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For $q=(q^0,\mathbf{q})$ with $\mathbf{q}\ne 0$, the Coulomb gauge propagator hasThe temporal component is instantaneous; the spatial part projects onto transverse photon polarizations. Although these components are not individually Lorentz covariant, their contraction with conserved external currents equals the covariant answer. Zero-spatial-momentum limits should be taken after combining the current contractions.

For $q\cdot J=q\cdot K=0$, the instantaneous term in the Coulomb-gauge photon propagator combines with its spatial longitudinal subtraction:The identity follows from $\mathbf{q}\cdot \mathbf{J}=q^0{J}^0$ and $q^2=(q^0{)}^2-|\mathbf{q}{|}^2$. It is equality after contraction with conserved currents, not equality of the uncontracted propagator tensors. It is especially useful in showing that tree scattering amplitudes computed in Coulomb gauge are Lorentz invariant.

The conjugate complex scalar field has the opposite charge. Thus the derivative on ${\varphi }^{\ast }$ in the kinetic term must mean $(D_{\mu }\varphi {)}^{\ast }=({\partial }_{\mu }-ie{A}_{\mu }){\varphi }^{\ast }$, while the derivative on $\varphi$ is $D_{\mu }={\partial }_{\mu }+ie{A}_{\mu }$. This is the conjugate gauge covariant derivative. Applying the same plus-charge differential operator to both fields would not give the stated gauge-invariant theory. For example, take $A=0$, $\varphi =x^0$ and $\alpha =c{x}^0$; the literal plus-charge kinetic product changes from $1$ to $1+2iec{x}^0$. The opposite-charge interpretation removes this term.

For an arbitrary real function $\alpha (x)$, takeThen $D_{\mu }^{\prime }{\varphi }^{\prime }=e^{-ie\alpha }{D}_{\mu }\varphi$ and the conjugate derivative transforms oppositely. Also $F_{\mu \nu }^{\prime }=F_{\mu \nu }$ because mixed derivatives commute. The kinetic contraction and $|\varphi {|}^2$ are unchanged, proving local U(1) gauge symmetry.

Expanding the covariant kinetic term identifies the interactions:The first term gives the scalar electrodynamics three-point vertex, and the second gives the seagull vertex. For scalar charge flow from an incoming particle of momentum $p$ to an outgoing one of momentum $p^{\prime }$, the factors areThe factor two in the second rule comes from the two identical photon fields. Equivalently, with all momenta incoming, the three-point rule is $ie(r-q{)}^{\mu }$ for a $\varphi$ leg of momentum $q$ and a ${\varphi }^{\ast }$ leg of momentum $r$. An antiparticle line has the opposite charge-flow sign. The vertices are shown with dashed scalar lines and wavy photon lines:

For particle-antiparticle scattering, use incoming $p_1,p_2$ and outgoing $p_3,p_4$, with $p_1,p_3$ particle momenta. At order $e^2$, there are exactly two tree-level Feynman diagrams: $t$-channel photon exchange and $s$-channel annihilation. The seagull vertex has only two scalar legs and cannot alone provide four external scalar legs.

Set $q_t=p_1-p_3=p_4-p_2$ and $q_s=p_1+p_2=p_3+p_4$. The external currents areAll are transverse to the corresponding exchanged momentum: for instance $q_t\cdot J_t=p_1^2-p_3^2=0$ and $q_s\cdot J_s=p_1^2-p_2^2=0$. This is the on-shell scalar quantum electrodynamics Ward identity. With the vertex convention above,The relative sign comes from the opposite particle/antiparticle charge in the exchange diagram and the two equal annihilation-vertex signs.

