Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 43 / 3

In the Polonyi model, the Kähler metric is $K_{z\bar{z}}=1$. The relevant Kähler covariant derivative of a superpotential isThe supergravity auxiliary field is $F^z=-e^{|z{|}^2/2}\overline{D_{z}W}$, up to an irrelevant common phase convention. Thus a constant vacuum preserves supersymmetry exactly when this auxiliary field vanishes. For $m=0$, the superpotential and scalar potential vanish identically, and every constant scalar value is a supersymmetric vacuum.

For $m\ne 0$, put $⟨z⟩=x+iy$. Its supersymmetry condition becomesSince $\beta >0$, it requires $y=0$ and $x^2+\beta x+1=0$. Hence the Polonyi supersymmetry branches areAt these points $W\ne 0$, so the supergravity F-term potential is negative, $V=-3e^K|W{|}^2$: these are supersymmetric Anti-de Sitter spacetime vacua, not zero-energy ones. The condition $D_{z}W=0$ also makes them stationary, as follows by differentiating the supergravity F-term potential.

For $0<\beta <2$ and $m\ne 0$, the auxiliary field cannot vanish anywhere, so any vacuum has supersymmetry breaking. For $\beta \ge 2$, a stationary vacuum at any other scalar value still breaks supersymmetry; the parameter condition alone does not determine which vacuum is selected. In particular, a nontrivial zero-energy vacuum cannot preserve supersymmetry: $D_{z}W=0$ and $V=0$ would also imply $W=0$, whereas $W=0$ gives $z=-\beta$ and $D_{z}W=m^2\ne 0$.

The canonical Kähler potential gives $e^K=e^{|z{|}^2}$ and inverse Kähler metric one. Substituting the superpotential and its Kähler covariant derivative of a superpotential into the supergravity F-term potential yieldsFor real $m$, the prefactor is simply $m^4$; the modulus form also covers a complex phase. To keep both scalar directions explicit, write $z=x+iy$ and $P=1+x^2+y^2+\beta x$. ThenThe negative term is essential: unlike the global F-term scalar potential, the supergravity F-term potential need not be nonnegative. This is why cancelling the cosmological constant does not force the auxiliary field to vanish.

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.