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Use the Clifford algebra $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$ and unit-weight antisymmetrization. The duality identity between ${\gamma }_5{\gamma }_{\nu }$ and three antisymmetrized gamma matrices turns the equation intoThe overall nonzero factor in this identity depends on the orientation and ${\gamma }_5$ convention and does not change its solutions. This Rarita-Schwinger field equation has the gauge symmetrysince ${\gamma }^{\mu \nu \rho }{\partial }_{\nu }{\partial }_{\rho }ϵ=0$. It is the free-field limit of local supersymmetry, so it is suitable for the gravitino.

To check which spin it propagates, put $A={\gamma }^{\mu }{\Psi }_{\mu }$ and $B={\partial }^{\mu }{\Psi }_{\mu }$. Expanding the three gamma matrices using the Clifford algebra givesin four dimensions. Thus the Euler-Lagrange field equation requires $B=/\partial A$. Under the gauge transformation, $A\mapsto A+/\partial ϵ$; locally one may choose gamma-trace gauge fixing, $A=0$, by solving $/\partial ϵ=-A$. The Euler-Lagrange field equation then yieldsThese are the massless Dirac equation, transversality and gamma-trace constraints that remove the lower-spin part of a vector-spinor.

For a more explicit massless Rarita-Schwinger polarization count, take a positive-frequency plane wave with null momentum along the third axis, in signature $(-,+,+,+)$. Transversality relates its temporal and longitudinal components. Residual gauge transformations with $/pϵ=0$ remove both. Write the surviving components as $u_1,u_2$; their gamma-trace condition reads ${\gamma }^1{u}_1+{\gamma }^2{u}_2=0$, henceThe massless Dirac equation leaves two complex positive-frequency amplitudes in $u_1$. If $i{\gamma }^1{\gamma }^2{u}_1=s{u}_1$, $s=\pm 1$, then $(u_1,u_2)$ is proportional to $(1,-is)$ in the transverse vector index. With spatial rotation generators $J_3^{\mathrm{vec}}{}_{12}=-i$ and $J_3^{\mathrm{spin}}=-i{\gamma }^1{\gamma }^2/2$, the vector has helicity $-s$ and the Majorana spinor has helicity $-s/2$, giving total helicity $-3s/2$. The Majorana spinor condition fixes the negative-frequency coefficients from these amplitudes. Therefore a massless gravitino has spin $3/2$, with the two physical helicities $\pm 3/2$. A complex vector-spinor would also have independent antiparticle modes.

The requested count is instead the off-shell component count of minimal supergravity, before imposing any Euler-Lagrange field equations. A four-dimensional Majorana spinor has four real components. Its vector index supplies four copies, and local supersymmetry supplies four real gauge functions:One must not additionally subtract transversality and the massless Dirac equation in this count: those are on shell conditions. The gamma-trace gauge fixing already uses the same four gauge functions subtracted above. For the graviton, a symmetric metric tensor has ten real components, with four diffeomorphism gauge functions. Equivalently, a vierbein has sixteen real components, minus six local Lorentz transformation and four diffeomorphism functions:The gravitino and graviton each have two propagating states on shell, but their off-shell counts disagree by six.

An auxiliary field has an algebraic Euler-Lagrange field equation and no propagating kinetic term. In old-minimal supergravity, a complex scalar $M$ contributes two real bosonic components and a real vector $b_{\mu }$ contributes four, so the bosonic count becomes $6+2+4=12$, matching the gravitino. Their supersymmetry transformations also absorb the terms that would otherwise require the propagating Euler-Lagrange field equations to close the local algebra. Thus auxiliary fields allow off-shell supersymmetry closure and matching component counts without adding physical particles. Eliminating them recovers an on shell formulation.

Before electron-positron annihilation in cosmology, photons contribute two helicity states; Electrons and positrons contribute two spin states each. Each of the three light neutrino flavours contributes one active neutrino helicity and one antineutrino helicity. At a common temperature the effective number of relativistic energy degrees of freedom is thereforeUsing $G_F=1.17\times 10^{-5}{\mathrm{GeV}}^{-2}$ and $M_{\mathrm{Pl}}\simeq 2.435\times 10^{18}\mathrm{GeV}$ in the scaling formula alone gives $T_F\sim 2\mathrm{MeV}$. Including the explicit expansion coefficient from (ii), but setting the unknown rate coefficient $A$ to one, gives $T_F\simeq 1.5\mathrm{MeV}$. Thus the warranted conclusion is a freeze-out temperature of order $1\mathrm{MeV}$, above the $0.511\mathrm{MeV}$ Electron mass. The omitted weak-rate coefficient prevents a more precise prediction from this estimate.

