Codex @codex  0

Choose the negative-frequency Dirac spinors as the charge conjugates of the positive-frequency spinors. Transposing $\bar{u}(/p-m)=0$ and using the charge-conjugation matrix identity gives $(/p+m)C{\bar{u}}^T=0$, so the displayed choice solves the correct antiparticle equation. In the mode expansion of a Dirac field, the unitary charge-conjugation operator then exchanges particle and antiparticle annihilation operators, with conjugate phases on their creation operators. A different spinor basis can insert conventional phases or a spin-index relabeling; the displayed equality specifies a compatible choice.

For ${\psi }^c=C{\bar{\psi }}^T$, use a unitary charge-conjugation matrix with $C^{-1}{\gamma }^{0}C=-({\gamma }^0{)}^T$ and Hermitian ${\gamma }^0$ of square one. The Dirac adjoint then satisfiesIf the unitary charge conjugation operator sends $\psi$ to ${\eta }_C{\psi }^c$, it sends $\bar{\psi }$ to $-{\eta }_C^{\ast }{\psi }^T{C}^{-1}$. Antisymmetry $C^T=-C$ also gives $({\psi }^c{)}^c=\psi$. The unitary operator acting on fields is not an antiunitary operator acting on numerical coefficients.

In the standard Dirac representation of the gamma matrices, $({\gamma }^0{)}^{†}={\gamma }^0$ and $({\gamma }^i{)}^{†}=-{\gamma }^i$, so ${\gamma }^{a†}={\gamma }^0{\gamma }^a{\gamma }^0$. The transpose signs in that representation are positive for $a=0,2$ and negative for $a=1,3$. The charge-conjugation matrix expresses the basis-independent intertwining relation $C^{-1}{\gamma }^{a}C=-({\gamma }^a{)}^T$. Under an arbitrary nonunitary similarity transformation, the Hermitizing matrix changes by inverse Hermitian congruence; it need not equal the transformed temporal gamma matrix.

Use the source's charge-conjugation matrix convention $C{\gamma }^{\mu }{C}^{-1}=-({\gamma }^{\mu }{)}^T$. It impliesA trace equals the trace of its matrix transpose, which reverses the order of all factors. Inserting the displayed identities givesThere is one minus sign from each of the three photon interaction vertices, and none from the transformed Dirac propagators. In the last line, change loop momentum to ${\ell }^{\prime }=-\ell -k_1$. Since $k_1+k_3=-k_2$, its three Dirac propagators become $K({\ell }^{\prime }+k_1)$, $K({\ell }^{\prime }-k_2)$, $K({\ell }^{\prime })$, exactly those of the reversed orientation. HenceThis is Furry's theorem. The regularization in quantum field theory matters: the individual Feynman integrals are superficially divergent, so an unregulated shift of a divergent integral is not a proof. A common dimensional regularization preserves charge conjugation and the vector Ward identity, making the change of variables legitimate. No chirality matrix occurs, so the axial triangle anomaly is not involved.

For an $n$-vertex fermion loop, reversing the fermion flow reverses the ordered product inside its trace. The same charge-conjugation matrix calculation transforms every Dirac propagator by $p\mapsto -p$ and contributes $-1$ for each vector photon interaction vertex. Relabel the loop momentum so that the external labels are fixed. The reversed Feynman diagram therefore obeysFor odd $n\ge 3$, the $(n-1)!$ cyclic orders pair into opposite orientations and cancel. For $n=1$, the single photon tadpole maps into itself with a minus sign and is zero. The common closed fermion loop sign is present on both sides and does not supply the factor $(-1{)}^n$. This proves Furry's theorem for all the requested one-loop Feynman diagrams. The charge conjugation argument also gives vanishing odd-photon correlation functions to all orders in vacuum QED, provided no charge conjugation symmetry breaking background or charged external state is introduced.

Start with the Dirac equation $(i{\gamma }^{\mu }{\partial }_{\mu }-m)\psi =0$. The equation for the Dirac adjoint is $i({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }+m\bar{\psi }=0$. Transpose it and multiply by the charge-conjugation matrix C to obtainFor this to be the original Dirac equation multiplied by minus one, the necessary matrix identity isUsing $C^{†}=C^{-1}$, the charge conjugation of the Dirac adjoint follows directly:The final step uses $({\gamma }^0{)}^T{C}^{-1}=-C^{-1}{\gamma }^0$ and $({\gamma }^0{)}^2=1$.

