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$2M$ Majorana zero modes define $M$ complex fermionic modes. Fixing total fermion parity leaves a $2^{M-1}$-dimensional ground space, encoding $M-1$ qubits.

A fermionic operator is odd under fermion parity: if $P=(-1{)}^N$ is the parity operator, then $PAP=-A$. A fermionic creation operator or fermionic annihilation operator is an example. Interchanging odd insertions in a time-ordered product contributes the corresponding fermionic sign. Odd parity alone does not imply that every pair of such operators has zero anticommutator.

For a paired non-self-conjugate momentum, rotate the Ising-chain Nambu spinor by an angle with $\cos \theta =(g-\cos k)/r_k$, $\sin \theta =-\sin k/r_k$ and $r_k=\sqrt{1+g^2-2g\cos k}$. Opposite angles at $k,-k$ preserve the canonical anticommutation relations. The result is $H={\sum }_k{ϵ}_k({\gamma }_k^{†}{\gamma }_k-1/2)$. Self-paired momenta have no pairing and are handled directly as number levels. Periodic spin-chain sectors additionally constrain fermion parity and allowed mode quantization.

The Jordan–Wigner transformation represents a one-dimensional spin one-half chain by fermions with a parity string preceding each site. With $P_n={\prod }_{m<n}(1-2n_m)$, one has $S_n^{+}=P_n{c}_n$ and $S_n^z=1/2-n_n$. The string converts the off-site canonical anticommutation relations into commuting spin operators. Adjacent exchange terms become quadratic fermion expressions, while a periodic end bond depends on global fermion parity.

An odd involution $A$ reverses fermion parity and is its own inverse, so it identifies the bosonic and fermionic subspaces. They have equal dimension in a finite-dimensional representation. For a supercharge $q$ with $q^2=0$ and $\{q,q^{†}\}=cI$, $c>0$, use $A=(q+q^{†})/\sqrt{c}$. This gives a direct proof of boson-fermion degeneracy in a supermultiplet at positive energy. A zero-energy supersymmetric vacuum need not have an accompanying fermionic vacuum.

At fixed nonzero four-momentum, take the finite physical supermultiplet and write $P_F=(-1{)}^F$ for fermion parity. Each supercharge is odd, so $P_F{Q}_{\alpha }=-Q_{\alpha }{P}_F$ and the same holds for its adjoint. By the cyclic property of the trace,Consequently $\mathrm{Tr}[P_F\{Q_{\alpha },Q_{\alpha }^{†}\}]=0$. Sum the two spinor components of the $\mathcal{N}=1$ Super-Poincaré algebra. Since ${\mathrm{Tr}}_{2\times 2}{\sigma }^i=0$, their sum is $4H$. On a fixed-energy representation, $H=EI$, yieldingFor a nonzero physical one-particle momentum, $E>0$, including the massless case. Hence the bosonic and fermionic physical state counts agree:Equivalently a nonzero normalized supercharge and its adjoint define an odd involution pairing the two subspaces. This is supertrace pairing at positive energy; it counts spin polarizations, not merely distinct names of particles.

The phrase “every supermultiplet” requires this positive-energy qualification. A zero-energy supersymmetric vacuum with all supercharges acting as zero is a one-dimensional bosonic representation with no fermionic partner. It is a counterexample to the unrestricted statement, because division by $E$ is then impossible. It does not contradict the requested equality for particle supermultiplets.

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.

Spontaneous supersymmetry breaking preserves the action and conserved supercharges, but the vacuum fails to be invariant. This differs from explicit breaking, which changes the symmetry of the action itself. In global four-dimensional N=1 supersymmetry, summing diagonal entries of the Super-Poincaré algebra writes the Hamiltonian as a sum of positive charge anticommutators. Its expectation is a sum of squared norms. Consequently a zero-energy vacuum is supersymmetric, whereas an existing vacuum with broken global supersymmetry has positive energy density. The energy zero is fixed by the algebra. Neither this positivity statement nor the particle-multiplet mass relations should be transferred without qualification to supergravity.

