Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 53 / Section B

Spontaneous supersymmetry breaking preserves the action and conserved supercharges, but the vacuum fails to be invariant. This differs from explicit breaking, which changes the symmetry of the action itself. In global four-dimensional N=1 supersymmetry, summing diagonal entries of the Super-Poincaré algebra writes the Hamiltonian as a sum of positive charge anticommutators. Its expectation is a sum of squared norms. Consequently a zero-energy vacuum is supersymmetric, whereas an existing vacuum with broken global supersymmetry has positive energy density. The energy zero is fixed by the algebra. Neither this positivity statement nor the particle-multiplet mass relations should be transferred without qualification to supergravity.

For supersymmetric quantum mechanics, let $H=\{q,q^{†}\}/2$, $q^2=0$, and let fermion parity anticommute with both charges. On a positive-energy eigenspace, $(q+q^{†})/\sqrt{2E}$ is an odd involution: its square is one, it preserves energy, and it reverses parity. Thus every positive-energy bosonic state has a fermionic partner and conversely. Their contributions cancel in the Witten indexFor a discrete spectrum with a well-defined trace, this proves independence of $\beta$. Continuous changes of parameters also preserve the index if the trace stays well defined and no state escapes to infinity or crosses an unregulated continuum threshold. A nonzero index guarantees an unbroken supersymmetric ground state; a zero index is inconclusive. For example, de Rham supersymmetric quantum mechanics on a circle has a bosonic constant zero-form and a fermionic constant one-form, both normalizable zero modes, so it is unbroken with index zero.

A more explicit one-dimensional family illustrates how normalizability determines the answer. DefinePositivity of these Hamiltonian operators shows that a zero mode must solve $A{\psi }_{-}=0$ or $A^{†}{\psi }_{+}=0$. Thus the only candidates areFor $w(x)=x$, the Gaussian minus-sector wavefunction is normalizable and its reciprocal is not, giving an unbroken model with $I=1$ when the minus sector is assigned even parity. For $w(x)=x^2+a$, $a>0$, the exponent $x^3/3+ax$ tends to opposite infinities at the two ends, so neither candidate is normalizable. Both partner potentials $[w^2\mp w^{\prime }]/2$ grow quartically; their spectra are discrete and their lowest energies cannot be zero. The lowest states are therefore paired at positive energy, $I=0$, and supersymmetry is spontaneously broken. This is supersymmetric factorization and zero-mode normalizability. In quantum mechanics the paired ground states illustrate the charge algebra; a relativistic massless-particle conclusion requires field-theoretic hypotheses.

In relativistic global field theory, Goldstone's theorem has a fermionic version. If a conserved supercurrent charge does not annihilate the vacuum, its Ward identity has a nonzero vacuum contact term. Fourier transforming, the divergence of the current correlator cannot tend to zero smoothly at zero momentum. Its spectral representation therefore contains a massless fermionic pole: the Goldstino, with spin one-half. This statement assumes a Poincaré-invariant vacuum and the usual local unitary field-theory conditions. It does not require a bosonic Goldstone particle for the fermionic symmetry. In supergravity the Goldstino can instead supply the longitudinal polarization of a massive gravitino.

For canonical chiral superfields ${\Phi }_i$, elimination of their auxiliary fields givesA supersymmetric vacuum solves every $W_i=0$. At a stationary broken vacuum,The same matrix $W_{ij}$ is the chiral-fermion mass matrix, so the nonzero vector $\overline{W_j}$ is a zero mode. This derives the Goldstino zero mode from vacuum stationarity. The transformation $\delta {\psi }_i=\sqrt{2}ϵF_i+\cdots$ independently identifies that direction through its inhomogeneous shift.

For one canonical chiral field, $W=fZ$, $f\ne 0$, has $U=|f{|}^2$ everywhere, a massless fermion and a classically flat scalar direction. It is flat F-term breaking with a linear superpotential, rather than an isolated vacuum. In contrast, a nonconstant polynomial $W^{\prime }$ of positive degree has a complex root and hence a supersymmetric vacuum; in particular a generic quadratic or cubic single-field superpotential cannot remove all such roots. This argument assumes canonical nonsingular kinetic terms and does not claim that every runaway or nonpolynomial model has a vacuum.