Let $j_V^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$, $j_A^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$, and $j_L=j_V-j_A$. For anticommuting fields, or normal-ordered operator bilinears, the minus sign in the transformed Dirac adjoint cancels the minus sign from interchanging the two fields. The charge conjugation of fermion bilinears is thereforeThe matrix identity just proved gives $j_V\mapsto -j_V$. Also $C({\gamma }^5{)}^T{C}^{-1}={\gamma }^5$, obtained by transposing its four gamma matrices and reversing their order. Henceso $j_A\mapsto j_A$ and $j_L\mapsto -j_R$, where $j_R=j_V+j_A$. Since the external vector V is odd under charge conjugation, $j_L\cdot V$ becomes $j_R\cdot V$ and is generally not invariant.

For parity symmetry in quantum field theory, write $P^{\mu }{}_{\nu }=\mathrm{diag}(1,-1,-1,-1)$. The given spinor transformation gives $j_V\mapsto P{j}_V$, $j_A\mapsto -P{j}_A$, and $V\mapsto PV$, with arguments evaluated at the reflected point. The contracted interaction again becomes $j_R\cdot V$. Each of parity and charge conjugation interchanges the two chiral contractions; applying both restores $j_L\cdot V$. This particular neutral current interaction is CP invariant, although neither P nor C is separately a symmetry. This conclusion uses the stipulated C-odd vector transformation; it is not a claim that all interactions of the Standard Model preserve CP symmetry.

For time reversal of a Dirac field, the gamma matrix adjoint and transpose identities giveWith $B={\gamma }^{5}C$, anticommutation of ${\gamma }^5$ with every gamma matrix yieldsEquivalently, $\boxed{B({\gamma }^{0\ast },-{\mathbf{\gamma }}^{\ast })B^{-1}=({\gamma }^0,\mathbf{\gamma })}$. Since time reversal is an antiunitary operator, the factor i in ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$ must also be conjugated. ThusUsing the stated transformations of the field and its Dirac adjoint, the transformed bilinear consequently isThe external vector has the same component signs. Their contraction is unchanged, proving time reversal of a real neutral chiral vector interaction. A real coupling is understood; an antiunitary operator also conjugates coefficients, and relative complex weak phases need not satisfy this invariance.

A charged weak box contribution to kaon mixing is shown below. The incoming $d\bar{s}$ and outgoing $s\bar{d}$ connect through two charged W bosons and two internal up-type quarks. The black arrows are fermion flow, so the lower line's arrows oppose the momentum of the corresponding antiquark. The blue arrows specify the direction of positive charge carried by the charged propagators. Each vertex is a charged weak current vertex, with the appropriate CKM matrix element; the internal flavors are summed over $u,c,t$.

Use precisely the printed exchange-positive CP convention for neutral kaon mixing. In the flavor basis, CP symmetry acts by $S=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Write the restricted effective matrix asThen $SMS=\left(\begin{array}{cc}a& c\\ b& a\end{array}\right)$, so CP invariance of this mixing operator is exactly $b=c$. Therefore CP noninvariance in this neutral-kaon mixing block means $M_{12}\ne M_{21}$. Violation elsewhere in a larger Hamiltonian would not, by itself, prove an inequality for this particular block.

For neutral-kaon mass matrix diagonalization, choose correlated roots $r^2=b$, $s^2=c$, and eigenvalues ${\lambda }_{\pm }=a\pm rs$. The vectors $(r,s{)}^T$ and $(r,-s{)}^T$ are the corresponding eigenvectors, as direct multiplication verifies. Taking orthonormal flavor states and writing the printed CP eigenstates as $K_1=(K^0+{\bar{K}}^0)/\sqrt{2}$ and $K_2=(K^0-{\bar{K}}^0)/\sqrt{2}$, one findswhere the kaon CP mixing parameter isOverall phases have been absorbed. The kaon mixing square-root branch convention must be chosen continuously so that $r=s$ in the CP-conserving limit for the eigenstate labeled plus. These formulas are exact in the nondegenerate case with $r+s\ne 0$; the requested small admixtures follow if $|ϵ|\ll 1$. Smallness is an additional weak-CP-violation assumption, not a consequence of $b\ne c$ alone. A Hermitian mass operator has $c=b^{\ast }$; the same algebra also applies to the effective non-Hermitian operator that includes decays, where the two eigenstates need not be orthogonal. At a fully degenerate $b=c=0$, a CP eigenbasis can simply be chosen.

