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Keep the mostly-plus Minkowski metric and the Fourier transform $e^{ip\cdot x}$. Write $/p={\gamma }^a{p}_a$, with $\{{\gamma }^a,{\gamma }^b\}=2{\eta }^{ab}$. To fix the otherwise unspecified phase of ${\gamma }^5$ and the Dirac adjoint, takewhere ${\Gamma }^a$ are ordinary mostly-minus gamma matrices. Then $({\gamma }^5{)}^2=-1$, $\{{\gamma }^5,{\gamma }^a\}=0$, and the fermionic time-derivative term is $i{\psi }^{†}{\partial }_t\psi$. In these conventions the interaction with real $g$ is Hermitian: in mostly-minus notation it is $ig{\bar{\psi }}_{\mathrm{std}}{\Gamma }^5\psi \varphi$. If one instead calls $i{\gamma }^5$ the square-one chirality matrix, its coefficient must be $-ig$ to represent the same interaction. These phase choices leave physical relativistic scattering cross-sections unchanged.

Expanding $e^{i{S}_{\mathrm{int}}}$ yields the following Feynman rules with relativistically normalized external states:

Strip the overall four-momentum conservation delta function when defining the scattering amplitude. There are no further bare interaction vertices, no gauge fixing and no Faddeev-Popov ghost fields in this theory. Renormalized higher-order calculations add the required counterterms; these are additional to the rules of the displayed classical Lagrangian density.

The Dirac field has the global symmetry $\psi \mapsto e^{i\alpha }\psi$, $\bar{\psi }\mapsto e^{-i\alpha }\bar{\psi }$, while the real scalar is unchanged. Both the free Dirac action and the pseudoscalar Yukawa interaction preserve this symmetry. Therefore Dirac fermion number conservation holds: its charge counts particles minus antiparticles.

The initial state has charge $+1$, whereas a final antiparticle and a neutral scalar have charge $-1$. Since the S-matrix commutes with that charge,This holds at every order, not just tree level. In the Feynman rules, a continuous fermion arrow cannot connect these specified external states. Replacing only the outgoing $\bar{u}$ by a $v$ in the previous expression would not give a physical amplitude. Changing both external fermions to antiparticles would instead produce an allowed process with reversed fermion flow and the appropriate $v$ spinors; moving a leg between initial and final states is a different operation governed by crossing symmetry.

Apply the Grassmann Gaussian integral to a regulated finite collection of field components. Up to a field-independent measure normalization and phase,In the continuum this is a formal functional determinant, including spinor and spacetime indices. Its meaningful definition requires a regulator and boundary conditions. A complex Dirac field supplies a determinant, not the inverse square root obtained for a real commuting field.

The comparison is clean after Wick rotation. A free real scalar field with positive Euclidean operator $P_E=-{\partial }_E^2+m^2$ hasTaking a logarithm gives $-\frac{1}{2}\mathrm{Tr}\log P_E$ for the scalar and $+\mathrm{Tr}\log K_E$ for the Dirac field. The opposite statistics sign is the determinant counterpart of the minus sign for a closed fermion loop. The magnitude also differs because a Dirac field has several independent spin and antiparticle degrees of freedom.

The vacuum energy sign of a fermionic oscillator makes the comparison explicit. A real scalar mode contributes $+E/2$; each independent fermionic oscillator contributes $-E/2$. There is one oscillator per scalar momentum, but a massive Dirac field has two particle and two antiparticle oscillators. Thus the vacuum energy densities are formallyEquivalently, the large Euclidean-time vacuum functional behaves as $\log Z_E\sim -\beta V{\rho }_{\mathrm{vac}}$. Both displayed vacuum energies are ultraviolet divergent; a common regularization in quantum field theory and the appropriate renormalization are needed before comparing them. The zero-point vacuum energies of bosons and fermions have opposite signs, but do not cancel without matching masses and degrees of freedom. Normal ordering removes an additive vacuum constant in nongravitating flat-space theory. Coupling to the metric in general relativity makes that constant contribute to the cosmological constant, so it cannot simply be discarded without a renormalization condition.

The effective number of relativistic energy degrees of freedom count two photon polarizations, four electron/positron spin states, and six neutrino/antineutrino states, with the Fermi-Dirac distribution factor $7/8$ for fermions. Treating these species as relativistic and at a common temperature gives $\boxed{g_{\ast }=2+(7/8)(4+6)=10.75}$.

