Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 307 (Supersymmetry and Extra Dimensions)

Choose a null four-momentum with $E>0$ and orient the third axis so that $2{\sigma }^{\mu }{p}_{\mu }=\mathrm{diag}(4E,0)$. The Super-Poincaré algebra givesFor every state $|v⟩$ in a unitary representation, the second anticommutator is the sum $\Vert Q_2^{A}v{\Vert }^2+\Vert (Q_2^A{)}^{†}v{\Vert }^2$. Thus both operators act trivially. A central charge in supersymmetry also vanishes in this massless supermultiplet: its mixed-spinor anticommutator contains the now-trivial $Q_2$. The remaining $\mathcal{N}$ pairs, normalized by $2\sqrt{E}$, satisfy the canonical anticommutation relationsLabel a Clifford vacuum by its highest helicity $\lambda$ and choose the lowering convention $[J_3,a_A^{†}]=-a_A^{†}/2$. Acting with distinct fermionic creation operators produces the fermionic Fock spaceThis proves the helicity spectrum of a massless supermultiplet and the $2^{\mathcal{N}}$ state count. If the internal charges or helicities do not already make the spectrum invariant under the CPT theorem, add its conjugate using the CPT completion of a supermultiplet.

The spin bound for massless supermultiplets follows directly from the width $\mathcal{N}/2$ of this helicity interval. Requiring $|h|\le 1$ gives $\mathcal{N}\le 4$, achieved by the four-dimensional N=4 super Yang-Mills theory vector multiplet with multiplicities $1,4,6,4,1$ at helicities $1,1/2,0,-1/2,-1$. These helicity bounds are necessary spin restrictions for the usual interacting renormalizable quantum field theory; they do not by themselves establish renormalizability for an arbitrary action. Requiring $|h|\le 2$ gives $\mathcal{N}\le 8$. The four-dimensional N=8 supergravity supergravity multiplet attains this bound with highest helicity $2$ and multiplicities $1,8,28,56,70,56,28,8,1$. There is one $+2$ state and one $-2$ state, the two polarizations of a single graviton.

The stronger restriction relevant to matter is the chirality constraint on extended supersymmetry. An $\mathcal{N}=1$ chiral superfield can contain a left-handed Weyl spinor in a complex gauge group representation $R$. By contrast, a full $\mathcal{N}=2$ hypermultiplet consists of $\mathcal{N}=1$ matter in $R\oplus \overline{R}$, and the gauginos of a vector multiplet are in the real adjoint group representation. Higher unbroken extended supersymmetry retains this pairing. A pseudoreal representation may allow a half-hypermultiplet, but it does not yield net chirality in a genuinely complex gauge group representation. Consequently the chiral gauge spectrum of the Standard Model permits at most unbroken $\mathcal{N}=1$ supersymmetry in its four-dimensional low-energy matter sector.

On shell, a massless $\mathcal{N}=1$ vector multiplet contains a gauge boson with two transverse polarizations and a gaugino with two fermionic states. A chiral superfield contains a complex scalar field, hence two real bosonic states, and a Weyl spinor with two fermionic states. Their total is $4$ bosonic and $4$ fermionic states.

Suppose charged scalar fields acquire vacuum expectation values at a supersymmetric vacuum, where F-flatness and D-flatness hold. For each broken generator of the gauge symmetry, the supersymmetric Higgs mechanism combines one massless vector multiplet and one chiral superfield into a massive N=1 vector multiplet. The gauge boson absorbs one real Goldstone boson, acquiring its third polarization. The other real scalar field remains physical. The gaugino and the relevant higgsino mix into the four fermionic states. Thus the spin content isUnbroken supersymmetry gives all these states the same mass through supersymmetric mass degeneracy. Remaining massive chiral superfields instead have spin content $(1/2,0,0)$, with two fermionic and two bosonic states of a common mass. The auxiliary fields ensure matching off shell but are not additional propagating particles in these counts.

