so no nonzero constant supersymmetry parameter leaves this vacuum invariant. Hence supersymmetry is spontaneously broken even though the vacuum energy vanishes: the positive auxiliary field contribution cancels the negative gravitational term. The matter fermion supplies the goldstino, which is absorbed by the gravitino in the super-Higgs mechanism.

Using the gravitino mass from a superpotential, and $z_0+\beta =m_p$, givesAn auxiliary field $F^z$ has mass dimension two. Defining the supersymmetry breaking scale by ${\Lambda }_{\mathrm{SUSY}}=\sqrt{|F^z|}$, one obtainsThus the gravitino mass is small compared with the supersymmetry breaking scale when that scale is sub-Planckian, as in the usual hierarchy $|\mu |\ll m_p$. This smallness is gravitational suppression; it is not a conclusion valid for arbitrary choices of the mass parameter.