The mass-equivalent radiation mass density is $\rho =({\pi }^2/30)g_{\ast }(k_{B}T{)}^4/({\hslash }^3{c}^5)$. Therefore the Friedmann equation givesAt $T_0=10^{10}\mathrm{K}$, $H_0\simeq 0.503{\mathrm{s}}^{-1}$. Equating the specified weak interaction rate to $H_0(T/T_0{)}^2$ gives $(T_{\ast }/T_0{)}^3=H_0/(1{\mathrm{s}}^{-1})$ and henceThe approximate equilibrium neutron-to-proton ratio, neglecting chemical potentials, is $r_{\ast }\simeq \exp [-(m_n-m_p)c^2/(k_B{T}_{\ast })]$, about $0.152$ using the mass difference $1.293\mathrm{MeV}$. This is a freeze-out estimate rather than a detailed reaction-network computation.

After weak cosmological weak freeze-out, free neutrons undergo beta-minus decay; neutrinos decouple, and electron-positron annihilation heats the photons relative to the neutrinos. Initially the abundant energetic photons destroy deuterium almost as soon as it forms. Once the deuterium bottleneck ends, nuclear reactions rapidly pass through deuterium, tritium and helium-three to helium-4. Reactions eventually become too slow compared with cosmic expansion; nearly all surviving neutrons are bound in helium-four. Its mass fraction is approximately $Y_4=2r/(1+r)$, where $r$ is evaluated when nuclear assembly begins.

Increasing $g_{\ast }$ increases the Hubble parameter, raises $T_{\ast }$, increases the freeze-out neutron-to-proton ratio, and leaves less time for free-neutron decay after freeze-out. Both effects increase $Y_4$; decreasing $g_{\ast }$ reverses them. Most deuterium is only an intermediate product and is burned into helium, so its residual abundance and its absolute change are tiny compared with the helium mass fraction. This does not imply a negligible fractional change in residual deuterium. Faster expansion also shortens the time available to burn it, and its relative abundance can be appreciably sensitive to extra radiation. The distinction matters when interpreting the final wording of the question; joint deuterium and helium abundances constrain extra relativistic species, as discussed in the Particle Data Group nucleosynthesis review.

A Majorana spinor equals its charge conjugation, ${\psi }^c=C{\bar{\psi }}^T=\psi$, with conventional phase choices absorbed into $C$. The two-component worldsheet Majorana fermion therefore has no independent complex conjugate components; in a Majorana representation its components can be real Grassmann variables. The same condition applies to the constant supersymmetry parameter. Grassmann statistics matter in the variation: replacing the spinors by commuting numerical vectors would give the wrong bilinear interchange signs.

The displayed rigid transformation is understood in flat conformal gauge. Take $h_{\mu \nu }={\eta }_{\mu \nu }=\mathrm{diag}(-1,1)$, $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$, and chooseHere $C={\gamma }^0$ is a valid invariant charge-conjugation form: $C^T=-C$ and $({\gamma }^{\mu }{)}^{T}C=-C{\gamma }^{\mu }$. These concrete matrices make all signs checkable; equivalent Majorana conventions give the same result with their consistently transformed bars.

Let $A=\sqrt{{\alpha }^{\prime }/2}$ and $B=1/\sqrt{2{\alpha }^{\prime }}$, so ${\alpha }^{\prime }B=A$. Suppressing the common factor $-1/(4\pi {\alpha }^{\prime })$ in the action, the flat kinetic density isThe rigid variations have $\delta X^a=iA\overline{ϵ}{\psi }^a$ and $\delta {\psi }^a=B{\gamma }^{\nu }{\partial }_{\nu }{X}^aϵ$. The supersymmetry variation is even, so its product rule has no extra graded sign. From the Majorana Grassmann bilinear interchange identities,For example the last identity follows by moving the Grassmann-odd parameter through $\psi$, then using $C^T=-C$ and the transpose relations twice. No equation of motion has been used.

The two kinetic variations areThe Clifford algebra turns the first fermionic term plus the bosonic variation into $iA{\partial }_{\nu }{X}_a\overline{ϵ}{\gamma }^{\mu }{\gamma }^{\nu }{\partial }_{\mu }{\psi }^a$. In the second term, commute the partial derivatives and use the same algebra: the antisymmetric gamma product drops out, leaving $iA\square X_a\overline{ϵ}{\psi }^a$. Together they are precisely the rigid worldsheet supersymmetry boundary term:Therefore $\delta S=-(4\pi {\alpha }^{\prime }{)}^{-1}{\int }_{\partial \Sigma }d{\Sigma }_{\mu }{J}^{\mu }$. It vanishes on a closed or periodic worldsheet, for compactly supported changes, or under compatible supersymmetric endpoint conditions. The local total-derivative identity is off shell; the boundary assumptions are needed to call the integrated action invariant.

