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If a differential operator $L$ is a linear map and has nonzero kernel of a linear map, the equation $LA=J$ determines $A$ only up to a homogeneous solution. Lorentz covariance of $L$ and $J$ therefore does not force independently selected solutions in different inertial frames to obey a particular tensor transformation law. A covariant Green function and consistently transformed homogeneous data do provide such a choice. For the four-potential in Lorenz gauge, $\square A^{\mu }=-{\mu }_0{J}^{\mu }$ still permits $A^{\mu }\mapsto A^{\mu }+{\partial }^{\mu }\chi$ with $\square \chi =0$, a residual Lorenz gauge symmetry. Constant spatial potentials for zero sources are a concrete example.

For the QED proper Feynman vertex with both external four-momenta zero, a massive internal fermion and a massless internal photon in Feynman gauge, the even numerator reduces under Lorentz covariance to ${\gamma }^{\mu }[\alpha k^2+\beta m^2]$. The gamma matrices identity ${\gamma }^{\nu }{\gamma }^{\mu }{\gamma }_{\nu }=(2-D){\gamma }^{\mu }$ and the integrated replacement $k_{\rho }{k}_{\sigma }\mapsto g_{\rho \sigma }{k}^2/D$ prove the coefficients. The term $m^2/[k^2(k^2-m^2+i0{)}^2]$ has neither an infrared divergence nor an ultraviolet divergence for $m>0$; its integral in four dimensions is $-i/(16{\pi }^2)$. Only the remaining squared massive denominator supplies the ultraviolet pole.

A local scalar nonlinearity preserves Lorentz covariance of the wave operator when the wave speed is one. If $f=F^{\prime }$, the equation follows from $\mathcal{L}=({\partial }_t\varphi {)}^2/2-({\partial }_x\varphi {)}^2/2+F(\varphi )$. The physical potential is $-F$, so its sign must be distinguished from the sign written in the Lagrangian.

Let $A_b^{\prime }=J^d{}_b{A}_d$, where $J^d{}_b=\partial x^d/\partial x^{\prime b}$. ThenThe second term prevents the ordinary derivative from being a generally covariant tensor. It vanishes for affine coordinate changes, including Cartesian Lorentz transformations, so the restricted claim of Lorentz covariance is valid. Antisymmetrization cancels the same term for arbitrary smooth coordinate changes and gives the exterior derivative of a one-form.

Lorentz-covariant tensor integration replaces $q_{\mu }{q}_{\nu }$ by $g_{\mu \nu }{q}^2/d$. The numerator consequently contributes $(d-1)k^2{q}^2/d$. The supplied master integral with powers $a=1,b=3$ givesInserting the corrected parameter integral givesThe pole uses $\Gamma (ϵ/2)=2/ϵ+O(1)$, $D^{-ϵ/2}=1+O(ϵ)$ and ${\int }_0^1(1-x)dx=1/2$. ThusThere is no momentum-independent pole at $\lambda =0$ in this transverse gauge after removing the photon tadpole. The two-point counterterm insertion is $-i(\delta Z_{\varphi }{k}^2+\delta M^2)$, so minimal subtraction requiresThese coefficients cancel the actual one-loop insertion derived from the action. Applying the misprinted weight in part e would give a wave-function pole four times too small. In a $d=4-2\epsilon$ convention the same answer is $\delta Z_{\varphi }=3e^2/(16{\pi }^2\epsilon )$. Finally, $\delta M^2=0$ does not mean that the bare mass parameter is unchanged: $Z_{\varphi }{m}_0^2=m^2$ gives $m_0^2-m^2=-m^2\delta Z_{\varphi }+O(e^4)$. Photon, charge and quartic counterterms require their respective Green functions and cannot be determined from this scalar two-point function alone.

Set the external four-momenta to zero after identifying the proper diagrams. In each box or triangle, at least one endpoint of the internal photon is a cubic vertex adjacent to an external scalar of zero momentum. That vertex is proportional to $l_{\rho }$, whereas the supplied numerator obeys $l^{\rho }{P}_{\rho \sigma }(l)=0$. All seven such graphs vanish algebraically. This use of the transverse numerator is stronger than merely saying that one has chosen a generic Lorenz gauge.