To see how the Coulomb gauge expression becomes covariant, let $J,K$ be either pair, $q=(q^0,\mathbf{q})$, and initially take $|\mathbf{q}|\ne 0$. The displayed photon propagator givesCurrent conservation gives $\mathbf{J}\cdot \mathbf{q}=q^0{J}^0$ and the same identity for $K$. Since $q^2=(q^0{)}^2-|\mathbf{q}{|}^2$, the temporal coefficient simplifies asThereforeThe common Feynman pole prescription is understood in these formulas. This Coulomb-gauge propagator between conserved currents identity proves equality of the physical amplitudes despite the noncovariant individual propagator components. It should be formed before taking limits such as $\mathbf{q}\to 0$: individual instantaneous and transverse terms can be undefined separately in that limit while their sum has a finite covariant limit away from a physical pole.

Substitution gives the final scattering amplitudeHere $s,t,u$ are the Mandelstam variables for the equal-mass external scalars. This is scalar particle-antiparticle tree scattering; overall external-state phases do not affect its relative channel sign.

A symmetry preserves the action, possibly up to a boundary term. In a classical field theory, this connects transformations of fields to conserved quantities. In a quantum field theory, it also constrains states, observables and scattering amplitudes. Continuous symmetries have Noether currents; discrete symmetries can be equally important without supplying a current through the same theorem.

For a local density depending on fields ${\Phi }_a$ and their first derivatives, define ${\Pi }_a^{\mu }=\partial \mathcal{L}/\partial ({\partial }_{\mu }{\Phi }_a)$ and ${\mathcal{E}}_a=\partial \mathcal{L}/\partial {\Phi }_a-{\partial }_{\mu }{\Pi }_a^{\mu }$. Suppose a one-parameter transformation has fixed-coordinate variation $\delta {\Phi }_a=\epsilon R_a$ and satisfies $\delta \mathcal{L}=\epsilon {\partial }_{\mu }{K}^{\mu }$. The Noether's first theorem asserts thatFor Grassmann-valued fields the derivatives and ordering are chosen consistently; the same integration-by-parts argument applies. To prove the theorem, expand the density variation and integrate its derivative term once algebraically:Comparison with the assumed divergence gives ${\partial }_{\mu }{j}^{\mu }=-{\sum }_a{\mathcal{E}}_a{R}_a$. On shell every Euler-Lagrange expression vanishes. Integrating this continuity equation over space gives the Noether charge $Q=\int j^0{d}^{3}x$ and $dQ/dt=-{\int }_{\partial {\mathbb{R}}^3}\mathbf{j}\cdot d\mathbf{S}$. Thus $Q$ is conserved whenever the boundary flux vanishes. Boundary conditions are part of the conservation statement; the local identity alone does not discard nonzero flux.

Translations illustrate the theorem in every relativistic field theory. For $R_a=a^{\nu }{\partial }_{\nu }{\Phi }_a$ and $K^{\mu }=a^{\mu }\mathcal{L}$, the current is $a^{\nu }{T}^{\mu }{}_{\nu }$, whereThe four charges are energy and momentum. For the Klein-Gordon field they reduce to the expressions derived in Question 1. Lorentz invariance gives angular momentum and boost currents. With a symmetric improved stress-energy tensor, these are $M^{\lambda \mu \nu }=x^{\mu }{T}^{\lambda \nu }-x^{\nu }{T}^{\lambda \mu }$. The canonical Dirac expression also carries an intrinsic spin current; improvement incorporates it into the symmetric tensor. Together translations and Lorentz transformations form the Poincare group.

Internal symmetries act on field components without moving spacetime points. The global phase symmetry of a complex scalar field, $\delta \varphi =-i\epsilon \varphi$, hasIts charge distinguishes particles and antiparticles. A Dirac phase transformation $\delta \psi =-i\epsilon \psi$ gives $j^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$. This is preserved by the real-scalar Yukawa interaction, because the phases of $\bar{\psi }$ and $\psi$ cancel; consequently fermion number is conserved even while scalar and fermion particles interact. A multiplet of real scalars with a potential depending only on ${\sum }_a{\varphi }_a^2$ has orthogonal internal rotations. Their antisymmetric currents can be written $j_{ab}^{\mu }={\varphi }_a{\partial }^{\mu }{\varphi }_b-{\varphi }_b{\partial }^{\mu }{\varphi }_a$.