A renormalizable three-field O'Raifeartaigh model gives a less degenerate illustration:The auxiliary equations require $W_2=m{\varphi }_1=0$, then $W_1=hX{\varphi }_1+m{\varphi }_2=0$, leaving $W_X=f\ne 0$. They cannot vanish simultaneously. For real positive $f,h,m$ with $m^2>fh$, the potential obeysTherefore its global tree-level minimum has ${\varphi }_1={\varphi }_2=0$, arbitrary $X$, and energy $f^2$. The massless ${\psi }_X$ is the Goldstino, while $X$ is a pseudomodulus. At $X=0$, the scalar mass-squared spectrum is $0,0,m^2,m^2,m^2+fh,m^2-fh$, and the fermion masses are $0,m,m$. This exhibits both the stability condition and the loss of ordinary supermultiplet mass degeneracy in a broken vacuum. Quantum corrections can lift the pseudomodulus and must be analysed rather than assumed away; the noncompact classical valley also prevents an unqualified finite-volume Witten index trace argument. The index, the vacuum equations and the Goldstino theorem consequently answer complementary questions.

Ordinary electromagnetism uses a one-form gauge potential, its gauge transformation and its curvature. On superspace the corresponding object is a superspace gauge connection, with both vector and spinor components. Write ${\nabla }_A=D_A+i{\mathcal{A}}_A$. Its graded commutator contains the flat-frame torsion plus supercurvature. The conventional constraints set the two undotted-spinor curvatures, the two dotted-spinor curvatures and the conventional mixed spinor curvature to zero; the mixed derivative algebra then determines the vector connection. The remaining independent gauge curvature is packaged into a chiral field-strength superfield. A real vector superfield $V=V^{†}$ is a prepotential solving these constraints, rather than four unrelated ordinary gauge potentials.

Use the left-derivative conventions of Question 5, with $D^2=D^{\alpha }{D}_{\alpha }$ and ${\bar{D}}^2={\bar{D}}_{\dot{\alpha }}{\bar{D}}^{\dot{\alpha }}$. An Abelian supergauge transformation has a chiral parameter $\Lambda$, and one consistent normalization isThe parameter is chiral so that the transformed matter is still a chiral superfield; its complex components remove more of the prepotential than an ordinary real gauge parameter. In Wess-Zumino gauge, one convenient expansion isIts components are the gauge potential $A_m$, the gaugino $\lambda$ and the real auxiliary field $D$. Wess-Zumino gauge removes the extra scalar and spinor components of a general real superfield. Supersymmetry transformations usually leave that gauge slice and require a compensating supergauge transformation. For a residual chiral parameter with real lowest component $a(x)$, the expansion gives $A_m^{\prime }=A_m+2{\partial }_{m}a$ and ${\varphi }^{\prime }=e^{-2iqa}\varphi$; thus $\alpha =2a$ is the ordinary Maxwell gauge parameter.

The Abelian field strength is the Abelian field-strength chiral projectionThree barred derivatives vanish because there are only two anticommuting dotted components, so ${\bar{D}}_{\dot{\beta }}{W}_{\alpha }=0$. Gauge invariance follows directly: $D_{\alpha }\bar{\Lambda }=0$ by antichirality, while the mixed derivative algebra gives ${\bar{D}}^2{D}_{\alpha }\Lambda =0$ for chiral $\Lambda$. In chiral coordinates its component structure iswhere ${\sigma }^{mn}=({\sigma }^m{\bar{\sigma }}^n-{\sigma }^n{\bar{\sigma }}^m)/4$ fixes the tensor normalization. Overall gaugino phase choices do not change the gauge-invariant field strength or its Bianchi constraint.

Reality of $V$ and the derivative algebra imply the superspace Yang-Mills Bianchi identityIndeed, the two sides are the same differential operator on $V$, because $D^{\alpha }{\bar{D}}^2{D}_{\alpha }={\bar{D}}_{\dot{\alpha }}{D}^2{\bar{D}}^{\dot{\alpha }}$. This is the superspace counterpart of $dF=0$, and its components contain ${\partial }_{[m}{F}_{np]}=0$. It is a defining curvature constraint, not the Maxwell equations. Varying the pure gauge action additionally sets the common real superfield $D^{\alpha }{W}_{\alpha }$ to zero; matter supplies a current instead.

