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Antiparticles are positive-energy quanta of relativistic quantum fields, not physical particles with negative energies. The negative-frequency part of a quantum field creates these quanta; the distinction between its classical frequency and a state's excitation energy is central to their interpretation.

For a real scalar field, Hermiticity identifies the two frequency branches:There is one species of excitation. Its particle is its own antiparticle; a single Hermitian scalar cannot carry the nonzero continuous phase charge of a complex field. This does not prevent self-conjugate particles from having other quantum numbers.

A complex scalar field is not Hermitian. Its two independent sets of oscillators giveThe canonical commutation relations make $a^{†}$ and $b^{†}$ create different positive-energy species. With charge convention fixed by $[Q,\Phi ]=-q\Phi$, normal ordering givesThus $a^{†}|0⟩$ and $b^{†}|0⟩$ have the same mass and spin but opposite charge. Although $b^{†}{e}^{ipx}$ has negative classical frequency, $[H,b^{†}]=E_p{b}^{†}$, so it creates a positive-energy excitation. Two real scalar fields can be combined into one complex field; the particle–antiparticle basis diagonalizes the conserved phase charge. Neutrality alone does not require a particle to be self-conjugate.

Locality explains why both branches occur. For the complex scalar field,At spacelike separation, Lorentz invariance of the mass shell measure permits a frame with equal times; reversing $\mathbf{p}$ makes the two terms cancel. This is microcausality. If the antiparticle creation term were omitted, the commutator would be just a Wightman function, generally nonzero at spacelike separation. For example, for a massless scalar at equal times and separation $r>0$, it is $1/(4{\pi }^2{r}^2)$. The cancellation is the role of antiparticle modes and spacelike commutativity. In a real scalar field its self-conjugate creation modes supply the same cancellation.

A nonrelativistic particle field obeys a first-order time equation and need not contain an antiparticle-creation branch. For example,allows an expansion using only $a_{\mathbf{p}}{e}^{-i{\mathbf{p}}^{2}t/(2m)+i\mathbf{p}\cdot \mathbf{x}}$ in $\Psi$; its adjoint creates those particles. The basic theory conserves particle number and does not impose a Lorentz-invariant light cone locality condition. It often arises from a relativistic quantum field by factoring out the rest-energy oscillation and restricting to a low-energy particle sector, where creation of particle–antiparticle pairs costs roughly $2m$. An antiparticle species can certainly also be treated nonrelativistically with its own field; antiparticles are simply not forced by this nonrelativistic quantum field equation.

Historically, the Dirac equation has one-particle energy branches $\pm \sqrt{{\mathbf{p}}^2+m^2}$. Treating both as ordinary electron energies would leave no lowest-energy one-particle vacuum: an electron could fall through arbitrarily negative levels. The Dirac sea interpretation fills every negative-energy electron level. The Pauli exclusion principle then blocks further occupation. A missing negative-energy electron is a hole: removing energy $-E$ increases energy by $E$, and removing charge $-e$ leaves charge $+e$. This predicts the positron's mass, spin and opposite electric charge.

The quantum field theory account quantizes a Dirac field instead:The independent particle and antiparticle operators obey canonical anticommutation relations. Before normal ordering, the negative-frequency contribution to the Dirac Hamiltonian has the form $-E_p{b}_r{b}_r^{†}$. Anticommutation rewrites it as $+E_p{b}_r^{†}{b}_r$ plus a vacuum constant. Removing the constant leaves positive particle and antiparticle energies. The charge has opposite signs on the two occupation numbers, and interactions permit pair creation and annihilation while conserving the relevant charges.

There is therefore no need for a literal infinitely populated material sea. The hole picture is a useful fermionic reinterpretation, but it cannot explain bosonic antiparticles through filled-level exclusion, since bosons can multiply occupy a state. Canonical quantization of a complex scalar field supplies them just as consistently. More generally charge conjugation relates field representations with opposite internal charges; real scalar fields and Majorana fermions are self-conjugate exceptions. The relativistic Fock space description handles these cases, statistics and changing particle number within one framework.