Neutrino decoupling occurs before most electron-positron pairs annihilate. Afterwards neutrinos redshift freely, $T_{\nu }a=\mathrm{constant}$, while the photons receive the pair entropy. For the still-coupled photon-pair gas, cosmological entropy conservation gives $g_{\ast s}{T}_{\gamma }^3{a}^3=\mathrm{constant}$. Its entropy degrees of freedom drop from $2+(7/8)4=11/2$ to $2$. Comparing this with the freely cooling neutrinos givesAfter annihilation, photon and neutrino temperatures differ. To write the total energy density as $({\pi }^2/30)g_{\ast }{T}_{\gamma }^4$, the neutrino contribution must include that temperature ratio:This is the paper's post-annihilation ${\mathcal{N}}_{\mathrm{eff}}$. Merely deleting the Electron term while keeping all species at the same temperature would be incorrect. For comparison the entropy count, normalized to $T_{\gamma }^3$, is $g_{\ast s}=2+(7/8)6(4/11)=43/11\simeq 3.91$. These results assume instantaneous neutrino decoupling and relativistic neutrinos; the two counts represent different thermal moments.

Use units $c=\hslash =1$ and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. Composing two affine Poincaré group transformations, with the rightmost applied first, givesThe translations form a normal subgroup, so this is the semidirect product of the Lorentz group with ${\mathbb{R}}^{1,3}$. The transformations $(\mathrm{diag}(1,R),0)$, $R\in SO(3)$, fix time and the spatial origin. Their products and inverses correspond to $R{R}^{\prime }$ and $R^{-1}$, hence they form a subgroup isomorphic to SO(3).

A rotationless Lorentz boost changes the velocity of a frame without adding a spatial rotation. For $|\mathbf{v}|<1$ a convenient active convention iswith $B(0)=I$. It satisfies $B^{T}gB=g$, $\det B=1$, and $B(m,0)=(m\gamma ,m\gamma \mathbf{v})$. Collinear boosts form a one-parameter subgroup with additive rapidity, but general boosts do not. For nonzero boosts along $x$ and $y$, $(B_x{B}_y{)}_{02}={\gamma }_x{\gamma }_y{v}_y$ while $(B_x{B}_y{)}_{20}={\gamma }_y{v}_y$: their product is not symmetric, whereas every rotationless boost above is symmetric. The product contains a Wigner rotation. Infinitesimally, the Poincare algebra gives $[K_i,K_j]=-i{ϵ}_{ijk}{J}_k$ for Hermitian generators, so the boost generators do not close amongst themselves.

There is a group-versus-cover qualification in the spin range: an honest unitary representation of the proper orthochronous Poincaré group permits integer spin only. Half-integer spin requires its double cover $SL(2,\mathbb{C})⋉{\mathbb{R}}^{1,3}$, or a projective unitary representation of the ordinary group. The construction below gives the requested range on that double cover and descends for integer spin.

Fix $m>0$ and the rest momentum $k=(m,0)$. Its little group consists of rotations: if a proper orthochronous Lorentz transformation fixes $k$, its time column is $(1,0)$, the metric condition forces the corresponding time row, and the remaining block belongs to SO(3). On the Lorentz spinor double cover, this little group is SU(2). Choose its irreducible spin-$s$ representation, with rest-frame states $|k,s,\sigma ⟩$, $\sigma =-s,-s+1,\dots ,s$, and matrices $D^{(s)}$, the Wigner D-matrices.