For supersymmetric quantum mechanics, let $H=\{q,q^{†}\}/2$, $q^2=0$, and let fermion parity anticommute with both charges. On a positive-energy eigenspace, $(q+q^{†})/\sqrt{2E}$ is an odd involution: its square is one, it preserves energy, and it reverses parity. Thus every positive-energy bosonic state has a fermionic partner and conversely. Their contributions cancel in the Witten indexFor a discrete spectrum with a well-defined trace, this proves independence of $\beta$. Continuous changes of parameters also preserve the index if the trace stays well defined and no state escapes to infinity or crosses an unregulated continuum threshold. A nonzero index guarantees an unbroken supersymmetric ground state; a zero index is inconclusive. For example, de Rham supersymmetric quantum mechanics on a circle has a bosonic constant zero-form and a fermionic constant one-form, both normalizable zero modes, so it is unbroken with index zero.

A more explicit one-dimensional family illustrates how normalizability determines the answer. DefinePositivity of these Hamiltonian operators shows that a zero mode must solve $A{\psi }_{-}=0$ or $A^{†}{\psi }_{+}=0$. Thus the only candidates areFor $w(x)=x$, the Gaussian minus-sector wavefunction is normalizable and its reciprocal is not, giving an unbroken model with $I=1$ when the minus sector is assigned even parity. For $w(x)=x^2+a$, $a>0$, the exponent $x^3/3+ax$ tends to opposite infinities at the two ends, so neither candidate is normalizable. Both partner potentials $[w^2\mp w^{\prime }]/2$ grow quartically; their spectra are discrete and their lowest energies cannot be zero. The lowest states are therefore paired at positive energy, $I=0$, and supersymmetry is spontaneously broken. This is supersymmetric factorization and zero-mode normalizability. In quantum mechanics the paired ground states illustrate the charge algebra; a relativistic massless-particle conclusion requires field-theoretic hypotheses.

In relativistic global field theory, Goldstone's theorem has a fermionic version. If a conserved supercurrent charge does not annihilate the vacuum, its Ward identity has a nonzero vacuum contact term. Fourier transforming, the divergence of the current correlator cannot tend to zero smoothly at zero momentum. Its spectral representation therefore contains a massless fermionic pole: the Goldstino, with spin one-half. This statement assumes a Poincaré-invariant vacuum and the usual local unitary field-theory conditions. It does not require a bosonic Goldstone particle for the fermionic symmetry. In supergravity the Goldstino can instead supply the longitudinal polarization of a massive gravitino.

For canonical chiral superfields ${\Phi }_i$, elimination of their auxiliary fields givesA supersymmetric vacuum solves every $W_i=0$. At a stationary broken vacuum,The same matrix $W_{ij}$ is the chiral-fermion mass matrix, so the nonzero vector $\overline{W_j}$ is a zero mode. This derives the Goldstino zero mode from vacuum stationarity. The transformation $\delta {\psi }_i=\sqrt{2}ϵF_i+\cdots$ independently identifies that direction through its inhomogeneous shift.

For one canonical chiral field, $W=fZ$, $f\ne 0$, has $U=|f{|}^2$ everywhere, a massless fermion and a classically flat scalar direction. It is flat F-term breaking with a linear superpotential, rather than an isolated vacuum. In contrast, a nonconstant polynomial $W^{\prime }$ of positive degree has a complex root and hence a supersymmetric vacuum; in particular a generic quadratic or cubic single-field superpotential cannot remove all such roots. This argument assumes canonical nonsingular kinetic terms and does not claim that every runaway or nonpolynomial model has a vacuum.