Consider positive-energy unitary irreducible representations of the proper orthochronous Poincaré group, allowing its double cover for physical half-integer spin. Use metric $(+---)$ and define the Pauli-Lubanski pseudovector byThis overall sign gives $\mathbf{W}=m\mathbf{J}$ at rest. The antisymmetric tensor and commuting translation generators give $P_{\mu }{W}^{\mu }=0$. The given commutators say $W$ commutes with translations and transforms as a Lorentz vector. Hence both $P^2$ and $W^2$ commute with the Poincare algebra; they are Casimir operators.

The commuting Hermitian translation generators admit generalized joint momentum eigenstates. A Lorentz transformation maps a state at $p$ to a state at $\Lambda p$, so an irreducible representation has momentum support on a single Lorentz orbit. Choose a standard momentum $k$. Its little group consists of the transformations fixing $k$ and acts on the internal states there. Transport those internal states over the orbit by chosen transformations $L(p)k=p$. This gives the construction underlying Wigner's classification.

For timelike future-directed momenta, $P^2=m^2$ with $m>0$ and $k=(m,0,0,0)$. The massive particle little group is SO(3), or SU(2) on the double cover. Write $J_i={ϵ}_{ijk}d(M^{jk})/2$. The rotation commutators give $[J_i,J_j]=i{ϵ}_{ijk}{J}_k$, so the irreducible little-group representation hasAt rest, $W^0=0$ and $\mathbf{W}=m\mathbf{J}$. Thus the massive Pauli-Lubanski Casimir isThe rest spin space has dimension $2s+1$. All $s=0,\frac{1}{2},1,\dots$ occur for the double cover; only integer spin descends to an ordinary single-valued representation of the proper orthochronous Poincare group.

For the detailed representation, choose a Lorentz boost $L(p)$ from $k$ to every point of the positive-energy mass shell. With invariant normalization, states $|p,\sigma ⟩=U(L(p))|k,\sigma ⟩$ transform asThe matrix $R$ fixes $k$, so it is the Wigner rotation in the little group. Its cocycle identityproves the representation law. Translations give phases; with the convention $U(a)|p,\sigma ⟩=e^{ip\cdot a}|p,\sigma ⟩$, this extends to the massive induced representation of the Poincare double cover onThe invariant measure and unitary spin matrices give unitarity. A closed invariant subspace restricts at fixed momentum to a little-group invariant spin subspace; irreducibility of that spin representation and transitivity over the mass shell force the full representation to be irreducible. Conversely the momentum-orbit construction recovers every massive positive-energy irreducible representation. The labels are therefore $(m,s)$.

For a nonzero future null momentum choose $k=(E,0,0,E)$. The massless particle little group is $ISO(2)$. With $K_i=d(M^{0i})$, its massless little-group generators arewith $[J_3,N_1]=i{N}_2$, $[J_3,N_2]=-i{N}_1$ and $[N_1,N_2]=0$. Evaluating the Pauli-Lubanski vector gives $W^2=-E^2(N_1^2+N_2^2)$.

In a finite-dimensional unitary little-group representation, the joint spectrum of the commuting Hermitian $N_1,N_2$ is finite and invariant under continuous planar rotations. Its only possible values are zero. Thus finite-dimensional unitary representations of the massless little group have trivial translations. Irreducibility then leaves a one-dimensional rotation character of helicity $h$, givingHelicity supplies an additional label that the two zero Casimirs cannot distinguish. It is integer for the ordinary group and may be half-integer on its physical double cover. Nontrivial little-group translations instead lead to an infinite-dimensional continuous-spin representation with $W^2<0$; the Poincare algebra by itself allows these too. This completes the timelike and nonzero null cases of physical interest.

Use $x^{\mu }=(ct,x,y,z)$ and metric signature $(+,-,-,-)$. Since $d\tau =dt/\gamma$, the four-velocity and four-momentum areThese use the proper time and Lorentz factor; the mass shell identity is $E^2-c^2|\mathbf{p}{|}^2=M^2{c}^4$.

Initially the parent four-momentum is $(Mc,0)$. By four-momentum conservation, the remaining single particle has energy $M{c}^2-E$ and three-momentum $-\mathbf{p}$. Applying its mass shell identity gives $m^2{c}^4=(M{c}^2-E{)}^2-c^2|\mathbf{p}{|}^2$, henceThe positive root is selected for the rest mass; physically admissible decay data also give nonnegative remaining energy.