The super-Higgs mechanism occurs when local supersymmetry is spontaneously broken in supergravity. Its spin-$3/2$ gauge field, the gravitino, absorbs the spin-$1/2$ goldstino:The extra states have helicities $\pm 1/2$, supplementing $\pm 3/2$. The ordinary supersymmetric Higgs mechanism can preserve supersymmetry while breaking the internal gauge symmetry; the super-Higgs mechanism breaks the local supersymmetry itself. Spontaneously broken global supersymmetry leaves a physical massless goldstino, because there is no dynamical gravitino to absorb it.

Take a canonical Kähler potential $K={\Phi }^{†}\Phi$ and the Wess–Zumino model superpotentialThe F-term component givesEliminating the auxiliary field by $F=-\overline{W^{\prime }(\varphi )}$ gives the F-term scalar potential $V=|W^{\prime }(\varphi ){|}^2$. ConsequentlyThis is the supersymmetric relation between quartic and Yukawa couplings: there is one holomorphic parameter $g$, with the quartic coefficient $|g{|}^2$ and the Yukawa coupling $g$ in the displayed normalization. For real $g$ the quartic coefficient is the $g^2$ of the question. They cannot be varied independently while retaining this canonical supersymmetry.

The relation produces the supersymmetric cancellation of quadratic divergences. Bosonic and fermionic Feynman diagrams correcting a scalar mass have equal ultraviolet quadratic coefficients and opposite signs, so the mass is not driven directly to the ultraviolet cutoff scale. In a theory with soft supersymmetry breaking, residual corrections instead have the schematic sizewith coefficients and logarithms depending on the spectrum. Thus superpartner masses near the electroweak scale can stabilize the Higgs boson mass against a much larger cutoff, addressing the radiative part of the hierarchy problem. Heavy sparticles leave a corresponding residual sensitivity; the cancellation does not by itself explain every mass scale.

In the conventions $\{Q_{\alpha },Q_{\beta }^{†}\}=2{\sigma }_{\alpha \dot{\beta }}^{\mu }{P}_{\mu }$, the trace over the two spinor components of the Super-Poincaré algebra givesHence, on a normalized state in the domain of the supercharges,This proves energy positivity in global supersymmetry. The zero of energy is fixed by the algebra; adding an arbitrary constant to $H$ while keeping that algebra unchanged is not allowed.

For a finite-dimensional supermultiplet at fixed positive energy and four-momentum, choose a nonzero component $q$ with $\{q,q^{†}\}=cI$, $c>0$, and $q^2=0$. The Hermitian odd operatorIt is therefore an invertible map from the even fermion parity subspace to the odd subspace and conversely. This proves boson-fermion degeneracy in a supermultiplet, $n_B=n_F$, for every positive-energy physical particle supermultiplet. Equivalently, the supertrace of the supercharge anticommutator vanishes because odd operators interchange these two subspaces. A zero-energy vacuum is an exception to the literal claim about every possible representation: it can be a one-dimensional bosonic singlet with all supercharges acting trivially.

For a vacuum $|\Omega ⟩$, the sum-of-squares formula gives $E_0=0$ exactly when all supercharges and their adjoints annihilate it. If global supersymmetry is spontaneously broken and a vacuum exists, at least one such norm is nonzero, so the vacuum energy is positive. In infinite volume this is the statement about the vacuum energy density, understood with a finite-volume regulator. Explicit supersymmetry breaking need not preserve the algebra used in the proof. The conclusion also does not extend to the supergravity F-term potential, whose negative $-3e^K|W{|}^2$ term permits, for example, the zero-energy broken vacua of no-scale supergravity.

Use left Grassmann derivatives and place the odd constant transformation parameters on the left. A scalar superfield transformation can be writtenEquivalently, pull back the scalar function under the infinitesimal superspace translationThe term involving $\overline{ϵ}$ has the displayed sign because interchanging it with $\theta$ reverses the sign in the Grassmann algebra. Overall factors of $i$ in the definition of the supercharges can be changed together without changing this coordinate transformation.

