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Choosing $N$ different complete one-particle states from an orthonormal basis of size $M$ gives $\left(\genfrac{}{}{0ex}{}{M}{N}\right)$ independent Slater determinants. This is the binomial coefficient because only the occupied subset matters: permutations change the sign of the Slater determinant, not the physical quantum state. Summing over particle numbers gives $2^M$ states in the finite fermionic Fock space.

For the diagram's two spatial orbitals $A,B$, take energies $0,\Delta$ with $\Delta >0$, each independent of spin. In the complete-state order $(A\uparrow ,A\downarrow ,B\uparrow ,B\downarrow )$, the six $N=2$ occupation number strings are $1100,1010,1001,0110,0101,0011$. Their total energies are $0,\Delta ,\Delta ,\Delta ,\Delta ,2\Delta$, giving energy multiplicities $1,4,1$. For $N=3$, the ground state energy is $\Delta$ with two independent ground states, according to which spin orbital in $B$ is occupied. Across $N=0,1,2,3,4$, the counts are $1,4,6,4,1$, summing to $16$. These are exact counts for this model, not measured atomic energies.

For a one-particle Hilbert space $H$, the bosonic or fermionic Fock space is ${\mathcal{F}}_{\pm }(H)={⨁}_{n=0}^{\infty }{H}^{{\otimes }_{\pm }n}$, using symmetric or antisymmetric tensor powers. Its zeroth summand is the one-dimensional vacuum sector. Creation operators and annihilation operators change particle number. The bosonic Fock space and fermionic Fock space implement the corresponding exchange statistics.

For a real smooth function $\chi$ on Euclidean space, put $A_a={\partial }_a-{\partial }_a\chi$ and $q={\sum }_a{\psi }_a^{†}{A}_a$. The adjoint operator for the usual $L^2$ inner product is $A_a^{†}=-{\partial }_a-{\partial }_a\chi$. Since $[A_a,A_b]=0$, the canonical anticommutation relations give $q^2=0$. Expanding $H=\{q,q^{†}\}$ and using $[A_a,A_b^{†}]=-2{\partial }_a{\partial }_b\chi$ givesIn the empty fermionic Fock space sector the last term vanishes; in the filled sector ${\psi }_a^{†}{\psi }_b={\delta }_{ab}$. Consequently the two scalar Hamiltonian operators are $H_0=-\Delta +|\nabla \chi {|}^2+\Delta \chi$ and $H_d=-\Delta +|\nabla \chi {|}^2-\Delta \chi$. On the compatible self-adjoint operator domain the Hamiltonian operator is nonnegative because $⟨v,Hv⟩=\Vert qv{\Vert }^2+\Vert q^{†}v{\Vert }^2$.

A massless supermultiplet is a positive-energy unitary irreducible representation of the Super-Poincaré algebra with null four-momentum. For finite helicity states in four dimensions, half the supercharges act trivially and the others form a fermionic Fock space.

Use $[L_m,L_n]=(m-n)L_{m+n}+\frac{c}{12}(m^3-m){\delta }_{m+n,0}$ and $L_n^{†}=L_{-n}$. A normalized highest-weight vector obeys $L_0|h⟩=h|h⟩$ and $L_n|h⟩=0$ for $n>0$. Its Virasoro descendants are products of negative modes. Their inner products are determined entirely by $c,h$ and the Virasoro algebra; for example $\Vert L_{-1}|h⟩{\Vert }^2=2h$, so unitarity requires $h\ge 0$. An irreducible unitary highest-weight Virasoro module is the Virasoro Verma module quotient by its null submodules, and its surviving inner product must be positive definite.

The classification for $c<1$ is the Unitary Virasoro discrete series:where $m$ is an integer at least $2$, $1\le r\le m-1$, and $1\le s\le m$. The identification $(r,s)\sim (m-r,m+1-s)$ leaves $m(m-1)/2$ distinct modules. The two generating singular-vector levels are $rs$ and $(m-r)(m+1-s)$; all their Virasoro descendants are removed in the irreducible quotient. For $m=2$ the only module is the trivial $c=h=0$ module; for $m\ge 3$ these are the nontrivial unitary discrete-series modules. There are no other irreducible unitary highest-weight representations with $c<1$. This is the classification theorem, rather than a condition inferred only from finitely many descendant norms.