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.

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.

For non-null four-momentum, the transverse projector of a vector field and the complementary longitudinal projector of a vector field areIndeed, $q_a{P}^a{}_b=0$, $P^2=P$, $L^2=L$, and $P+L=I$. Acting on an arbitrary four-vector,So $P$ removes the unwanted component, whereas $L$ extracts it. On the massive mass shell, $q^2=-m^2$ and $P^a{}_b={\delta }_b^a+q^a{q}_b/m^2$. This on-shell form should not be used as an off-shell linear projection: away from $q^2=-m^2$ it is not idempotent. The non-null hypothesis matters; this decomposition is undefined at $q^2=0$.

Keep the mostly-plus Minkowski metric and the Fourier transform $e^{ip\cdot x}$. Write $/p={\gamma }^a{p}_a$, with $\{{\gamma }^a,{\gamma }^b\}=2{\eta }^{ab}$. To fix the otherwise unspecified phase of ${\gamma }^5$ and the Dirac adjoint, takewhere ${\Gamma }^a$ are ordinary mostly-minus gamma matrices. Then $({\gamma }^5{)}^2=-1$, $\{{\gamma }^5,{\gamma }^a\}=0$, and the fermionic time-derivative term is $i{\psi }^{†}{\partial }_t\psi$. In these conventions the interaction with real $g$ is Hermitian: in mostly-minus notation it is $ig{\bar{\psi }}_{\mathrm{std}}{\Gamma }^5\psi \varphi$. If one instead calls $i{\gamma }^5$ the square-one chirality matrix, its coefficient must be $-ig$ to represent the same interaction. These phase choices leave physical relativistic scattering cross-sections unchanged.

Expanding $e^{i{S}_{\mathrm{int}}}$ yields the following Feynman rules with relativistically normalized external states:

Strip the overall four-momentum conservation delta function when defining the scattering amplitude. There are no further bare interaction vertices, no gauge fixing and no Faddeev-Popov ghost fields in this theory. Renormalized higher-order calculations add the required counterterms; these are additional to the rules of the displayed classical Lagrangian density.

Use the physical rapidity strip $0<\mathrm{Im}\vartheta <\pi$ for poles of the two-body S-matrix, and write $\vartheta$ for scattering rapidity to distinguish it from the theta angle in question 2. A denominator in the kink-antikink product vanishes atNo numerator cancels these poles. For two equal-mass constituents with rapidities $\chi \pm i{u}_k/2$, their four-momentum vectors sum toThis gives the relativistic bound-state mass from a rapidity pole. The ordered breather spectrum isIt increases strictly with $k$. The hypothetical $k=n$ state would lie at the two-kink threshold $2M$ and is not a bound state; it is absent from the pole product. At these couplings $M=mn/\pi$ and ${\beta }^2=8\pi /(n+1)$. In particular $m_1\to m$ at weak coupling $n\to \infty$. This is the Sine-Gordon breather spectrum at reflectionless couplings. At $n=1$ there are no breathers; the subsequent processes involving a physical ${\mathcal{B}}_2$ require $n\ge 3$.

For two identical neutral ${\mathcal{B}}_1$ particles, exchanging the two outgoing labels does not produce a distinguishable channel. In one spatial dimension the elastic final momenta are the incoming pair, up to interchange. Thus there is one scalar identical-particle amplitude, rather than separately observable transmission and reflection amplitudes.

Put $a=\pi /(2n)$, so the basic amplitude uses $\sin (2a)$. Its poles in the physical strip occur at $\vartheta =2ia$ and $\vartheta =i(\pi -2a)$. The first is the direct bound-state pole. Choosing constituent rapidities $\chi \pm ia$ gives real total energy-momentumSince $m_1=2M\sin a$,Both energy and momentum therefore match an on shell ${\mathcal{B}}_2$ with rapidity $\chi$. The complementary pole is its crossed-channel partner. At $n=2$ the would-be ${\mathcal{B}}_2$ is a threshold state, so this physical fusion interpretation must not be imposed there.