For two even scalar superfields $S,T$, the odd generators obey the graded Leibniz rule, while the full parameter-weighted variation $\delta$ is even. ThusThis is precisely the transformation of a product of scalar superfields. The finite version is even clearer: if $S^{\prime }(z)=S(z^{\prime })$ and $T^{\prime }(z)=T(z^{\prime })$ for the same transformed superspace point, then $(ST{)}^{\prime }(z)=S(z^{\prime })T(z^{\prime })=(ST)(z^{\prime })$. No new representation law is required.

The expression $S(x,\theta ,\bar{\theta })=\varphi (x)$ needs a distinction between a configuration and a closed field content. As a function on superspace, it is a valid special configuration of an unconstrained superfield. But its variation isFor nonconstant $\varphi$, this generates nonzero components depending on Grassmann variables. Therefore the pure-scalar truncation is not a supersymmetric multiplet: an isolated spacetime scalar field with all its partners permanently set to zero is not closed under supersymmetry. This is the pure-scalar truncation of a superfield. A constant gauge-singlet $\varphi$ is the trivial invariant exception. Calling $\varphi (x)$ a special superfield configuration is consistent; claiming that its restricted field content remains a superfield representation is not.

All derivatives below are left Grassmann derivatives. Decompose the supersymmetric covariant derivatives into odd piecesWhen these operators act on an arbitrary superfield, the graded Leibniz rule givesThe anticommutator of the two pure Grassmann derivatives vanishes. Also $\{A_{\alpha },B_{\dot{\beta }}\}=0$: the spacetime derivatives commute, while the $\theta$ and $\bar{\theta }$ coefficients anticommute. Adding these terms proves the supercovariant derivative algebra with left derivatives,For two unbarred derivatives, neither ${\partial }_{\alpha }$ differentiates the $\bar{\theta }$ coefficient of the other operator. Their coefficient products anticommute, so $\boxed{\{D_{\alpha },D_{\beta }\}=0}$. The same argument with barred and unbarred coordinates exchanged gives $\boxed{\{{\bar{D}}_{\dot{\alpha }},{\bar{D}}_{\dot{\beta }}\}=0}$. These signs follow the derivatives printed in this paper; a convention replacing $\bar{D}$ by $-\bar{D}$ reverses the mixed sign as well.

Set $y^{\mu }=x^{\mu }+i{\theta }^{\alpha }{\sigma }_{\alpha \dot{\beta }}^{\mu }{\bar{\theta }}^{\dot{\beta }}$. Left differentiation givesThe chain rule therefore makes ${\bar{D}}_{\dot{\alpha }}$ simply ${\bar{\partial }}_{\dot{\alpha }}$ at fixed $y,\theta$. The chiral superfield constraint ${\bar{D}}_{\dot{\alpha }}\Phi =0$ means that $\Phi$ is independent of $\bar{\theta }$ in these coordinates. Since there are only two independent entries of $\theta$, its Grassmann algebra expansion terminates at degree two. Naming the coefficients gives the chiral-superfield component expansionHere $\varphi$ is a complex scalar field, $\psi$ a Weyl spinor, and $F$ a nonpropagating auxiliary field; the $\sqrt{2}$ is their conventional normalization.

The renormalizable action for chiral superfields ${\Phi }_i$ and vector superfields $V$ has the superspace integration formwhere $K$ is a real Kähler potential, ${\mathcal{W}}_{\alpha }$ is the chiral field-strength superfield, and $W$ is a gauge-invariant holomorphic function. For the renormalizable theory choose canonical positive kinetic terms, a polynomial $W$ of degree at most three, and field-independent gauge kinetic coefficients $f_{ab}$. We prove the non-renormalization theorem for the local Wilsonian effective action, with a nonzero infrared cutoff and a regulator preserving supersymmetry. The elementary chiral superfields are retained; this is not a claim that eliminating whole massive fields leaves the same polynomial.

