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

Take $\{{\gamma }^a,{\gamma }^b\}=2{\eta }^{ab}$ and ${\gamma }^{ab}={\gamma }^{[a}{\gamma }^{b]}=[{\gamma }^a,{\gamma }^b]/2$, with unit-weight antisymmetrization. The Clifford algebra first givesExpand the two antisymmetrized gamma matrices using ${\gamma }_{ab}={\gamma }_a{\gamma }_b-{\eta }_{ab}$:Anticommuting ${\gamma }^m$ past the next contracted factor givesSubtracting the remaining $D{\gamma }^m$ gives the gamma-matrix vector sandwichIn particular it vanishes in four dimensions. The result is independent of the choice of Lorentz signature when all indices are contracted consistently. Defining the bivector without its factor $1/2$ would multiply $\alpha$ by four.

Choose the charge conjugation matrix so that $C^T=-C$ and $({\gamma }^m{)}^T=-C{\gamma }^m{C}^{-1}$, and write $\bar{\lambda }={\lambda }^{T}C$. Since the components are odd Grassmann variables,Thus a symmetric matrix gives zero, whereas a nonzero antisymmetric matrix can give a nonzero bilinear. The four-dimensional Majorana bilinear vanishing criterion follows by transposing an antisymmetrized product and reversing its $n$ factors:For $n=0,1,2,3$, the symmetry types are respectively antisymmetric, symmetric, symmetric and antisymmetric. These are the same-spinor selection rules of the Majorana spinor; they would be reversed for a commuting spinor. The vanishing choices are (a) and (c).

The matrix $C{\gamma }^m$ is symmetric. Its contraction with two identical odd Grassmann variables is antisymmetric under their exchange, soThis is the vector case of the four-dimensional Majorana bilinear vanishing criterion.

The matrix $C$ is antisymmetric, so the scalar Majorana spinor bilinear is not identically zero. For example, choose two independent odd Grassmann variables ${\lambda }_{\alpha },{\lambda }_{\beta }$ at a pair of indices with $C_{\alpha \beta }\ne 0$. Their contribution is $2C_{\alpha \beta }{\lambda }_{\alpha }{\lambda }_{\beta }$, which is nonzero as a Grassmann polynomial. This is why a Majorana mass term is possible.

Four-dimensional Hodge duality identifies ${\gamma }^{mn}{\gamma }_5$ with a nonzero convention-dependent multiple of ${ϵ}^{mnpq}{\gamma }_{pq}$. Every $C{\gamma }_{pq}$ is symmetric by the four-dimensional Majorana bilinear vanishing criterion; taking this linear combination preserves symmetry. HenceThe phase in the duality formula depends on the definition of ${\gamma }_5$ and orientation, but does not affect the vanishing result.

The three-gamma matrix $C{\gamma }^{mnp}$ is antisymmetric, and for three distinct indices it is nonzero. A pair of nonzero Grassmann components therefore produces a nonzero term exactly as in the scalar case. Consequently this three-form bilinear is not identically zero. Equivalently it is dual to the axial-vector Majorana spinor bilinear, which is allowed; it must not be confused with the vanishing vector in part (a).

Here a supersymmetry-preserving combination means a real combination of Hermitian Majorana spinor supercharges, in a positive-norm unitary representation of the ordinary Super-Poincaré algebra. This reality qualification is essential. Let $|\psi ⟩$ be normalized and in the charge domains, and define the real symmetric matrixFor real $u$, the Hermitian operator $q_u=u^a{Q}_a$ satisfiesTherefore $M$ is a positive semidefinite matrix, andThe reverse implication uses positivity: diagonalize $M$, so a zero quadratic form has zero component in every positive-eigenvalue direction. This converts the problem into nullity of real N=1 supercharges, not just a condition on the determinant of an operator-valued expression.