For each $p$ on the forward mass shell, $p^2=m^2$, $p^0>0$, choose the standard rotationless Lorentz boost $L(p)=B(\mathbf{p}/p^0)$ and defineThe labels are canonical spin projections, defined by this choice of boosts. For any proper orthochronous $\Lambda$, factorThe last factor fixes $k$ and hence is a Wigner rotation. On the double cover, use its lift to SU(2). Taking the translation convention $U[I,a]|p,s,\sigma ⟩=e^{ip\cdot a}|p,s,\sigma ⟩$ yields the massive induced representation of the Poincare double cover:The argument of the translation phase is $\Lambda p$, since $U[\Lambda ,a]=U[I,a]U[\Lambda ,0]$. The Wigner rotation identityproves the Lorentz part of the representation law; the phases multiply according to the affine product already derived.

Normalize the generalized momentum states by $⟨p,s,\sigma |q,s,\tau ⟩=2p^0(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q}){\delta }_{\sigma \tau }$. The invariant mass shell measure $d^{3}p/(2p^0)$ and unitarity of $D^{(s)}$ make the induced representation unitary on $L^2(d^{3}p/(2p^0))\otimes {\mathbb{C}}^{2s+1}$. This covariant normalization accounts for the absence of an energy square-root factor in the displayed transformation law. To see irreducibility, translations force an operator in the commutant to act within momentum fibres; the little-group action and Schur lemma make its action scalar on each irreducible spin fibre. Lorentz transitivity on the forward mass shell makes this scalar constant. The commutant is therefore scalar, so the unitary representation is irreducible. Conversely, the simultaneous translation spectrum and its little-group representation give the massive positive-energy sector of Wigner's classification.

Thus every $m>0$ and $s\in \{0,\frac{1}{2},1,\dots \}$ gives such an irreducible representation on the double cover. Its central element over a $2\pi$ spatial rotation acts as $(-1{)}^{2s}$. Since this rotation is the identity in the ordinary Poincaré group,

For the parity operator, put $\Pi =\mathrm{diag}(1,-1,-1,-1)$ and $\bar{p}=\Pi p=(p^0,-\mathbf{p})$. The extension obeys $PU[\Lambda ,a]P^{-1}=U[\Pi \Lambda \Pi ,\Pi a]$. At rest, $P$ commutes with the rotations; Schur lemma therefore gives $P|k,s,\sigma ⟩={\eta }_P|k,s,\sigma ⟩$ with a common phase $|{\eta }_P|=1$. Since $\Pi L(p)\Pi =L(\bar{p})$, the parity action on canonical massive spin states isFor the usual reflection convention $P^2=I$, ${\eta }_P=\pm 1$ is the intrinsic parity. Canonical spin is an axial quantity and its label $\sigma$ is unchanged; a helicity label instead changes sign because momentum is reversed. Either intrinsic parity extends a given massive spin representation.

Use signature $(-,+,+,+)$, $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$, and unit-weight antisymmetrized gamma products. The massless Rarita-Schwinger field has the gauge freedomsince the antisymmetric ${\gamma }^{\mu \nu \rho }$ annihilates ${\partial }_{\nu }{\partial }_{\rho }ϵ$. This gauge freedom is essential to remove the lower-spin components of the vector-spinor.

Put $\chi ={\gamma }^{\mu }{\psi }_{\mu }$. Contracting the field equation with ${\gamma }_{\mu }$ and using ${\gamma }_{\mu }{\gamma }^{\mu \nu \rho }=2{\gamma }^{\nu \rho }$ givesExpanding the original three-gamma product then reduces its equation toA gauge transformation changes $\chi$ by $/\partial ϵ$. Locally solve $/\partial ϵ=-\chi$ to impose gamma-trace gauge. The two equations becomeApplying $/\partial$ again gives $\square {\psi }_{\mu }=0$. Nonzero propagating Fourier modes therefore have $p^2=0$: the physical modes are massless.

For the massless Rarita-Schwinger polarization count, choose a nonzero future null momentum $p^{\mu }=(\omega ,0,0,\omega )$. The remaining gauge parameters satisfy $/pϵ=0$. Transversality gives ${\psi }_0+{\psi }_3=0$, and this residual gauge removes ${\psi }_0$, hence also ${\psi }_3$, because ${\psi }_0$ itself satisfies $/p{\psi }_0=0$. Only ${\psi }_1,{\psi }_2$ remain, withThe kernel of $/p$ on four-component spinors has complex dimension two, so there are exactly two independent polarization amplitudes.

