Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 55 (Supersymmetry and Extra Dimensions)

For null four-momentum, $\{Q^I,{\bar{Q}}^J\}=2{\delta }^{IJ}\sigma \cdot p$ has rank one in its spinor indices. Positivity makes the null combinations act trivially; the remaining $N$ normalized pairs obey $\{a_I,a_J^{†}\}={\delta }_{IJ}$. Acting with distinct lowering operators on a highest-helicity state gives the helicity spectrum of a massless supermultiplet:A CPT completion of a supermultiplet adds the opposite helicities if needed. Interpreting the physical spin bounds as $|\lambda |\le s$, the width $N/2$ fits inside $[-s,s]$ only if $N\le 4s$. A literal one-sided bound on helicity without CPT would not give this maximum. The maximal spectra areandThus the maxima are (N=4) for spin at most one and (N=8) for spin at most two, with 16 and 256 states. Dimensional reduction of ten-dimensional super Yang-Mills theory on a six-torus gives the first: the vector yields one four-dimensional vector and six real scalar fields, and its Majorana-Weyl spinor yields four Weyl spinors. Reducing eleven-dimensional supergravity on a seven-torus gives the second: metric components supply one graviton, seven vectors and 28 scalars; the three-form supplies 21 vectors, 35 scalars and seven dualized two-form scalars; the gravitino supplies eight gravitini and 56 spin-half fields. These total 28 vectors and 70 scalars, with 128 boson and 128 fermion states.

R-parity is the multiplicative ${\mathbb{Z}}_2$ quantum numberwhere $B,L,s$ are baryon number, lepton number and spin. Ordinary Standard Model particles have $R_p=+1$ and their superpartners have $R_p=-1$. Gauge invariance alone allows MSSM superpotential terms $LL{E}^c$, $LQ{D}^c$, $U^c{D}^c{D}^c$ and $L{H}_u$. Simultaneous baryon- and lepton-number violation can induce rapid proton decay. Exact R-parity forbids these renormalizable terms, since the associated matter parity is odd; it does not forbid every higher-dimensional proton-decay operator, for example $QQQL$.

Two consequences are superpartner production in pairs from ordinary-particle initial states and stability of the lightest supersymmetric particle: a single parity-odd particle cannot decay only into parity-even particles. Cascades therefore end in the lightest supersymmetric particle; if it is neutral and weakly interacting, it produces missing-energy signatures.

A continuous R-symmetry rotates $\theta$ and the supercharges. With $R(\theta )=1$, the superpotential has R-charge two and the fermion in a chiral superfield has one less charge than its scalar. R-parity is only a discrete restriction, represented by combining matter parity with $\theta \mapsto -\theta$, and need not preserve a full continuous R-symmetry. For example, a gaugino Majorana mass term violates an unbroken continuous R-symmetry but preserves R-parity. Thus proton protection by parity and continuous R-charge selection rules are distinct requirements.

In conventional Kaluza-Klein theory, bulk fields have a Kaluza-Klein tower with masses of order $1/R$. If ordinary matter propagates in the extra dimension, the absence of accessible excited modes requires a small $R$: a TeV compactification scale corresponds to roughly $10^{-19}$ m; Planck-scale compactification would be much smaller, of order the Planck length. This is a scale estimate, not a universal radius limit independent of the fields and spectrum.

An unwarped brane-world scenario localizes ordinary fields on a brane while gravity explores a large compact bulk. The volume relationmakes gravity weak without a fundamental Planck-scale cutoff. In the original large-extra-dimension model, $M_{\ast }\sim \mathrm{TeV}$ permits approximately millimetre/submillimetre dimensions for $n=2$; more dimensions are smaller. Ordinary-field KK bounds no longer limit these gravity-only radii, although gravitational and other constraints do. This replaces the large fundamental scale hierarchy by a large compact volume needing stabilization.