In the equivalent Weyl spinor form, $\{q_{\alpha },q_{\beta }^{†}\}=2{\sigma }_{\alpha \dot{\beta }}^m{P}_m$. After a fixed invertible change of real charge basis, $M$ is the realification of the Hermitian two-by-two matrixup to the harmless spatial-sign convention. A complex Hermitian matrix has the same eigenvalues in its realification, each repeated twice. The Pauli matrix multiplication law gives $(\mathbf{p}\cdot \mathbf{\sigma }{)}^2=|\mathbf{p}{|}^{2}I$, so these eigenvalues are $E+|\mathbf{p}|$ and $E-|\mathbf{p}|$, each twice. Positivity forces $E\ge |\mathbf{p}|$. If one nonzero real $u$ annihilates the state, $M$ has a kernel: either $E=|\mathbf{p}|>0$, with exactly two zero eigenvalues, or $E=|\mathbf{p}|=0$, with four. ThusFor a momentum eigenstate these are the null-momentum and zero-momentum cases; the expectation-matrix proof also covers states not assumed to have definite momentum. There are no massive states preserving a real charge in this unextended algebra.

If “linear combination” instead allows arbitrary complex coefficients, the unrestricted claim is false. In a massive fermionic Fock space with two lowering operators $a_1,a_2$ and a spin-half Clifford vacuum spectator, takeThen $a_1\psi =0$. The three vectors $a_2\psi$, $a_2^{†}\psi$ and $a_1^{†}\psi$ are linearly independent, by their occupation numbers and spectator spin labels. The complex annihilator is therefore only one-dimensional. Such a lowering operator is not a Hermitian generator of preserved real supersymmetry. The boxed conclusion applies to the physical real-charge interpretation; it also assumes no domain-wall charge in N=1 supersymmetry.

In a massive four-dimensional N=1 supersymmetry representation, the two normalized raising supercharges transform as a spin-half doublet. Start with a Clifford vacuum of superspin $j=1/2$. At occupation levels zero, one and two, the ordinary spins areThe two-creation state has a spin-singlet oscillator factor because the creation operators anticommute. Thus the massive N=1 vector multiplet containsThere are four bosonic polarizations, $3+1$, and four fermionic polarizations, $2+2$, as required by boson-fermion degeneracy in a supermultiplet.

In the massless limit the helicity multiplicities become $+1,-1$, two copies each of $+1/2,-1/2$, and two zero-helicity states. They separate into the four irreducible helicity pairsWith the lower-helicity convention for superhelicity, a pair $(\kappa ,\kappa +1/2)$ has labels $\kappa =1/2,0,-1/2,-1$, respectively. A midpoint convention shifts each label by $1/4$; the explicit helicity pairs remove any naming ambiguity.

Pairing under the CPT theorem, the first and last pairs form a massless vector multiplet, and the middle two form a massless chiral multiplet. This is the massless limit of a massive N=1 vector multiplet. The longitudinal vector polarization supplies one of the chiral multiplet's two real scalar modes; the original spin-zero state supplies the other. Simply discarding the longitudinal polarization would lose a physical degree of freedom and give the wrong decomposition.

A scalar superfield transformation is a pullback under a supertranslation. Since that transformation mixes $x$ and $\theta$, an odd-coordinate derivative acquires an extra spacetime-derivative term under the coordinate Jacobian. It therefore does not transform as a spinor superfield merely by differentiating the scalar transformation law. This is partial derivatives and superfield covariance.

For explicit signs, split the Majorana spinor coordinates into ${\theta }^{\alpha },{\bar{\theta }}^{\dot{\alpha }}$ and use left Grassmann derivatives. Choose the differential generatorsFor example, $\{{\partial }_{\alpha },{\bar{Q}}_{\dot{\beta }}\}=i{\sigma }_{\alpha \dot{\beta }}^m{\partial }_m$, which is nonzero. Commuting the even supersymmetry variation through ${\partial }_{\alpha }$ consequently gives the unwanted translation term.