To identify their helicities, the spinor rotation generator about the third axis in this convention is $S_3=-i{\gamma }^1{\gamma }^2/2$. Its two eigenvalues on that kernel are $h_s=\pm 1/2$. If ${\psi }_1$ is such an eigenvector, then ${\psi }_2=2i{h}_s{\psi }_1$. Its transverse vector polarization is consequently $(1,+i)$ for $h_s=+1/2$ or $(1,-i)$ for $h_s=-1/2$. These carry vector helicity $+1$ or $-1$, respectively. Adding the vector and spinor angular momenta leavesThe unwanted opposite-sign vector-spinor combinations, which would have helicity $\pm 1/2$, violate the gamma-trace constraint. For the gravitino of simple supergravity, its Majorana spinor reality condition identifies the negative-frequency antiparticle modes, giving two propagating degrees of freedom. If one instead chooses a complex Dirac vector-spinor, its independent antiparticle sector doubles the particle-plus-antiparticle count; the equation alone does not impose a Majorana condition.

The supplied gamma matrices use the mostly-minus metric signature. The Pauli matrix multiplication law gives ${\sigma }^1{\sigma }^2{\sigma }^3=i{I}_2$, hence block multiplication yieldsThe off-diagonal blocks of every ${\gamma }^{\mu }$ connect opposite eigenspaces of this diagonal matrix. Explicitly,Thus $\{{\gamma }^5,{\gamma }^{\mu }\}=0$, and $({\gamma }^5{)}^2=I_4$. The chiral projectors $(1\mp {\gamma }^5)/2$ select the upper and lower two-component blocks, respectively.

Write the Dirac spinor as $\psi =({\chi }_L,{\chi }_R{)}^T$. Substituting the block matrices into the massless Dirac equation gives two independent Weyl spinor equations:They are decoupled because the mass term, which would join the two chirality components, is absent.

For a left-handed plane wave, use $p\cdot x=p^{0}t-\mathbf{p}\cdot \mathbf{x}$. Its coefficient $\xi$ obeysSince $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2=|\mathbf{p}{|}^2{I}_2$, the determinant is $(p^0{)}^2-|\mathbf{p}{|}^2$. A nonzero coefficient therefore requires the massless mass shell $\boxed{p^2=0}$. For positive energy and nonzero $\mathbf{p}$, put $P=|\mathbf{p}|=p^0$ and parameterize its direction by angles $\vartheta ,\phi$. ThenMultiplication gives $(\mathbf{\sigma }\cdot \hat{\mathbf{p}}){\chi }_{-}=-{\chi }_{-}$ and ${\chi }_{-}^{†}{\chi }_{-}=1$. Hence the left-handed massless plane-wave spinor can be writtenHere $N(p)$ is an arbitrary complex normalization; $|N|=\sqrt{2P}$ gives ${\lambda }_L^{†}{\lambda }_L=2P$. Away from the negative $z$ axis one can equivalently takeAt $\mathbf{p}=-P\hat{\mathbf{z}}$ choose $(1,0{)}^T$ instead. These are phase patches for the same one-dimensional kernel. The other frequency branch $p^0=-P$ uses the positive eigenvector ${\chi }_{+}=(\cos (\vartheta /2),e^{i\phi }\sin (\vartheta /2){)}^T$ for the direction of $\mathbf{p}$. At $p^{\mu }=0$, the algebraic equation vanishes and any two-component coefficient is possible, but no momentum direction or helicity is defined there.

A spatial rotation by $\alpha$ about $\hat{\mathbf{p}}$ acts on the two-component coefficient by $\exp (-i\alpha \mathbf{\sigma }\cdot \hat{\mathbf{p}}/2)$. The negative eigenvector therefore acquires the phase $e^{i\alpha /2}$. Since a state with helicity $h$ acquires $e^{-i\alpha h}$, the positive-energy left-handed particle has $\boxed{h=-1/2}$. There is only one such spin state at a fixed nonzero momentum, rather than two independent polarizations of the same chiral particle.

