Massive and Massless Unitary Supermultiplets

A unitary particle supermultiplet is obtained by evaluating the supercharge anticommutator on a fixed-momentum Wigner representation and retaining its positive oscillator modes. In four-dimensional $\mathcal N=1$ supersymmetry, a massive sector has two complex oscillators, whereas a finite-helicity massless sector has only one. This rank change explains both the state counts and the helicity ladders; CPT then decides whether the resulting irreducible multiplet can occur by itself in a local relativistic theory.

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.

Positive one-particle representation space

This page classifies the finite-dimensional polarization fiber of a sharp-mass, positive-energy Wigner one-particle sector with unbroken Poincaré supersymmetry. A sharp momentum ket $|p,\lambda\rangle$ is a generalized, distributional state; normalizable one-particle states are wave packets over the same fiber. In a gauge theory, the count is made on the positive physical quotient—equivalently, on gauge-equivalence classes or the appropriate BRST cohomology—not on the covariant field-component space.

The construction does not cover continuous-spin representations, infraparticles without a sharp mass shell, confined colored excitations, unstable resonances, or the exceptional orbit $p^\mu=0$. Those cases do not possess the finite little-group fiber assumed below.

Because $[P_\mu,Q_\alpha]=0$, a supercharge preserves momentum and mass. At a chosen $p$, one therefore:

1. evaluates the Hermitian matrix $G_{AB}=\{Q_A,Q_B^\dagger\}$;
2. uses positivity to quotient every zero-eigenvalue direction;
3. rescales the positive eigenmodes to canonical fermionic oscillators;
4. chooses a state annihilated by all oscillator lowerers;
5. forms its finite exterior-algebra orbit; and
6. restores little-group covariance and, when necessary, the CPT-conjugate sector.

The state in step 4 is called a Clifford vacuum. It is not the zero-particle vacuum of the QFT: it is a state at the same nonzero momentum as the rest of the multiplet, selected only because all oscillator annihilators kill it. One complex oscillator means one annihilation–creation pair, not two independent degrees of freedom. This fixed-momentum oscillator construction is reviewed in Sohnius 1985, § 3.

The shortening and recombination map and its comparison table place this kinematic construction beside BPS and superconformal shortening without identifying the three mechanisms.

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 that

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

The two creators transform as a spin-$\tfrac12$ doublet of the massive little group $SU(2)$. If $|\Omega;j,m_j\rangle$ has spin $j$, definite fermion parity, and $a_\alpha|\Omega\rangle=0$, its levels are

Level	Exterior 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 when $j=0$. Including every magnetic sublevel,

$$\dim\mathcal H_j=(2j+1)\dim\Lambda^\bullet\mathbb C^2 =4(2j+1), \qquad N_B=N_F=2(2j+1).$$

Here a spin-$\tfrac12$ representation counts as two spin states. This distinction between an irreducible spin representation and its dimension prevents a common factor-of-two error.

For a bosonic scalar Clifford vacuum, the content collapses to

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

namely two bosonic and two fermionic states. For a fermionic spin-$\tfrac12$ Clifford vacuum, levels 0 and 2 give two spin-$\tfrac12$ fermion representations, while level 1 gives the bosonic decomposition $\tfrac12\otimes\tfrac12=1\oplus0$. The result is the massive vector content

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

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

These counts describe one connected super-Poincaré irrep in a fixed internal sector. If its Clifford vacuum transforms in a complex representation $R$, CPT adds the conjugate $\bar R$ multiplet and doubles the displayed count.

More generally, for $n>0$ non-null complex oscillators,

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

Thus bosonic and fermionic states balance in each finite multiplet. The argument concerns the fixed-momentum polarization fiber; it does not apply without modification to the zero-momentum vacuum or to continuum states.

Massless rank reduction

For a finite-helicity massless representation, the little group is $ISO(2)$ and its translation subgroup acts trivially. This is precisely the assumption that excludes continuous-spin representations. Choose

$$p^\mu=(E,0,0,E), \qquad p_\mu=(E,0,0,-E)$$

in the mostly-minus convention. Then

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

For any physical state $|\psi\rangle$,

$$0=\langle\psi|\{Q_1,Q_1^\dagger\}|\psi\rangle =\lVert Q_1|\psi\rangle\rVert^2 +\lVert Q_1^\dagger|\psi\rangle\rVert^2.$$

Positivity therefore makes both $Q_1$ and $Q_1^\dagger$ act trivially in the irreducible physical representation. This is a supersymmetry-null direction caused by the rank of $\sigma\cdot p$; it is not the same statement as a covariant state being pure gauge or BRST exact.

