Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 307 / 1

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.