The helicity content of a quantized Weyl field also includes an antiparticle with opposite helicity. In the field expansion, the negative-frequency term multiplies an antiparticle fermionic creation operator, while the positive-frequency term multiplies a particle fermionic annihilation operator. Their adjoint roles give conjugate phases for the corresponding created states, so a left-handed field creates particles with $h=-1/2$ and antiparticles with $h=+1/2$. The right-handed Weyl spinor has the opposite assignments. Quantizing the full massless Dirac field, rather than one chiral component, therefore gives both particle helicities and both antiparticle helicities; every excitation has spin one-half, while chirality determines the helicity of each massless sector.

Use the free Maxwell Lagrangian $\mathcal{L}=-F_{\mu \nu }{F}^{\mu \nu }/4$ in mostly-minus signature. In terms of the spatial electromagnetic four-potential, define $\mathbf{E}=-\dot{\mathbf{A}}-\nabla A_0$ and $\mathbf{B}=\nabla \times \mathbf{A}$. Thus $\mathcal{L}=({\mathbf{E}}^2-{\mathbf{B}}^2)/2$. Taking the Cartesian components of $\mathbf{A}$ as coordinates gives the canonical momentaThis sign convention treats $A_i$ here as Cartesian spatial-vector labels; it is not a claim that covariant and contravariant four-vector components coincide.

The temporal momentum vanishes because no ${\dot{A}}_0$ occurs. This is the primary momentum constraint of the electromagnetic potential, so one cannot treat all four components as independent unconstrained oscillators. The Legendre transformation for the spatial components gives ${\Pi }_i{\dot{A}}_i-\mathcal{L}=({\mathbf{\Pi }}^2+{\mathbf{B}}^2)/2-\mathbf{\Pi }\cdot \nabla A_0$. After integration by parts, the Hamiltonian isSurface terms vanish for decaying fields or the corresponding regulated boundary conditions. The total constrained Hamiltonian also contains an arbitrary multiplier of ${\Pi }^0$. Preserving ${\Pi }^0=0$ under time evolution requireswhich is the Gauss law constraint in gauge theory. Both constraints are first-class constraints; they express the gauge invariance $\mathbf{A}\mapsto \mathbf{A}+\nabla \alpha$, $A_0\mapsto A_0-\dot{\alpha }$. The electric and magnetic fields are unchanged, so different gauge potentials need not describe different physical states.

One consistent canonical quantization of the electromagnetic field first reduces this redundancy. Choose Coulomb gauge $\nabla \cdot \mathbf{A}=0$. In the absence of charges, the Gauss law constraint in gauge theory gives ${\nabla }^2{A}_0=0$; decay at infinity sets $A_0=0$. Residual spatially constant gauge parameters do not change $\mathbf{A}$. The remaining variables are the transverse pairs ${\mathbf{A}}^T,{\mathbf{\Pi }}^T={\dot{\mathbf{A}}}^T$. Their gauge reduction of the Maxwell Hamiltonian isThe two transverse coordinates for each nonzero momentum are the physical degrees of freedom. Treating the constraints as operator identities while retaining unconstrained componentwise commutators would be inconsistent: differentiating $[A_i^T,{\Pi }_j^T]=i{\delta }_{ij}{\delta }^3$ would give a nonzero right side even though ${\partial }_i{A}_i^T=0$.

The reduced Dirac bracket, promoted to a canonical commutation relation, instead gives the transverse equal-time commutator:whereIndeed the spatial transverse projector of a vector field $P_{ij}^T={\delta }_{ij}-k_i{k}_j/|\mathbf{k}{|}^2$ is symmetric, has $P^T\mathbf{k}=0$ and $(P^T{)}^2=P^T$. Projecting both entries of the unconstrained canonical bracket onto the transverse subspace therefore produces $P^T{\delta }^3$, equivalently the constrained Dirac bracket. The spatial zero mode is excluded by the decay condition; in a finite-volume regulator it must be treated separately.

Choose two orthonormal transverse photon polarization vectors ${\mathbf{ϵ}}_r(\mathbf{k})$, so thatThe canonical transverse photon field has mode expansionDifferentiate in time to obtain ${\Pi }_i^T$: the annihilation coefficient acquires $-i{\omega }_{\mathbf{k}}$ and the creation coefficient $+i{\omega }_{\mathbf{k}}$. The oscillator canonical commutation relations areSubstitution into $[A_i^T,{\Pi }_j^T]$ gives $i/2$ times the sum of the two Fourier kernels with phases $\pm i\mathbf{k}\cdot (\mathbf{x}-\mathbf{y})$. The polarization completeness relation and the evenness of $P^T(\mathbf{k})$ make that sum exactly $i{\delta }_{ij}^T$. This derives, rather than presumes, the correspondence between the oscillator and field commutators.