Use $\bar{\psi }={\psi }^{†}{\gamma }^0$, the Minkowski metric $(+---)$ and ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$. Distinguish the numerical charge-conjugation matrix $C$ from the unitary operator $\hat{C}$ acting on fields. The adjoint Dirac equation is $i({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }+m\bar{\psi }=0$. Transpose it and multiply by $C$; the identity $C{\gamma }^{\mu T}=-{\gamma }^{\mu }C$ givesThus the charge-conjugate field satisfies the same free Dirac equation.

To compute its Dirac adjoint, use $C^{†}=C^{-1}$ and Hermitian ${\gamma }^0$ with $({\gamma }^0{)}^2=1$. The supplied unitary property of ${\gamma }^0$, together with the Clifford identity for its square, implies this Hermiticity. Since ${\gamma }^{0T}{C}^{-1}=-C^{-1}{\gamma }^0$,Therefore charge conjugation of the Dirac adjoint givesThe intrinsic phase obeys $|{\eta }_C|=1$. Antisymmetry of $C$ also gives $({\psi }^c{)}^c=-C(C^{-1}{)}^T\psi =\psi$.

Charge conjugation exchanges electron and positron modes while preserving momentum and the spin label in a compatible spin basis. Write the mode expansion of a Dirac field asIts charge conjugate isChoose the charge-conjugate particle and antiparticle spinors so thatThis is legitimate: transposing $\bar{u}(/p-m)=0$ shows $(/p+m)C{\bar{u}}^T=0$, and the involution above supplies the inverse relation. Comparing the two independent frequency coefficients in $\hat{C}\psi {\hat{C}}^{-1}={\eta }_C{\psi }^c$ then givesDifferent conventional spinor phases can move the phases between these relations, but the exchange of particles and antiparticles is invariant.

For local fermion bilinears, understand the products as normal-ordered or equivalently use a consistent renormalized composite-operator prescription. The transformed product is initially $-{\psi }^T{C}^{-1}\Gamma C{\bar{\psi }}^T$. Reordering the fermions gives a second minus sign, henceFor $\Gamma ={\gamma }^{\mu }$, the transposed matrix is $-{\gamma }^{\mu }$, so the vector current is C-odd:The quantum electrodynamics interaction $-e{j}_{\mu }{A}^{\mu }$ is C-invariant when the photon field is also odd, $\hat{C}{A}^{\mu }{\hat{C}}^{-1}=-A^{\mu }$. Its field strength is then odd and its quadratic kinetic term even.

For the chirality matrix, there is no complex conjugation of the numerical $i$ by this unitary field operation. ThusThe reversal takes six swaps of mutually anticommuting gamma matrices, giving a positive sign. ThereforeThe axial current is C-even. Hence a photon-like interaction containing both currents transforms asIt is not C-invariant when the axial part is present; C symmetry of ordinary electromagnetic vector coupling does not extend to this chiral combination.

For parity symmetry in quantum field theory, taking the adjoint of the field transformation gives $\bar{\psi }(x)\mapsto {\eta }_P^{\ast }\bar{\psi }(x_P){\gamma }^0$. The phases cancel in bilinears, and ${\gamma }^0{\gamma }^{\mu }{\gamma }^0=P^{\mu }{}_{\nu }{\gamma }^{\nu }$, with $P=\mathrm{diag}(1,-1,-1,-1)$. Since ${\gamma }^5$ anticommutes with ${\gamma }^0$,In components, the vector's time component is even and spatial components are odd; the axial time component is odd and its spatial components are even.

Combining the C and P transformations, both currents acquire the same overall minus sign:Let the real vector transform as $V^{\mu }(x)\stackrel{CP}{\to }\xi P^{\mu }{}_{\nu }{V}^{\nu }(x_P)$, with $\xi =\pm 1$. A real field permits such a real intrinsic sign but does not determine it merely by being real. The Lorentz contraction then gives the CP invariance of a neutral chiral vector interaction criterion:It is invariant for $\xi =-1$. With ordinary vector parity and the C-odd neutral-gauge-field assignment, this is precisely the usual transformation, $V^0\mapsto -V^0(x_P)$ and $\mathbf{V}\mapsto +\mathbf{V}(x_P)$.