A warped brane-world scenario instead uses a position-dependent metric factor. In the two-brane Randall–Sundrum model, $m_{\mathrm{IR}}=e^{-kL}{m}_0$, so $kL\sim 35$ generates a weak scale from a Planck-sized input without a millimetre proper separation; for $k$ of Planck order, $L$ is tens of inverse-Planck lengths. The single-brane version even allows infinite proper size through gravity localization in a noncompact warped dimension; it is not by itself the two-brane redshift hierarchy construction. Thus warped proper size can be unbounded, while finite hierarchy models need specified warping and radius stabilization.

Supersymmetry addresses the hierarchy problem differently: boson/fermion cancellations protect scalar masses against quadratic sensitivity to high scales, with residual corrections controlled by soft supersymmetry breaking. It stabilizes a chosen hierarchy rather than explaining its initial size. Volume dilution and the warped redshift mechanism explain scale separation geometrically; neither automatically determines its moduli or removes all tuning.

For a chiral superfield of mass dimension one, a renormalizable quantum field theory has a quadratic Kähler potential and a superpotential of degree at most three. Before canonical normalization one may writewith real $k_0$, andThe purely holomorphic terms in $K$, their conjugates and its constant have vanishing full superspace integration; they are Kähler transformations of the global theory. Rescaling the chiral superfield sets $Z=1$, so the physically general kinetic choice is $K={\Phi }^{†}\Phi$ up to these terms. The constant superpotential in global supersymmetry also has no component effect. The dimensions of $\ell ,m,g$ are respectively two, one and zero.

To eliminate the linear term, translate the complex chiral superfield as $\Phi ={\Phi }^{\prime }+s$. The shifted linear coefficient isIf $g\ne 0$, this quadratic has a root over the complex numbers; if $g=0$ but $m\ne 0$, take $s=-\ell /m$. The resulting linear term vanishes and the new quadratic coefficient is $m+2gs$. The canonical Kähler potential changes only by holomorphic, antiholomorphic and constant terms, which do not change the global kinetic action. This is the superpotential critical-point shift.

The claim needs an exception: if $m=g=0$ but $\ell \ne 0$, the superpotential is purely linear and no invertible translation can remove it. More generally an invertible regular field redefinition cannot turn a nowhere-zero derivative into a critical point. The theory with $W=\ell \Phi$ has $F=-\bar{\ell }$, a constant positive scalar potential $|\ell {|}^2$, and flat F-term breaking with a linear superpotential. Below, $m$ and $g$ denote the quadratic/cubic coefficients after the valid shift, as intended by the question.

Expand $\Phi =\varphi +\sqrt{2}\theta \psi +{\theta }^{2}F$ in chiral coordinates. The highest component of $W(\Phi )$ is $W^{\prime }(\varphi )F-\frac{1}{2}{W}^{\prime \prime }(\varphi )\psi \psi$, while the full-superspace kinetic term contains $F^{\ast }F$. Thus the auxiliary field part of the component Lagrangian isVarying $F$ and $F^{\ast }$ independently givesSubstitution gives ${\mathcal{L}}_F=-|m\varphi +g{\varphi }^2{|}^2$, and hence the F-term scalar potentialIt is nonnegative and vanishes at $\varphi =0$, and also at $\varphi =-m/g$ when $g\ne 0$. These are supersymmetric vacua, since the auxiliary field vanishes. Thus the quadratic/cubic model does not spontaneously break supersymmetry. If $m=0$ but $g\ne 0$, zero is the sole critical point; if both vanish, every $\varphi$ is a zero-energy vacuum. Nonzero $m,g$ give the usual two vacua, coinciding when $m=0$.

For the loop argument, promote the coefficients to background chiral spurions. One convenient pair of independent formal $U(1)$ charge assignments isBoth symmetries act nontrivially on all three of $\Phi ,m,g$; the second is an R-symmetry. The superspace integration measure $d^2\theta$ has R-charge $-2$, so both tree monomials give invariant actions. These are spurionic selection rules, not necessarily actual symmetries after fixing numerical couplings. The R-row is an ordinary-charge redefinition of the usual Wess–Zumino spurion charge assignment.