The bound-state fusion of factorized S-matrices treats a bound particle as its on shell constituents with analytically continued rapidities. If equal-mass particles $A,A$ fuse to $C$ at relative rapidity $iu$, use constituent rapidities $\chi \pm iu/2$. To scatter a third particle $D$ off $C$, multiply its scattering amplitudes with each constituent and take the bound-state residue or projection in the constituent channel. For a scalar amplitude this givesFor particles with internal indices, the product is projected using the bound-state coupling tensors. The heuristic reason is factorized scattering: conserved higher charges prevent particle production and fix the rapidity data, so the third particle scatters through the constituents by successive two-body processes. Consistency of different orders is the Yang-Baxter equation. Without integrability, an independent three-body interaction would invalidate this simple bootstrap product.

For ${\mathcal{B}}_2=({\mathcal{B}}_1{\mathcal{B}}_1)$, the constituent shifts are $\pm ia$, givingIt is useful to write this Sine-Gordon breather fusion amplitude in explicitly factorized form:To check the reduction, put $z=\sinh \vartheta$. Multiplying the shifted factors gives numerator and denominator $z^2\pm 2iz\sin (2a)\cos a+{\sin }^{2}a-{\sin }^2(2a)$. Use $\sin a+\sin 3a=2\sin 2a\cos a$ and $\sin a\sin 3a={\sin }^{2}2a-{\sin }^{2}a$ to factor them as $(z\pm i\sin a)(z\pm i\sin 3a)$. The product tends to one at large positive real rapidity, fixing its overall phase in this bootstrap convention.

For $n\ge 3$, the nearest pole to the real axis is $\vartheta =ia$. It is simple and comes from the first factor. In the crossed, or t-channel, the momentum carried between the external particles is their difference. With the $+,-$ metric its invariant isAt the pole, substitute the breather masses:The trigonometric identity is applied with angles $2a$ and $a$. Thus the exchanged one-particle state is the lightest breather ${\mathcal{B}}_1$, on its mass shell. This is crossed-channel lightest-breather exchange. The external momenta at a bound-state pole are analytically continued; on shell here means the invariant mass relation and conservation of the continued energy-momentum, not a pole at real physical rapidity. At $n=3$ the more distant central factor has a double pole, but the nearest $ia$ pole and its ${\mathcal{B}}_1$ interpretation remain unchanged.

For this question use the mostly-plus Dirac convention: the Minkowski metric is ${\eta }_{ab}=\mathrm{diag}(-1,1,1,1)$ andThis convention matches the printed plane-wave phase and final identity. It is related to the usual mostly-minus Dirac equation by reversing the metric and taking the negatives of the usual gamma matrices. In particular, the resulting Dirac action and its Dirac adjoint describe the same physical massive field. The Dirac gamma matrices are four complex $4\times 4$ matrices representing the spacetime Clifford algebra with quadratic form $-\eta$. The irreducible complex representation has dimension four. A convenient explicit choice is the negative of the standard Dirac representation of the gamma matrices:where ${\sigma }^i$ are the Pauli matrices. Their multiplication law verifies the displayed anticommutator.

In this representation the gamma matrix adjoint and transpose identities areandThe invariant way to express the latter pattern uses the charge-conjugation matrix:Individual transpose signs depend on the basis. More generally, a similarity transformation ${\gamma }^{\prime a}=M{\gamma }^a{M}^{-1}$ changes the Hermitizing matrix to $H^{\prime }=M^{-†}{\gamma }^0{M}^{-1}$ and the charge-conjugation matrix to $C^{\prime }=MC{M}^T$. Then ${\gamma }^{\prime a†}=H^{\prime }{\gamma }^{\prime a}{H}^{\prime -1}$ and $C^{\prime -1}{\gamma }^{\prime a}{C}^{\prime }=-{\gamma }^{\prime aT}$. Thus the simple formula with ${\gamma }^{\prime 0}$ itself presumes a compatible Hermitian basis, rather than an arbitrary nonunitary similarity transformation.

