Massive and Massless Unitary Supermultiplets

A unitary supermultiplet is built at fixed momentum by turning the positive supercharge anticommutator into a finite set of fermionic oscillators. A massive four-dimensional $\mathcal N=1$ representation has two complex oscillators and therefore four states for a scalar Clifford vacuum. A massless representation has only one active oscillator because $\sigma\cdot p$ has rank one; it contains a helicity pair and must often be joined to its CPT conjugate.

Required background. The four-dimensional N=1 algebra fixes the anticommutator and adjoint. One-particle states, mass, spin, and normalization supplies Wigner’s massive and massless little groups.

Helpful background. Representations and intertwiners supplies tensor-product decomposition, while multiplets and selection rules distinguishes symmetry organization from dynamics.

Physical-state multiplets

This page counts normalizable, gauge-invariant one-particle states in a positive Hilbert space. It does not count field components before constraints, gauge redundancy, or auxiliary fields. Those are treated in component multiplets and closure.

Because $[P_\mu,Q_\alpha]=0$, every state obtained by acting with a supercharge has the same momentum and mass. The construction at fixed $p$ is therefore:

1. evaluate the Hermitian matrix $\{Q_A,Q_B^\dagger\}$;
2. quotient its null directions;
3. rescale positive eigenmodes to canonical fermionic oscillators;
4. choose a Clifford vacuum annihilated by the lowering operators;
5. take its finite exterior-algebra orbit; and
6. restore little-group covariance and, when required, CPT.

Unitarity enters at step 2: a vector with zero norm is set to zero in the physical Hilbert space. Merely finding a zero eigenvalue in a formal component space is not shortening.

Massive N=1 multiplets

For $P_\mu=(m,\mathbf0)$ with $m>0$,

$$\{Q_\alpha,Q_\beta^\dagger\}=2m\,\delta_{\alpha\beta}.$$

Define

$$a_\alpha=\frac{Q_\alpha}{\sqrt{2m}}, \qquad a_\alpha^\dagger=\frac{Q_\alpha^\dagger}{\sqrt{2m}},$$

so

$$\{a_\alpha,a_\beta^\dagger\}=\delta_{\alpha\beta}, \qquad \{a_\alpha,a_\beta\}=0.$$

The two creation operators transform as a spin-$\tfrac12$ doublet of the massive little group $SU(2)$. If the Clifford vacuum $|\Omega;j,m_j\rangle$ has spin $j$ and is annihilated by both $a_\alpha$, its Fock levels are

Level	Oscillator factor	Little-group content
0	$1$	$j$
1	$a_\alpha^\dagger$	$j\otimes\tfrac12=(j+\tfrac12)\oplus(j-\tfrac12)$
2	$a_1^\dagger a_2^\dagger$	$j$

The $j-\tfrac12$ term is absent for $j=0$. Thus a generic $j>0$ multiplet has spins $j+\tfrac12$, two copies of $j$, and $j-\tfrac12$. The collapsed $j=0$ case has two spin-zero states and one spin-$\tfrac12$ representation:

$$2\times(0)\oplus\left(\frac12\right).$$

It contains two bosonic and two fermionic spin states. Choosing a spin-$\tfrac12$ Clifford vacuum instead gives the massive vector content

$$(1)\oplus(0)\oplus2\times\left(\frac12\right),$$

with four bosonic and four fermionic states. The complete Clebsch–Gordan construction is given in Weinberg 2000, § 25.5, pp. 48–51.

For $n$ non-null complex oscillators, the Fock factor has dimension $2^n$. Boson–fermion equality follows without inspecting spins:

$$\operatorname{Tr}_{\Lambda^\bullet\mathbb C^n}(-1)^F =\sum_{k=0}^n(-1)^k\binom nk =(1-1)^n=0$$

for $n>0$. The statement applies at fixed nonzero momentum in a finite unitary multiplet; vacuum and continuum subtleties require separate care.

Massless rank reduction

Take $p^\mu=(E,0,0,E)$, so with the mostly-minus metric $p_\mu=(E,0,0,-E)$. Then

$$2\sigma^\mu p_\mu =2E(\mathbf1-\sigma^3) =\begin{pmatrix}0&0\\0&4E\end{pmatrix}.$$

The $Q_1$ direction has zero anticommutator with its adjoint. Positivity therefore makes it act trivially on the irreducible physical representation. The surviving pair may be normalized as

$$b=\frac{Q_2^\dagger}{2\sqrt E}, \qquad b^\dagger=\frac{Q_2}{2\sqrt E}, \qquad \{b,b^\dagger\}=1,$$

where this naming chooses $b^\dagger$ to lower helicity by $\tfrac12$. Starting from a highest-helicity state $|h\rangle$ gives

$$|h\rangle, \qquad b^\dagger|h\rangle=|h-\tfrac12\rangle.$$

This two-state set is an irreducible representation of the connected super-Poincaré algebra. It need not be CPT invariant. CPT reverses helicity and conjugates all internal charges, so it supplies a second pair

$$|-h+\tfrac12\rangle, \qquad |-h\rangle.$$

The rank-one construction and normalization are derived in Weinberg 2000, § 25.4, pp. 43–47.