At $c=1/2$, $m=3$ and the possible weights areConstruct them with a single free chiral Majorana fermion:The free chiral Majorana fermion conformal field theory has $T=-:\psi \partial \psi :/2$. In the Neveu–Schwarz sector, $r\in \mathbb{Z}+1/2$, while in the Ramond sector, $r\in \mathbb{Z}$. With positive modes moved to the right and the zero-mode product antisymmetrized, the corresponding generators areThe oscillator relation gives $[L_n,b_r]=-(r+n/2)b_{n+r}$. The stress-tensor operator product expansion, or the reordering of the quadratic generators, then gives the Virasoro algebra with $c=1/2$. One can check both constants directly in the ground states. In the NS sector,In the R sector, $b_0^2=1/2$ and $L_{-1}|R⟩=\frac{1}{2}{b}_{-1}{b}_0|R⟩$, whose norm squared is $1/8$. The relation $[L_1,L_{-1}]=2L_0$ requires $2h_R=1/8$, so $h_R=a=1/16$. Thus the Ramond sector zero-mode shift is essential.

The three constructions in Ising Virasoro modules from a Majorana fermion are explicit. The NS vacuum is a highest-weight vector of weight $0$. The state $b_{-1/2}|0⟩$ is a highest-weight vector of weight $1/2$: applying any positive $L_n$ gives $-(n/2-1/2)b_{n-1/2}|0⟩=0$, including the zero coefficient for $n=1$. Since $L_n$ is quadratic in fermions, the two cyclic modules lie in the even and odd fermion parity sectors, respectively. For R, choose a normalized ground state annihilated by all $b_n$ with $n>0$ and with $b_0|R⟩=\pm |R⟩/\sqrt{2}$. All positive $L_n$ annihilate it and its weight is $1/16$. The two zero-mode sign choices give isomorphic Virasoro modules; one choice suffices to construct the third module.

The oscillator inner products are positive, and each cyclic highest-weight representation just constructed is irreducible. Indeed any nonzero proper invariant graded submodule would have a lowest-energy vector $w$ of weight greater than the cyclic ground weight. Every positive $L_n$ annihilates $w$; thus $w$ is orthogonal to every negative-mode descendant of the original ground vector, by moving the modes across the inner product. It is also orthogonal to that ground vector by its different $L_0$ eigenvalue. Since those Virasoro descendants span the cyclic module, $w$ would have zero norm, a contradiction. This proves irreducibility directly inside the positive fermionic Fock space.

In fact the cyclic modules fill the entire even NS, odd NS and chosen R Fock spaces. Otherwise an orthogonal remaining lowest vector would be the ground vector of a further unitary highest-weight Virasoro module; its weight would lie in ${\mathbb{Z}}_{>0}$, $1/2+{\mathbb{Z}}_{>0}$ or $1/16+{\mathbb{Z}}_{>0}$, respectively. None belongs to the classified list $\{0,1/2,1/16\}$. As a check, their Virasoro characters, defined by ${\chi }_h(q)=\mathrm{Tr}{q}^{L_0-c/24}$, areThe NS signs project onto even and odd occupation number; the R product uses the single chosen ground state. Null vectors of the abstract Virasoro Verma modules are automatically zero in this construction. In particular $L_{-1}|0⟩=0$, and the weight-$1/2$ and weight-$1/16$ ground states obey, respectively, $(L_{-2}-\frac{3}{4}{L}_{-1}^2)|1/2⟩=0$ and $(L_{-2}-\frac{4}{3}{L}_{-1}^2)|1/16⟩=0$, as is also verified by their level-two Gram matrices.

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.

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 the printed interaction, so the Dyson series contributes $-ig$ per vertex before its spinor matrix. The momentum-space Feynman rules areEach interaction vertex has one pseudoscalar line and one incoming and one outgoing fermion-number line. Conserve four-momentum at every vertex, integrate each independent loop momentum with $d^4\ell /(2\pi {)}^4$, and divide by any Feynman-diagram symmetry factor. A closed fermion loop contributes an extra minus sign. The external wavefunctions are $u$ for incoming fermions, $\bar{u}$ for outgoing fermions, $\bar{v}$ for incoming antifermions and $v$ for outgoing antifermions; spinor matrices are multiplied in order along the fermion line. An external pseudoscalar has unit wavefunction factor in standard relativistic normalization. Relative fermionic signs also occur when the external contractions are permuted.

The leading scattering amplitudes are of order $g^2$. Label the incoming momenta $p_1,p_2$ and outgoing momenta $p_3,p_4$, all with positive energy and $p_i^2=m^2$. Use the Mandelstam variables $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$ and $u=(p_1-p_4{)}^2$. The tree scattering with pseudoscalar exchange is shown below; dashed edges are pseudoscalar propagators, and arrows indicate fermion-number flow, opposite to the motion of an antifermion.

