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For the stated matrix exponential parametrization in the connected proper orthochronous Lorentz group, use the corresponding Spinor representation of the Lorentz group:The spacetime argument is inverse-transformed because this is an active transformation of the field at a fixed coordinate $x$. Equivalently, with $x^{\prime }=\Lambda x$, ${\psi }^{\prime }(x^{\prime })=D(\Lambda )\psi (x)$. The identity $[S^{\mu \nu },{\gamma }^{\rho }]={\eta }^{\nu \rho }{\gamma }^{\mu }-{\eta }^{\mu \rho }{\gamma }^{\nu }$ integrates to $D^{-1}{\gamma }^{\mu }D={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$, which ensures covariance of the Dirac equation.

Strictly, $D$ depends on a lift to the Spin group, not just on the final Lorentz transformation: two lifts differ by sign. The specified generator exponential chooses a lift along its continuous path from the identity. This qualification is essential for the full-rotation comparison in part (g).

For fixed $(P,Q)\ne (0,0)$ the effective Fermi interaction predicts $d\sigma /d\Omega \propto G_F^{2}s$. The angular odd term integrates to zero, while $\int d\Omega (1+{\cos }^2\theta )=16\pi /3$, soThe corresponding fixed-spin scattering amplitudes grow as $G_{F}s$, eventually conflicting with partial-wave unitarity. This is a failure of extrapolating the effective field theory, whose contact approximation requires $s\ll M_W^2$, not a failure of the Standard Model.

Restoring the virtual W boson Feynman propagator replaces its constant low-energy denominator by $M_W^2/(M_W^2-s)$, away from the resonance. The massless external Dirac equations give zero contraction of each current with the exchanged momentum, so the longitudinal numerator does not contribute. Its squared modulus multiplies the effective cross-section. For $s\gg M_W^2$ the resulting tree-level rate decreases as $1/s$ rather than growing as $s$. Near the pole one must include the W boson decay width. These statements assume the same massless conserved external currents; the contact expression alone is not valid there.

Use the Minkowski metric $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ throughout. Taking the Hermitian conjugation of the massive Dirac equation and using the Dirac adjoint givesTaking the matrix transpose and multiplying by the charge conjugation matrix $C$ yieldsThus the charge conjugation of a Dirac field preserves the free massive Dirac equation:The unitarity of $C$ and ${\gamma }^0{C}^{-1}{\gamma }^0=-C^{-1}{\gamma }^{0T}$ also giveConsequentlyHere $\hat{C}$ is the unitary operator acting on the relativistic quantum field, whereas $C$ acts on Dirac spinor indices.

The tree amplitude is, up to an overall convention-dependent sign,The Clifford algebra gives the momentum-space Dirac equation and its adjoint:ThereforeAlthough $\mu >2m$ permits the relativistic two-body decay kinematically, the matrix element vanishes. Thus

Multiplying the QED Feynman rules in fermion-line order gives, up to the common convention for $i\mathcal{M}$,The two terms are both required by the Ward identity; replacing either external polarization by its photon momentum makes their sum vanish after using momentum conservation and the external Dirac equations.

Write ${\psi }^P(x)={\eta }_P{\gamma }^0\psi (x_P)$. The chain rule gives ${\partial }_0\psi (x_P)=({\partial }_0\psi )(x_P)$ and ${\partial }_i\psi (x_P)=-({\partial }_i\psi )(x_P)$. Using the gamma matrix relations $({\gamma }^0{)}^2=1$ and ${\gamma }^i{\gamma }^0=-{\gamma }^0{\gamma }^i$,Thus the Dirac equation is invariant under parity symmetry in quantum field theory.

For a massless Dirac fermion,The local gauge symmetry gives the vector current $J^{\mu }=e\bar{\psi }{\gamma }^{\mu }\psi$. The global chiral transformation $\psi \mapsto e^{i\beta {\gamma }^5}\psi$ gives the classical axial currentUsing the massless Dirac equation and $\{{\gamma }^5,{\gamma }^{\mu }\}=0$ gives ${\partial }_{\mu }{J}_{\mathrm{ax}}^{\mu }=0$ classically.

Quantum mechanically, a gauge-invariant regulator for the fermion measure is not invariant under the axial rotation. In the Fujikawa form, its infinitesimal Jacobian containsThe first nonzero term in the heat kernel expansion is quadratic in $F_{\mu \nu }$. Usingand the Gaussian momentum integral gives the chiral anomalyThe overall sign follows the gamma-matrix, charge, and Levi-Civita conventions stated in the question. The same coefficient is obtained from the one-loop axial-vector-vector triangle diagram.

The Wigner classification labels a massive one-particle state by $|p;j,m_s⟩$, where $p^2=m^2>0$, $p^0>0$, $j$ is the spin representation of the $SO(3)$ little group, and $m_s=-j,\dots ,j$. A finite-spin massless state is $|p;\lambda ⟩$, where $p^2=0$, $p^0>0$, and the helicity $\lambda$ labels a one-dimensional representation of the rotational part of the $ISO(2)$ little group. A parity-invariant theory pairs nonzero helicities $\lambda$ and $-\lambda$.