The two surviving seagull vertex bubbles have vertex factors $2i{e}_D^2{g}^{\mu \rho }$ and $2i{e}_D^2{g}^{\sigma \nu }$. Combining their two propagators givesThis scalar-QED four-point vertex pole in transverse gauge calculation determines the proportionality constant as well as the required tensor structure. Keep $m>0$ while extracting the ultraviolet pole. Lorentz covariance inside the regulated integral replaces $l^{\mu }{l}^{\nu }$ by $g^{\mu \nu }{l}^2/D$, giving the factor $(D-1)g^{\mu \nu }/D$.

For its scalar integral, Feynman parameter givesThere is no infrared divergence at $l=0$ with nonzero $m$. At pole order the two photon assignments and the factor $3/4$ therefore yieldThe common engineering scale factor is included in $e_D^2$; finite logarithms are immaterial to the pole. Adding $2i{e}_D^2{\delta }_4{g}^{\mu \nu }$ cancels it whenThis is minimal subtraction scheme, up to optional finite terms. It agrees with the transverse-gauge scalar QED wave-function pole and $Z_4=Z_2$. Setting $m=0$ at the outset would instead make the zero-momentum integral scaleless and obscure its separate ultraviolet divergence.

Keep $m>0$ and set $p=p^{\prime }=0$, which is an off shell choice. The numerator isTerms odd in $k$ integrate to zero. Lorentz covariance and dimensional regularization give, inside this integral,Applying the gamma matrices identity from part (ii) once more yieldsConsequentlyThis massive zero-momentum QED vertex integral uses an integrated tensor identity for the replacement of $N^{\mu }$, not an equality for each fixed $k$. The physical photon remains massless. The retained fermion mass supplies a scale and makes this zero-external-four-momentum calculation suitable for extracting the ultraviolet pole without a scaleless integral.

The integral printed in the text can be evaluated without reconstructing Figure 1. Put $A=l^2-m^2+i0$ and $B=(l+k{)}^2-m^2+i0$, and let $I^{\mu }$ denote that integral without its factor $\xi$. The numerator contraction responsible for its covariant gauge dependence isWith the scalar electrodynamics three-point vertex $-ie(p+p^{\prime }{)}^{\mu }$, three such vertices, two scalar propagators and one photon propagator have overall factor $(-ie{)}^3{i}^2(-i)=-e^3$. Thus the supplied integrand gives $V_{\mathrm{supplied}}^{\mu }=-e^3\xi I^{\mu }$.

First isolate the numerator term $l^2$. Its contribution isUnder $l\mapsto -l-k$ the two denominators exchange and the numerator changes sign. A shift-invariant dimensional regularization therefore gives $\boxed{I_0^{\mu }=0}$, including the finite part.

For the remaining numerator $2(l\cdot k)(2l+k{)}^{\mu }$, only its term $4l^{\mu }(l\cdot k)$ contributes to the ultraviolet pole. Corrections from expanding the denominators, and the numerator $2k^{\mu }(l\cdot k)$, are ultraviolet finite by power counting. Use auxiliary-mass extraction of ultraviolet poles with any $M>0$ rather than literally integrating a massless scaleless integral. Lorentz covariance then givesTODO (source gap): The algebra of the supplied expression is complete, but its graphical identification with Figure 1(b), including the routing at each internal vertex, cannot be visually verified from the unreadable panel.

The complete renormalization constant can nevertheless be obtained independently. Here is a short scalar QED wave-function pole calculation, to avoid assuming either missing panel's value. For an external four-momentum $p$, the exchange contribution to the scalar self-energy has numeratorIts denominator is $l^2[(p-l{)}^2-m^2+i0]$. The necessary ultraviolet poles, with the measure $d^{d}l/(2\pi {)}^d$, are respectivelyThe first two follow by a Feynman parameter and a shift of $l$; the third is a tadpole integral; the last follows by the same large-$l$ extraction used above. The seagull vertex contributes only a photon tadpole integral, which is a scaleless integral, and in any case cannot affect the coefficient of $p^2$. Multiplication by the exchange graph's factor $-e^2$ gives the scalar self-energy insertionThe kinetic counterterm $i{\delta }_2{p}^2$ therefore requires ${\delta }_2=e^2(3-\xi )C_{ϵ}$. The scalar QED Ward identity sets ${\delta }_1={\delta }_2$, soIndeed the full unrenormalized scalar electrodynamics three-point vertex has ultraviolet pole $i{e}^3(3-\xi )C_{ϵ}{k}^{\mu }$, which is cancelled by $-ie{\delta }_1{k}^{\mu }$. Subtracting the supplied contribution gives the gauge-independent sum stated in part (ii), without assigning either unreadable panel an integrand.