A local Wilsonian superpotential is a holomorphic function of the chiral fields and chiral spurions. The ratio $z=g\Phi /m$ is neutral under both charge assignments and has dimension zero, whereas $m{\Phi }^2$ has exactly the charges and dimension of $W$. The general symmetry-allowed form, on a patch with $m\ne 0$ and up to a physically irrelevant global constant, is thereforewith $f$ holomorphic where that description is regular. Symmetry by itself does not determine $f$.

For a regular perturbative local Wilsonian effective action retaining the same elementary field and an infrared cutoff, expand $f(z)={\sum }_n{c}_n{z}^n$. Its term of order $n$ is $c_n{g}^n{m}^{1-n}{\Phi }^{n+2}$. Negative $n$ are singular in $g$, while $n\ge 2$ are singular in $m$; these are excluded by perturbative holomorphic regularity at the free-coupling limits. Thus only the quadratic and cubic terms survive. Their regular holomorphic coefficients cannot acquire coupling-dependent invariant corrections. At $g=0$ the quadratic theory is Gaussian, fixing $c_0=1/2$ exactly. A contribution linear in $g$ to the cubic vertex has only one cubic interaction: three external lines exhaust that vertex, leaving no contraction to form a loop. Thus $c_1=1/3$ as well in the present normalization. This supplies the holomorphy argument for superpotential non-renormalization rather than concluding non-renormalization from symmetry alone.

The Kähler potential can still undergo wave-function renormalization. If its quadratic coefficient becomes $Z$, canonical normalization gives $m_c=m/Z$ and $g_c=g/Z^{3/2}$, so physical masses and interactions can run despite superpotential non-renormalization. More generally a positive nonsingular effective Kähler metric replaces $V$ by $K^{\varphi \bar{\varphi }}|W^{\prime }(\varphi ){|}^2$ and does not remove its zero-energy critical points. Infrared-singular nonlocal one-particle-irreducible terms and eliminating the entire massive field are outside this Wilsonian statement.

Use a flat five-dimensional metric of signature $(+----)$ and a circle coordinate $y\sim y+2\pi r$. Start with the canonically normalized Maxwell actionExpand the real Maxwell field in a Fourier series:The circle integral makes different modes orthogonal. Writing $a^{(n)}=A_5^{(n)}$ and $k_n=n/r$, the resulting four-dimensional action isThe scalar sign is positive because the fifth direction is spacelike in the chosen mostly-minus metric. This expression follows directly by separating the $\mu \nu$ and the two identical $\mu 5$ contributions in $F_{MN}{F}^{MN}$.

A five-dimensional gauge transformation $A_M\mapsto A_M+{\partial }_M\Lambda$ acts on its modes byFor $n\ne 0$, define the gauge-invariant vectorThen $F_{\mu 5}^{(n)}=-i{k}_n{B}_{\mu }^{(n)}$, so its term is a Proca field mass term $+\frac{1}{2}{k}_n^2{B}_{\mu }^{(n)}{B}^{\mu (-n)}$. Equivalently, a periodic mode-dependent gauge transformation sets $a^{(n)}=0$. The nonzero modes are massive vectors whose longitudinal polarization is supplied by $a^{(n)}$ through the Stueckelberg mechanism:For each positive $n$ there is one complex vector, equivalently the two real sine/cosine vectors, each with three physical polarizations. The negative mode is its conjugate, not another independent complex field. This is the infinite Kaluza-Klein tower of the Maxwell reduction on a circle.

At $n=0$, $\delta a^{(0)}=0$ under periodic infinitesimal gauge transformations. The zero modes are a massless vector with two polarizations and a real massless scalar with one. The scalar is the circle component, or Wilson line degree of freedom; a generic constant value cannot be removed by a periodic gauge parameter. Both have $m_0=0$. At energies $E\ll 1/r$, the free theory's low-energy effective action is thereforeThere is no scalar potential in this free Maxwell theory. If the original action instead has coefficient $1/g_5^2$ and unrescaled constant zero-mode fields are used, integrating the circle gives $g_4^2=g_5^2/(2\pi r)$; canonical field rescaling recovers the normalization displayed above.