The flat-gauge qualification is substantive. On an arbitrary curved worldsheet, spinors need a zweibein and spin connection and a constant spinor parameter need not exist. A fully covariant locally supersymmetric action also involves the worldsheet gravitino; the isolated ordinary-derivative expression does not prove rigid invariance for arbitrary $h_{\mu \nu }$. We have proved the intended rigid symmetry of its flat gauge-fixed action, with the fermionic term inside the same integral and overall normalization. If the displayed last term were read as outside the integral, it would not even define an action.

For phenomenology, bosonic string theory has no spacetime fermions and its usual vacuum contains a tachyon. The spinning string has a Ramond sector, whose fermionic zero modes form a spacetime Clifford algebra and give spacetime spinor states, as well as a Neveu–Schwarz sector. In suitable consistent theories the GSO projection removes the tachyonic NS ground state and keeps the appropriate Ramond chirality. The critical dimension of the RNS superstring is 10 rather than 26: $D$ bosons and $D$ real fermions have matter central charge $3D/2$ per chiral sector, while the reparameterization and superconformal ghosts contribute $-26+11=-15$, so the total anomaly cancels at $D=10$.

Suitable GSO-projected superstrings admit spacetime fermions and a tachyon-free spectrum, making them more promising for particle physics than the bosonic string. Rigid worldsheet supersymmetry alone does not establish a tachyon-free spacetime theory or realistic phenomenology. The projection, consistent sectors and compactification in string theory are additional input; spacetime supersymmetry and a realistic four-dimensional spectrum are not automatic consequences of this classical action.

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.

For noninteracting fermions, each single-particle level has occupation $n_i\in \{0,1\}$ by the Pauli exclusion principle. A microstate has $E={\sum }_i{ϵ}_i{n}_i$, $N={\sum }_i{n}_i$, so its grand-canonical weight is the product of its independent level weights. The grand potential is $\Omega =-kT\log \mathcal{Z}$; the particle number, entropy and pressure follow from $N=-{\partial }_{\mu }\Omega$, $S=-{\partial }_T\Omega$ and $P=-{\partial }_V\Omega$. This occupation description is the basis for the product, continuum and adiabatic calculations.

Summing separately over the two allowed occupations of every level givesThusThe factorization uses the absence of interactions; each level, including any degeneracy labels, is counted separately.

The scalar potential can be written asIts vacuum manifold is the sphere ${\varphi }_a{\varphi }_a=v^2$. The Adjoint double cover from SU(2) to SO(3) rotates any nonzero vacuum expectation value to $(0,0,v)$, with $v>0$. A local unitary gauge removes the two angular fluctuations, leaving $\varphi =(0,0,v+\eta )$. This choice holds in a neighbourhood of the nonzero vacuum; a single global rotation cannot align arbitrary position-dependent fields, and topologically nontrivial configurations can obstruct a global unitary gauge.

The generator $t^3$ fixes the vacuum expectation value, while $t^1,t^2$ do not. Thus the Higgs mechanism in this SU(2) gauge theory with an adjoint Higgs field givesThe two angular Goldstone bosons become the longitudinal polarizations of two massive gauge bosons. Define physical fieldsThese labels describe the fields of this model; they do not identify it with the Standard Model. With the stated adjoint covariant derivative,To express all Yang-Mills theory terms in physical fields, introduce the gauge field strengthsThen $F_{\mu \nu }^3=f_{\mu \nu }+ig{Q}_{\mu \nu }$ and $(F_{\mu \nu }^1\mp i{F}_{\mu \nu }^2)/\sqrt{2}={\mathcal{W}}_{\mu \nu }^{\pm }$. Dropping the constant vacuum energy, the Lagrangian density isTaking the positive gauge coupling convention $g>0$, the canonically normalized mass terms giveThe residual U(1) gauge symmetry makes $W^{\pm }$ oppositely charged. The Yang-Mills theory terms supply interactions of $A$ with $W^{\pm }$ and four-vector interactions. The Higgs mode has cubic and quartic self-interactions, together with $2g^{2}v\eta W^{+}\cdot W^{-}$ and $g^2{\eta }^2{W}^{+}\cdot W^{-}$ couplings; it is neutral under the surviving U(1) gauge symmetry.

