Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 307 (Supersymmetry)

In the chiral coordinate $y^{\mu }=x^{\mu }+i\bar{\theta }{\bar{\sigma }}^{\mu }\theta$, a chiral superfield isHere $\varphi$ is a complex scalar, ${\psi }_{\alpha }$ is a two-component Weyl spinor, and $F$ is a complex auxiliary field. Off shell, $\varphi$ and $F$ contribute two real bosonic components each, while $\psi$ has two complex Grassmann components. Thus there are four real bosonic and four real fermionic degrees of freedom.

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The eight Gell-Mann matrices generate $G=SU(3)$, and the three superfields form one fundamental triplet. A renormalizable superpotential may have degree at most three. There is no invariant vector or symmetric rank-two tensor in the fundamental. The only apparent cubic singlet isbut chiral superfields are Grassmann-even and commute, so this contraction of a symmetric product with ${ϵ}^{ijk}$ vanishes identically. Consequently the most general invariant superpotential as the question is written is only a constant,which has no effect in global supersymmetry. In particular, $W=\lambda {\Phi }_1{\Phi }_2{\Phi }_3$ is not invariant under a general $SU(3)$ rotation.

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The canonical Kähler potential isUnder $\Phi \mapsto U\Phi$ with $U\in SU(3)$, it becomes ${\Phi }^{†}{U}^{†}U\Phi =K$, so it respects $G$.

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The F-term scalar potential isSince the invariant superpotential is constant, all derivatives vanish andThere are no gauge fields in the stated model, so there is no D-term potential.

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Soft supersymmetry breaking means adding supersymmetry-breaking operators with positive-mass-dimension coefficients that do not create new ultraviolet quadratic divergences. Standard examples are scalar masses, holomorphic bilinear and trilinear scalar couplings, gaugino masses, and singlet tadpoles when the field content and symmetries allow them.

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For one $SU(3)$ triplet and no gauge multiplet, the only nonzero $G$-invariant soft term is the universal scalar mass,An invariant bilinear does not exist, and the formal trilinear $A{ϵ}^{ijk}{\varphi }_i{\varphi }_j{\varphi }_k$ vanishes because the scalar fields commute. Thus the phrase “general soft terms” does not generate further interactions for the field content stated in the question.

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There are no scalar interaction vertices in the theory as written. The canonical Kähler term supplies kinetic terms, the constant superpotential supplies none, and the soft term is only a two-point mass insertion with Feynman rule $-i{m}_s^2$ in Minkowski conventions. Hence no interaction diagram involving only scalar components exists.

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The Yukawa matrix is $W_{ij}={\partial }_i{\partial }_{j}W$, which vanishes for constant $W$. Therefore there is no scalar-fermion interaction vertex and no corresponding Feynman diagram. The requests in parts (vi)(a) and (vi)(b) would become nontrivial only after changing the field content or reducing the symmetry so that a nonzero invariant superpotential exists.

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The three Weyl components form one fundamental representation of the global $SU(3)$, so coupling $G$ to background gauge fields reveals a nonzero $SU(3{)}^3$ 't Hooft anomaly. This does not destroy the global symmetry in flat space or make the theory inconsistent; it constrains any infrared description through 't Hooft anomaly matching. If $G$ were instead intended as a dynamical gauge symmetry, the same uncancelled gauge anomaly would make the model inconsistent and additional chiral matter would be required. Ordinary quantum corrections cannot generate the forbidden superpotential, and the non-renormalization theorem protects its absence.

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A vacuum preserves supersymmetry exactly whenA nonzero expectation value alone therefore does not decide the issue. On the real slice this condition is $1+{\varphi }^2+\beta \varphi =0$, which has real solutions only when $\beta \ge 2$. The sub-Planckian Minkowski solution found below has $\beta =2-\sqrt{3}<2$, so $D_{\varphi }W\ne 0$ there and supersymmetry is spontaneously broken.

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For $K=|\varphi {|}^2$,and the supergravity F-term potential isThe potential is invariant under complex conjugation. Writing $\varphi =x+iy$, its derivative with respect to $y$ vanishes at $y=0$, so the extremisation equations consistently admit a real solution. On that slice,

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At a real Minkowski vacuum, $V=0$ giveswhere the branch relevant to the sub-Planckian solution has been chosen. Since the bracket multiplying $e^{x^2}$ vanishes, stationarity givesCombining the equations yields $2x+\beta =\sqrt{3}$. Solving givesThus $X=3$, $Y=2$, $U=-1$, and $Z=3$.

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At $x_0=\sqrt{3}-1$ and $\beta =2-\sqrt{3}$,The stationary point is therefore a local minimum along the real direction. In fact $V(x)$ is positive away from this point on the real axis, tends to $+\infty$ as $x\to \pm \infty$, decreases to the zero-valued minimum at $x_0\approx 0.732$, and then rises again. Its sketch is a single well tangent to the $V=0$ axis at $x_0$.

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With canonical Kähler potential, the Wess–Zumino model action isA spurion is a nondynamical background superfield assigned charges so that a coupling transforms covariantly and the action can be treated as formally symmetric.

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Take an ordinary $U(1)$ and an $R$-symmetry, assigning $q(\Phi )=R(\Phi )=1$. The superpotential has chargesbecause $d^2\theta$ is neutral under the first symmetry and has $R$-charge $-2$. Formal invariance of each term then assigns

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The ratiois neutral under both $U(1)$ symmetries. The unique monomial with mass dimension three and charges $(0,2)$ isHolomorphy and the spurion symmetries therefore requireExpanding a term givesRegularity as $g\to 0$ requires $n\ge 2$, while regularity as $m\to 0$ requires $n\le 3$. Hence only $n=2,3$ can occur.

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Weak-coupling matching givesbecauseThus $W_{\mathrm{eff}}=W_{\mathrm{tree}}$. The holomorphic superpotential receives neither perturbative nor nonperturbative corrections compatible with the limits and symmetries. The Kähler potential and wavefunction normalization can still be renormalized, so canonically normalized couplings can run through anomalous dimensions.

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Eliminating the auxiliary field givesExplicitly,The two supersymmetric vacua for $g\ne 0$ are $\varphi =0$ and $\varphi =-m/g$.

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The component interactions include the Yukawa vertexand the quartic scalar vertex $-|g{|}^2|\varphi {|}^4$, with cubic scalar interactions fixed by the same $m$ and $g$. At one loop, the scalar propagator has a bosonic tadpole from the quartic vertex and a closed fermion loop with two Yukawa vertices. The loop's fermionic minus sign makes their quadratic cutoff dependence equal and opposite; diagrams involving the dimensionful cubic scalar vertex are at most logarithmically divergent. This is the supersymmetric cancellation of quadratic divergences, and it relies on the coupling relations imposed by supersymmetry.

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