Normalize the surviving pair by

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

The spinor transformation law gives

$$[J_3,Q_2]=-\frac12Q_2, \qquad [J_3,Q_2^\dagger]=+\frac12Q_2^\dagger.$$

Hence $b^\dagger$ lowers helicity by $\tfrac12$. If $|h\rangle$ is killed by $b$, the connected super-Poincaré irrep is

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

This two-state set need not be CPT self-conjugate. CPT reverses helicity and conjugates the internal representation $R\mapsto\bar R$, supplying

$$|-h+\tfrac12;\bar R\rangle, \qquad |-h;\bar R\rangle.$$

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

Chiral and vector examples

The familiar CPT-complete massless $\mathcal N=1$ representations are

Multiplet	Irreducible pair in $R$	CPT pair in $\bar R$	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 excitations are the particle and antiparticle of one complex scalar—equivalently, two real scalar degrees of freedom. The helicities $\pm\tfrac12$ are the corresponding Weyl-fermion excitations. For a vector multiplet, the helicities $\pm1$ are transverse gauge-boson polarizations and the helicities $\pm\tfrac12$ are the gaugino. Longitudinal modes, ghosts, and auxiliary fields are absent from the physical count. In the usual Yang–Mills example the adjoint internal representation is real, so the CPT partner lies in the same representation even though the helicities reverse.

A raw pair may be displayed without its conjugate when a complex charge sector and its conjugate are being listed separately. Under the usual CPT-theorem hypotheses—local Lorentz-covariant fields, a positive Hilbert space, the spectrum condition, and a stable vacuum—the complete particle spectrum is closed under CPT. This conclusion uses more than the connected super-Poincaré algebra alone; parity is a separate transformation.

Extended massless multiplets

Let $\mathcal N$ denote $\mathcal N$ Weyl supercharges, or $4\mathcal N$ real supercharges. The massless null relation holds for every $I$. It also forces the scalar central charges to act trivially:

$$Q_1^I=0 \quad\Longrightarrow\quad \{Q_1^I,Q_2^J\} =2\epsilon_{12}Z^{IJ}=0$$

on the representation. Thus the vanishing central charge in a finite-helicity massless multiplet follows from positivity and the algebra rather than being an unrelated assumption; compare extended supersymmetry and central charges.

One complex oscillator survives for each $I$. In the site’s convention the active creators transform in the fundamental $\mathbf{\mathcal N}$, so level $k$ contains

$$\Lambda^k\mathbf{\mathcal N}, \qquad \dim\Lambda^k\mathbf{\mathcal N}=\binom{\mathcal N}{k}, \qquad h_k=h_{\max}-\frac{k}{2}.$$

There are $2^{\mathcal N}$ states before any separate CPT completion. For the $\mathcal N=4$ vector multiplet, the entire ladder is visible at once:

$k$	Helicity	$SU(4)_R$ representation	Multiplicity
0	$+1$	$\mathbf1$	1
1	$+\tfrac12$	$\mathbf4$	4
2	$0$	$\mathbf6$	6
3	$-\tfrac12$	$\bar{\mathbf4}$	4
4	$-1$	$\mathbf1$	1

The helicity range is $h_{\min}=h_{\max}-\mathcal N/2$. Endpoint symmetry therefore requires $h_{\max}=\mathcal N/4$ for a multiplet to be CPT self-conjugate, but this is only necessary: the internal representations and Clifford-vacuum quantum numbers must also map back to their conjugates. The $\mathcal N=4$ vector multiplet satisfies the full test and has $16$ states. The $\mathcal N=8$ gravity multiplet similarly runs from helicity $+2$ to $-2$ and has $256$ states.

These are representation-theoretic statements. Whether a given multiplet occurs as an asymptotic particle spectrum also depends on locality, interactions, gauge symmetry, infrared behavior, and the existence of a sharp massless pole.