A renormalizable three-field O'Raifeartaigh model gives a less degenerate illustration:The auxiliary equations require $W_2=m{\varphi }_1=0$, then $W_1=hX{\varphi }_1+m{\varphi }_2=0$, leaving $W_X=f\ne 0$. They cannot vanish simultaneously. For real positive $f,h,m$ with $m^2>fh$, the potential obeysTherefore its global tree-level minimum has ${\varphi }_1={\varphi }_2=0$, arbitrary $X$, and energy $f^2$. The massless ${\psi }_X$ is the Goldstino, while $X$ is a pseudomodulus. At $X=0$, the scalar mass-squared spectrum is $0,0,m^2,m^2,m^2+fh,m^2-fh$, and the fermion masses are $0,m,m$. This exhibits both the stability condition and the loss of ordinary supermultiplet mass degeneracy in a broken vacuum. Quantum corrections can lift the pseudomodulus and must be analysed rather than assumed away; the noncompact classical valley also prevents an unqualified finite-volume Witten index trace argument. The index, the vacuum equations and the Goldstino theorem consequently answer complementary questions.

For global four-dimensional $\mathcal{N}=1$ supersymmetry, useLet $\Pi =(-1{)}^F$ be fermion parity. On a finite one-particle supermultiplet at fixed positive-energy momentum, the supercharges act within the multiplet and $\Pi Q_{\alpha }=-Q_{\alpha }\Pi$. Cyclicity of the finite-dimensional trace givesThus the trace with $\Pi$ of every displayed diagonal anticommutator vanishes. Summing the two components, ${\sum }_{\alpha }\{Q_{\alpha },Q_{\alpha }^{†}\}=4E1$, soSince $E>0$, this proves $N_B=N_F$, for both massive and massless one-particle multiplets. The restriction matters: a zero-momentum supersymmetric vacuum singlet, annihilated by every supercharge, need not contain a fermionic partner. The algebra does not imply equal counts for such a trivial zero-energy representation or for an arbitrary unregulated infinite-dimensional trace.

Consider a physical supermultiplet at fixed four-momentum with positive energy $E$ and finitely many polarization states. Let $\Pi =(-1{)}^F$ be fermion parity, equal to $+1$ on bosons and $-1$ on fermions. Every supercharge is odd, so $\Pi Q_{\alpha }=-Q_{\alpha }\Pi$ and likewise for its adjoint.

Trace cyclicity gives, for each component,Thus the supertrace pairing at positive energy yieldswhere the middle equality uses $\mathrm{tr}{\sigma }^0=2$ and $\mathrm{tr}{\sigma }^i=0$. Since $E>0$,This proves boson-fermion degeneracy in a supermultiplet for both massive and massless particle multiplets, counting physical states rather than unphysical gauge components or auxiliary fields. The positive-energy qualification is essential: a zero-energy supersymmetric vacuum can be a one-dimensional bosonic singlet with all supercharges zero. It need not have a fermionic partner, so the printed assertion cannot literally include arbitrary zero-energy representations.

Use ${\eta }_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, ${\sigma }^{\mu }=(I,\mathbf{\sigma })$ and ${\bar{\sigma }}^{\mu }=(I,-\mathbf{\sigma })$, where $\mathbf{\sigma }$ are the Pauli matrices. The translation and Lorentz generators obey the Poincare algebraOne can derive the additional relations by realizing the supercharges as differential operators on superspace. With left Grassmann derivatives and $P_{\mu }=i{\partial }_{\mu }$, takeUsing $\{{\partial }_{{\theta }^{\alpha }},{\theta }^{\beta }\}={\delta }_{\alpha }{}^{\beta }$, the two mixed derivative-coordinate terms each give $i{\sigma }_{\alpha \dot{\beta }}^{\mu }{\partial }_{\mu }$. The remaining terms cancel by Grassmann anticommutation. Equal-chirality terms cancel in the same way. HenceAll coefficients are independent of $x$, so direct differentiation gives $[P_{\mu },Q_{\alpha }]=[P_{\mu },{\bar{Q}}_{\dot{\alpha }}]=0$. This also shows that supercharges commute with translations and preserve the four-momentum of a state.