For the gravitational part, allow a circle radius field $R(x)$ with vacuum value $r$, and use a dimensionless angle $\chi \sim \chi +2\pi$ in the Kaluza-Klein theory ansatzThe vector comes from off-diagonal metric components; $R$ is the radion. Perform the coordinate transformation ${\chi }^{\prime }=\chi -\kappa \lambda (x)$, leaving $x$ unchanged. Since $d\chi =d{\chi }^{\prime }+\kappa {\partial }_{\mu }\lambda d{x}^{\mu }$, the metric has the same form withThus a higher-dimensional diffeomorphism becomes an Abelian gauge transformation in the reduced theory. Changing the sign convention for the internal coordinate shift reverses the displayed sign without changing its content.

More generally, continuous internal isometries give lower-dimensional gauge symmetries, with vector fields from metric components along the corresponding Killing vector fields. The circle translation group gives $U(1)$. A scalar mode ${\psi }_n(x)e^{in\chi }$ transforms as ${\psi }_n^{\prime }=e^{in\kappa \lambda }{\psi }_n$, so $D_{\mu }{\psi }_n=({\partial }_{\mu }-in\kappa A_{\mu }){\psi }_n$ transforms covariantly. This derives Kaluza-Klein charge quantization in the chosen vector normalization. The metric vector is a graviphoton; if a five-dimensional Maxwell field is present as well, its zero-mode photon is a separate field originating from that Maxwell gauge symmetry.

Take canonical kinetic terms, gauge coupling $g>0$, and a nonzero charge $q$. Define $e=gq$ and choose the convention in which the Fayet–Iliopoulos term contributes $\xi D$ to the Lagrangian. With one charged chiral superfield, gauge invariance permits no nonconstant polynomial superpotential; an irrelevant global constant gives $F=0$. The real auxiliary field part isIts equation of motion is $D=-(\xi +e|\varphi {|}^2)$. Eliminating it gives the D-term scalar potentialThe gaugino transformation has the form $\delta \lambda =iϵD+$ a field-strength term, with a convention-dependent overall phase. In a translationally invariant vacuum with zero field strength, a nonzero $⟨D⟩$ therefore gives an inhomogeneous shift of $\lambda$: no nonzero supersymmetry parameter leaves the vacuum invariant. The massless Goldstino is consequently the gaugino, not the matter fermion, since $F=0$.

Minimize the potential over $u=|\varphi {|}^2\ge 0$. If $q\xi <0$, the allowed value $u=-\xi /e$ sets $D=0$, so supersymmetry is unbroken while the internal gauge symmetry is Higgsed. If $\xi =0$, the origin also has $D=0$. For the broken branch, $q\xi >0$ and the formal zero of $D$ would require negative $u$. The actual minimum isThis is single charged-field D-term breaking. Nonzero FI parameter alone does not force breaking if a scalar can cancel it. The degenerate case $q=0$ is different: any $\xi \ne 0$ breaks supersymmetry in the vector sector while the neutral scalar remains flat; the sign test above assumes an actually charged field.

Expand around the broken vacuum:For $\varphi =({\varphi }_R+i{\varphi }_I)/\sqrt{2}$, both canonically normalized real scalars have squared mass $e\xi =gq\xi >0$. The matter Weyl spinor has no superpotential mass. The gauge boson remains massless because $⟨\varphi ⟩=0$. The gauge Yukawa coupling, proportional to ${\varphi }^{\ast }\lambda \psi$, gives no fermion mixing or mass at this vacuum, so the gaugino is also massless. The tree-level spectrum is thereforeand the chiral-multiplet splitting isThe vector and its gaugino happen to remain degenerate at zero mass even though supersymmetry is broken. If the sign convention for the FI contribution is reversed, the sign criterion changes accordingly; the invariant condition is that the scalar condensate cannot cancel the auxiliary source. This is the classical toy-model spectrum; a quantum completion of the single charged-fermion theory also has to cancel its gauge anomalies.

