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For $d$ fermionic creation operators with the canonical anticommutation relations, the Hilbert space $L^2({\mathbb{R}}^d)\otimes {\Lambda }^{\bullet }{\mathbb{C}}^d$ splits into ${\mathcal{H}}_n=L^2({\mathbb{R}}^d)\otimes {\Lambda }^n{\mathbb{C}}^d$, $0\le n\le d$. The fermion number operator $N={\sum }_a{\psi }_a^{†}{\psi }_a$ has value $n$ on ${\mathcal{H}}_n$; its internal multiplicity is $\left(\genfrac{}{}{0ex}{}{d}{n}\right)$. An odd supercharge $Q=q+q^{†}$, with $q$ raising $n$ and $q^2=0$, gives $H=\{q,q^{†}\}$ and preserves these sectors.

On an eigenspace with $E>0$, $Q/\sqrt{E}$ is an odd involution pairing bosonic and fermionic states. More precisely,gives an orthogonal decomposition of each state into the images of $q$ and $q^{†}$: their inner product vanishes because $(q^{†}{)}^2=0$. If $u\in \mathrm{im}{q}^{†}$, then $q^{†}u=0$, $q^{†}qu=Eu$ and $\Vert qu{\Vert }^2=E\Vert u{\Vert }^2$. Thus $q/\sqrt{E}$ pairs such a state in degree $n$ with one in degree $n+1$, and $q^{†}/\sqrt{E}$ is the inverse. States with zero energy are killed by both operators and may remain unpaired. The pairing applies to positive spectral subspaces as well; discreteness is not implied.

The canonical anticommutation relations realize the internal states as a fermionic Fock space ${\Lambda }^{\bullet }{\mathbb{C}}^3$. Choosing a Clifford vacuum $|0⟩$ annihilated by all ${\psi }_a$, the states are functions multiplying $|0⟩$, ${\psi }_a^{†}|0⟩$, ${\psi }_a^{†}{\psi }_b^{†}|0⟩$ with $a<b$, or ${\psi }_1^{†}{\psi }_2^{†}{\psi }_3^{†}|0⟩$. Thus the four fermion-number sectors in supersymmetric quantum mechanics arewith internal dimensions $1,3,3,1$. The fermion number operator $N={\sum }_a{\psi }_a^{†}{\psi }_a$ has value $n$ on ${\mathcal{H}}_n$. The sectors $n=0,2$ have even fermion parity, and $n=1,3$ have odd fermion parity.

Write $q={\sum }_a{\psi }_a^{†}{A}_a$, so that the supercharge is $Q=q+q^{†}$. The commutator $[A_a,A_b]$ is zero because the two mixed partial derivatives of $\chi$ agree. The canonical anticommutation relations then imply $q^2=(q^{†}{)}^2=0$ andThe Hamiltonian operator preserves each sector, while $q$ raises and $q^{†}$ lowers its degree by one. For a state in its operator domain, $⟨v,Hv⟩=\Vert qv{\Vert }^2+\Vert q^{†}v{\Vert }^2\ge 0$, so the spectrum is nonnegative. If $Hv=Ev$ with $E>0$, then $Qv/\sqrt{E}$ is a state of the same energy and opposite fermion parity, and $Q(Qv/\sqrt{E})=\sqrt{E}v$. This is an odd involution pairing bosonic and fermionic states and proves equal even and odd multiplicities at every positive eigenvalue.

There is also a more precise adjacent-sector pairing. On the positive-energy eigenspace,The two summands are orthogonal since $q^2=(q^{†}{)}^2=0$. A state $u$ in the image of $q^{†}$ obeys $q^{†}u=0$, $q^{†}qu=Eu$, so $q/\sqrt{E}$ maps it to a normalized partner in the next sector, with inverse $q^{†}/\sqrt{E}$. Thus positive-energy doublets lie in ${\mathcal{H}}_0\oplus {\mathcal{H}}_1$, ${\mathcal{H}}_1\oplus {\mathcal{H}}_2$, or ${\mathcal{H}}_2\oplus {\mathcal{H}}_3$. In particular, every positive-energy state of ${\mathcal{H}}_0$ has a partner in ${\mathcal{H}}_1$, and every one of ${\mathcal{H}}_3$ has a partner in ${\mathcal{H}}_2$; the remaining middle-sector states pair with each other. There is no requirement that all four individual sector spectra coincide. Zero-energy states satisfy $qv=q^{†}v=Qv=0$ and can be unpaired. No confinement assumption was given: a continuous spectrum is possible, with the same even/odd pairing on its positive spectral subspaces.