The physical polarization counts areA scalar obeys the Klein-Gordon equation. A spinor obeys the Dirac equation, and a massless irreducible spinor additionally has a fixed chirality. A massive vector obeys the Proca equation, whose divergence gives ${\partial }_{\mu }{A}^{\mu }=0$ and leaves three polarizations. A massless vector instead has the gauge redundancy $A_{\mu }\sim A_{\mu }+{\partial }_{\mu }\alpha$, leaving two transverse polarizations. A massive symmetric tensor obeys the Fierz-Pauli equations: symmetry, transversality, and tracelessness leave five components. A massless tensor has linearized diffeomorphism redundancy $h_{\mu \nu }\sim h_{\mu \nu }+{\partial }_{\mu }{\xi }_{\nu }+{\partial }_{\nu }{\xi }_{\mu }$, leaving helicities $\pm 2$. A redundancy identifies field configurations representing the same physical state; gauge invariance is invariance under that identification.

The Dirac field Lagrangian isChoose the Majorana spinorSubstitution shows that its two chiral kinetic terms agree after integration by parts and thatThus ${\mathcal{L}}_D[{\psi }_M]$ equals twice the displayed one-Weyl-field Lagrangian, up to a total derivative.

The lower chiral component of the Dirac equation isPutting $u_R=i{\sigma }^2{u}_L^{\ast }$ gives exactly $i{\bar{\sigma }}^{\mu }{\partial }_{\mu }{u}_L-im{\sigma }^2{u}_L^{\ast }=0$; the upper component is its complex conjugate.

The translation changes the Lagrangian density by $\delta {\mathcal{L}}_0={ϵ}^{\nu }{\partial }_{\nu }{\mathcal{L}}_0={\partial }_{\mu }({ϵ}^{\mu }{\mathcal{L}}_0)$. The Noether current is thereforeThis is the canonical stress-energy tensor. Directly,up to the analogous adjoint-field term, so the Euler-Lagrange field equations imply ${\partial }_{\mu }{T}^{\mu \nu }=0$ on shell. The Dirac equation also gives ${\mathcal{L}}_0=0$ on shell.

After using the Clifford algebra, integrations by parts and the Dirac equation, every allowed Lorentz-covariant symmetric rank-two expression with one derivative reduces, modulo terms vanishing on shell and identically conserved improvements, to an overall multiple ofThus the general nontrivial tensor in the stated class isIt is manifestly symmetric. Differentiating it, commuting partial derivatives and usingmakes the terms cancel pairwise, proving ${\partial }_{\mu }{\hat{T}}^{\mu \nu }=0$ on shell. The normalization $C$ remains free at this stage.

Choose $C=i/4$. Expandingwith the Clifford algebra, applying the product rule and then using the Dirac equation and its adjoint givesThis is the Belinfante-Rosenfeld stress-energy tensor written as an improvement of the canonical tensor.

The translation charges areFor spatial $\nu$, their difference is the integral of ${\partial }_i{S}^{i0\nu }-{\partial }^{\nu }(\bar{\psi }{\gamma }^0\psi )$. For $\nu =0$, use conservation of the Dirac current to replace ${\partial }^0(\bar{\psi }{\gamma }^0\psi )$ by a spatial divergence; also $S^{00\nu }=0$. Hence ${\hat{P}}^{\nu }-P^{\nu }$ is always a spatial boundary integral. Under the usual decay boundary condition it vanishes, so both currents generate the same four-momentum.

Define ${\sigma }^{\mu \nu }=\frac{i}{2}[{\gamma }^{\mu },{\gamma }^{\nu }]$ using the gamma matrix Clifford algebra. Moving ${\gamma }^{\rho }$ through the two factors with $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }$ proves the displayed operator commutator. For real antisymmetric ${\omega }_{\mu \nu }$ and $S=1-\frac{i}{4}{\sigma }^{\mu \nu }{\omega }_{\mu \nu }$, it gives $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }+O({\omega }^2)$. This is the infinitesimal Lorentz covariance identity required by the Dirac equation.

For tree Electron-Positron annihilation into three photons, all six orders along the fermion line must be summed. Write $D(r)=/r-m$ and $S(r)=D(r{)}^{-1}=(/r+m)/(r^2-m^2)$. Contracting an internal photon with its momentum gives $S(R)/qS(r)=S(R)-S(r)$ when $q=r-R$. Contracting an end photon removes its adjacent Dirac propagator by the external Dirac equation, with a minus sign at the electron end and a plus sign at the positron end. For either fixed order of the two uncontracted photons, the three insertion positions therefore telescope to zero. Summing their two orders proves the Ward identity. The mass terms in both remaining Dirac propagators are essential when $m\ne 0$. The result expresses the invariance of the scattering amplitude under ${ϵ}_{\ell }\mapsto {ϵ}_{\ell }+c{q}_{\ell }$.

The two photon orders in a tree-level charged-fermion amplitude must be summed. Contracting one photon vertex with its momentum writes the contracted slash as a difference of inverse Dirac propagators. The external Dirac equations remove the on shell inverse factors; the two remaining terms have equal size and opposite sign. Thus the sum vanishes under the momentum replacement although individual diagrams generally do not. This implements the gauge equivalence of photon polarization vectors.

In signature $(-,+)$, choose real gamma matrices ${\gamma }^0=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$, ${\gamma }^1=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and ${\gamma }_{\ast }=-{\gamma }^0{\gamma }^1$. The Dirac equation then gives $({\partial }_{\tau }+{\partial }_{\sigma }){\psi }_{+}=0$ and $({\partial }_{\tau }-{\partial }_{\sigma }){\psi }_{-}=0$. Positive chirality is right-moving and negative chirality is left-moving in this orientation. Reversing the definition of ${\gamma }_{\ast }$ exchanges the labels without changing the physics.