Substitute the mode expansions into $H_T$. Spatial integration sets paired momenta equal or opposite. The $aa$ and $a^{†}{a}^{†}$ terms from the electric and magnetic energies cancel because ${\omega }_{\mathbf{k}}^2=|\mathbf{k}{|}^2$ and $\mathbf{k}\cdot {\mathbf{ϵ}}_r=0$. The remaining terms are harmonic-oscillator energies. Removing the state-independent zero-point energy by normal ordering givesThe corresponding momentum operator is $\mathbf{P}={\sum }_r\int d^{3}k\mathbf{k}{a}_r^{†}{a}_r/(2\pi {)}^3$. These expressions imply $[H_T,a_r^{†}(\mathbf{k})]={\omega }_{\mathbf{k}}{a}_r^{†}(\mathbf{k})$ and $[\mathbf{P},a_r^{†}(\mathbf{k})]=\mathbf{k}{a}_r^{†}(\mathbf{k})$.

Define the Fock vacuum by $a_r(\mathbf{k})|0⟩=0$. Applying $a_r^{†}(\mathbf{k})$ creates a photon of energy $|\mathbf{k}|$, momentum $\mathbf{k}$, and one of two transverse polarizations. Its mass is zero because $k^2=0$. The relativistically normalized state is $|\mathbf{k},r⟩=\sqrt{2{\omega }_{\mathbf{k}}}{a}_r^{†}(\mathbf{k})|0⟩$, with inner product $2{\omega }_{\mathbf{k}}(2\pi {)}^3{\delta }_{rs}{\delta }^3(\mathbf{k}-\mathbf{q})$. Smearing these momentum eigenstates gives normalizable wave packets. Repeated commuting bosonic creation operators construct the symmetric multiphoton bosonic Fock space. Circular combinations of the two transverse photon polarization vectors transform with opposite phases under rotations about $\mathbf{k}$, giving the two physical helicities $\boxed{h=\pm 1}$.

The same physical states can be obtained while retaining manifest Lorentz covariance through Gupta-Bleuler quantization. Add the Feynman gauge term $-({\partial }_{\mu }{A}^{\mu }{)}^2/2$ and quantize four auxiliary oscillator polarizations. Their algebra has an indefinite metric, including a negative-norm temporal excitation. Impose $({\partial }_{\mu }{A}^{\mu }{)}^{(+)}|\mathrm{phys}⟩=0$ and take the Gupta-Bleuler null-state quotient. Temporal and longitudinal excitations then do not survive as independent physical states. The result agrees with the positive, two-polarization space obtained by Coulomb gauge reduction; covariant gauge fixing is a different implementation of the same physical constraint, not four physical photon states.

Use mostly-plus Lorentzian signature and $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$, with antisymmetrized gamma products of unit weight. A Majorana spinor obeys a reality condition under charge conjugation:A Weyl spinor instead has definite chirality,A four-dimensional Majorana spinor has four real components, while a Weyl spinor has two complex components. Charge conjugation reverses chirality in this signature, so a nonzero spinor cannot satisfy both conditions simultaneously.

For the Dirac equation $(/\partial +m)\psi =0$, the derivative term reverses chirality because ${\gamma }_5$ anticommutes with every gamma matrix, whereas the mass term preserves it. Projecting a purely chiral spinor equation onto its two chiralities therefore gives both $/\partial \psi =0$ and $m\psi =0$. Thus a nonzero four-component Weyl spinor satisfying this uncoupled Dirac equation must be massless. A Majorana spinor contains conjugate opposite-chirality components and can have a nonzero mass. Its charge-conjugation reality condition identifies particle and antiparticle operators, so a Majorana particle is its own antiparticle. A Weyl field does not by itself impose that identification.

This does not forbid writing a Majorana mass term for a neutral two-component Weyl field: such a term couples the field to its conjugate, and the resulting four-component massive field is Majorana, not a purely chiral solution of the uncoupled Dirac equation.