Define the scattering amplitude by $⟨f|S-1|i⟩=i(2\pi {)}^4{\delta }^4(p_3+p_4-p_1-p_2)\mathcal{M}$. Use standard relativistic normalization for external states and spinors. To fix even the overall signs, suppress the positive state-normalization factors and write the creator order of the two-fermion states as $|i⟩=b_1^{†}{b}_2^{†}|0⟩$, $|f⟩=b_3^{†}{b}_4^{†}|0⟩$, and the fermion-antifermion states as $|i⟩=b_1^{†}{d}_2^{†}|0⟩$, $|f⟩=b_3^{†}{d}_4^{†}|0⟩$. Spin labels on the $u_i,v_i$ are implicit.

For two fermions, the direct $t$ graph connects $1\to 3$ and $2\to 4$, while the exchanged $u$ graph connects $1\to 4$ and $2\to 3$. The canonical anticommutation relations give signs $+$ and $-$ for these contractions in the chosen state order. Multiplication of the two vertices and the scalar propagator gives $(-ig{)}^{2}i=-i{g}^2$. HenceInterchanging the labels of either pair of identical external fermions reverses this amplitude, as required for the antisymmetric states of fermionic Fock space.

For a fermion and an antifermion, the graphs are $t$ exchange and $s$ annihilation. Their signs can be derived without guessing an analogy with the first process. The normal-ordered product of the pseudoscalar bilinear contains terms of the formsThe minus in the second term is the antifermion sign of a normal-ordered bilinear. Between the chosen external states, exchange has sign $-$ and pair annihilation followed by pair creation has sign $+$. ThereforeChanging an external-state phase can change the common overall sign; it cannot change the relative minus between the two channels. Both amplitudes use the coefficient exactly as printed. For the Hermitian convention $g=i{g}_P$ discussed in part (a), substitute $g^2=-g_P^2$ throughout, or equivalently use vertex $g_P{\gamma }^5$ from the outset.

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.

Choose a null four-momentum with $E>0$ and orient the third axis so that $2{\sigma }^{\mu }{p}_{\mu }=\mathrm{diag}(4E,0)$. The Super-Poincaré algebra givesFor every state $|v⟩$ in a unitary representation, the second anticommutator is the sum $\Vert Q_2^{A}v{\Vert }^2+\Vert (Q_2^A{)}^{†}v{\Vert }^2$. Thus both operators act trivially. A central charge in supersymmetry also vanishes in this massless supermultiplet: its mixed-spinor anticommutator contains the now-trivial $Q_2$. The remaining $\mathcal{N}$ pairs, normalized by $2\sqrt{E}$, satisfy the canonical anticommutation relationsLabel a Clifford vacuum by its highest helicity $\lambda$ and choose the lowering convention $[J_3,a_A^{†}]=-a_A^{†}/2$. Acting with distinct fermionic creation operators produces the fermionic Fock spaceThis proves the helicity spectrum of a massless supermultiplet and the $2^{\mathcal{N}}$ state count. If the internal charges or helicities do not already make the spectrum invariant under the CPT theorem, add its conjugate using the CPT completion of a supermultiplet.

The spin bound for massless supermultiplets follows directly from the width $\mathcal{N}/2$ of this helicity interval. Requiring $|h|\le 1$ gives $\mathcal{N}\le 4$, achieved by the four-dimensional N=4 super Yang-Mills theory vector multiplet with multiplicities $1,4,6,4,1$ at helicities $1,1/2,0,-1/2,-1$. These helicity bounds are necessary spin restrictions for the usual interacting renormalizable quantum field theory; they do not by themselves establish renormalizability for an arbitrary action. Requiring $|h|\le 2$ gives $\mathcal{N}\le 8$. The four-dimensional N=8 supergravity supergravity multiplet attains this bound with highest helicity $2$ and multiplicities $1,8,28,56,70,56,28,8,1$. There is one $+2$ state and one $-2$ state, the two polarizations of a single graviton.

The stronger restriction relevant to matter is the chirality constraint on extended supersymmetry. An $\mathcal{N}=1$ chiral superfield can contain a left-handed Weyl spinor in a complex gauge group representation $R$. By contrast, a full $\mathcal{N}=2$ hypermultiplet consists of $\mathcal{N}=1$ matter in $R\oplus \overline{R}$, and the gauginos of a vector multiplet are in the real adjoint group representation. Higher unbroken extended supersymmetry retains this pairing. A pseudoreal representation may allow a half-hypermultiplet, but it does not yield net chirality in a genuinely complex gauge group representation. Consequently the chiral gauge spectrum of the Standard Model permits at most unbroken $\mathcal{N}=1$ supersymmetry in its four-dimensional low-energy matter sector.

