Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 43 (Supersymmetry)

Use the hypercharge normalization $Q_{\mathrm{electric}}=T^3+Y$. All matter chiral superfields are written with left-handed Weyl spinors, so the fields denoted by a superscript $c$ contain the charge conjugates of the usual right-handed Standard Model fermions. The MSSM superfield representations areEach fermion generation contributes the first five chiral superfields. Their complex scalar field partners are squarks for the quarks and sleptons for the leptons. Each Higgs chiral doublet contains a Higgs complex scalar field and a higgsino. Every chiral superfield also has a complex auxiliary field. The vector superfields areThey contain the corresponding gauge bosons, gauginos in the Adjoint representation, and real auxiliary fields. A right-handed neutrino chiral superfield is not part of the minimal field content.

A gauge anomaly is a quantum obstruction to a classical gauge symmetry. For hypercharge, triangle diagrams with left-handed Weyl spinors can violate the Ward identity of the gauge boson; an uncancelled gauge anomaly makes the gauge theory inconsistent. Anomaly cancellation sums over every component, including colour and weak multiplicities. Complex scalar fields do not contribute to these chiral gauge anomalies. For one fermion generation, the cubic hypercharge coefficient isThe other coefficients involving a hypercharge gauge boson vanish too. With the fundamental index $T(\mathbf{N})=1/2$,The last line is the mixed gauge-gravitational anomaly. Coefficients with one non-Abelian generator and two hypercharge generators vanish by tracelessness. For completeness, the purely colour cubic gauge anomaly cancels between the two fundamental quark components and the two antifundamentals; the weak group has no perturbative cubic gauge anomaly. Its four left-handed doublets per fermion generation also avoid the Witten SU(2) anomaly. Thus each family is separately anomaly-free, not merely their sum.

The gauginos do not spoil this result: their hypercharge is zero and their Adjoint representation is real. One extra Higgs chiral doublet is different because its higgsino is chiral. For $H_u$, its contributions areThey have no compensating contribution from the Higgs complex scalar field. The higgsino in $H_d$ supplies precisely the negative of each coefficient. Opposite-hypercharge Higgs chiral doublets restore anomaly cancellation. The same pair restores an even number of weak fermion doublets, so the Witten SU(2) anomaly provides an additional check of the higgsino anomaly cancellation.

The independent reason is the holomorphic need for two Higgs chiral doublets. A superpotential is a holomorphic function of chiral superfields, so it cannot use a conjugate Higgs superfield to generate the missing Yukawa couplings. The ordinary Standard Model can use a Higgs scalar and its conjugate, but the MSSM needs distinct chiral superfields of both hypercharges. For example,where the dot contracts weak indices with the antisymmetric tensor. All three terms are gauge-invariant holomorphic functions. $H_u$ supplies up-type masses, while $H_d$ supplies down-type and charged-lepton masses. Replacing either by the conjugate of the other would violate the holomorphic closure of chiral superfields.

Use left Grassmann derivatives and the usual contractions ${\partial }^2={\partial }^{\alpha }{\partial }_{\alpha }$ and ${\bar{\partial }}^2={\bar{\partial }}_{\dot{\alpha }}{\bar{\partial }}^{\dot{\alpha }}$. Set $a={\theta }^1,b={\theta }^2,c={\bar{\theta }}^{\dot{1}},d={\bar{\theta }}^{\dot{2}}$. The printed epsilon convention givesThe reversal of the barred contraction is essential. The left Grassmann derivative obeys the graded Leibniz rule, so ${\partial }_a(ab)=b$ but ${\partial }_b(ab)=-a$. With the index-raising convention,Applying these to the displayed Grassmann algebra monomials gives $A=B=-4$. Since every antisymmetric two-index product is proportional to epsilon, the $12$ and $\dot{1}\dot{2}$ components giveThus $C=-2$ and $D=2$.

In the remaining contraction, moving the first barred Grassmann variable past the second unbarred one introduces a minus sign. Inserting the two spinor identities then givesThe trace normalization is the one explicitly supplied in the paper. These two-component superspace contraction signs therefore giveAs a direct check in the mostly-minus convention, ${\sigma }^0=I$ gives $(ac+bd{)}^2=-2abcd$, whereas $(\theta \theta )(\bar{\theta }\bar{\theta })=-4abcd$. Their ratio is $+1/2$. A convention with $\mathrm{Tr}({\sigma }^{\mu }{\bar{\sigma }}^{\nu })=-2{\eta }^{\mu \nu }$ would change this last metric-relative sign; it is not the printed trace convention.

The defining restriction is the reality conditionFor a non-Abelian gauge group, write the vector superfield in a Hermitian generator basis with real superfield coefficients. This is a condition on the full superfield, not a chirality constraint. It relates conjugate component coefficients and makes the vector and $D$ auxiliary field real. Wess-Zumino gauge is a further supergauge transformation choice removing redundant components; it is not the defining condition for a vector superfield.

Introduce $y^{\mu }=x^{\mu }+i\theta {\sigma }^{\mu }\bar{\theta }$ and retain the same Grassmann variables. The supersymmetric derivatives in chiral coordinates follow from the left Grassmann derivative chain rule. Differentiating $y$ givesThe minus sign in the second relation comes from moving the odd Grassmann derivative past $\theta$. Hence, as operators on a superfield expressed in $(y,\theta ,\bar{\theta })$,Substitution into the two supersymmetric covariant derivatives adds the two unbarred spacetime terms and cancels the two barred ones:These operator equalities prove both requested actions on $V$. In particular, a chiral superfield becomes independent of $\bar{\theta }$ at fixed $y$. The TeX aid corrupts the second formula by replacing its ordinary barred derivative with a covariant one; the original PDF has the ordinary $-{\bar{\partial }}_{\dot{\alpha }}$.

The Abelian field-strength chiral projection is linear in the vector superfield, so isolate the terms containing the gaugino $\lambda$ and the $D$ auxiliary field. Replacing $x$ by $y-i\theta \sigma \bar{\theta }$ leaves these terms unchanged: the shift of the gaugino term would contain three barred Grassmann variables, and the shift of the $D$ term would contain three of each chirality, hence both vanish. Their contribution is thereforeThe $2i({\sigma }^{\mu }\bar{\theta }{)}_{\alpha }{\partial }_{\mu }$ part of the supersymmetric covariant derivative also adds a third barred factor, so it vanishes on these two terms. The ordinary left Grassmann derivative, with ${\partial }_{\alpha }(\theta \theta )=2{\theta }_{\alpha }$, givesAt fixed $y$, $\bar{D}=-\bar{\partial }$ and ${\bar{D}}^2(\bar{\theta }\bar{\theta })=-4$. Thus the chiral field-strength superfield hasThe remaining components come from the other independent terms in the vector superfield; linearity ensures they cannot alter the two coefficients just calculated. This proves the requested components without computing the omitted terms. The gaugino phase and the sign of the vector component are those in the printed Wess-Zumino gauge expansion.

$W_{\alpha }$ is a fermionic chiral spinor superfield. Its chirality follows directly from the Abelian field-strength chiral projection:There are only two independent barred supersymmetric covariant derivatives, and their equal-chirality anticommutators vanish. Every product of three barred derivatives therefore vanishes. Equivalently, the previous calculation has no independent barred Grassmann variable at fixed $y$. The free undotted index makes this a chiral spinor superfield, rather than a scalar chiral superfield; its lowest component is the odd gaugino ${\lambda }_{\alpha }$.

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.