For real $\chi$, the adjoint operator is $A_a^{†}=-{\partial }_a-{\partial }_a\chi$. In ${\mathcal{H}}_0$, every ${\psi }_a^{†}{\psi }_b$ vanishes; in the one-dimensional internal space of ${\mathcal{H}}_3$, it equals ${\delta }_{ab}$. The two scalar Hamiltonian operators are consequentlyThese are the extreme sectors of the gradient superpotential Hamiltonian in supersymmetric quantum mechanics.

For radial $\chi$ and a normalized state $v_0=f(r)|0⟩$ with $E>0$, $q^{†}{v}_0=0$ and $A_{a}f=(x_a/r)(f^{\prime }-{\chi }^{\prime }f)$. The radial superpartner in three-dimensional supersymmetric quantum mechanics therefore has the simple formIt belongs to ${\mathcal{H}}_1$; $H{v}_1=E{v}_1$ follows from $[H,Q]=0$, and its wavefunction normalization follows from $\Vert Q{v}_0{\Vert }^2=⟨v_0,H{v}_0⟩=E$. Also $Q{v}_1=\sqrt{E}{v}_0$. At the origin the formula is interpreted by regular extension; its degree-one components carry the radial unit vector rather than being three independent radial scalars.

For a zero mode in the fermion-vacuum sector, the factorization of $H_0$ gives $0={\sum }_a\Vert A_{a}f{\Vert }^2$. Hence ${\partial }_{a}f=({\partial }_a\chi )f$, or ${\partial }_a(e^{-\chi }f)=0$. The requested wavefunction isA nonzero normalizable wavefunction exists only if this integral is finite. The positive sign in the exponential is fixed by the given convention $A_a={\partial }_a-{\partial }_a\chi$.

Use a positive-energy unitary representation with finite helicity states, excluding continuous-spin representations. In the specified frame the Pauli matrices giveFor every $A$ and every state $|\chi ⟩$,Positivity therefore gives $Q_2^A={\bar{Q}}_{\dot{2}A}=0$ on this massless supermultiplet. The equal-chirality anticommutator $\{Q_1^A,Q_2^B\}={ϵ}_{12}{Z}^{AB}$ then shows that every central charge in supersymmetry vanishes on this superalgebra representation.

Normalize the remaining supercharges as fermionic raising and lowering operators:They obey the canonical anticommutation relationsChoose a Clifford vacuum $|\lambda ⟩$ of helicity $\lambda \in \frac{1}{2}\mathbb{Z}$ annihilated by every $a_A$. Such a vector exists because the commuting fermion number operators $a_A^{†}{a}_A$ have eigenvalues $0$ and $1$, and lowering every occupied mode produces a nonzero empty state. In an irreducible representation this empty space has dimension one; multiple empty states produce a direct sum of copies. The states areThe canonical anticommutation relations make these states orthonormal for a normalized Clifford vacuum. In the helicity convention $[J_3,Q_1^A]=\frac{1}{2}{Q}_1^A$, its adjoint has $[J_3,a_A^{†}]=-\frac{1}{2}{a}_A^{†}$. Thus the Clifford vacuum is the highest-helicity state andReversing the rotation convention reverses the displayed ordering and leaves all counts unchanged. The level-$k$ states form the exterior power ${\Lambda }^k{\mathbb{C}}^{\mathcal{N}}$, which explains the binomial coefficient. Summing the binomial coefficients gives the helicity spectrum of a massless supermultiplet:Each application of a fermionic raising and lowering operator reverses fermion parity. The even and odd levels each contain $2^{\mathcal{N}-1}$ states, so a physical massless supermultiplet has $2^{\mathcal{N}-1}$ boson and $2^{\mathcal{N}-1}$ fermion states.

