Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 307 (Supersymmetry and Duality)

The Kähler potential $K(\Phi ,{\Phi }^{†})$ is a real function integrated over all four fermionic coordinates, $\int d^4\theta K$. Its complex Hessian gives the scalar Kähler metric and therefore the kinetic terms. The superpotential $W(\Phi )$ is holomorphic and is integrated over chiral superspace, $\int d^2\theta W+h.c.$; its derivatives determine Yukawa couplings and the F-term scalar potential.

For the canonical Kähler potential $K={\Phi }^{†}\Phi$, extracting the ${\theta }^2{\bar{\theta }}^2$ component of the supplied chiral-superfield component expansion and integrating by parts gives the bosonic actionin the mostly-plus convention. Extracting the ${\theta }^2$ component of a holomorphic function givesso its bosonic term is $F{W}^{\prime }(\varphi )$, with the Hermitian conjugate understood in a real action. For several fields, eliminating each algebraic auxiliary field by $F_i^{\ast }=-W_i$ produces $V_F={\sum }_i|W_i{|}^2$.

For $W=X^{2}Z$, the F-term scalar potential isBoth terms vanish exactly when $X=0$, while $Z$ is arbitrary. Thus the supersymmetric vacua form one complex line,

For $W=XYZ$,The F-flatness equations require at least two of $X,Y,Z$ to vanish. The vacuum space is therefore the union of the three coordinate axes in ${\mathbb{C}}^3$,and every point on it has zero vacuum energy.

Forthe three F-terms areThe equation $W_Z=0$ forces $X=0$, after which $W_Y=\alpha \ne 0$; hence no supersymmetric vacuum exists. Minimizing over $Z$ sets $Z=-2\beta XY/\gamma$, leavingThe two real quadratic eigenvalues about $X=0$ are $|\gamma {|}^2\pm 2|\alpha \beta |$, so the stated hierarchy makes $X=0$ stable. The vacua areThis is an O'Raifeartaigh model: supersymmetry is spontaneously broken and $Y$ is a classical pseudomodulus.

Write the scalar components as ${\varphi }_i,{\tilde{\varphi }}_i,x$. The superpotential givesChoosing the sign of the Fayet–Iliopoulos term so that positive $\zeta$ favors the positively charged fields, the full Supersymmetric quantum electrodynamics potential isChanging the FI sign convention exchanges $\varphi$ with $\tilde{\varphi }$ in the descriptions below.

Let $m_i=0$ and first take $\zeta >0$. The mass terms and D-flatness require $x=0$ and ${\sum }_i|{\varphi }_i{|}^2-{\sum }_i|{\tilde{\varphi }}_i{|}^2=\zeta$, while F-flatness adds ${\sum }_i{\tilde{\varphi }}_i{\varphi }_i=0$. Dividing by the U(1) gauge action gives the Higgs branchwith complex dimension $2N-2$. For $\zeta <0$, the same result follows after exchanging $\varphi$ and $\tilde{\varphi }$.

Suppose $\zeta >0$ and all $m_i$ are distinct. A nonzero ${\varphi }_j$ forces $x=m_j$. Distinctness then makes every other charged field vanish, and ${\sum }_i{\tilde{\varphi }}_i{\varphi }_i=0$ forces ${\tilde{\varphi }}_j=0$. The D-flatness equation fixes $|{\varphi }_j{|}^2=\zeta$, while its phase is removed by the gauge group. There is therefore one isolated supersymmetric vacuum for each flavor,for a total of $N$ vacua. Negative $\zeta$ gives the analogous vacua with ${\tilde{\varphi }}_j$ nonzero.

For $\zeta =0$ and distinct masses, any nonzero charged field can occur only at one $x=m_j$. At such a point, D-flatness gives $|{\varphi }_j|=|{\tilde{\varphi }}_j|$, whereas F-flatness gives ${\tilde{\varphi }}_j{\varphi }_j=0$; together they force both fields to vanish. Thus only the Coulomb branch remains,of complex dimension one. The points $x=m_i$ are special because the corresponding charged multiplets become massless there.

When every $m_i$ and $\zeta$ vanishes, the Coulomb branch again has ${\varphi }_i={\tilde{\varphi }}_i=0$ and arbitrary $x$. A Higgs branch also occurs at $x=0$: it is the zero-level quotient ofby U(1), and has complex dimension $2N-2$ for $N>1$. The two branches meet at the origin. For $N=1$, the F- and D-flat equations force both charged fields to vanish, so the Higgs branch collapses to that intersection point.

There is a conflict in the printed question: with dynamical $X$ and the displayed superpotential, $F_X=\tilde{\varphi }\varphi =M$ forces $M=0$, so the $N=1$ theory has no vacuum branch parametrized by nonzero $M$. The natural intended calculation is the Kähler quotient of the $N=1$ D-flat SQED matter fields before imposing that extra F-term.

On this quotient, $|\varphi |=|\tilde{\varphi }|$ and the gauge-invariant coordinate is $M=\tilde{\varphi }\varphi$. Hence $|\varphi {|}^2=|\tilde{\varphi }{|}^2=|M|$, and the canonical Kähler potential descends toThe associated Kähler metric isIt is smooth for $M\ne 0$ and has a conical singularity at $M=0$, locally $\mathbb{C}/{\mathbb{Z}}_2$. Physically, the charged fields become massless and the U(1) gauge symmetry is restored there, so integrating out the vector multiplet to obtain a sigma-model metric ceases to be valid.