Discrete examples include $\varphi \mapsto -\varphi$ for an even real-scalar potential, parity symmetry in quantum field theory, and charge conjugation, which interchanges a complex scalar with its conjugate and reverses the electromagnetic potential. Discrete transformations are not generated by an infinitesimal continuous parameter, so Noether theorem does not attach a local conserved current to each of them. Nevertheless they forbid interaction terms and relate physical processes.

A global symmetry in field theory uses parameters constant across spacetime and can act nontrivially on physical states. A gauge redundancy allows arbitrary spacetime-dependent parameters and relates descriptions of the same physical configuration. In scalar electrodynamics, replacing the scalar's global phase by a local phase requires the compensating gauge transformation of $A_{\mu }$ and the gauge covariant derivative; the invariant contraction uses the opposite representation for ${\varphi }^{\ast }$. In Yang-Mills theory, matter transforms in a representation of a non-Abelian gauge group and the gauge potential transforms so that covariant derivatives and field strengths transform covariantly. Gauge fixing, such as Coulomb gauge, chooses representatives of these descriptions rather than changing physical predictions. The current-contracted propagator identity of Question 3 illustrates why a noncovariant gauge choice leaves scattering Lorentz invariant.

The arbitrary local parameter produces a stronger statement than one independent conserved charge for each function. For the abelian scalar transformation, vary the action with an arbitrary compactly supported $\alpha (x)$ and integrate its derivative term:ThereforeThis consequence of Noether second theorem, the Noether identity for abelian scalar gauge symmetry, holds off shell: gauge invariance makes the field equations dependent. In Hamiltonian language the same redundancy is reflected by the Gauss law constraint in gauge theory and the elimination of unphysical gauge degrees of freedom. Gauge transformations that vanish at the boundary are redundancies; transformations with nontrivial boundary behaviour can instead carry physical surface charges. This is why identifying every gauge transformation with a zero-charge operation would be too strong.

Finally, symmetry of the action need not imply symmetry of a chosen vacuum. For a complex scalar with a symmetry-preserving potential whose minima occur at nonzero $|\varphi |$, different constant phases label degenerate vacua. Choosing one breaks the global phase symmetry of the state; the Goldstone theorem supplies a massless mode under its usual relativistic assumptions. When the same phase is gauged, the Higgs mechanism describes the phase degree of freedom becoming the longitudinal polarization of a massive gauge boson; the local redundancy is still present in the underlying description. At the quantum level a classical symmetry can also fail through a quantum anomaly, so a classical Noether derivation alone does not guarantee an exact quantum Ward identity. Conserved charges, physical global transformations and local gauge redundancy are related but distinct consequences of symmetry.

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.

For the commonly used covariant form, add the gauge fixing term $-({\partial }_{\mu }{A}^{\mu }{)}^2/(2\xi )$. The resulting Fourier-space kinetic operator isThe inversion of the gauge-fixed Maxwell kinetic operator gives $K^{\mu \nu }{D}_{\nu \rho }^F=i{\delta }^{\mu }{}_{\rho }$. In Feynman gauge, $\xi =1$, the photon propagator isEquivalently, $\square D_{\mu \nu }^F=i{\eta }_{\mu \nu }{\delta }^4(x-y)$ with the vacuum pole prescription. For general $\xi$, the momentum-space numerator is ${\eta }_{\mu \nu }-(1-\xi )k_{\mu }{k}_{\nu }/(k^2+i0)$.

A covariant canonical realization uses four polarization oscillators with $[a_r,a_s^{†}]=-(2\pi {)}^3{\eta }_{rs}{\delta }^3(\mathbf{k}-{\mathbf{k}}^{\prime })$. The resulting indefinite inner product is auxiliary. In Gupta-Bleuler quantization, impose $({\partial }_{\mu }{A}^{\mu }{)}^{(+)}|\mathrm{phys}⟩=0$ and take the Gupta-Bleuler null-state quotient. The scalar-longitudinal combination is thereby removed from the physical state space, leaving the same two transverse photon states. Thus the four-component Feynman-gauge numerator does not imply four physical polarization states.