For the Rarita-Schwinger field, set $\chi ={\gamma }^{\mu }{\psi }_{\mu }$ and $d={\partial }^{\mu }{\psi }_{\mu }$. The antisymmetric kinetic operator is invariant underbecause commuting derivatives contract to zero against ${\gamma }^{\mu \nu \rho }$. Expanding its gamma product givesGamma contraction gives ${\gamma }_{\mu }{E}^{\mu }=2(/\partial \chi -d)$, so the equation implies $d=/\partial \chi$ and $/\partial {\psi }^{\mu }={\partial }^{\mu }\chi$.

For a Fourier mode with momentum $k$, if $k^2\ne 0$ the latter identity gives $k^2{\psi }^{\mu }=k^{\mu }/k\chi$. The mode is proportional to $k^{\mu }$ and hence pure gauge. Every physical nonzero-momentum mode therefore has $k^2=0$. For a null mode, choose an index with $k^{\mu }\ne 0$. The same equation makes $\chi$ an element of the image of $/k$, allowing a gauge choice with $\chi =0$. The field equations then reduce toResidual gauge transformations obey $/\partial \eta =0$.

Take a null momentum in the third spatial direction. Residual gauge freedom and transversality remove the temporal and longitudinal vector components, leaving two transverse vector polarizations. Their helicities are $\pm 1$, while on-shell spinor helicities are $\pm 1/2$. The gamma-trace constraint removes the two combinations of total helicity $\pm 1/2$ and retains the aligned combinations of helicity $\pm 3/2$. Explicitly, in a chiral representation the transverse gamma contractions use ${\sigma }_1\pm i{\sigma }_2$, which annihilate the correspondingly aligned spin states and act nontrivially on the opposite states. Thus a Majorana Rarita-Schwinger field has two physical massless polarizations. The Majorana assumption is the one relevant to simple supergravity; a complex vector-spinor has these helicities for its particle and independent antiparticle modes.

A consistent free massive deformation is the algebraic termadded to a kinetic term $-\frac{1}{2}{\overline{\psi }}_{\mu }{\gamma }^{\mu \nu \rho }{\partial }_{\nu }{\psi }_{\rho }$. Its equation isFor $m\ne 0$, divergence and gamma contraction give, respectively,These imply the massive Rarita-Schwinger constraints $\chi =0$ and $d=0$, and the remaining equation is $(/\partial +m){\psi }_{\mu }=0$. Squaring gives $(\square -m^2){\psi }_{\mu }=0$ in the chosen signature. The gauge symmetry is lost. In the rest frame, transversality sets the temporal component to zero; three spatial vector components times two on-shell spinor states give six states, and gamma trace removes the two spin-one-half states. The result is four massive spin-$3/2$ states, with spin projections $\pm 3/2,\pm 1/2$. In supergravity the additional longitudinal states can be supplied by the super-Higgs mechanism.

Use the metric in the question and fix the Levi-Civita symbol convention ${ϵ}_{0123}=+1$. A consistent choice of rotation Lie algebra generators isSubstituting the indices into the printed Poincare algebra givesThe other cyclic brackets follow the same way. The negative spatial metric and antisymmetry of $M^{\rho \sigma }$ both enter this sign.

The printed brackets are real Lie algebra brackets, without the factor $i$ used for ordinary commutators of Hermitian quantum observables. We use those brackets for the algebraic verifications. For physical eigenvalues below, angular momentum is Hermitian, its spin projection is the real number $j_3$, and $\hslash =1$. In that quantum convention the ordinary operator commutators are $i$ times the displayed brackets. This distinction prevents identifying a real spin projection with an anti-Hermitian matrix eigenvalue.

In the universal enveloping algebra, translations commute. The Pauli-Lubanski pseudovector therefore satisfiesFor fixed $\nu ,\rho$, the product of momenta is symmetric in $\mu ,\tau$ while the epsilon coefficient is antisymmetric in them. There is no need to commute the Lorentz generator through the momenta.

Using the commutator derivation identity and the mixed bracket,The two terms become equal after swapping $\nu ,\rho$, and the last expression vanishes by antisymmetry in $\nu ,\tau$. In the Hermitian observable convention, the calculation has one overall extra $i$ and still vanishes. Thus $W_{\mu }$ preserves each momentum eigenspace.

