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The covariant free-photon oscillator construction uses four polarization vectors and $[a^{\lambda },a^{{\lambda }^{\prime }†}]=-{\eta }^{\lambda {\lambda }^{\prime }}$. Starting from a positive Fock vacuum, it induces an indefinite Hermitian form on the multiparticle state space: the temporal oscillator has negative norm while the three spatial oscillators have positive norm. It is therefore not the physical positive Hilbert space. The Gupta-Bleuler null-state quotient selects a positive physical space with two transverse photon polarizations. Continuum momentum oscillators and their states are understood after smearing or finite-volume regularization.

Choose $C_{\mathbf{p}}=a_{\mathbf{p}}^0-a_{\mathbf{p}}^3$ with the standard temporal and longitudinal polarization vectors. The physical pre-space is the common kernel of all $C_{\mathbf{p}}$. In each regulated mode, $a^0$ acts on creator polynomials as $-{\partial }_{a^{0†}}$ and $a^3$ as ${\partial }_{a^{3†}}$, so the constraint kernel consists of transverse creator polynomials and polynomials in $C^{†}=a^{0†}-a^{3†}$. Since $[C,C^{†}]=0$, these latter excitations remain constrained. They are orthogonal to every constrained state because $⟨\Phi |C^{†}=⟨C\Phi |=0$. Quotienting this radical of a Hermitian form leaves only the transverse bosonic Fock space with a positive inner product. The constraint alone gives a positive semidefinite Hermitian form; quotienting removes its null directions. If starting from finite-particle creator polynomials, take the Hilbert space completion of this positive quotient to obtain the physical Hilbert space.

For $x=m_W^2/m_H^2<1/4$, the on-shell longitudinal pair contributes the displayed tree-level partial decay width. In the Higgs boson rest frame, its polarization vectors have contraction $(m_H^2/2-m_W^2)/m_W^2$. Multiplying by the Higgs boson coupling to W bosons gives $|{\mathcal{M}}_{LL}{|}^2=m_H^4(1-2x{)}^2/v^2$; the two-body Lorentz-invariant phase space then gives the width. The sum over transverse polarizations adds $8x^2$ to the longitudinal polynomial. Below threshold, physical decays through virtual W bosons require a different multiparticle calculation.

A free Proca field is consistent and has three physical particle polarizations. The difficulty for generic interactions is the ultraviolet behavior of its Proca propagator: the longitudinal numerator makes part of it approach order $1/m^2$, rather than decay as $1/k^2$. Longitudinal external polarization vectors also behave as ${\epsilon }_L^{\mu }=k^{\mu }/m+O(m/E)$. Without suitable identities and cancellations, amplitudes grow with energy and perturbative counterterms of increasing dimension are needed. Conserved-current Abelian models and the Stueckelberg mechanism are important exceptions; a massive interacting vector is not intrinsically inconsistent as an effective field theory.

For a Dirac field annihilating with its antiparticle into two massless gauge bosons, the two ordered fermion-exchange diagrams leave a colour matrix commutator when one outgoing momentum contracts the scattering amplitude. The three-boson exchange diagram cancels this current term but leaves a term proportional to the other outgoing momentum. With $D=\partial -igA$, $J_c^{\lambda }=\bar{v}{\gamma }^{\lambda }{t}^{c}u$, $q=k_1+k_2$, and the source's positive-sign three-boson vertex, the tree result is $k_{2\nu }{\mathcal{M}}^{ab\mu \nu }=i{g}^2{f}^{abc}(J_c\cdot k_2)k_1^{\mu }/q^2$. It vanishes after contraction with a physical polarization vector on leg 1, although the open-index tensor is generally nonzero.

Set $s=(p_1+p_2{)}^2=m_h^2$, so $p_1^2=p_2^2=0$ and $p_1\cdot p_2=s/2$. The physical polarization vectors obey ${ϵ}_1\cdot p_1={ϵ}_2\cdot p_2=0$. Write $X_{\mu \nu }\simeq Y_{\mu \nu }$ when they agree after contraction with these polarization vectors; terms proportional to $p_{1,\mu }$ or $p_{2,\nu }$ may then be discarded.