Set $x=T+\bar{T}>0$, the domain in which the Kähler potential is defined, and use Planck units. The nonzero entries of the Kähler metric and its inverse areThe Kähler covariant derivatives of a superpotential areDenote $W=C^3+B$ and $S=3C^2+\bar{C}W$. The supergravity F-term potential isHere the $T$ contribution is exactly $3|W{|}^2$, cancelling the negative term. This is the no-scale supergravity identity, givingThe cancellation is important: replacing the covariant derivatives by ordinary derivatives would miss both the matter dependence and the possibility of zero vacuum energy with supersymmetry breaking.

In the convention $F^i=-e^{K/2}{K}^{i\bar{j}}\overline{D_{j}W}$, the chiral supergravity auxiliary fields areAn overall auxiliary phase or sign convention does not change their vanishing conditions. These are upper-index fields, including the inverse Kähler metric, rather than just the covariant derivatives.

Since $V$ is nonnegative and vanishes at $C=0$, its minimum energy isThere are additional minima that should not be discarded by examining only real $C$. For $C=\rho e^{i\alpha }\ne 0$, the condition $S=0$ becomesIf $B\ne 0$, the magnitude equationhas exactly one positive root: its left side is strictly increasing from zero to infinity. There are three phases, differing by $2\pi /3$, determined by $e^{3i\alpha }=-B/|B|$. Thus for nonzero $B$ the finite zero-energy matter vacua consist of $C=0$ and these three nonzero phase-related values, each with arbitrary $T$ in the allowed half-plane. At a nonzero matter vacuum,At $C=0$, $W=B$.

For $B\ne 0$, every such minimum has $F^C=0$ but $F^T\ne 0$, so supersymmetry is spontaneously broken despite zero vacuum energy. Indeed no finite point anywhere is fully supersymmetric: $D_{T}W=0$ would force $W=0$, after which $D_{C}W=3C^2=0$ forces $C=0$ and then $B=0$. This proves breaking for the specified nonzero-constant case without relying just on the positivity of the potential.

For the allowed exceptional value (B=0), the requested breaking statement is false. Then $S=C^2(3+|C{|}^2)$ vanishes only at $C=0$, where $W=F^T=F^C=0$. Every finite allowed $T$ gives an unbroken supersymmetric vacuum with zero energy. This is an explicit counterexample to unqualified breaking for an arbitrary complex $B$.

Both real components of $T$ are flat directions of a scalar potential along every minimum: the imaginary component never appears in $V$, and when $S=0$ the entire $x$ dependence is multiplied by zero. The matter vacua are isolated in the $C$ plane; there is no continuous matter vacuum direction. At $B=0$ the potential begins quartically in $C$, so zero quadratic mass is not a flat vacuum manifold. Away from $S=0$, the real $T$ direction instead has $V\propto x^{-3}$ and a decompactification-type runaway towards zero as $x\to \infty$, not a further finite positive-energy minimum.

The gravitino mass from a superpotential isAt the $C=0$ branch it is $|B|/x^{3/2}$; at each nonzero branch it is $3\rho e^{{\rho }^2/2}/x^{3/2}$. Its value is not fixed because the $T$ modulus is flat. For $B=0$ it vanishes at the supersymmetric minimum. For nonzero $B$, the super-Higgs mechanism absorbs the Goldstino associated with $F^T$ into the massive gravitino. The zero-energy relation is explicitly satisfied:which balances the auxiliary contribution against the gravitational negative term.