Adopt the Minkowski metric $\eta =\mathrm{diag}(-1,+1)$ and the Clifford algebra convention $\{{\gamma }_{\mu },{\gamma }_{\nu }\}=2{\eta }_{\mu \nu }I$. The source's Pauli matrices giveThey satisfy ${\gamma }_0^2=-I$, ${\gamma }_1^2=I$ and ${\gamma }_0{\gamma }_1=-{\gamma }_1{\gamma }_0=-{\sigma }_3$, which verifies the required Clifford algebra. Raising the indices gives ${\gamma }^0=-{\gamma }_0$ and ${\gamma }^1={\gamma }_1$.

Fix the sign of the rapidity $\theta$ by taking the Lorentz boostThe corresponding Spinor representation of the Lorentz group isFor example $S^{-1}{\gamma }^{0}S={\gamma }^0\cosh \theta -{\gamma }^1\sinh \theta$ and $S^{-1}{\gamma }^{1}S={\gamma }^1\cosh \theta -{\gamma }^0\sinh \theta$, establishing Lorentz covariance by $S^{-1}{\gamma }^{\mu }S={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$.

Thus the Dirac spinor transforms as ${\psi }^{\prime }(x^{\prime })=S(\theta )\psi (x)$, with infinitesimal lawThese variations compare the fields at corresponding spacetime points. At a fixed coordinate one adds the usual orbital term $-\delta x^{\mu }{\partial }_{\mu }\psi$. Reversing the rapidity convention reverses both boost signs consistently. Infinitesimal transformations describe the proper orthochronous Lorentz group.

Use natural units and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. Multiplication of the block gamma matrices givesThe Pauli matrices obey ${\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j=2{\delta }_{jk}{I}_2$, so the spatial anticommutators are $-2{\delta }_{jk}{I}_4$, the mixed ones are zero, and the temporal one is $2I_4$. ThusBecause partial derivatives commute, only the symmetric part of ${\gamma }^{\mu }{\gamma }^{\nu }$ survives in their contraction. Multiplying the Dirac equation on the left by its opposite-mass factor therefore gives the flat-space Dirac factorizationHence every component satisfies $\boxed{(\square +m^2)\psi =0}$, the Klein-Gordon equation. This implication does not remove the first-order spinor constraints imposed by the Dirac equation.

For Lorentz covariance, transform coordinates by $x^{\prime }=Lx$ and the Dirac spinor by ${\psi }^{\prime }(x^{\prime })=S(L)\psi (x)$. The chain rule gives ${\partial }_{\mu }^{\prime }=(L^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. ConsequentlyThe cancellation uses $L^{\mu }{}_{\rho }(L^{-1}{)}^{\nu }{}_{\mu }={\delta }^{\nu }{}_{\rho }$. Thus the Lorentz transformation maps every solution to another, establishing the Lorentz covariance of the Dirac operator.

The infinitesimal identity follows directly from the Clifford algebra. Moving ${\gamma }^{\rho }$ past the two factors givesSubtracting the same identity with $\mu ,\nu$ interchanged proves the spinor Lorentz generator commutatorContracting with the antisymmetric tensor ${\omega }_{\mu \nu }$ yieldsIt also verifies $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }+O({\omega }^2)$ for the infinitesimal Spinor representation of the Lorentz group.