The gamma matrix trace identities kill the terms with an odd number of gamma matrices. The remaining three two-momentum terms and the term with three factors of $m$ areIn the $g_{\mu \nu }$ coefficient the terms linear in $l\cdot p_i$ cancel, leaving $m^2-l^2-s/2$. Among the other terms, the surviving mixed products of $l$ and $p_i$ also cancel. ConsequentlyIn the second Feynman diagram, exchange the external labels and change the dummy loop momentum $l\mapsto -l$. Its projected numerator and denominator become identical to those of the first Feynman diagram. HenceWith the common factor $8m$ extracted, the primitive integer coefficients areEquivalently, extracting only $m$ gives $(A,B,C)=(32,8,8)$ and prefactor $i{N}_c{\lambda }_t{e}^2{e}_t^2{m}_t/\sqrt{2}$. A common rescaling of all three coefficients changes the prefactor inversely. The physical scattering amplitude is ${ϵ}_1^{\mu }{ϵ}_2^{\nu }{\mathcal{M}}_{\mu \nu }^{\prime }$; the projected integrand is not an identity for an arbitrary uncontracted off-shell tensor.

Use Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$ and the convention $S_{fi}={\delta }_{fi}+i(2\pi {)}^4{\delta }^{(4)}(p_f-p_i){\mathcal{M}}_{fi}$. Put $m=m_t$, $y={\lambda }_t/\sqrt{2}$, $Q=e_t=2/3$ andThe Feynman i-epsilon prescription is understood in subsequent denominators. Contracting the spinor indices around the fermion loop gives the traceHere $/p={\gamma }^{\rho }{p}_{\rho }$ is Feynman slash notation. The closed fermion loop sign is $-1$, the three interaction vertices contribute $(-iy)(-ieQ{)}^2$, and the three Dirac propagators contribute $i^3$. Their product, including the color charge trace, is $-N_{c}y{e}^2{Q}^2$. Thus one ordering hasMultiplying by the incoming polarization vectors and adding the exchanged Feynman diagram givesThe displayed four-dimensional Feynman integral is formal until regularized; subsequent manipulations use dimensional regularization, which permits the required loop momentum shifts.

Continue the loop momentum to $D=4-\epsilon$ dimensions, keeping $\mathrm{tr}1=4$ and the external physical polarization vectors four dimensional. Introduce Feynman parameters $x,y\ge 0$, $x+y\le 1$, denoting this integration triangle by $\mathcal{T}$. Completing the square givesOdd terms in $k$ integrate to zero. Lorentz invariance of dimensional regularization gives $k_{\mu }{k}_{\nu }\mapsto g_{\mu \nu }{k}^2/D$. The general projected numerator therefore becomes, under the Feynman integral,Only the $k^2$ term can give an ultraviolet divergence. Define, with the conventional scale factor implicit,The gamma function formula givesThus the pole of the numerator integral, including $2{\int }_{\mathcal{T}}dxdy=1$, isThe coefficients $(4,1,1)$ satisfy this condition. A nonzero local divergence would have to be removed by a counterterm. The Ward identity forbids a local $h{A}_{\mu }{A}^{\mu }$ term, and the first electromagnetic gauge-invariant coupling with one Higgs boson and two photons has the form $h{F}_{\mu \nu }{F}^{\mu \nu }$, of engineering dimension five. It is not an independent counterterm of the Standard Model, a renormalizable quantum field theory. The top quark contribution is separately electromagnetic gauge-invariant; no additional W boson loop Feynman diagram is needed to cancel this pole.

The finite remainder must still be retained. For $A=4C$,Putting $D=4$ before evaluating the divergent Feynman integral would incorrectly set this finite contribution to zero.

