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The Grassmann field $c$ is odd and the adjoint scalar field $\varphi$ is even. The two printed signs are consistent with a right-acting BRST differential, whose graded Leibniz rule isThe bracket between two odd fields is graded, so $[c,c]=2c^2$. Applying this rule to the ordinary Lie bracket $[c,\varphi ]=c\varphi -\varphi c$ givesThe second bracket here is graded because both its entries are odd. The graded Jacobi identity gives $[c,[c,\varphi ]]=\frac{1}{2}[[c,c],\varphi ]$. ThereforeThe side of the odd derivation is essential. With the usual left graded Leibniz rule and the same two printed signs, the result would be $-[[c,c],\varphi ]$, which is generally nonzero. A left-acting convention must reverse one of those signs. All subsequent BRST symmetry formulas here use the right-acting convention.

Choose an anti-Hermitian basis $T^a$ with $[T^b,T^c]=f^{abc}{T}^a$, an invariant positive invariant bilinear form on a Lie algebra, and the adjoint covariant derivative $D_{\mu }X={\partial }_{\mu }X+[A_{\mu },X]$. Couplings are absorbed into this convention; restoring $D_{\mu }={\partial }_{\mu }+g[A_{\mu },\cdot ]$ multiplies each ghost interaction vertex below by $g$. The associated BRST charge acts as $Q{A}_{\mu }=-D_{\mu }c$, $Q\bar{c}=ih$, and $Qh=0$.

Write $F^a=\partial \cdot A^a+(n\cdot \partial {\varphi }^a)/\sqrt{2}$. For the gauge-fixing fermion $\Psi =\int {\bar{c}}^a(F^a-i{h}^a/2)$, the right graded Leibniz rule givesThe Gaussian functional integral over the Nakanishi-Lautrup field produces the positive gauge fixing term $F^a{F}^a/2$. The ghost operator is $-\partial \cdot D-(n\cdot \partial ){\mathrm{ad}}_{\varphi }/\sqrt{2}$.

Now use the canonical free kinetic terms $F_{\mu \nu }^a{F}_{\mu \nu }^a/4+(\partial {\varphi }^a{)}^2/2$. At nonzero momentum in Euclidean space $p$, put $\eta =n\cdot p$. The quadratic kernel for $(A_{\mu }^a,{\varphi }^a)$, per color, isThe transverse gauge-field kernel plus the gauge-fixing longitudinal term has become $p^2{\delta }_{\mu \nu }$. The scalar quantum field theory propagator is the inverse Schur complement, not merely the inverse of the scalar diagonal entry:Hence the free adjoint-scalar propagator in scalar-dependent gauge fixing isFor completeness the mixed quantum field theory propagator is $⟨A_{\mu }^a{\varphi }^b{⟩}_0=-{\delta }^{ab}\eta p_{\mu }/(\sqrt{2}{p}^4)$; ignoring this mixing would give an incorrect scalar answer.

The scalar propagator is independent of the gauge vector $n$, not of the momentum component parallel to $n$. For a unit $n$, $p^2=p_{\perp }^2+p_{\parallel }^2$, and the free scalar quantum field theory propagator still depends on $p_{\parallel }$. Thus the literal momentum-independence clause in the PDF is false for the standard minimally coupled massless scalar action; the cancellation above establishes the natural gauge-vector-independence statement. The usual massless zero mode in field theory at $p=0$ needs a separate infrared prescription.

Finally, integration by parts gives the ghost action in an unambiguous convention:Use $e^{ip\cdot x}$ for the Fourier transform of every field, with all momenta incoming. Let the antighost carry color $a$ and momentum $q$, the boson color $b$ and momentum $k$, and the ghost color $c$ and momentum $p$, so $q+k+p=0$. Expansion of $e^{-S_{\mathrm{gh}}}$ gives the ghost vertices in scalar-dependent gauge fixingThe free Faddeev-Popov ghost field propagator is ${\delta }^{ab}/p^2$ and every closed ghost loop contributes a minus sign. There are no further ghost interaction vertices in this gauge. Factors of $i$ and an overall ghost-vertex sign depend on the Fourier transform and ghost-ordering conventions; the displayed ghost action fixes both here.

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.