The N=1 chiral and vector examples

The familiar CPT-complete massless representations are:

Multiplet	One helicity pair	CPT-conjugate pair	Physical count
chiral	$+\tfrac12,0$	$0,-\tfrac12$	$2$ bosonic + $2$ fermionic
vector	$+1,+\tfrac12$	$-\tfrac12,-1$	$2$ bosonic + $2$ fermionic

For a chiral multiplet the two helicity-zero states are a complex scalar and its antiparticle; the helicity $\pm\tfrac12$ states form a Weyl fermion and its antiparticle. For a vector multiplet the helicity $\pm1$ states are the transverse gauge boson polarizations and the helicity $\pm\tfrac12$ states are the gaugino. Longitudinal gauge modes, ghosts, and auxiliary fields are not one-particle states in this table.

A pair can be displayed without its CPT conjugate when one is deliberately discussing a chiral charge sector or a complex representation and the conjugate sector is understood separately. A local, unitary, CPT-invariant QFT must contain the full conjugate spectrum. For massive charged representations, CPT similarly maps the charge-$q$ multiplet to charge $-q$; it need not act within one irreducible charge sector.

Extended massless multiplets

With $\mathcal N$ four-dimensional supercharges and no central charge acting on a massless representation, one complex oscillator survives for each $I$. Acting with $k$ distinct lowering operators gives

$$\binom{\mathcal N}{k} \quad\text{states of helicity}\quad h_{\max}-\frac{k}{2},$$

so there are $2^{\mathcal N}$ states before a separate CPT completion. The states at level $k$ transform in $\Lambda^k\mathbf{\mathcal N}$ of the $SU(\mathcal N)$ part of R-symmetry.

The helicity range is

$$h_{\min}=h_{\max}-\frac{\mathcal N}{2}.$$

A single multiplet is CPT self-conjugate when its helicities and internal representations map back to themselves. At the level of endpoints this requires $h_{\max}=\mathcal N/4$. Two important cases are:

• $\mathcal N=4$, $h_{\max}=1$: the vector multiplet runs from $+1$ to $-1$ and has $16$ states;
• $\mathcal N=8$, $h_{\max}=2$: the gravity multiplet runs from $+2$ to $-2$ and has $256$ states.

This count is algebraic. The statement that interacting massless particles of spin above one or two are obstructed uses additional soft-theorem and locality assumptions, not the oscillator algebra alone.

A reusable multiplet check

For any proposed table, verify:

1. Representation space: physical one-particle states or field components?
2. Momentum orbit: massive $SU(2)$ little group or massless helicity?
3. Positive matrix: how many non-null complex supercharge modes remain?
4. State count: does the Fock dimension equal $2^n$ times the Clifford-vacuum degeneracy?
5. Statistics: do bosonic and fermionic physical states balance?
6. Gauge quotient: have longitudinal and pure-gauge states been removed?
7. CPT: is the displayed set self-conjugate; if not, where is its conjugate?
8. Shortening: is a missing state caused by a null supercharge, or merely by an equation of motion or gauge choice?

Common pitfalls

Counting fields instead of states. An off-shell vector field has four components, a massless gauge particle has two helicities, and an auxiliary field has no one-particle excitation. These counts answer different questions.

Dropping null charges without using positivity. The implication $\{Q,Q^\dagger\}=0\Rightarrow Q=0$ on physical states uses a positive-definite Hilbert norm. It fails in an unphysical gauge-fixed state space.

Assuming every helicity pair is CPT complete. The pair $(h,h-\tfrac12)$ is generally mapped to a distinct pair. Check charges and helicities rather than the total number of states.

Check your understanding

Construct the massive $\mathcal N=1$ multiplet over a scalar Clifford vacuum and verify its state count.

Answer

There are two fermionic creators. Levels $0$ and $2$ are rotational singlets; level $1$ is a spin-$\tfrac12$ doublet. Hence the representation contains two scalar states and two spin states of one spin-$\tfrac12$ particle: two bosonic and two fermionic states, four in total.

References

• Steven Weinberg, The Quantum Theory of Fields, Volume III: Supersymmetry, Cambridge University Press (2000), §§ 25.4–25.5, DOI.