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For an unpolarized electromagnetic target, Lorentz covariance and parity symmetry in quantum field theory permit a symmetric hadronic tensor built from $g^{\mu \nu }$, $P^{\mu }$ and $q^{\mu }$, where $P=P_H$. Its surviving general form isThe totally antisymmetric tensor structure ${ϵ}^{\mu \nu \rho \sigma }{P}_{\rho }{q}_{\sigma }$ is excluded by electromagnetic parity symmetry in quantum field theory for this unpolarized target. The other possible antisymmetric structure, $P^{\mu }{q}^{\nu }-q^{\mu }{P}^{\nu }$, cannot be transverse for generic scattering kinematics and hence is excluded by current conservation. Current conservation gives the Ward identities $q_{\mu }{W}^{\mu \nu }=q_{\nu }{W}^{\mu \nu }=0$, imposingTwo scalar functions remain. Taking ${\tilde{P}}^{\mu }=P^{\mu }-\nu q^{\mu }/q^2$ gives the transverse basisThese are the deep-inelastic structure functions; $x$ is Bjorken x. At fixed target mass, the two independent Lorentz scalar invariants can equivalently be chosen as $x,q^2$. In this paper $\nu =P\cdot q$ has dimensions of mass squared, rather than the alternative convention $\nu =P\cdot q/M$ used for energy transfer.

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 the Minkowski metric $\eta =\mathrm{diag}(-1,1,1,1)$ and define the electromagnetic field tensor byAs a rank-two Lorentz tensor, it transforms under a Lorentz transformation asTwo independent quadratic electromagnetic field invariants are

With $A^{\mu }=(\varphi /c,\mathbf{A})$, $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$, and $\mathbf{B}=\nabla \times \mathbf{A}$, the components areDirect contraction giveswhere the sign of $I_2$ changes if the opposite spacetime orientation is chosen. Since both are complete tensor contractions, $S$ and $T$ are Lorentz scalars.

For a constant uniform field, the relativistic Lorentz force is a constant-coefficient linear ordinary differential equation. Its solution is the matrix exponential

Now $S=T=0$ implies that the field is a null electromagnetic field:The stated axes allow us to choosePut $\Omega =qB/m$. On the $(ct,x,y)$ coordinates, the mixed tensor divided by $B$ isThe particle starts with four-velocity $u(0)=(c,0,0,0)$, so the exponential terminates after its quadratic term and givesIntegrating with the particle initially at the origin gives the relativistic trajectory in a constant null crossed fieldEliminating $\tau$ yieldsIf the original positive $x$ direction is opposite to $\mathbf{E}$, both the sign of $x$ and the displayed signed coefficient reverse.

In the Weyl representation of the gamma matrices,A Lorentz transformation acts as ${\psi }^{\prime }(x^{\prime })=S(\Lambda )\psi (x)$ with $S(\Lambda )=\exp [-\frac{i}{4}{\omega }_{\mu \nu }{\sigma }^{\mu \nu }]$. In this basis a boost is block diagonal and gives $u_L\mapsto e^{\mathbf{\chi }\cdot \mathbf{\sigma }/2}{u}_L$ and $u_R\mapsto e^{-\mathbf{\chi }\cdot \mathbf{\sigma }/2}{u}_R$.

For boosts, the two exponentials cancel in $u_L^{†}{u}_R$; for rotations, the unitary rotation and its inverse cancel. Hence this bilinear is a Lorentz scalar. The Pauli matrices obey the identityshows that $i{\sigma }^2{u}_L^{\ast }$ acquires the right-handed boost matrix and the usual rotation matrix, so it is right-handed.

For $x^{\prime }=\Lambda x$, a Dirac spinor and vector field transform aswhere $S^{-1}{\gamma }^{\mu }S={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$. Hence the spinor terms and $F_{\mu \nu }{F}^{\mu \nu }$ are Lorentz scalars. The Maxwell term is real. Integration by parts and $({\gamma }^{\mu }{)}^{†}={\gamma }^0{\gamma }^{\mu }{\gamma }^0$ show that the adjoint of $i\bar{\psi }{\gamma }^{\mu }{D}_{\mu }\psi$ differs from it by $-i{\partial }_{\mu }(\bar{\psi }{\gamma }^{\mu }\psi )$. Finally, $i[{\gamma }^{\mu },{\gamma }^{\nu }]$ is Dirac-Hermitian, so the real coefficient $\lambda$ makes the Pauli bilinear real. Thus the action is real.

The antisymmetric $C_{\mu \nu }$ transforms as a Lorentz tensor, so its complete contractionis a Lorentz scalar. HenceParts ii and v give $[P_{\rho },{\tilde{C}}_2]=[Q_{\alpha },{\tilde{C}}_2]=[{\bar{Q}}_{\dot{\alpha }},{\tilde{C}}_2]=0$. It therefore commutes with every generator of the Super-Poincaré algebra and is a Casimir element, called the superspin Casimir.

A scalar field theory describes one or more Lorentz scalar fields through an action. The displayed Lagrangian density is a common classical normalization for a real scalar field; a complex scalar field has two real components. Canonical quantization turns these classical degrees of freedom into operators. The free real scalar field gives spin-zero particles, while interactions arise from the potential $V$.

In light-cone coordinates $x^{\pm }$ on two-dimensional Minkowski spacetime, add two real odd Grassmann variables ${\theta }^{\pm }$. With left Grassmann derivatives, the supercharges and supersymmetric covariant derivatives can be represented asThe identity $\{{\partial }_{\theta },\theta \}=1$ gives $Q_{\pm }^2=i{\partial }_{\pm }$, $D_{\pm }^2=-i{\partial }_{\pm }$, $\{Q_{+},Q_{-}\}=\{D_{+},D_{-}\}=0$, and $\{Q_s,D_t\}=0$ for every $s,t$. A Lorentz boost acts as $x^{\pm }\mapsto e^{\pm \omega }{x}^{\pm }$, ${\theta }^{\pm }\mapsto e^{\pm \omega /2}{\theta }^{\pm }$. Thus $D_{\pm }$ have boost weights $\mp 1/2$, and $D_{-}{D}_{+}$ is a Lorentz scalar. Since it commutes with both supercharges, an equation $i{D}_{-}{D}_{+}\Phi =W(\Phi )$ for an even scalar superfield is Lorentz invariant and supersymmetry covariant.