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Both the D meson and kaon are spin-zero pseudoscalars. The strong interaction states obey parity symmetry in quantum field theory. Consequently an axial current matrix element between them would have to be a pseudovector formed from only $p$ and $k$, which is impossible: an totally antisymmetric tensor would require more linearly independent vectors. ThusLorentz covariance then leaves two linearly independent vectors for the vector current matrix element, giving the pseudoscalar-to-pseudoscalar form factorsThe vanishing axial current here follows from strong interaction parity symmetry in quantum field theory, not parity symmetry in quantum field theory of the weak interaction.

To obtain the requested decay formula, use naive factorization of a nonleptonic meson decay: approximate the four-quark matrix element by the product of the ${\bar{D}}^0\to K^{+}$ current matrix element and the vacuum-to-pion matrix element. This is an additional hadronic approximation; tree-level weak vertices alone do not establish it, and nonfactorizable Quantum chromodynamics effects can change the result. The vacuum-to-pion vector current matrix element vanishes by parity symmetry in quantum field theory, while the specified pion decay constant normalization givesThe $\sqrt{2}$ cancels the $1/\sqrt{2}$ in the four-fermion interaction. Up to an irrelevant overall phase, the scattering amplitude is thereforeFor $m_{\pi }=0$, $q^2=0$, so only $f_{+}(0)$ survives. Integrating the two-body decay phase space in the D meson rest frame givesHenceThis coefficient uses exactly the stated pion decay constant convention and the stated massless-pion approximation.

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.

For $m>0$, the three physical polarization vectors are transverse to $k$ and have norm $-1$ in metric $(+---)$. In the rest frame their sum is $\mathrm{diag}(0,1,1,1)$; Lorentz covariance then gives the displayed expression for $k^2=m^2$. This includes the longitudinal mode. It cannot be specialized to $m=0$ by setting its denominator to zero.

Between spin-zero pseudoscalars, Lorentz covariance decomposes a vector current matrix element into $(p+k{)}^{\mu }{f}_{+}(q^2)+(p-k{)}^{\mu }{f}_{-}(q^2)$, with $q=p-k$. Strong interaction parity symmetry in quantum field theory makes the corresponding axial current matrix element vanish.