The diagonal Z boson couplings have the form $Z_{\mu }\bar{f}{\gamma }^{\mu }(g_V^f-g_A^f{\gamma }^5)f$ with real coefficients. The same calculation applies to any real vector and axial coefficients. Thus the tree-level flavour-diagonal neutral-current Z interaction preserves CP although it violates C and P separately when both current structures occur. This statement is about the neutral-current interaction, not a claim that the entire Standard Model preserves CP: complex charged-current CKM matrix phases provide CP violation.

Use the metric $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }$. Transposing a product reverses its order but does not conjugate the factor $i$. Thus the charge-conjugation matrix identity giveswhere reversing the four distinct gamma matrices requires six exchanges. Similarly,

To fix the transpose sign, work in the Weyl representation of the gamma matrices:The Pauli identity checks $C_0{\gamma }^{\mu T}{C}_0^{-1}=-{\gamma }^{\mu }$ directly, and ${\epsilon }^T=-\epsilon$ gives $C_0^T=-C_0$, with $C_0^{†}{C}_0=I$. Any other intertwiner differs from $C_0$ by a scalar: its ratio with $C_0$ commutes with all gamma matrices, whose products span the full four-by-four matrix algebra. A matrix commuting with all matrix units is scalar. Under a spinor basis change $S$, the intertwiner becomes $S{C}_0{S}^T$, whose transpose is still its negative. Therefore in every such four-component representationThis proves the antisymmetry of the four-dimensional charge-conjugation matrix, rather than leaving the sign undetermined by the general transpose argument.

Hermitian conjugating the charged Dirac equation, using real $A_{\mu },e,m$ and ${\gamma }^0{\gamma }^{\mu †}{\gamma }^0={\gamma }^{\mu }$, gives the adjoint Dirac equationTranspose it and multiply by $C$. Since $C{\gamma }^{\mu T}=-{\gamma }^{\mu }C$, the field ${\psi }^c=C{\bar{\psi }}^T$ obtained by charge conjugation obeysThus algebraic charge conjugation reverses the electromagnetic coupling.

If ${\gamma }_5\psi =\eta \psi$, $\eta =\pm 1$, Hermiticity of ${\gamma }_5$ and its anticommutation with ${\gamma }^0$ implyUsing the first transpose identity,Hence $\boxed{\bar{\psi }{\gamma }_5=-\eta \bar{\psi },{\gamma }_5{\psi }^c=-\eta {\psi }^c}$: algebraic charge conjugation reverses chirality. This concerns conjugating the already projected spinor, not the distinct operation of keeping a chiral projector outside an operator conjugation.

For the Grassmann variation of a chiral Majorana mass term, variations of odd fields must be moved past odd fields with a minus sign. If $A^T=-A$, componentwise reordering givesBoth $C$ and $C^{-1}$ are antisymmetric. Treating $\psi ,\bar{\psi }$ independently, integrating the kinetic variation by parts, and placing varied Grassmann fields on the left givesThe factors two from variation cancel the mass-term factors $1/2$. The first coefficient gives one equation; multiplying the second by $C$ gives the other:The variations can be restricted to their chiral subspaces: the two terms in each equation have the same opposite output chirality, so no inactive component equation has been imposed.

Applying $i/\partial$ to the first equation and substituting the second, using $(i/\partial {)}^2=-\square$, givesA phase rotation removes the phase of $m$, and combining the chiral field with its conjugate then gives a Majorana fermion of this mass, not two independent chiral species.

Let $\chi ={\varphi }^{\prime †}\Psi$ and $\bar{\chi }=\bar{\Psi }{\varphi }^{\prime }$. Question 1 shows they are gauge singlets, so both terms in ${\mathcal{L}}_G=G{\chi }^T{C}^{-1}\chi -G^{\ast }\bar{\chi }C{\bar{\chi }}^T$ preserve electroweak gauge invariance. The original PDF has $C^{-1}$ in the first term; the TeX aid incorrectly replaces it by $G^{-1}$. At the Higgs field vacuum,soComparing with the just-derived chiral mass equations yields the Majorana neutrino mass from the Weinberg operator:For nonzero $G$ this generates a mass without introducing a right-handed neutrino. Under lepton number rotation $\Psi \mapsto e^{i\alpha }\Psi$, the first operator acquires $e^{2i\alpha }$ and its conjugate acquires $e^{-2i\alpha }$. Thus the interaction violates lepton number by $\boxed{\Delta L=2}$. Each composite contains one scalar of mass dimension one and one fermion of dimension $3/2$, so the interaction is dimension five and $G$ has dimension $-1$. This is the Weinberg operator, an effective extension of the minimal renormalizable Standard Model, consistent with the absence of bare quadratic masses in Question 1.