For the Lorentz generators, define the left-handed spinor matricesThe identity ${\sigma }^{\mu }{\bar{\sigma }}^{\nu }+{\sigma }^{\nu }{\bar{\sigma }}^{\mu }=2{\eta }^{\mu \nu }I$ gives their Lorentz commutation relations. Acting on the spinor superspace coordinates and their dual derivatives then gives, in a column-component convention,Here dotted index values are identified with the conjugate undotted values, and ${\bar{Q}}_{\dot{\alpha }}=Q_{\alpha }^{†}$. The second formula also follows directly by taking the adjoint of the first, using Hermitian $M_{\mu \nu }$. For example, $S_{12}={\sigma }^3/2$, so $[J_3,Q_1]=-Q_1/2$ and $[J_3,Q_2]=Q_2/2$ for $J_3=M_{12}$. This explicit component convention avoids reversing signs by mixing state matrices with operator-component matrices. Covariance of the mixed anticommutator follows by applying these matrices to ${\sigma }^{\rho }{P}_{\rho }$.

These are the minimal Super-Poincaré algebra relations. In the ordinary particle algebra a possible scalar term in $\{Q_{\alpha },Q_{\beta }\}$ would be proportional to ${ϵ}_{\alpha \beta }$, but an anticommutator is symmetric in its two labels; with only one supersymmetry such a central term vanishes. Extended algebras can instead use an antisymmetric internal index tensor. Tensor charges associated with extended objects are not part of this minimal particle algebra.

For a positive-energy massless supermultiplet, choose $p^{\mu }=(E,0,0,E)$ with $E>0$, so $p_{\mu }=(E,0,0,-E)$ andPositivity of the Hilbert-space norm implies that both $Q_1$ and $Q_1^{†}$ annihilate every state: their squared norms sum to zero. SetFor momentum along the positive third axis, helicity is $h=J_3$. The Lorentz commutators give $[h,a]=a/2$ and $[h,a^{†}]=-a^{†}/2$. Choose a highest-helicity Clifford vacuum $|\lambda ⟩$ with $a|\lambda ⟩=0$. Its partner $a^{†}|\lambda ⟩$ has unit norm if the first state does and has helicity $\lambda -1/2$. No further state is generated because $(a^{†}{)}^2=0$. ThereforeThe two states have opposite fermion parity. A CPT completion of a supermultiplet adds helicities $(-\lambda ,-\lambda +1/2)$ and conjugate internal charges when necessary. Taking $\lambda =1/2$ produces a complex scalar and a Weyl spinor after CPT completion; taking $\lambda =1$ produces the massless vector multiplet. This construction assumes ordinary finite-helicity particle representations, rather than continuous-spin representations.

Let $\Gamma =(-1{)}^F$ be fermion parity. It acts as $+1$ on bosons and $-1$ on fermions, and anticommutes with every odd supercharge. In a finite physical supermultiplet at fixed positive energy $E$, trace cyclicity givesIndeed, cycling $Q_{\alpha }$ in the second term and using $Q_{\alpha }\Gamma =-\Gamma Q_{\alpha }$ cancels the first term. Summing diagonal spinor components of the Super-Poincaré algebra gives ${\sum }_{\alpha }\{Q_{\alpha },Q_{\alpha }^{†}\}=4H$, since the Pauli matrices are traceless. Hence supertrace pairing at positive energy yieldsThis is boson-fermion degeneracy in a supermultiplet. Equivalently, a nonzero charge combination with positive anticommutator gives an invertible parity-changing map between the two sectors. The positive-energy qualification is important: a zero-energy supersymmetric vacuum may be a one-state bosonic representation and need not have a fermionic vacuum partner.