Applying $(i{\gamma }^a{\partial }_a-m)$ to the Dirac equation gives $(\square -m^2)\Psi =0$. Its mass shell is $p^2=-m^2$, so the frequencies are $p^0=\pm \sqrt{{\mathbf{p}}^2+m^2}$. The Dirac spinor transforms in the four-component Spinor representation of the Lorentz group. Under spatial rotations, the two upper and the two lower components each transform as a two-component spin-$1/2$ representation: the spin angular momentum matrices are $\mathrm{diag}({\sigma }^i,{\sigma }^i)/2$. At rest the positive-energy equation selects the upper two components, giving two independent spin polarizations, and the negative-frequency equation selects the lower two.

In the quantum theory a mode expansion isWith the mostly-plus Dirac convention, $e^{ip\cdot x}=e^{-iEt+i\mathbf{p}\cdot \mathbf{x}}$ is positive frequency. The negative-frequency coefficient obeys $(/p+m)v_s(p)=0$. The fermionic annihilation operators $b_s$ and $d_s$ satisfy the canonical anticommutation relations, with their respective fermionic creation operators. The $b_s^{†}$ excitations are particles; the $d_s^{†}$ excitations are antiparticles with the same positive energy, mass and spin-$1/2$ but opposite charge. After normal ordering, the Hamiltonian operator contains positive multiples of $b_s^{†}{b}_s+d_s^{†}{d}_s$. Reinterpreting the negative-frequency part as antiparticle creation supplies a spectrum bounded below rather than a physical tower of negative-energy particles.

For the printed $e^{ip\cdot x}$ wave, ${\partial }_a\Psi =i{p}_a\Psi$. Substitution givesThe spin label $s$ indexes the two states of a spin-$1/2$ particle, rather than varying the particle's total spin. For real on-shell $p$, Hermitian conjugation and ${\gamma }^{a†}={\gamma }^0{\gamma }^a{\gamma }^0$ giveThese are right and left null-vector equations for the same on-shell matrix.

To obtain the Gordon identity, take both external Dirac spinors to have the same real mass $m$. Their two equations implyDefine ${\gamma }^{ab}=[{\gamma }^a,{\gamma }^b]/2$. The Clifford algebra relation yieldsThereforeMultiplying the previous null-vector equation by $-1$ proves the required formula exactly:For $m\ne 0$ it can be solved for the vector-current matrix element, separating a momentum term from the antisymmetric Dirac spinor term. The identity before division also holds at $m=0$. The signs depend jointly on the metric, Clifford relation, Dirac mass term and plane-wave phase. In the mostly-minus convention of Questions 1 and 3, the printed phase instead gives $(/p+m)u=0$; the corresponding identity uses $(p_b-p_b^{\prime })$ in the antisymmetric term. Mixing that convention with the formula proved here would produce an apparent sign error.

Return to the mostly-minus Minkowski metric of Question 3, and write $f={\partial }_a{A}^a$. The field $\xi (x)$ is assumed real and nonzero wherever its inverse occurs. Under the usual Abelian gauge transformation $A_a\mapsto A_a+{\partial }_a\lambda$, with $\xi$ inert, the electromagnetic field tensor is invariant but $f\mapsto f+\square \lambda$. The change in the added density isThis is not generally a total derivative. The action is not invariant under arbitrary gauge transformations. It retains residual Lorenz gauge symmetry for transformations satisfying $\square \lambda =0$, with the same boundary conditions imposed before and after the transformation. The action itself is undefined at $\xi =0$; that value can only be considered as a limiting gauge.

To vary the electromagnetic four-potential, use $\delta F_{ab}={\partial }_a\delta A_b-{\partial }_b\delta A_a$ and $\delta f={\partial }_a\delta A^a$. The integration by parts of both terms giveswith surface terms removed by the variational boundary conditions. HenceFor variable $\xi$, the derivative must act on $1/\xi$ as well as on $f$:Since $\xi$ enters algebraically, its Euler-Lagrange equation isFor a real field this gives $f=0$. Thus a dynamical gauge-fixing parameter imposes a constraint equation in field theory; it is not an ordinary propagating scalar. On configurations satisfying both equations, ${\partial }_a{F}^{ab}=0$, ${\partial }_a{A}^a=0$, and therefore $\square A^b=0$. There is no independent kinetic equation that determines $\xi$.