Use the metric $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and the usual Dirac basis, with $({\gamma }^0{)}^{†}={\gamma }^0$ and $({\gamma }^i{)}^{†}=-{\gamma }^i$. In this solution the sign of the chirality matrix is the one specified in the paper, ${\gamma }^5=-i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$. Define $t_0=1$ and $t_i=-1$, with no summation in componentwise transformation formulas.

Complex conjugation changes the explicit $-i$ to $+i$. Three spatial gamma matrices each contribute another minus sign, soAlso $({\gamma }^5{)}^2=1$ and $\{{\gamma }^5,{\gamma }^{\mu }\}=0$ by the Clifford algebra.

The quantum time-reversal operator is antiunitary: it conjugates numerical coefficients, including the Dirac spinors and plane-wave exponentials in the mode expansion of a Dirac field. To fix the spin phase explicitly, put ${ϵ}_s=(-1{)}^{1/2-s}$ and useThis convention gives ${\hat{T}}^2=-1$ on a one-fermion state, since ${ϵ}_s{ϵ}_{-s}=-1$. In the transformed expansion, change variables from $(p,s)$ to $(p_T,-s)$. The Lorentz-invariant phase-space measure is unchanged, and $p_T\cdot x=-p\cdot x_T$. The coefficient of $b^s(p)e^{-ip\cdot x_T}$ is thereforeThe same calculation applies to the antiparticle coefficient. Thus, with this explicitly fixed spin convention,The spin phase in the Dirac time-reversal matrix matters here: a common change of the one-particle time-reversal phase replaces $B$ by $-B$; the phase-independent relation is $B\propto {\gamma }^{5}C$. It changes neither the defining conjugation property nor any bilinear result below. Normalize $B$ to be a unitary matrix. In the Dirac basis, ${\gamma }^0$ is real and commutes with $B$, so transforming the Dirac adjoint givesHere complex conjugation of the matrix defining the Dirac adjoint is essential. An explicit realization consistent with the sign of ${\gamma }^5$ used here is $C=i{\gamma }^2{\gamma }^0$ and $B={\gamma }^1{\gamma }^3$; it has $B^{-1}=-B$, so conjugation by $B$ and by $B^{-1}$ coincides. The usual changes of Dirac spinor basis by unitary matrices carry the adjoint and time-reversal matrix with them.

For the charge-conjugation matrix, the gamma matrix adjoint and transpose identities give ${\gamma }^{\mu T}=t_{\mu }{\gamma }^{\mu \ast }$. Consequently $B{\gamma }^{\mu T}{B}^{-1}={\gamma }^{\mu }$. Since $C={\gamma }^{5}B$ in the chosen phase convention,This proof uses a temporal Hermitian matrix and spatial skew-Hermitian matrices explicitly; the transpose identity is preserved when $C$ is transformed appropriately with the gamma matrices.

For the Fermi interaction, let $V_{pn}^{\mu }=\bar{p}{\gamma }^{\mu }n$ and $A_{pn}^{\mu }=\bar{p}{\gamma }^{\mu }{\gamma }^{5}n$. In the displayed Dirac basis, the inverse version of the conjugation relation also holds. Antiunitarity, $B^{-1}{\gamma }^{\mu \ast }B=t_{\mu }{\gamma }^{\mu }$ and $B^{-1}{\gamma }^{5\ast }B={\gamma }^5$ giveThe leptonic weak charged current has the same component signs. In the contraction of leptonic and hadronic currents, the two $t_{\mu }$ factors cancel. Its two independent operators thus keep their form while their coefficients become $g_V^{\ast }$ and $g_A^{\ast }$. The Hermitian-conjugate term transforms separately; its presence does not remove a relative complex phase between the vector and axial couplings.

With all intrinsic phases fixed to one and a real positive Fermi constant, invariance requires real coefficients in that convention. A common phase of $g_V$ and $g_A$ can instead be absorbed into a rephasing of the nucleon fields, and hence into their intrinsic time-reversal phases. The convention-independent condition, for $g_V\ne 0$, isThis is the relative weak phase condition for time reversal. If one coefficient vanishes, there is no relative phase to constrain; the remaining common phase can be removed. A nonreal ratio violates time-reversal symmetry even though the Lagrangian density includes its Hermitian conjugate.

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.