For any normalizable state $|v⟩$, the same algebra gives energy positivity in global supersymmetry:It vanishes exactly when every supercharge and its adjoint annihilates the state. In particular, for an existing translation-invariant vacuum, this provesIn infinite volume, interpret the statement as one about vacuum energy density, using finite-volume normalization first. The energy zero is fixed by the supersymmetry algebra; adding an arbitrary constant to the Hamiltonian while leaving that algebra unchanged is not allowed. For a canonical matter and gauge action, the same positivity is seen from $V={\sum }_i|F_i{|}^2+\frac{1}{2}{\sum }_a{D}_a^2$, so a nonzero auxiliary-field expectation breaks the symmetry. This global Minkowski statement does not apply unchanged to the supergravity F-term potential, which can have negative-energy supersymmetric vacua.

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 mostly-minus metric signature, $\eta =\mathrm{diag}(1,-1,-1,-1)$. The Weyl sigma matrices are ${\sigma }^{\mu }=(I,\mathbf{\sigma })$ and ${\bar{\sigma }}^{\mu }=(I,-\mathbf{\sigma })$, with the three Pauli matricesThe Pauli matrix multiplication law ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$ immediately gives ${\sigma }^{\mu }{\bar{\sigma }}^{\nu }+{\sigma }^{\nu }{\bar{\sigma }}^{\mu }=2{\eta }^{\mu \nu }I$: the $00$ case is $2I$, the mixed cases cancel, and the spatial cases give $-2{\delta }_{ij}I$. The spinor indices of ${\sigma }_{\alpha \dot{\beta }}^{\mu }$ express the correspondence between a Lorentz four-vector and a Hermitian $2\times 2$ matrix. For $X=x^{0}I+x^i{\sigma }_i$, $\det X=(x^0{)}^2-|\mathbf{x}{|}^2$; $X\mapsto MX{M}^{†}$ preserves this determinant for $M\in \mathrm{SL}(2,\mathbb{C})$, yielding the Lorentz spinor double cover.

In four-dimensional N=1 supersymmetry, the odd supercharges $Q_{\alpha }$ and ${\bar{Q}}_{\dot{\alpha }}=Q_{\alpha }^{†}$ transform as conjugate Weyl spinors. Their Super-Poincaré algebra isThe ordinary translation and Lorentz generators satisfy the Poincare algebra, and the Lorentz generators act on the supercharges by the corresponding group representations on Weyl spinors. Tracing the mixed anticommutator gives $P_0=\frac{1}{4}{\sum }_{\alpha }\{Q_{\alpha },Q_{\alpha }^{†}\}$, proving energy positivity in global supersymmetry. For a massive rest-frame state, $P_{\mu }=(M,0,0,0)$, the rescaled charges $Q_{\alpha }/\sqrt{2M}$ obey the canonical anticommutation relations. Acting with their adjoints builds a supermultiplet containing states of alternating fermion parity.

Use the usual normalization of the Super-Poincaré algebra, with ${\sigma }^{\mu }=(1,\mathbf{\sigma })$ and ${\bar{Q}}_{\dot{\alpha }}=Q_{\alpha }^{†}$ under the corresponding index convention:The supercharges are odd operators: they turn bosonic states into fermionic states and vice versa. If $\Pi =(-1{)}^F$ is fermion parity, this statement is $\Pi Q_{\alpha }{\Pi }^{-1}=-Q_{\alpha }$, soThe same relation holds for the conjugate supercharges.

For a finite-dimensional physical supermultiplet at fixed four-momentum with energy $E>0$, let $n_B,n_F$ count physical bosonic and fermionic states. Cyclicity of the ordinary trace and the parity anticommutation implySumming over the two spinor indices givesThis supertrace pairing at positive energy proves boson-fermion degeneracy in a supermultiplet for massive as well as massless positive-energy representations. It counts on-shell polarizations, not merely the names of fields. The requirement $E>0$ matters: a zero-energy supersymmetric vacuum can be a bosonic singlet without a paired fermionic vacuum.