Replace the ordinary odd-coordinate derivatives by the supersymmetric covariant derivativesThe graded product rule cancels the mixed coordinate/derivative terms, giving $\{D,Q\}=\{D,\bar{Q}\}=\{\bar{D},Q\}=\{\bar{D},\bar{Q}\}=0$. Thus $D_{\alpha }V$ and ${\bar{D}}_{\dot{\alpha }}V$ transform as spinor superfields. Their own mixed algebra is $\{D_{\alpha },{\bar{D}}_{\dot{\beta }}\}=-2i{\sigma }_{\alpha \dot{\beta }}^m{\partial }_m$, with equal-chirality anticommutators zero. This covariant differentiation is also what makes a chiral superfield constraint ${\bar{D}}_{\dot{\alpha }}Z=0$ supersymmetry invariant. Negating every barred derivative is another convention, but changes the displayed mixed-algebra sign.

In the ordinary local two-derivative matter class, the general ungauged supersymmetric action for a chiral superfield isThe Kähler potential $K$ is real, and the superpotential $W$ is a holomorphic function. Adding $f(Z)+\bar{f}(\bar{Z})$ to $K$ changes the action only by a boundary term: full superspace integration of a chiral or antichiral function vanishes modulo that term. This is a Kähler transformation. A positive Kähler metric $G=K_{z\bar{z}}$ is required for a nonsingular positive-norm kinetic term, not just for derivative counting.

To see the differential order, the bosonic component Lagrangian in mostly-plus signature containsThe auxiliary field equation is algebraic: $F=-G^{-1}\overline{W_z}$ on a purely bosonic configuration. It gives the scalar potential $U=G^{-1}|W_z{|}^2$. The scalar equations have at most two spacetime derivatives, the fermion equations from the sigma-model kinetic term have one, and the remaining interaction terms do not raise this order. The fermionic corrections to the auxiliary equation are still algebraic.

For four-dimensional renormalizability, assign engineering dimensions $[Z]=1$, $[d^4\theta ]=2$, and $[d^2\theta ]=1$. Hence, up to a Kähler transformation and a constant field normalization, the restrictions areA mixed higher-degree term in $K$ or a term of degree above three in $W$ introduces a negative-mass-dimension coupling. The constant $w_0$ has no local effect in global supersymmetry, as in the constant superpotential in global supersymmetry.

The boxed general-action statement uses the conventional two-derivative matter interpretation. It is not a classification of all derivative-dependent effective superfield actions by the order of their bosonic equations alone. For example, a scalar Lagrangian quartic in first derivatives still yields a second-order Euler–Lagrange equation, and supersymmetric derivative-dependent interactions can contain such terms. Their auxiliary and fermionic equations require separate checking. This distinction is explicit in the original construction of supersymmetric derivative-dependent scalar actions. Such actions are outside the ordinary $K+W$ class used for the renormalizable answer here.

Use the supertranslation-invariant superspace coframeIt is a basis because the change from $(dX,d\theta )$ is triangular with identity diagonal blocks. Under a rigid supertranslation, $\delta \theta =ϵ$, $\delta X^m=-i\overline{ϵ}{\gamma }^m\theta$ and $dϵ=0$. Therefore $\delta {\Pi }^m=-i\overline{ϵ}{\gamma }^{m}d\theta +i\overline{ϵ}{\gamma }^{m}d\theta =0$, and $\delta d\theta =0$.

Writing wedge products explicitly, the three-form isBoth factors are supertranslation invariant. Under a Lorentz transformation, ${\Pi }^m$ and the bilinear in $d\theta$ transform as vectors, so their contraction is invariant. Thus $H_3$ is invariant under the Super-Poincaré group.

The Maurer-Cartan equation is $d{\Pi }^m=id\bar{\theta }{\gamma }^{m}d\theta$. Since the other factor is closed,The required four-dimensional Fierz rearrangement identity, which may be stated without proof, isDifferentials of odd Grassmann variables commute in the graded exterior algebra. Thus the coefficient of the product of four $d\theta$ factors is the fully symmetric spinor-index part appearing in this identity. It vanishes, provingThis is the closed three-form on four-dimensional superspace; it should not be confused with the different closed four-form used for a membrane coupling in Question 10.