The identity divergence of a curl is zero follows by the symmetry of second partial derivatives and antisymmetry of the Levi-Civita symbol. ThereforeIf $\mathbf{B}=\nabla \times \mathbf{A}$ and $\mathbf{A}$ is in Coulomb gauge, the curl of the curl identity gives $\mathbf{J}=-{\nabla }^2\mathbf{A}$. Apply the supplied scalar Poisson equation representation to each component of the decaying vector potential. In the units of this problem the Coulomb-gauge vector potential of a localized steady current isHere one needs $\mathbf{A}=O(|\mathbf{x}{|}^{-1})$, $\nabla \mathbf{A}=O(|\mathbf{x}{|}^{-2})$ and enough localization of $\mathbf{J}$ for the stated representation and differentiations. A smooth compactly supported current density is a sufficient convenient case. The local differential equations and gauge alone do not impose these conditions: $\mathbf{A}=(-y,x,0)$, $\mathbf{B}=(0,0,2)$, $\mathbf{J}=0$ satisfy them but do not satisfy the displayed integral formula. The formula selects the decaying solution, excluding nondecaying homogeneous harmonic functions.

With $\mathbf{r}=\mathbf{x}-\mathbf{y}$, the gradient of the kernel is ${\nabla }_x(1/|\mathbf{r}|)=-\mathbf{r}/|\mathbf{r}{|}^3$. Since $\mathbf{J}(\mathbf{y})$ is independent of $\mathbf{x}$,Thus taking the curl gives the Biot-Savart law in the problem's normalization:Finally, change variables $\mathbf{z}=\mathbf{x}-\mathbf{y}$ in the integral for $\mathbf{A}$. This keeps the singular kernel independent of the differentiation variable:For smooth localized $\mathbf{J}$, differentiation under the integral sign is legitimate: $1/|\mathbf{z}|$ is locally integrable in three dimensions, and on a bounded neighbourhood of $\mathbf{x}$ the translated sources and their derivatives have common compact support. The integral indeed satisfies Coulomb gauge.

For a time-independent current, the Maxwell equations reduce to magnetostatics:Write $\mathbf{B}=\nabla \times \mathbf{A}$ using a vector potential and choose the Coulomb gauge $\nabla \cdot \mathbf{A}=0$. The vector identity for the curl of a curl gives the Poisson equationThe free-space Green function therefore givesTaking the curl with respect to $\mathbf{r}$, moving it under the integral, and using $\nabla (1/|\mathbf{r}-{\mathbf{r}}^{\prime }|)=-(\mathbf{r}-{\mathbf{r}}^{\prime })/|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3$ gives the Biot-Savart law

The magnetostatic Ampère-Maxwell equation is $\nabla \times B={\mu }_{0}J$. Substituting $B=\nabla \times A$ and using the Coulomb gauge $\nabla \cdot A=0$ givesThe free-space Green function of the Laplacian therefore givesFor the thin wire current stated in the question this becomes

The magnetic field is $B=\nabla \times A$. The magnetostatic Ampère-Maxwell equation is $\nabla \times B={\mu }_{0}J$, so the curl of the curl identity givesA gauge transformation $A\mapsto A+\nabla \chi$ does not change $B$. We may therefore choose the Coulomb gauge $\nabla \cdot A=0$ by solving the Poisson equation ${\nabla }^2\chi =-\nabla \cdot A$. It follows component by component in a Cartesian coordinate system thatThe free-space Green function of the Laplacian then givesFor a localized steady current density, conservation of electric charge gives ${\nabla }^{\prime }\cdot J=0$; integration by parts confirms directly that this integral has zero divergence and hence obeys the Coulomb gauge.

For a free electromagnetic field with suitable boundary conditions, combine Coulomb gauge ${\partial }_i{A}_i=0$ with $A_0=0$. Gauss's law makes the second condition consistent in the source-free problem. The remaining field operator is spatially transverse and describes the two radiative photon polarizations. A constrained scalar potential must be restored when charges are present.