For a positive-energy four-momentum, put $\psi =u(p)e^{-ip\cdot x}$, $E=\sqrt{{\mathbf{p}}^2+m^2}>0$. The Dirac equation becomes $(/p-m)u=0$, where the Feynman slash notation uses $/p={\gamma }^{\mu }{p}_{\mu }=E{\gamma }^0-\mathbf{\gamma }\cdot \mathbf{p}$. Splitting $u=(\xi ,\eta {)}^T$ givesThe second equation sets $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\xi /(E+m)$, and the first is then the mass shell, since $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2={\mathbf{p}}^2{I}_2$. Choosing two orthonormal two-spinors ${\chi }_s$ gives the Dirac plane waves in the standard representation:The negative-frequency solutions are independentlywith orthonormal ${\eta }_s$. They are the antiparticle modes after field quantization. The normalization gives $u_r^{†}{u}_s=v_r^{†}{v}_s=2E{\delta }_{rs}$ and ${\overline{u}}_r{u}_s=2m{\delta }_{rs}$, ${\overline{v}}_r{v}_s=-2m{\delta }_{rs}$, using the Dirac adjoint $\overline{u}=u^{†}{\gamma }^0$.

For a spatial rotation, ${\sigma }^{ij}={ϵ}_{ijk}\mathrm{diag}({\sigma }_k,{\sigma }_k)$. Hence the intrinsic rotation generators areFor a massive particle in its rest frame, the two independent spinors therefore carry the two-dimensional spin one-half representation, with $J_3$ eigenvalues $\pm 1/2$ and Casimir $s(s+1)=3/4$. A $2\pi$ rotation acts as $-I$, the spinor sign under a full spatial rotation. Thus the particle has spin $\boxed{s=1/2}$; four spinor components encode particle and antiparticle sectors, not four spin states of one particle. In the massless limit the corresponding physical labels are helicities $\pm 1/2$ rather than a rest-frame spin basis.

Vary the Maxwell Lagrangian with respect to $A_{\nu }$. Since $\delta F_{\mu \nu }={\partial }_{\mu }\delta A_{\nu }-{\partial }_{\nu }\delta A_{\mu }$ and $F$ is antisymmetric,After integration by parts, the Euler-Lagrange field equation isThe gauge transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$ leaves $F_{\mu \nu }$ unchanged because partial derivatives commute. This gauge invariance means that potentials related in this way represent the same electromagnetic field; all four potential components are not independent physical photon degrees of freedom. In particular, the ungauged kinetic operator is singular and its temporal canonical momentum vanishes.

Choose the Lorenz gauge ${\partial }_{\mu }{A}^{\mu }=0$. The field equation then becomes $\boxed{\square A_{\mu }=0}$. Residual gauge transformations with $\square \chi =0$ preserve this condition. The correctly named Lorenz gauge condition concerns the divergence of the potential; Lorentz covariance is the separate transformation property of the equations.

For the alternative quadratic Lagrangian density, direct differentiation givesso its Euler-Lagrange field equation is also $\square A^{\nu }=0$. This density is the Feynman-gauge Maxwell kinetic density after a boundary-term subtraction. Indeed, with $D={\partial }_{\mu }{A}^{\mu }$,The divergence identity follows by expanding ${\partial }_{\mu }{K}^{\mu }$ and commuting its second derivatives. The quadratic density alone describes four wave equations; the gauge constraint must still be incorporated in the physical states. It is not an unconstrained four-positive-polarization replacement for Maxwell equations.

Using this specified boundary-term convention, the canonical momentum conjugate to $A_{\nu }$ isThe canonical commutation relations are thereforewith field-field and momentum-momentum brackets zero. Equivalently $[A_{\mu },{\dot{A}}_{\nu }]=-i{g}_{\mu \nu }{\delta }^3$. The temporal component has the opposite sign to the spatial components; replacing the metric by a positive Euclidean one would destroy these brackets.

The wave equation gives massless frequencies ${\omega }_{\mathbf{k}}=|\mathbf{k}|$. Hermiticity pairs its positive- and negative-frequency solutions, so the Heisenberg picture field has the expansionThis is a real field expansion for each component, but its oscillator metric is indefinite. The assumed brackets, with the dagger's vector index restored, areThey verify the photon oscillator completeness and canonical brackets. Inserting ${\Pi }^{\nu }=-{\dot{A}}^{\nu }$ gives at equal timesThe other brackets cancel between the two frequency contributions. Thus both the $2\omega$ normalization and the sign $-g_{\mu \nu }$ are needed.

