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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.

For an ideal infinitely long solenoid with interior magnetic field $B_0\hat{\mathbf{z}}$ and radius $R$, an axisymmetric Coulomb gauge representative of the magnetic vector potential isThis follows from $2\pi r{A}_{\varphi }=\Phi (r)$ by Stokes theorem. The exterior has zero curl but nonzero circulation around the enclosed magnetic flux. The Coulomb gauge condition alone does not eliminate every residual gauge transformation; symmetry and regularity select this customary representative.

For a spatial potential $\mathbf{A}$ with $\mathbf{E}=-\dot{\mathbf{A}}-\nabla A_0$, the Maxwell Lagrangian gives $\mathbf{\Pi }=-\mathbf{E}$ and ${\Pi }^0=0$. Its Hamiltonian is $\int [({\mathbf{\Pi }}^2+{\mathbf{B}}^2)/2+A_0\nabla \cdot \mathbf{\Pi }],d^{3}x$. Preserving the temporal momentum first-class constraint gives the Gauss law constraint in gauge theory. In free space, Coulomb gauge and decay at spatial infinity set $A_0=0$ and leave only transverse modes. The reduced Dirac bracket uses the projector ${\delta }_{ij}-k_i{k}_j/|\mathbf{k}{|}^2$, so quantization gives the transverse equal-time commutator and the positive Hamiltonian displayed above.

For a localized quantum harmonic oscillator along $z$, let the transverse magnetic vector potential be $A_z(t)=2A_0\cos (\theta -\omega t)$ for $t>0$, with zero potential before the switch. The Coulomb gauge form of minimal coupling gives $H_1=e{A}_z{p}_z/m$. The creation operator matrix element $⟨1|p_z|0⟩=i\sqrt{m\hslash {\omega }_0/2}$ means that only $|1⟩$ is reached at first order from the ground state. Its interaction picture amplitude isIn particular $F_0=t$, so resonant finite-time expressions are nonsingular. The induced electric dipole moment is $-2e\sqrt{\hslash /(2m{\omega }_0)}\mathrm{Re}(e^{-i{\omega }_{0}t}{b}_1)$. It contains a driven component with the dynamic electric polarizability and free-frequency transients required by the initial conditions. The discontinuity in $A_z$ gives an impulsive electric field and an initial mechanical velocity $e{A}_z(0^{+})/m$ even though the initial canonical momentum has zero mean.

The components of the momentum operator do not generally commute with a position-dependent magnetic vector potential. Expanding minimal coupling with the ordering intact givesActing on a wavefunction shows that $\mathbf{p}\cdot \mathbf{A}-\mathbf{A}\cdot \mathbf{p}=-i\hslash \nabla \cdot \mathbf{A}$. In the Coulomb gauge the divergence vanishes, so the two cross terms agree. Discarding the quadratic term in the weak radiation amplitude givesThe equality of the two operator orderings follows from the gauge condition, not from treating $\mathbf{p}$ and $\mathbf{A}$ as commuting numbers component by component.

For the requested oscillator response, treat $y$ as the fixed transverse position of a one-dimensional quantum harmonic oscillator along $z$, and put $\theta =ky$. The supplied radiation is transverse and already satisfies the Coulomb gauge. Its perturbation isThe annihilation operator and creation operator giveThus the only first-order transition out of the ground state is to $|1⟩$. More generally, the oscillator electric-dipole selection rule is $\Delta n=\pm 1$. This is a rule for matrix elements; exact energy conservation is replaced by a finite-time resonance profile during a switched finite-duration interaction.

To display both the resonant and nonresonant terms without spurious singularities, defineThe interaction picture coefficient in the given expansion is thenThe full Schrödinger picture coefficient of $|1⟩$ is $e^{-3i{\omega }_{0}t/2}{b}_1(t)$, while that of $|0⟩$ is $e^{-i{\omega }_{0}t/2}$ at this order. The normalization changes only at order $A_0^2$. In particular, at $\omega ={\omega }_0$ the first term is $C{e}^{i\theta }t$, finite for every finite time, and the perturbative result requires $|b_1|\ll 1$.

The fixed-$y$ interpretation is needed for the displayed target formulae. If $p^2$ were instead a three-dimensional kinetic operator with no transverse confinement, that model would have no normalizable transverse ground state, and $e^{\pm iky}$ would transfer transverse momentum $\pm \hslash k$. A treatment of that different model needs specified transverse states and recoil energies; they are absent from the question. The following parts use the intended localized-oscillator response.