Write $W={\sum }_A{\lambda }_A{\mathcal{O}}_A(\Phi )$, treating masses and interaction coefficients alike as background chiral spurions. A local F-term is chiral even when these backgrounds vary, so its coefficient depends holomorphically on $\Phi ,{\lambda }_A$ and the holomorphic gauge couplings. Dependence on ${\bar{\lambda }}_A$ would not be chiral. Such dependence can enter D-terms or terms with supercovariant derivatives instead. With the infrared cutoff kept fixed, the perturbative local coefficients are regular power series around vanishing masses and interactions. This removes the inverse-mass and infrared singular terms that would spoil the following spurion selection rule for perturbative non-renormalization.

Assign every elementary ${\Phi }_i$ R-charge zero, every ${\lambda }_A$ R-charge two, and $\theta$ R-charge one. The superpotential must have R-charge two because $d^2\theta$ has R-charge minus two. Holomorphy excludes the conjugate spurions, so regularity and this formal R-symmetry require exactly one factor of a superpotential spurion in any perturbative candidate. Thus $W_{\mathrm{eff}}$ is linear in the ${\lambda }_A$, with the ordinary flavor spurion charges fixing the permitted field monomials. An anomalous formal R-symmetry is implemented by also transforming the holomorphic gauge coupling; its anomaly must not simply be ignored.

Gauge interactions cannot provide a perturbative coefficient modifying this conclusion. With $\tau =\vartheta /(2\pi )+4\pi i/g^2$, perturbative diagrams are independent of the topological angle $\vartheta$. A holomorphic coefficient invariant under continuous real shifts of $\tau$ is constant, by the Cauchy-Riemann equations. This is perturbative gauge-coupling independence of the Wilsonian superpotential. Contributions involving $e^{2\pi i\tau }$ are nonperturbative and are outside this argument. For an Abelian factor the same angle-independent holomorphy argument applies. Hence a putative perturbative coefficient linear in ${\lambda }_A$ is independent of the gauge coupling and can be evaluated as that coupling tends to zero.

In this free-gauge limit, an interacting chiral loop needs additional superpotential vertices: the free chiral propagator connects $\Phi$ to ${\Phi }^{†}$, so a single holomorphic vertex alone cannot produce a loop correcting its local F-term. Extra vertices introduce extra spurions or their conjugates, both excluded by the preceding charge and holomorphy argument. The surviving coefficient is its tree value. Therefore

For an explicit charge check, take $W=m{\Phi }^2/2+g{\Phi }^3/3$ in the Wess–Zumino model. Give $(\Phi ,m,g)$ ordinary charges $(1,-2,-3)$ and R-charges $(1,0,-1)$. A holomorphic monomial $m^a{g}^b{\Phi }^n$ of the required charges satisfies $n-2a-3b=0$ and $n-b=2$, giving $a=1-b$, $n=b+2$. Perturbative regularity requires $a,b\ge 0$, hence only $(a,b,n)=(1,0,2)$ and $(0,1,3)$ survive. The free limit fixes their coefficients to $1/2$ and $1/3$; there is no room for a coupling-dependent loop correction. This illustrates the holomorphy argument for superpotential non-renormalization explicitly.

The Kähler potential can acquire wave-function renormalization. Restoring canonical field normalization changes physical masses and Yukawa couplings, so their running does not contradict the boxed statement about holomorphic coordinates. Gauge kinetic F-terms have their own renormalization and are not the superpotential. Infrared effects in the one-particle-irreducible quantum effective action, tree-level elimination of retained fields, and genuine nonperturbative effects must likewise be distinguished from this perturbative Wilsonian effective action theorem.