These formulas describe one irreducible representation of the massless Super-Poincaré algebra at fixed four-momentum. A reducible superalgebra representation can contain several such supermultiplets. Also, the CPT theorem sends helicity $h$ to $-h$ and conjugates internal charges. CPT completion of a supermultiplet may therefore require a second $2^{\mathcal{N}}$-state supermultiplet, with highest helicity $\mathcal{N}/2-\lambda$. A necessary condition for the original helicity spectrum of a massless supermultiplet to be self-conjugate is $\lambda =\mathcal{N}/4$; internal charges must also admit the conjugation in a realization compatible with the Spin-statistics theorem. Thus this helicity condition alone is not sufficient for CPT completion of a supermultiplet without doubling. In particular, the original irreducible count is not automatically the count after CPT completion of a supermultiplet.

For the conventional bounds on extended supersymmetry, suppose all massless states satisfy $|h|\le s_{\max }$. The helicity spectrum of a massless supermultiplet occupies an interval of width $\mathcal{N}/2$, whereas the available interval has width $2s_{\max }$. HenceAn ordinary four-dimensional renormalizable quantum field theory without gravity uses interacting massless fields of spin angular momentum at most one. Massless spin angular momentum $3/2$ requires supergravity, whose gravitational coupling has negative mass dimension and is a nonrenormalizable interaction by power counting in quantum field theory. Setting $s_{\max }=1$ gives $\mathcal{N}\le 4$ for conventional renormalizable theories. The bound is attained by four-dimensional N=4 super Yang-Mills theory: choosing $\lambda =1$ givesfor its supersymmetric vector multiplet.

Allowing gravity permits $s_{\max }=2$, so $\mathcal{N}\le 8$ for conventional theories including gravity. The four-dimensional N=8 supergravity supergravity multiplet attains the bound with $\lambda =2$:It has $256$ states, $128$ boson and $128$ fermion states. The conventional claim about a maximum “in general” assumes local interacting four-dimensional theories in Minkowski spacetime with a finite spectrum of fields and no massless spin angular momentum above two. It is not a theorem excluding arbitrary free higher-spin angular momentum constructions or all theories in other spacetime settings.

Finally, a chiral gauge spectrum means that the independent left-handed Weyl spinors occur in complex gauge group representations without obligatorily paired conjugate left-handed Weyl spinors. In four-dimensional N=1 supersymmetry, an irreducible matter supermultiplet with helicities $(0,-\frac{1}{2})$ can carry a complex gauge group representation $R$. Its CPT completion of a supermultiplet supplies the right-handed antiparticle in $\bar{R}$, as every left-handed Weyl spinor requires; it does not supply a second independent left-handed Weyl spinor in $\bar{R}$. Consequently a chiral gauge spectrum is possible.

For $\mathcal{N}=2$, the matter hypermultiplet has helicities $(\frac{1}{2},0,0,-\frac{1}{2})$ before internal charge conjugation. If it carries a complex gauge group representation $R$, its CPT completion of a supermultiplet adds the conjugate hypermultiplet. Equivalently, the resulting full hypermultiplet contains two $\mathcal{N}=1$ chiral superfields, one in $R$ and one in $\bar{R}$. These give a vectorlike gauge spectrum. Half-hypermultiplets can occur in pseudoreal representations, which also do not produce genuinely complex chiral gauge spectra. A supersymmetric vector multiplet has Adjoint representation, hence real gauge quantum numbers. Since every $\mathcal{N}\ge 2$ algebra contains an $\mathcal{N}=2$ subalgebra, the same obstruction applies to greater extended supersymmetry.

Therefore a conventional chiral gauge spectrum is possible only for $\mathcal{N}=0$ or $\mathcal{N}=1$. Here $\mathcal{N}=0$ means no supersymmetry; among theories with actual supersymmetry, only $\mathcal{N}=1$ qualifies. These are necessary structural conditions, not a guarantee that a chosen chiral gauge spectrum satisfies anomaly cancellation.

The Hilbert space is the fermionic Fock spaceIn a polarization where ${\hat{\eta }}^a$ acts by exterior multiplication and ${\hat{\bar{\eta }}}_a$ by contraction, a general state isThe fermion number operator $N={\hat{\eta }}^a{\hat{\bar{\eta }}}_a$ returns the exterior degree, and therefore