Coupling fermions permits a massless electromagnetic gauge boson and massive charged mediators of the weak interaction; fermion couplings with chirality can produce parity-violating weak charged currents. However, this model has no massive neutral Z boson, and its only charge generator is $t^3$. A fundamental doublet has opposite charges, so it cannot reproduce the observed doublet charge assignments through an independent hypercharge. The Standard Model instead has $SU(2{)}_L\times U(1{)}_Y\to U(1{)}_Q$, a complex Higgs doublet, three absorbed Goldstone bosons, a massive Z boson and a weak mixing angle.

Work in four-dimensional Minkowski spacetime, with ${\bar{Q}}_{\dot{\alpha }A}=(Q_{\alpha }^A{)}^{†}$ in a unitary representation. The usual Super-Poincaré algebra has the Poincare algebra and possible internal Lorentz scalars as its even generators, and only the stated Weyl spinors as its odd generators. This closure assumption is essential: Lorentz covariance alone would also permit additional tensor-valued generators. It is the setting selected by the Haag–Łopuszański–Sohnius theorem for ordinary interacting relativistic theories.

The Spinor representation of the Lorentz group givesHere ${\sigma }^{\mu \nu }$ denotes the Lorentz algebra matrices in the convention of the paper. Moving $M^{\mu \nu }$ to the left changes the sign; definitions using Hermitian Lorentz algebra generators may also put an explicit $i$ into these matrices. The conjugate relation acts on the dotted Weyl spinor indices.

The tensor product of group representationsis a Lorentz four-vector. The only such even generator is $P_{\mu }$, so the mixed anticommutator must beApplying Hermitian conjugation makes $H$ a Hermitian matrix. Positivity of $\{S,S^{†}\}$ for every linear combination $S$ of supercharges makes $H$ a positive semidefinite matrix on a positive-energy physical superalgebra representation. Remove any null supercharges, which act trivially, and use a unitary diagonalization of a normal matrix followed by rescaling to obtain $H^A{}_B=2{\delta }^A{}_B$. Thus $\mathcal{N}$ counts the independent nontrivial supercharges.

For equal chirality, the tensor product of group representations isConsequently the tentative equal-chirality anticommutator can contain ${ϵ}_{\alpha \beta }{Z}^{AB}$ and a symmetric spinor tensor $T_{\alpha \beta }^{\mu \nu }{M}_{\mu \nu }{Y}^{AB}$, where $T$ is the intertwining operator for Lorentz covariance onto $(1,0)$. Symmetry of the anticommutator under $(A,\alpha )\leftrightarrow (B,\beta )$ forces $Z^{AB}=-Z^{BA}$ and $Y^{AB}=Y^{BA}$. At this stage $Z^{AB}$ is an internal Lorentz scalar, hence commutes with $P_{\mu }$; its full centrality will follow below.

To determine the commutator with translations, Lorentz covariance and the absence of additional odd generators leave onlyand the relation obtained by Hermitian conjugation. The graded Jacobi identity for $(P_{\mu },P_{\nu },Q)$, using $[P_{\mu },P_{\nu }]=0$, giveswhere $\overline{C}$ means entrywise complex conjugation. This equation alone would not imply $C=0$. Contract the graded Jacobi identitywith ${ϵ}^{\alpha \beta }$. The $M$ term drops out because its spinor tensor is symmetric, and $[P_{\mu },Z^{AB}]=0$ makes the remaining left side vanish. Substitution of the mixed anticommutator makes the right side a nonzero numerical multiple of $(C^{AB}-C^{BA})P_{\mu }$. Thus $C=C^T$. It follows that $\overline{C}=C^{†}$ and $C{C}^{†}=0$, whence $C=0$. Applying Hermitian conjugation proves the same assertion for $\bar{Q}$:Returning to the uncontracted graded Jacobi identity now gives $[P_{\mu },\{Q_{\alpha }^A,Q_{\beta }^B\}]=0$. Since the Lorentz algebra generators do not commute with translations, the tentative $M$ term must have $Y^{AB}=0$. HenceIn particular $Z^{11}=0$ for $\mathcal{N}=1$. The dotted-dotted anticommutator follows by Hermitian conjugation applied to the second relation.