If supersymmetry-breaking operators are explicitly added to the Lagrangian, the original supercharges generally no longer commute with the full Hamiltonian. They are not conserved symmetries generating finite fixed-energy physical supermultiplets; their original anticommutator does not equal the full translation generator with the breaking terms included. Thus the step replacing the parity-weighted anticommutator by $4E\Pi$ on a closed physical representation fails. The odd parity relation alone does not force energy degeneracy or an equal number of physical states at each mass.

This is explicit versus spontaneous supersymmetry breaking. In spontaneous breaking the action still has conserved supercharges, but the vacuum is not annihilated by them. Acting on particle excitations about that vacuum involves the broken-vacuum/Goldstino sector, so an ordinary finite particle multiplet above an invariant vacuum is no longer the correct pairing argument. The vacuum-energy statements below refer to an exact globally supersymmetric Hamiltonian, including the spontaneously broken case; they are not positivity claims for an arbitrary explicitly broken Hamiltonian.

Put $n_n=c_n^{†}{c}_n$ and $P_n={\prod }_{m<n}(1-2n_m)$. The Jordan–Wigner transformation is $S_n^{+}=P_n{c}_n$, $S_n^{-}=P_n{c}_n^{†}$ and $S_n^z=1/2-n_n$. Since ${\sigma }_n^x=P_n(c_n+c_n^{†})$, $P_{n+1}=P_n(1-2n_n)$ and $(c_n+c_n^{†})(1-2n_n)=c_n^{†}-c_n$, adjacent bonds becomeThus, apart from the end bond,The end bond contains the global fermion parity and sets the sector-dependent periodic or antiperiodic fermion modes. It contributes an order-one boundary term, which is negligible for the thermodynamic energy density; it is not identically zero for the finite spin chain.

Choose the discrete Fourier transform convention $c_n=N^{-1/2}{\sum }_k{e}^{-ikn}{c}_k$. Hopping gives $-2J\cos k{c}_k^{†}{c}_k$. Opposite-momentum pairing gives $iJ\sin k(c_k^{†}{c}_{-k}^{†}+c_k{c}_{-k})$, using the canonical anticommutation relations to antisymmetrize the coefficient. ThereforeFor the Ising-chain Nambu spinor ${\Psi }_k=(c_k,-i{c}_{-k}^{†}{)}^T$, expansion ofrecovers every term: the diagonal contributes $J(g-\cos k)(n_k+n_{-k}-1)$, and the two off-diagonal entries supply the pairing. Since ${\sum }_k\cos k=0$, the constant is $-JgN$. Reversing the Fourier sign changes the pairing convention; the specified sign makes the displayed matrix agree directly.

In the conventions $\{Q_{\alpha },Q_{\beta }^{†}\}=2{\sigma }_{\alpha \dot{\beta }}^{\mu }{P}_{\mu }$, the trace over the two spinor components of the Super-Poincaré algebra givesHence, on a normalized state in the domain of the supercharges,This proves energy positivity in global supersymmetry. The zero of energy is fixed by the algebra; adding an arbitrary constant to $H$ while keeping that algebra unchanged is not allowed.

For a finite-dimensional supermultiplet at fixed positive energy and four-momentum, choose a nonzero component $q$ with $\{q,q^{†}\}=cI$, $c>0$, and $q^2=0$. The Hermitian odd operatorIt is therefore an invertible map from the even fermion parity subspace to the odd subspace and conversely. This proves boson-fermion degeneracy in a supermultiplet, $n_B=n_F$, for every positive-energy physical particle supermultiplet. Equivalently, the supertrace of the supercharge anticommutator vanishes because odd operators interchange these two subspaces. A zero-energy vacuum is an exception to the literal claim about every possible representation: it can be a one-dimensional bosonic singlet with all supercharges acting trivially.