For static fields, the Maxwell equations give $\nabla \times B={\mu }_{0}J$ and $\nabla \cdot B=0$. Write $B=\nabla \times A$. A gauge transformation $A\mapsto A+\nabla \chi$ leaves $B$ unchanged; choose $\Delta \chi =-\nabla \cdot A$ to impose the Coulomb gauge. ThenFor a localized conserved current with the potential chosen to vanish at infinity, the supplied fundamental solution gives the magnetic vector potentialIts divergence vanishes by integration by parts and $\nabla \cdot J=0$. For a filamentary loop the current integral becomes $I{\oint }_{L}dr$, soFor $R=|x|$ much larger than the fixed loop size, expand $|x-r{|}^{-1}=R^{-1}+(x\cdot r)R^{-3}+O(R^{-3})$. The leading constant term integrates to zero because $\oint dr=0$. ThereforeThe vector triple-product identity gives $x\times (r\times dr)=r(x\cdot dr)-(x\cdot r)dr$. Integrating the derivative of $(x\cdot r)r$ around the closed loop shows $\oint r(x\cdot dr)=-\oint (x\cdot r)dr$. ConsequentlyEquivalently, $A={\mu }_{0}m\times x/(4\pi R^3)+O(R^{-3})$, where $m=(I/2)\oint r\times dr$ is the magnetic dipole moment. The leading displayed potential is of order $R^{-2}$; the remainder is smaller by a factor $R^{-1}$.

For steady electromagnetic fields, Maxwell's equations give $\nabla \times \mathbf{B}={\mu }_0\mathbf{j}$ and $\nabla \cdot \mathbf{B}=0$. Introduce a magnetic vector potential with $\mathbf{B}=\nabla \times \mathbf{A}$ and the permitted Coulomb gauge $\nabla \cdot \mathbf{A}=0$. The vector identity for the double curl then gives $-\Delta \mathbf{A}={\mu }_0\mathbf{j}$. For localized currents and the decaying solution,Here $-\Delta (1/(4\pi |\mathbf{r}|))=\delta$ follows from harmonicity away from zero and the inward flux of the radial derivative through a small sphere, whose magnitude is one. It verifies the vector Poisson solution componentwise. Moreover steady electric charge conservation gives ${\nabla }^{\prime }\cdot \mathbf{j}=0$; integration by parts confirms the stated potential is divergence-free. Taking its curl, using $\nabla |\mathbf{r}-{\mathbf{r}}^{\prime }{|}^{-1}=-(\mathbf{r}-{\mathbf{r}}^{\prime })/|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3$, proves the Biot-Savart law:The zero-field condition at infinity selects the magnetic field produced by these currents, excluding an externally imposed homogeneous magnetic field.

At the origin write the source point as $r{\mathbf{e}}_r+z{\mathbf{e}}_z$. Its current is $kr{\mathbf{e}}_{\varphi }$, so the numerator of the Biot-Savart law is $k{r}^2{\mathbf{e}}_z-krz{\mathbf{e}}_r$. The radial term cancels after angular integration. Using $d{V}^{\prime }=rdrd\varphi dz$ givesThe inner integral is $2h/[r^2\sqrt{h^2+r^2}]$, as is verified by differentiating $z/[r^2\sqrt{r^2+z^2}]$. HenceFor $b,h>0$, its magnitude is ${\mu }_0|k|h(\sqrt{h^2+b^2}-h)$; it points along $+z$ for positive azimuthal current $k>0$, and along $-z$ for $k<0$. The limit $h\to \infty$ is ${\mu }_{0}k{b}^2/2$, agreeing with direct integration of the infinitely long cylindrical current distribution.

A steady localized electric current density satisfies $\nabla \cdot \mathbf{j}=0$, from charge conservation or the divergence of Ampère's equation. Choose an enclosing region whose boundary lies outside its support; extending the integration region does not change the source integrals. The divergence theorem applied componentwise givesTheir boundary fluxes vanish, proving the localized steady-current moment identitiesFor discontinuous but localized currents the same argument uses the distributional steady conservation law and an outer enclosing surface.

The magnetic vector potential is a vector field with $\mathbf{B}=\nabla \times \mathbf{A}$, consistent with $\nabla \cdot \mathbf{B}=0$. It is defined up to $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$. The steady Maxwell equation $\nabla \times \mathbf{B}={\mu }_0\mathbf{j}$ givesIn the Coulomb gauge $\nabla \cdot \mathbf{A}=0$, the equation is ${\nabla }^2\mathbf{A}=-{\mu }_0\mathbf{j}$, whose decaying Green-function solution is the one supplied in the source.

Let $r=|\mathbf{x}|$ and expand its kernel uniformly over the bounded source region:The first term integrates to zero. Put $T_{ki}=\int x_k^{\prime }{j}_i({\mathbf{x}}^{\prime })d{V}^{\prime }$, an antisymmetric tensor by the earlier identity. Define the magnetic dipole moment $m_{\ell }=\frac{1}{2}{ϵ}_{\ell ki}{T}_{ki}$, so $T_{ki}={ϵ}_{ki\ell }{m}_{\ell }$. ConsequentlyorThe displayed dipole term has size $r^{-2}$. If $\mathbf{m}=0$, it vanishes and a higher multipole determines the leading magnetic field instead.

For $r>0$, $\nabla \cdot (\mathbf{x}/r^3)=0$ andTherefore the far magnetic field is

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

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

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

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

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

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

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

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

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 Poincaré 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.