First assume global four-dimensional N=1 supersymmetry, canonical positive Kähler potential, and tree level. Write $W_i={\partial }_{i}W$ and $m_{ij}=W_{ij}$ at the stationary vacuum. The chiral-superfield fermion mass matrix is the symmetric matrix $m$, and the sum of squared Weyl spinor masses is $\mathrm{tr}(m^{†}m)$. From the F-term scalar potential $V={\sum }_k|W_k{|}^2$, its mixed and holomorphic Hessian matrix blocks areFor canonically normalized real and imaginary parts of the scalar fields, the real scalar mass matrix has trace twice the mixed trace. Its holomorphic blocks can split the two real scalar masses but do not change their sum. HenceA Weyl spinor has two spin states, so this proves the tree-level supertrace mass sum rule,The paper defines its supertrace with $(-1{)}^{2j+1}$, the negative of the conventional boson-minus-fermion supertrace used here. The asserted zero is the same in either convention. The result requires the stated canonical tree-level hypotheses; a noncanonical Kähler metric, supergravity, or radiative corrections can change the sum rule.

If vector multiplets are also present, the gauge contributions to the F-term supertrace cancel rather than being omitted. At $D_a=0$, define $C={\sum }_a{g}_a^2\Vert T_{a}v{\Vert }^2$ for the scalar expectation vector $v$. Differentiating $\frac{1}{2}{\sum }_a{D}_a^2$ adds $2C$ to the real-scalar trace. The symmetric gaugino-matter mass matrix has off-diagonal entries $\sqrt{2}{g}_a{T}_{a}v$, giving an extra $4C$ in its fermion squared-mass trace. The covariant scalar kinetic term gives vector squared-mass trace $2C$. Thus their contribution is $2C-2(4C)+3(2C)=0$. The gauge-theory result uses all spin states, including the vector weight three; it is not obtained by dropping massive gauge partners.

The MSSM tree-level sfermion mass constraint explains the phenomenological difficulty with direct visible-sector breaking. If F-term breaking is neutral under an unbroken electric and color gauge symmetry, and no D-term shifts are present, the same trace argument applies within each conserved-charge fermion block. Its corresponding scalar partners cannot all have squared masses above the mean of the light fermion squared masses. Thus canonical tree-level visible-sector supersymmetry breaking alone cannot make every squark and other scalar partner heavy while leaving the observed fermions light. The usual effective soft supersymmetry breaking terms arise after communicating breaking from a separate sector; integrating out that sector, noncanonical interactions and radiative effects evade the hypotheses. The global trace identity alone would not identify a particular light squark without the conserved-charge block argument.

For the specified O'Raifeartaigh model, phases may be chosen so that $\lambda ,\mu ,M$ are real and positive. Its superpotential derivatives areand its F-term scalar potential isFor fixed $x,z$, choose $y=-2\lambda xz/M$ to minimize the final square. Since $|z^2-{\mu }^2{|}^2\ge (|z{|}^2-{\mu }^2{)}^2$,Thus $M^2>2{\lambda }^2{\mu }^2$ makes $z=y=0$ a global minimum with arbitrary $x$. The hierarchy $M\gg \mu$ ensures this for fixed perturbative $\lambda$, rather than for arbitrarily large $\lambda$. The origin is one member of this classically flat family, with $V_0={\lambda }^2{\mu }^4>0$ and $F_X=\lambda {\mu }^2$. The complex field $x$ is a pseudomodulus.

At the origin the chiral-superfield fermion mass matrix, in the $({\psi }_X,{\psi }_Y,{\psi }_Z)$ basis, isIts physical fermion masses are $0,M,M$; the two massive Weyl spinors form one massive four-component fermion. The massless ${\psi }_X$ is the goldstino. More generally, stationarity gives $W_{ij}\overline{W_j}=0$, so the nonzero auxiliary field direction is a null vector of $m$, proving the goldstino zero mode from vacuum stationarity.

Writing $z=(z_R+i{z}_I)/\sqrt{2}$ and similarly for $x,y$, the quadratic scalar potential isThe mass spectrum of the quadratic-cubic O'Raifeartaigh model is thereforeThe scalar squared-mass sum is $4M^2$, while twice the fermion squared-mass sum is also $4M^2$, so the supertrace is zero, as required. The two massless real $x$ modes are tree-level pseudomoduli, which can be lifted by a quantum effective potential; their tree-level masslessness is not the exact symmetry protection enjoyed by the goldstino.