Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 45 / 1

It is essential to distinguish a finite-dimensional spinor matrix from its action on a family of operator components. Take the Pauli matrices and ${\bar{\sigma }}^{\mu \nu }$ in the supplied convention, with $\eta =\mathrm{diag}(1,-1,-1,-1)$. They giveIf both displayed transformations are read literally as left multiplication on the column ${\bar{Q}}^{\dot{\alpha }}$, expanding the operator conjugation and comparing its linear term gives the formal answer $[M^{\mu \nu },\bar{Q}]=+{\bar{\sigma }}^{\mu \nu }\bar{Q}$.

There is a sign inconsistency in that literal reading. With the usual angular momentum commutation relations, coefficient matrices $R_i$ in $[J_i,q_a]=(R_i{)}_{ab}{q}_b$ must satisfyIndeed, the Jacobi identity gives $[J_i,[J_j,q]]-[J_j,[J_i,q]]=(R_j{R}_i-R_i{R}_j)q$, whereas $[[J_i,J_j],q]=i{ϵ}_{ijk}{R}_{k}q$. The formal $R_i=+{\sigma }_i/2$ violates this identity. For example it would give $[J_3,q_1]=q_1/2$, $[J_{+},q_1]=q_2$ and $[J_3,q_2]=-q_2/2$; the two sides of the Jacobi identity for $J_3,J_{+},q_1$ are then $-q_2/2$ and $3q_2/2$.

A consistent canonical column-operator convention uses the inverse adjoint action for the displayed spinor transformation, equivalently the contragredient operator-component prescription. It givesThis is the convention used for the physical states below. Reversing the adjoint action is the minimum needed sign correction in the literal column reading; a dual row notation can encode the same correction by its index contractions. The formal printed-sign coefficients are also recorded below, so the distinction is explicit rather than silently changing a convention. The representation matrices themselves remain $J_i={\sigma }_i/2$ and $K_i=i{\sigma }_i/2$ in Question 3: the minus sign here concerns commutators of operator components.

A supercharge is an odd Weyl spinor generator. The mixed anticommutator has one undotted and one dotted index, hence transforms as a Lorentz four-vector. In the minimal Super-Poincaré algebra, its only spacetime generator is the four-momentum, so covariance fixes it to $c{\sigma }_{\alpha \dot{\beta }}^{\mu }{P}_{\mu }$. Normalizing the charges canonically fixes $c=2$:The number two is a normalization, not something Lorentz covariance alone could fix under $Q\mapsto aQ$.

For an explicit closure calculation, use left Grassmann derivatives in superspace and the differential translation convention $P_{\mu }=i{\partial }_{\mu }$:Each of the two derivative-coordinate cross terms contributes $i{\sigma }^{\mu }{\partial }_{\mu }$; the other terms cancel by anticommutativity. This proves the displayed coefficient. The same calculation gives $\{Q,Q\}=\{\bar{Q},\bar{Q}\}=0$ and $[P_{\mu },Q]=[P_{\mu },\bar{Q}]=0$. There is no scalar central charge for a single supersymmetry label: symmetry of the anticommutator is incompatible with a term proportional only to the antisymmetric ${ϵ}_{\alpha \beta }$.

In the massive rest frame $P_{\mu }=(m,0,0,0)$, the mixed relation is $\{Q_{\alpha },Q_{\beta }^{†}\}=2m{\delta }_{\alpha \beta }$. These positive canonical anticommutation relations supply the input for the normalization and finite fermionic Fock space in the question-level continuation.

Write $q_1={\bar{Q}}^{\dot{1}}$ and $q_2={\bar{Q}}^{\dot{2}}$. In the Jacobi-consistent convention just fixed,Applying the commutator product rule twice givesHere ${\sum }_i{\sigma }_i^2=3I$ produces the constant term; moving the $J_i$ to the right accounts for it. ConsequentlyThus the physically consistent numerical coefficients areIf one mechanically retains the literal positive-sign column transformation instead, the coefficient tuple would be $(\frac{1}{2},\frac{3}{4},1,-i,1,\frac{3}{4},1,i,-1)$. It cannot describe a unitary multiplet with the supplied rotation algebra, by the explicit Jacobi counterexample in part (a).

Take the spin-zero Clifford vacuum $|\Omega ⟩$, normalized to one and annihilated by the two annihilation charges. This is a lowest Fock state of a positive-mass representation, not a zero-energy vacuum annihilated by every charge. For the first one-charge state,The ladder commutators are $[J_{+},q_1]=-q_2$, $[J_{-},q_1]=0$, $[J_{+},q_2]=0$, $[J_{-},q_2]=-q_1$. Therefore these two states are the $j=1/2$ pair. A phase choice $|\frac{1}{2},+\frac{1}{2}⟩=-N{q}_2|\Omega ⟩$ makes the usual positive ladder coefficient hold.

Raising the dotted index only changes the oscillator basis by the unitary antisymmetric epsilon matrix, so $\{q_a,q_b^{†}\}=2m{\delta }_{ab}$. Since $q_a^{†}|\Omega ⟩=0$,The positive convention would formally label this state with $j_3=+1/2$ while $J_{+}$ acts nontrivially on it, contradicting the highest-weight ladder rule. This supplies an independent state-level check of the needed sign correction.

The two-charge state is a rotation singlet. Indeed, using $q_1^2=q_2^2=0$ and $q_2{q}_1=-q_1{q}_2$, the product rule gives $[J_i,q_1{q}_2]=0$ for every $i$: only the sum of the two diagonal spinor coefficients survives, and their trace is zero. Hence the requested commutators and quantum numbers areIts squared norm is $(2m{)}^2$ before multiplication by $N^4$, by the two oscillator anticommutators.

The spin-zero massive N=1 supermultiplet therefore consists of $|\Omega ⟩$, $N{q}_1|\Omega ⟩$, $N{q}_2|\Omega ⟩$ and $N^2{q}_1{q}_2|\Omega ⟩$. Their spins are $0,1/2,1/2,0$, all with the same mass $m$. Starting with an even vacuum, the two spin-zero states are bosonic and the two one-charge states fermionic. They are the two real scalar degrees of freedom and the two on-shell fermion polarizations of a massive chiral multiplet. Any product of three creation charges repeats an index and vanishes; annihilation charges reduce to these four states by the canonical anticommutation relations. There are exactly four independent states.