The Supersymmetric Yang-Mills action and minimal matter coupling areThe chiral phase factors cancel in $\bar{\Phi }{e}^{2qV}\Phi$. The component matter derivative is $({\partial }_m+iq{A}_m)\varphi$, which transforms covariantly under the residual gauge transformation just derived. The same superspace term generates the associated matter–gaugino Yukawa coupling and $qD|\varphi {|}^2$ auxiliary coupling, so minimal coupling is not only replacing an ordinary derivative in the scalar action. The gauge action contains $-F_{mn}{F}^{mn}/(4g^2)$, the gaugino kinetic term and $D^2/(2g^2)$. Rescaling the vector multiplet to canonical kinetic normalization gives the physical coupling $qg$. The holomorphic superpotential must be gauge invariant; a single field of nonzero continuous Abelian charge admits no nonconstant polynomial superpotential without additional fields. For several fields, invariance requires ${\sum }_i{q}_i{\Phi }_i{\mathcal{W}}_i=0$. A Fayet–Iliopoulos term is additionally possible for an Abelian factor, shifting the algebraic $D$ equation.

For a non-Abelian gauge group, $V$ and $\Lambda$ are Lie-algebra-valued, and multiplication cannot be reordered. For matter in a specified representation, setThe same matter quadratic form remains invariant. Its curvature isAt first order in an Abelian $V$, this reduces to $-{\bar{D}}^2{D}_{\alpha }V/4$, fixing the factor $1/8$. In the chiral gauge frame, $\bar{\nabla }=\bar{D}$ and ${\nabla }_{\alpha }=e^{-2V}{D}_{\alpha }{e}^{2V}$ as an operator. The gauge-covariant Bianchi identity is ${\nabla }^{\alpha }{W}_{\alpha }={\bar{\nabla }}_{\dot{\alpha }}{\tilde{\bar{W}}}^{\dot{\alpha }}$, where $\tilde{\bar{W}}=e^{-2V}\bar{W}{e}^{2V}$ is the conjugate curvature expressed in that same frame. It follows from the graded Jacobi identity of the covariant derivatives. Taking a trace of $W^{\alpha }{W}_{\alpha }$ makes the chiral gauge action invariant. Its bosonic curvature becomes the ordinary Yang-Mills curvature, and the gaugino transforms in the Adjoint representation. A constant Fayet–Iliopoulos coefficient cannot be inserted for a simple non-Abelian group, since it would need an invariant linear functional annihilating every Lie bracket. These constructions connect the constrained superspace connection, its prepotential and its component gauge multiplet.

A canonical Wess–Zumino model with a cubic superpotential already supplies a microscopic membrane-like object. Choose positive real $\mu ,g$ andThere are two isolated supersymmetric vacua, ${\varphi }_{\pm }=\pm \mu /g$. A planar domain wall has a static scalar profile depending on the transverse coordinate $z=X^3$, with one vacuum at either end. Its energy per unit area isChoosing $\alpha =\arg \Delta W$ gives $T\ge 2|\Delta W|$. The real increasing solution saturates the bound:The tension follows either from integrating $2({\varphi }^{\prime }{)}^2$ or from $W(\mu /g)-W(-\mu /g)=4{\mu }^3/(3g^2)$. This is a supersymmetric domain wall, and the first-order equation also implies the second-order scalar equation. Its thickness is of order $1/\mu$.

The wall breaks transverse translations: changing $z_0$ gives a normalizable scalar zero mode proportional to ${\varphi }^{\prime }(z)$. It also preserves only half the four real supercharges. In the chiral-fermion transformation, $F=-\overline{W^{\prime }}=-{\varphi }^{\prime }$ and the derivative term combine into a rank-half condition on the constant supersymmetry parameter, for example $ϵ=i{\sigma }^3\overline{ϵ}$ in a compatible Weyl convention. The orthogonal two real parameters are broken and generate localized fermion zero modes. One such profile satisfies $({\partial }_z+2g\varphi )f=0$, giving $f\propto {\mathrm{sech}}^2(\mu (z-z_0))$. The scalar and a three-dimensional Majorana fermion therefore form the low-energy wall multiplet. The domain-wall charge in N=1 supersymmetry is a boundary/tensor charge contributing to the extended algebra; the preserved combinations saturate $T=2|\Delta W|$. This avoids falsely applying the unextended particle-algebra argument of Question 3 to a wall with nonzero tension.