The Gupta-Bleuler formalism imposes the Lorenz gauge condition weakly on states:Here $(+)$ is the annihilation, positive-frequency part. Setting the full divergence to zero as an operator would conflict with the four-component canonical commutation relations. The subsidiary condition instead selects a physical pre-space inside the covariant photon Fock space, followed by a quotient of its null states.

Apply it to $|k,ϵ⟩={ϵ}^{\nu }{a}_{\nu }^{†}(k)|0⟩$. Commuting $k^{\mu }{a}_{\mu }$ through the creator gives a multiple of $-k_{\nu }{ϵ}^{\nu }|0⟩$. Hence the allowed photon polarization vectors obeyTake $k=(\omega ,0,0,\omega )$. The complete solution iswith complex coefficients. The last two vectors are transverse linear photon polarizations, and $(0,1,\pm i,0)/\sqrt{2}$ gives the two circular photon polarizations. The first vector is $k/\omega$, the allowed temporal-longitudinal combination. A purely spatial longitudinal vector $(0,0,0,1)$ by itself does not satisfy the subsidiary condition.

The oscillator brackets give, apart from the positive continuum normalization factor,Momentum eigenstates are distributionally normalized; a box or a wave packet makes the norm statement literal. The allowed longitudinal combination has zero norm because $k^2=0$, and it is orthogonal to every constrained state since $k\cdot ϵ=0$. It also gives a pure-gauge field wave: the field strength of $A_{\mu }\propto k_{\mu }{e}^{-ik\cdot x}$ vanishes. Thus $ϵ$ and $ϵ+ck$ represent the same physical photon polarization. The Gupta-Bleuler null-state quotient removes this null direction and leaves two positive-norm transverse photon photon polarizations. The negative-norm temporal oscillator of the unrestricted space is not an extra physical state. This is the transverse one-photon physical quotient.

Finally verify the two-photon fermion Ward identity directly for electron-positron annihilation into two photons. Define $D(r)=/r-m$ and the reduced Dirac propagatorThe equality follows from the Clifford algebra, since $(/r-m)(/r+m)=(r^2-m^2)I$. The incoming Dirac spinors obey $D(p)u(p)=0$ and $\overline{v}(q)(/q+m)=0$. On replacing one inserted photon polarization vector by $k$, the spinor factor in the two diagrams isAt the electron end, $/k=D(p)-D(p-k)$, soFor the other diagram, four-momentum conservation gives $p-k^{\prime }=k-q$. HenceThe two contributions therefore cancel exactly:Individual diagrams need not vanish; their sum implements the Ward identity. Linearity then gives invariance under $ϵ\mapsto ϵ+ck$, as required by gauge invariance. In particular, the null longitudinal photon polarization proportional to $k$ has zero physical production amplitude. Physical scattering produces the transverse photon classes, not an independent longitudinal massless photon.

Use the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. Multiplying the block gamma matrices and using the Pauli matrix multiplication law givesThus these matrices satisfy the Clifford algebra relation $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }{I}_4$. Since partial derivatives commute, the antisymmetric part of ${\gamma }^{\mu }{\gamma }^{\nu }$ drops out of its contraction with ${\partial }_{\mu }{\partial }_{\nu }$. Multiplying the Dirac equation on the left by $i{\gamma }^{\mu }{\partial }_{\mu }+m$ givesTherefore every component of the Dirac spinor obeys the Klein-Gordon equation with mass $m$.

For Lorentz covariance, take $x^{\prime }=Lx$ and ${\psi }^{\prime }(x^{\prime })=S(L)\psi (x)$, with constant $L$ and $S$. The chain rule gives ${\partial }_{\mu }^{\prime }=(L^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. The given intertwining relation then impliesThis proves invariance of the Dirac equation under the joint coordinate and spinor transformation.