Fix a positive-mass unitary irreducible representation at rest, $P^{\mu }=(M,0,0,0)$ with $M>0$. Use the central charge in supersymmetry conventionThe conjugate relation is $\{{\bar{Q}}_{\dot{\alpha }A},{\bar{Q}}_{\dot{\beta }B}\}={ϵ}_{\dot{\alpha }\dot{\beta }}\overline{Z^{AB}}$, with the dotted epsilon tensor the conjugate of the undotted one. The supercharges commute with $P_{\mu }$ and the central charges. Their Lorentz transformation is that of a Weyl spinor; together with the ordinary Poincare algebra, these relations specify the Super-Poincaré algebra. Take ${ϵ}_{12}=1$ and write $Z^{12}=2z$. This makes the normalization of the mass bound explicit: a convention that writes $2{ϵ}_{\alpha \beta }{Z}^{AB}$ instead has $z=Z^{12}$.

When $Z=0$, define four fermionic annihilation operators and their fermionic creation operators byIn the rest frame they obey the canonical anticommutation relations $\{a_i,a_j^{†}\}={\delta }_{ij}$ and $\{a_i,a_j\}=0$. Choose a spin-zero Clifford vacuum annihilated by all four $a_i$. Acting with each distinct creator at most once constructs the fermionic Fock spaceThe five levels have the binomial coefficient multiplicities $1,4,6,4,1$, so the massive supermultiplet contains $2^4=16$ states. The four creators are two copies of the spin-one-half SU(2) representation. Their second exterior power isHere the first factor carries physical spin and the second counts the two supersymmetries. Thus level two contains one spin-one triplet and three spin-zero singlets; levels one and three each contain two spin-one-half doublets. The final spin content isEven levels give eight boson states and odd levels eight fermion states, exhibiting boson-fermion degeneracy in a supermultiplet.

For nonzero $z=|z|e^{i\phi }$, pair the two supersymmetries before normalizing. DefineThe rest-frame algebra gives $\{q_{\alpha },r_{\beta }^{†}\}=2z{\delta }_{\alpha \beta }$ and consequentlyFor $M>|z|$, divide $b^{\pm }$ by $\sqrt{2(M\pm |z|)}$ to recover four normalized oscillators. Positivity of the Hilbert space norm proves the BPS bound in supersymmetry, sinceThereforeFor a BPS state saturating this bound, both $b_{\alpha }^{-}$ and their adjoints annihilate every state of the representation. Four of the eight real supercharges act trivially. Only the two $b^{+}$ oscillators remain, so a scalar Clifford vacuum gives a shortened massive supermultiplet with two spin-zero states and one spin-one-half doublet:This is the irreducible count at fixed central charge. A CPT completion of a supermultiplet can require the charge-conjugate multiplet as well, doubling it to eight states; the full hypermultiplet count must not be confused with this four-state count. Saturation is a representation-theoretic relation between mass and conserved charge: shortening persists while the state remains BPS, although its existence can change across parameter space.

For even $\mathcal{N}=2n$, unitary skew-diagonalization of an antisymmetric matrix brings the central-charge matrix to blocks $2z_r\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, $r=1,\dots ,n$. Repeat the oscillator construction separately for each block. Every block supplies two oscillators with norm $2(M+|z_r|)$ and two with norm $2(M-|z_r|)$, givingIf $k$ blocks saturate, $2k$ complex oscillators disappear, leaving $2\mathcal{N}-2k$. Starting with spin $j_0$, the irreducible state count and preserved fraction areThe largest massive shortening, $k=n$, preserves one-half. The ordinary massive representation has $k=0$. This rest-frame discussion excludes $M=0$, where the massless supermultiplet construction uses a null momentum instead.

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

The occupation formula from part (b) givesConsequentlyThe independent occupation numbers $n_1,n_2$ are each zero or one. The Hamiltonian is independent of $n_1$, so its four states haveEach energy admits $n_1=0$ or $n_1=1$; these states differ by occupancy of the zero-energy mode. The ground energy is $-|t|$ for either sign of $t$. These degeneracies refer to the full two-mode fermionic Fock space. If one separately restricts to a fixed global fermion parity, only one state at each energy remains in that parity sector.