Work with metric $\eta =\mathrm{diag}(1,-1,-1,-1)$, ${\sigma }^{\mu }=(I,\mathbf{\sigma })$ and ${\bar{\sigma }}^{\mu }=(I,-\mathbf{\sigma })$, where $\mathbf{\sigma }$ are the Pauli matrices. The supercharges of four-dimensional N=1 supersymmetry form one Weyl spinor $Q_{\alpha }$ and its conjugate ${\bar{Q}}_{\dot{\alpha }}$. Dotted and undotted indices cannot be interchanged; the conjugate charge in the printed mixed relation has a dotted index.

A direct derivation starts from superspace coordinates $x^{\mu },{\theta }^{\alpha },{\bar{\theta }}^{\dot{\alpha }}$, with the four odd coordinates generating a Grassmann algebra. Use left Grassmann derivatives, soOdd derivatives anticommute with one another, and multiplication by distinct odd coordinates also anticommutes. The operators specified by the question can be writtenTheir coefficients do not depend on $x$, hence $[{\mathcal{Q}}_{\alpha },{\mathcal{P}}_{\mu }]=[{\bar{\mathcal{Q}}}_{\dot{\alpha }},{\mathcal{P}}_{\mu }]=0$. Within $\{{\mathcal{Q}}_{\alpha },{\mathcal{Q}}_{\beta }\}$, the two odd derivative terms anticommute; the derivatives of the barred-coordinate coefficients vanish; and the two barred-coordinate multiplication terms cancel because the coordinates anticommute while the spacetime derivatives commute. ThusThis realizes the minimal algebra with no extra tensorial charges.

For the mixed anticommutator, split each operator into its odd derivative and odd-coordinate parts. The derivative/derivative anticommutator is zero. The coordinate/coordinate terms also cancel, since $\theta \bar{\theta }=-\bar{\theta }\theta$. The two remaining pieces areAdding gives the superspace differential realization of N=1 supercharges:In particular the differentiated-coordinate contributions have the same sign, not opposite signs. The graded product rule is the essential step; treating these as ordinary commuting-coordinate derivatives would give an incorrect algebra.

To obtain the Lorentz part, let $x$ transform as a four-vector and $\theta ,\bar{\theta }$ in conjugate spinor representations. DefineThe Pauli matrix multiplication law gives the intertwining identityIt shows that the two terms in ${\mathcal{Q}}_{\alpha }$ transform together as a covariant Weyl spinor: the coordinate derivative transforms in the dual spinor representation, while the barred-coordinate/spacetime-derivative term transforms in that same representation through this identity. Infinitesimally, with $U=\exp [-\frac{i}{2}{\omega }_{\mu \nu }{M}^{\mu \nu }]$, its transformation is $U{Q}_{\alpha }{U}^{-1}=Q_{\alpha }-\frac{1}{2}{\omega }_{\mu \nu }({\Sigma }^{\mu \nu }{)}_{\alpha }{}^{\beta }{Q}_{\beta }$. Comparing with the first-order expansion of $U$ givesIf the paper's spin matrix is denoted ${\sigma }^{\mu \nu }=i{\Sigma }^{\mu \nu }$, this is exactly its first displayed relation. Defining that spin matrix without the factor $i$ instead requires an explicit $i$ in the commutator; this is a generator convention, not a different algebra. The conjugate relation follows by Hermitian conjugation.

Combining the Lorentz transformation with the differential computations gives the Super-Poincaré algebra in the question's notation:The even generators retain the ordinary Poincare algebra. As a closure check, the even superspace variation $\delta =i(ϵ\mathcal{Q}-\overline{ϵ}\bar{\mathcal{Q}})$ gives $\delta \theta =ϵ$, $\delta \bar{\theta }=\overline{ϵ}$ and $\delta x^{\mu }=i\theta {\sigma }^{\mu }\overline{ϵ}-iϵ{\sigma }^{\mu }\bar{\theta }$. Two such variations commute to a translation by $2i({ϵ}_1{\sigma }^{\mu }{\overline{ϵ}}_2-{ϵ}_2{\sigma }^{\mu }{\overline{ϵ}}_1)$, with zero commutator on the odd coordinates, reproducing the same normalization and showing why the anticommutator produces momentum.