The infinitesimal identity follows directly from the spinor Lorentz generator commutator. To derive it, first move ${\gamma }^{\rho }$ through a two-matrix product using the Clifford algebra:Subtract the expression with $\mu$ and $\nu$ interchanged and multiply by $i/2$. ThusWriting $S=1-\frac{i}{4}{\sigma }^{\mu \nu }{\omega }_{\mu \nu }$ gives $S^{-1}=1+\frac{i}{4}{\sigma }^{\mu \nu }{\omega }_{\mu \nu }+O({\omega }^2)$, and henceThe last step uses ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$, including the metric signs in raising the first index.

For parity symmetry in quantum field theory, set $y=x_P=(t,-\mathbf{x})$. The time derivative of $\psi (y)$ is unchanged and each spatial derivative changes sign. Also ${\gamma }^0$ commutes with itself and anticommutes with every ${\gamma }^j$. ConsequentlyThis verifies the specified parity symmetry in quantum field theory action explicitly.

For charge conjugation, put $B=i{\gamma }^2$, so ${\psi }_C=B{\psi }^{\ast }$. In the Dirac representation of the gamma matrices, ${\gamma }^0,{\gamma }^1,{\gamma }^3$ are real and ${\gamma }^2$ is imaginary. The Clifford algebra therefore gives $B{\gamma }^{\mu \ast }{B}^{-1}=-{\gamma }^{\mu }$ for all four indices. Complex conjugation of the Dirac equation, followed by multiplication by $B$, yieldsThis is the antilinear charge-conjugation intertwiner property. No electromagnetic background is being held fixed in this free-field calculation.

Apply the specified parity symmetry in quantum field theory action before charge conjugation. Since ${\gamma }^0$ is real and $B{\gamma }^0=-{\gamma }^{0}B$,Thus the induced parity action on the charge-conjugate spinor has the opposite sign. In particular, if a spinor has intrinsic parity $\eta =\pm 1$, its charge conjugate has intrinsic parity $-\eta$. Equivalently, at rest the positive-frequency Dirac spinors occupy the upper two components and have ${\gamma }^0$ eigenvalue $+1$, whereas the negative-frequency Dirac spinors occupy the lower two components and have eigenvalue $-1$. The corresponding particle and antiparticle creation operators inherit these opposite signs under the specified parity convention. Hence electrons and positrons have opposite intrinsic parities; conventionally one assigns $+1$ to the electron and $-1$ to the positron. The common real overall parity sign may be reversed without changing this relative sign, as in opposite intrinsic parities of a Dirac particle and antiparticle.

The derivative convention uses hypercharge Y with $Q=T_3+Y$. For each lepton family $l=e,\mu ,\tau$, the left-handed fields form an SU(2) doublet $L_l=({\nu }_l,l{)}_L^T$, with $Y=-1/2$. Their weak isospin values are $T_3=+1/2,-1/2$ and their electric charges are $0,-1$. The right-handed charged lepton $l_R$ is an SU(2) singlet, with $T_3=0$, $Y=-1$ and charge $-1$. The minimal model used here has no right-handed neutrino. A sterile right-handed neutrino, if added as a singlet with $Y=0$, would have no charged weak current interaction. Antiparticles carry the conjugate representations and opposite electric charges.

Expanding ${\bar{L}}_{l}i{\gamma }^{\mu }{D}_{\mu }{L}_l$ supplies $-g{\bar{L}}_l{\gamma }^{\mu }{A}_{\mu }^a{\sigma }_a{L}_l/2$. The off-diagonal entries of $A^a{\sigma }_a$ are $\sqrt{2}{W}^{+}$ and $\sqrt{2}{W}^{-}$. With the chiral projector $P_L=(1-{\gamma }^5)/2$, one has ${\bar{\nu }}_L{\gamma }^{\mu }{l}_L=\bar{\nu }{\gamma }^{\mu }{P}_{L}l$. Thus the two W boson interactions areFor quarks, separate rotations diagonalize the up-type and down-type mass terms. If their left-handed rotations are $U_{uL}$ and $U_{dL}$, the charged interaction contains the CKM matrix $V=U_{uL}^{†}{U}_{dL}$:There is an implicit sum over color charge. This quark mixing permits transitions between different families, rather than just the paired fields in a single doublet.