The graded Jacobi identity for $(Q^A,Q^B,{\bar{Q}}_C)$, together with $[P,Q]=0$, gives $[Z^{AB},{\bar{Q}}_C]=0$. As a Lorentz scalar, $Z^{AB}$ commutes with $M_{\mu \nu }$, and it already commutes with $P_{\mu }$. Closure and Lorentz covariance make $[Z^{AB},Q_{\alpha }^C]=D^{AB,C}{}_D{Q}_{\alpha }^D$ for some coefficients. Apply the graded Jacobi identity once more:Thus $D=0$. Since every $Z$ and $Z^{†}$ is itself an anticommutator of supercharges, the $Z$'s also commute with each other and their conjugates. Therefore $Z^{AB}$ are central charges in supersymmetry of the displayed algebra. Additional R-symmetry automorphisms are not among the generators specified here.

For boson-fermion degeneracy in a supermultiplet, fix a physical four-momentum $p$ with $p_0=E>0$, and take the finite internal space of states at that four-momentum. Let $\Gamma =(-1{)}^F$ be fermion parity. It anticommutes with each supercharge and its adjoint. For any fixed $A$, summing the diagonal spinor indices givesbecause ${\sigma }^0=1$ and the Pauli matrices are traceless. Cyclicity of the operator trace and $\Gamma Q=-Q\Gamma$ implyThe supertrace of the preceding summed anticommutator therefore vanishes:This counts physical boson and fermion degrees of freedom, including polarizations. The phrase “any representation” needs a qualification: a one-dimensional even supersymmetric vacuum with $P=Q=Z=0$ has one boson and no fermion. Thus the equality applies to positive-energy physical supermultiplets with finite state counts at fixed four-momentum, not to arbitrary abstract superalgebra representations or unregulated infinite-dimensional operator traces.

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.

Use the canonical Kähler potential and normalize superspace integration by $\int d^2\theta {\theta }^2=1$. The superspace Lagrangian density isThe full superspace integration produces a D-term, while the chiral superfield integral produces an F-term. Work in chiral coordinates for the chiral-superfield component expansion, so the omitted spacetime-derivative terms do not enter the extraction of $F$.

Set $\delta \Phi =\sqrt{2}\theta \psi +{\theta }^{2}F$. A Taylor expansion of the superpotential givesHigher terms vanish because there are only two Grassmann variables ${\theta }^{\alpha }$. With $\theta \psi ={\theta }^{\alpha }{\psi }_{\alpha }$ and $\psi \psi ={\psi }^{\alpha }{\psi }_{\alpha }$, their anticommutation gives $(\theta \psi {)}^2=-\frac{1}{2}{\theta }^2\psi \psi$. ThusThe canonical Kähler potential contributes $F^{\ast }F$. Consequently the part of the Lagrangian density containing the auxiliary field isThe fermion terms coming from the superpotential areThe Weyl spinor kinetic term and the complex scalar field kinetic term are canonical because $K={\Phi }^{†}\Phi$.

Treat $F$ and $F^{\ast }$ as independent variables in their algebraic Euler-Lagrange equations. They giveEquivalently,After eliminating the auxiliary field, the Lagrangian density contains $-V$, so the F-term scalar potential isIt is nonnegative for arbitrary complex $m,g,\phi$, since it is a modulus squared. A supersymmetric vacuum satisfies F-flatness, so its zeros are:If $m=0$ and $g\ne 0$ these coincide; if $g=0$ and $m\ne 0$ only the origin remains; if $m=g=0$ every constant $\phi$ is a supersymmetric vacuum.

To compare masses, expand around a supersymmetric vacuum $v$ satisfying $W^{\prime }(v)=0$, writing $\phi =v+\eta$. With $M=W^{\prime \prime }(v)=m+2gv$,Writing $\eta =(A+iB)/\sqrt{2}$ gives the scalar field mass term $-\frac{1}{2}|M{|}^2(A^2+B^2)$; the Weyl spinor has physical mass $|M|$. Therefore the supersymmetric mass degeneracy isIn particular, at $v=0$ the boson and fermion masses are both $|m|$. At the second supersymmetric vacuum $v=-m/g$ the sign of $M$ reverses, but the physical mass is again $|m|$. Equality here concerns fluctuations about a supersymmetric vacuum, not arbitrary backgrounds with $W^{\prime }(v)\ne 0$.