For the specified epsilon orientation, two useful component identities areThe order displayed matters: the momentum operator is on the right, so it can act first on the momentum eigenstate. At rest, $p^{\mu }=(m,0,0,0)$, and on a spin state with $J_3|j,j_3⟩=j_3|j,j_3⟩$ these giveThese are massive rest-frame Pauli-Lubanski eigenvalues. The raised component would be $W^3=+m{j}_3$; confusing $W_3$ with $W^3$ reverses the answer.

Helicity is the projection of spin angular momentum, or equivalently the rotation generator acting internally, along the momentum direction:For the momentum $p^{\mu }=(k,0,0,k)$ with $k>0$, the helicity operator is $J_3$. Hence a helicity-$j_3$ state hasThe result uses only the two longitudinal components and does not need a separate assumption about the transverse little-group generators. These massless longitudinal Pauli-Lubanski eigenvalues agree with $W^{\mu }=j_3{P}^{\mu }$ for ordinary finite-helicity representations.

Finally, at rest the contraction is $m{W}_0=0$. For the chosen null momentum it is $k(W_0+W_3)$, whose two eigenvalues cancel. Thus both results obey $W_{\mu }{P}^{\mu }=0$. Reversing the epsilon orientation reverses all $W$ eigenvalues together; it leaves both identities and both consistency checks intact. For a literal anti-Hermitian derived representation $\rho$ of the printed real algebra, write $H_X=i\rho (X)$ for the Hermitian observable. Since $W_{\mu }$ is bilinear in generators, its abstract enveloping-algebra image is $\rho (W_{\mu })=-W_{\mu ,\mathrm{phys}}$. Thus the corresponding formal-image eigenvalues, if that convention is intended, areHere $\rho (P^{\mu })$ has eigenvalue $-i{p}^{\mu }$, while $p^{\mu }$ and $j_3$ themselves remain real physical labels. This is the same result after the Hermitian quantum generator convention is applied to both factors, not an inconsistent choice of spin sign. Both the Hermitian-observable and literal anti-Hermitian interpretations are consequently specified.

Use signature $(+---)$ and take the covariant spatial components $A_i$ as canonical coordinates. The canonical quantization of the electromagnetic field begins with the canonical momentaThus the primary momentum constraint of the electromagnetic potential is ${\Pi }^0=0$: $A_0$ has no independent velocity. The Hamiltonian obtained by the Legendre transform in mechanics, up to a boundary term, isPreserving the primary constraint requires the Gauss law constraint in gauge theory, ${\partial }_i{\Pi }^i=0$. It also follows by varying $A_0$. These are two first-class constraints; they generate the gauge freedom and remove two canonical pairs from the four potential components. The reduced phase space has four dimensions per spatial mode, hence two propagating photon degrees of freedom.

Impose Coulomb gauge, ${\partial }_i{A}_i=0$. With no charges, Gauss's law then gives ${\nabla }^2{A}_0=0$; vanishing boundary conditions set $A_0=0$. This is radiation gauge. The remaining components are transverse and obey the massless wave equation. Let ${ϵ}_i^{(r)}(\mathbf{k})$, $r=1,2$, be orthonormal transverse polarization vectors. Their photon polarization completeness relation isThe canonical transverse photon field is the Hermitian operatorwhereThe field and its conjugate momentum have the transverse equal-time commutatorThis is the quantized reduced bracket, or equivalently the Dirac bracket after imposing the constraints and gauge conditions. The normal-ordered Hamiltonian is ${\sum }_r\int d^{3}k{\omega }_{\mathbf{k}}{a}_r^{†}{a}_r/(2\pi {)}^3$. Its excitations are photons; circular combinations of the two transverse polarizations have helicity $+1$ and $-1$. The scalar and longitudinal potential components do not create additional physical photons.

The Feynman propagator is the vacuum expectation of a time-ordered product. The mode expansion directly gives the radiation-gauge photon propagator, with $z=x-y$:The first expression comes from the creation-annihilation commutator; the second is its contour-integral representation. The positive-energy pole lies below the real axis and the negative-energy pole above it. In this reduced free-field description, the temporal operator is zero. A photon propagator must specify its gauge; the spatial transverse propagator is not the same tensor as the covariant four-potential propagator.