Promote the translation collective coordinate to a slowly varying worldvolume field $Y({\xi }^0,{\xi }^1,{\xi }^2)$. At wavelengths long compared with the wall thickness, its leading relativistic geometric action is the Dirac membrane actionIt is invariant under worldvolume reparametrizations and the target Poincaré group. In static gauge $X^i={\xi }^i$, $X^3=Y$, the matrix determinant lemma givesDirectly substituting the leading collective-coordinate profile into the scalar kinetic term gives the same coefficient, since $\int dz({\varphi }^{\prime }{)}^2=T/2$. Target Lorentz symmetry fixes the leading first-derivative nonlinear completion to the invariant area; finite-thickness physics can add higher-curvature terms. The area action is therefore a long-wavelength effective field theory, not an exact replacement for the microscopic wall at arbitrary momenta. Its embedding equation is ${\partial }_i(\sqrt{-h}{h}^{ij}{\partial }_j{X}^m)=0$.

For the four-dimensional supermembrane, extend the embedding to $(X^m(\xi ),{\theta }^{\alpha }(\xi ))$ with a four-real-component Majorana spinor. Pull back the invariant coframe of Question 7:Replacing $\partial X$ by $\Pi$ makes the area term invariant under rigid target supersymmetry, but does not yet give the appropriate fermionic gauge symmetry. Add a Wess-Zumino brane coupling to a superspace three-form potential $C_3$. In one orientation and matching normalization,The supermembrane closed four-form uses the distinct four-dimensional Fierz identity $(C{\gamma }_{mn}{)}_{(\alpha \beta }(C{\gamma }^n{)}_{\gamma \delta )}=0$. Differentiating $G_4$ reduces its coefficient to this symmetric tensor, so $d{G}_4=0$. A local potential therefore exists. Since the curvature is supertranslation invariant, the potential can vary by an exact form; its pullback integral is invariant modulo boundary terms. This is why the three-form $H_3$ of Question 7 is not itself the membrane field strength: integrating a membrane potential requires three-form degree and hence four-form curvature.

The required additional local fermionic gauge invariance is kappa symmetry. Define the worldvolume gamma matrices ${\gamma }_i={\Pi }_i^m{\gamma }_m$ andWith mostly-plus target signature, a local orthonormal worldvolume frame gives ${\Gamma }_{\kappa }={\gamma }_0{\gamma }_1{\gamma }_2$, so ${\Gamma }_{\kappa }^2=1$ and $\mathrm{tr}{\Gamma }_{\kappa }=0$. Thus $(1\pm {\Gamma }_{\kappa })/2$ are rank-two kappa symmetry projectors. The embedding variation gives $\delta {\Pi }_i^m=2i\delta \bar{\theta }{\gamma }^m{\partial }_i\theta$. The area variation is consequently proportional to $-2iT\sqrt{-h}\delta \bar{\theta }{\gamma }^i{\partial }_i\theta$. Contracting the displayed $G_4$ with the fermionic variation and using ${ϵ}^{ijk}{\gamma }_{ij}=2\sqrt{-h}{\Gamma }_{\kappa }{\gamma }^k$ gives the opposite Wess-Zumino contribution with an extra ${\Gamma }_{\kappa }$. Their sum is proportional toFor the boxed kappa variation, $\delta \bar{\theta }=\bar{\kappa }(1+{\Gamma }_{\kappa })$, so it vanishes by $(1+{\Gamma }_{\kappa })(1-{\Gamma }_{\kappa })=0$. Reversing membrane orientation reverses the Wess-Zumino sign and the matching projector. The cancellation fixes the relative charge magnitude to the tension; rigid supersymmetry alone would not fix it.

Worldvolume reparametrizations remove three of the four embedding scalars, leaving one transverse mode. Kappa symmetry removes half the four real embedding spinor components, leaving a two-component three-dimensional Majorana field, whose first-order equation has one on-shell fermionic polarization. The physical worldvolume modes are one scalar and one Majorana fermion, matching the wall zero modes. Without kappa symmetry the fermionic degrees would be doubled. A planar configuration preserves the target supersymmetries that can be compensated by the kappa projector, reproducing the wall's half-supersymmetry condition. This is the four-dimensional effective domain-wall interpretation of a supermembrane, rather than the eleven-dimensional M2-brane particle-content problem.