A reusable multiplet check

For any proposed particle table, ask:

1. Is it counting a Wigner polarization fiber, physical wave packets, or field components?
2. Is the orbit massive with little group $SU(2)$, or finite-helicity massless with trivial $ISO(2)$ translations?
3. What is the spectrum of the positive matrix $\{Q,Q^\dagger\}$?
4. How many complex creation modes remain after the positive-null quotient?
5. Does $2^n$ times the Clifford-vacuum dimension reproduce the spin-state count and boson–fermion balance?
6. Has the gauge or BRST quotient already been taken?
7. Is the internal representation self-conjugate under CPT; if not, where is the conjugate sector?
8. Is the missing state caused by kinematic rank loss, BPS saturation, or a field equation? These are different mechanisms.

Common pitfalls

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

Treating a sharp momentum ket as normalizable. The finite oscillator calculation is performed fiber by fiber. A normalizable particle state is obtained only after smearing over momentum with the invariant one-particle measure.

Dropping a null charge without positivity. The implication $\{Q,Q^\dagger\}=0\Rightarrow Q=Q^\dagger=0$ uses the positive physical inner product. It need not hold in an unquotiented gauge-fixed space.

Checking CPT only by helicity endpoints. CPT also conjugates internal quantum numbers. A symmetric helicity range can still fail to be a self-conjugate representation.

Exercises

1. Massive spin counting. Starting from a Clifford vacuum of spin $j$ and definite fermion parity, derive the dimension and boson–fermion counts. Specialize to a bosonic $j=0$ vacuum and a fermionic $j=\tfrac12$ vacuum.

Solution

The two creators span $\Lambda^\bullet\mathbb C^2$, of dimension $1+2+1=4$, while the vacuum spin representation has dimension $2j+1$. Hence $\dim\mathcal H_j=4(2j+1)$. Even exterior levels have total dimension $2(2j+1)$ and the odd level has the same dimension, so $N_B=N_F=2(2j+1)$ after assigning the vacuum parity. For bosonic $j=0$ this gives two scalars and one spin-$\tfrac12$ representation. For fermionic $j=\tfrac12$, the even levels give two fermionic spin-$\tfrac12$ representations and the odd level gives bosonic spin $1\oplus0$.

2. Null action and CPT. For $p^\mu=(E,0,0,E)$, diagonalize $2\sigma\cdot p$, prove that $Q_1$ and $Q_1^\dagger$ vanish on the positive physical representation, and CPT-complete the pairs with $h=\tfrac12$ and $h=1$.

Solution

The eigenvalues are $0$ and $4E$. The zero expectation value of $\{Q_1,Q_1^\dagger\}$ is the sum $\lVert Q_1\psi\rVert^2+\lVert Q_1^\dagger\psi\rVert^2$, so both terms vanish. Since $Q_2$ lowers helicity by $\tfrac12$, the raw pairs are $(\tfrac12,0)$ and $(1,\tfrac12)$. CPT adds $(0,-\tfrac12)$ and $(-\tfrac12,-1)$, respectively, while conjugating the internal representation.

3. The N=4 ladder. Use the exterior powers of $\mathbf4$ to recover the multiplicities $1,4,6,4,1$ and verify CPT self-conjugacy.

Solution

The five levels are $\Lambda^k\mathbf4$ with dimensions $\binom4k$. Their representations are $\mathbf1$, $\mathbf4$, $\mathbf6$, $\bar{\mathbf4}$, and $\mathbf1$, while their helicities are $1$, $\tfrac12$, $0$, $-\tfrac12$, and $-1$. CPT pairs levels $k$ and $4-k$ because their helicities reverse and their $SU(4)_R$ representations are conjugate. The middle $\mathbf6$ is real. The total count is $1+4+6+4+1=16$.

References

• Martin F. Sohnius, “Introducing Supersymmetry,” Physics Reports 128 (1985), 39–204, especially § 3, DOI.
• Steven Weinberg, The Quantum Theory of Fields, Volume III: Supersymmetry, Cambridge University Press (2000), §§ 25.4–25.5, DOI.

Further reading

• David Tong, Supersymmetric Field Theory, §§ 2.3.2–2.4.1, lecture notes.