The transformation of the holomorphic strong-coupling scale is fixed by the mixed $SU(2{)}^{2}U(1)$ anomaly. The baryon-number contributions of $\Phi$ and $\tilde{\Phi }$ cancel, so ${\Lambda }^{b_0}$ is neutral under $U(1{)}_B$. Under $U(1{)}_A$, all $2N_f$ doublets have charge one and Dynkin index one, giving axial charge $2N_f$. For the R-symmetry, each matter fermion has charge $(1-2/N_f)-1=-2/N_f$, so the matter contribution is $-4$; the gaugino has R-charge one and contributes $I(\mathrm{adj})=4$, giving zero total anomaly. Thus

For $N_f=1$, antisymmetry makes the would-be baryons vanish, and the only generator of the chiral ring of a supersymmetric gauge theory is the meson operator in supersymmetric quantum chromodynamics $M=\tilde{\Phi }\Phi$. It obeys no classical constraint. Holomorphy, mass dimension and all three U(1) charges uniquely permit the Affleck–Dine–Seiberg superpotentialup to a nonzero constant absorbed into $\Lambda$. Since ${\partial }_{M}W=-{\Lambda }^5/M^2$ never vanishes at finite $M$, there is no finite supersymmetric ground state. Instead the potential approaches zero as $|M|\to \infty$: the theory has a supersymmetric runaway vacuum at infinity.

For $N_f=2$, the gauge-invariant chiral fields are the four mesons $M_i{}^j={\tilde{\Phi }}_i{\Phi }^j$ and the baryonsClassically they obeyup to the sign chosen in the definitions. No ordinary superpotential has the required R-charge. Nonperturbative dynamics instead produces the quantum-deformed moduli spaceThe right-hand side has exactly the axial charge, dimension and other symmetry properties required to deform the classical relation.

For $N_f=3$, the infrared fields at the origin are the nine mesons $M_i{}^j$ and three baryons plus three antibaryons, for fifteen chiral multiplets. Every elementary quark superfield has R-charge $1/3$, so its fermion has charge $-2/3$. In the ultraviolet there are twelve quark Weyl fermions and three adjoint gauginos. Hence the two anomalies areEach meson or baryon superfield contains two quarks and has R-charge $2/3$, so each composite fermion has charge $-1/3$. The unconstrained infrared fields therefore giveBoth results agree, establishing the requested 't Hooft anomaly matching. This smooth composite description is the $SU(2)$, $N_f=3$ example of s-confinement.

Two theories are Seiberg duals when they are distinct ultraviolet gauge theories that flow to the same infrared quantum field theory. Their gauge-invariant operator spectra, global symmetries, 't Hooft anomalies and responses to deformations agree, even though one description may be strongly coupled where the other is weakly coupled.

Quantum consistency requires cancellation of the $SU(N{)}^3$ gauge anomaly. The two-index symmetric square has cubic anomaly coefficient $A(S)=N+4$, whereas each antifundamental has coefficient $-1$. Therefore

The mixed $SU(N{)}^{2}U(1{)}_R$ anomaly receives $I(\mathrm{adj})=2N$ from the gaugino. A chiral multiplet of R-charge $r$ contains a fermion of charge $r-1$. Withthe symmetric-tensor fermion contributeswhile every antifundamental fermion contributes zero. The total is $2N-2N=0$, so this is a nonanomalous R-symmetry of the quantum theory.

Using the supplied Dynkin indices and $p=N+4$, the one-loop beta function of a supersymmetric gauge theory isThus $b_0>0$ for every nontrivial $SU(N)$ in this family, and the electric theory is asymptotically free.

In the proposed Spin(8) theory, the gaugino contributes $I(\mathrm{adj})=6$ to the mixed gauge-R anomaly. The $p$ fields $q$ have superfield R-charge one, so their fermions contribute zero. The spinor $t$ has fermion R-charge $-5-1=-6$ and index one, giving $-6$. The singlets $M$ and $U$ do not enter the gauge anomaly. Henceand the proposed R-symmetry survives quantization.

Gauge invariance, $SU(p)$ flavor symmetry and R-charge two permitwith the Spin(8) vector and spinor indices contracted by their invariant bilinear forms. Indeed, $R[Mqq]=0+1+1=2$ and $R[Utt]=12-5-5=2$. Powers of the matching scale can be inserted to give fields whichever canonical mass dimensions are chosen.

Only the $p$ Spin(8) vectors $q$ and the one spinor $t$ contribute to the magnetic gauge beta function. The supplied indices giveThe magnetic theory is infrared free when $b_0^{\mathrm{magnetic}}<0$, namelyAt $N=13$ the one-loop coefficient vanishes and higher-order dynamics decides the flow. For $N>13$, Seiberg duality says that the strongly coupled low-energy limit of the asymptotically free electric $SU(N)$ theory is described by the weakly coupled Spin(8) fields and the superpotential from part v.

Choose the flavor orientation in which each electric $\tilde{\Phi }$ is a fundamental of $SU(p)$. Its $N$ color components then giveIn the magnetic theory, the eight Spin(8) components of $q$ transform in the antifundamental and contribute $-8$. The singlet $M$ transforms in the symmetric representation of $SU(p)$, whose cubic anomaly coefficient is $p+4$, while $t$ and $U$ are flavor singlets. Thereforewhere $p=N+4$ was used. The two $SU(p{)}^3$ 't Hooft anomalies match, as required by Seiberg duality.