Use the Feynman parameters and $J_3$ defined in Solution. With $A=4C$, the finite term just found is $-C\Delta J_3$. Combining it with the remaining $g_{\mu \nu }$ coefficient givesThe other coefficient is $(B-4Cxy)J_3$. Thus, writing ${\mathcal{I}}_{\mu \nu }$ for the Feynman integral without the overall prefactor,For $T_{\mu \nu }=\frac{s}{2}{g}_{\mu \nu }-p_{2,\mu }{p}_{1,\nu }$, the two coefficients must have the corresponding ratio. The mismatch is simply $(B-C)J_3$, independent of $xy$. Therefore the required relation iswhich is satisfied. In the primitive normalization the result isand the full top quark scattering amplitude, in our stated $S$ convention, isIndeed $p_1^{\mu }{T}_{\mu \nu }=p_2^{\nu }{T}_{\mu \nu }=0$: replacing either polarization vector by its photon momentum makes the scattering amplitude vanish. This is the Ward identity, expressing electromagnetic gauge invariance. The result also is symmetric under exchange of the two photons. The finite term in Solution is essential for this Ward identity.

As an independent low-energy check, for $s/m^2\to 0$ the Feynman parameter integral isThe coefficient is consequently $-N_{c}y{e}^2{Q}^2/(6{\pi }^{2}m)$ at leading order. It has the inverse-mass dimension required for an effective $h{F}_{\mu \nu }{F}^{\mu \nu }$ coupling. If $y=m/v$, this coefficient stays finite in the large-$m$ limit at fixed $v$; that limit increases the Yukawa coupling, so it is not a fixed-coupling heavy-field decoupling limit.

For the outgoing massless gauge bosons, take $k_1^2=k_2^2=0$. Work away from internal poles, with the Feynman i-epsilon prescription understood. The Dirac equations and four-momentum conservation giveMultiplying by the neighboring Dirac propagator numerator cancels its denominator. The two fermion-exchange diagrams therefore yield the matrix commutatorThe annihilation vector current obeys $q_{\lambda }{J}_c^{\lambda }=\bar{v}(/p_1+/p_2)t^{c}u=0$. Direct contraction of the three-boson tensor givesThusThe terms proportional to $J_c^{\mu }$ cancel, leaving the non-Abelian Ward identity for fermion annihilationThe open-index tensor is generally nonzero: neither ${ϵ}^{abc}$ nor $J_c\cdot k_2$ is constrained to vanish. For a concrete nonzero configuration, take centre-of-momentum frame kinematics $p_1=(E,0,0,p)$, $p_2=(E,0,0,-p)$, $p=\sqrt{E^2-m^2}$, $k_1=(E,E,0,0)$, $k_2=(E,-E,0,0)$. Standard Dirac spinors with two-component spin labels $\chi =(1,0{)}^T$ for $u$ and $\eta =(0,1{)}^T$ for $v$ give the spinor current $(0,2E,2iE,0)$ and its contraction with $k_2$ is $2E^2$. Choosing both colour spinors $(1,0{)}^T$ and $(a,b,c)=(1,2,3)$ multiplies this by $1/2$, still nonzero. This explicitly rules out an unconditional tensor identity $k_{2\nu }{\mathcal{M}}^{ab\mu \nu }=0$.

For a physical polarization vector ${\epsilon }_{1\mu }^{\ast }$, however, $k_1\cdot {\epsilon }_1^{\ast }=0$, and thereforeThis is the required decoupling of a pure-gauge polarization vector on leg 2 when the other leg is physical. Interchanging the legs gives the corresponding identity for leg 1. The role of the three-boson diagram is explicit: it cancels the noncommuting colour-current term left by the two fermion exchanges.

The non-Abelian counterpart of the photon Ward identity is a Slavnov-Taylor identity, derived from BRST symmetry. A scattering amplitude with all other external states physical vanishes when one external massless gauge boson polarization vector is replaced by its four-momentum. An amputated tensor with the remaining indices open need not be transverse: part d gives its explicit nonzero longitudinal remainder.

Consequently, amplitudes involving several unphysical covariant polarization vectors need not vanish. For example, contracting the result of part d with a timelike vector $n_{\mu }$ gives $i{g}^2{ϵ}^{abc}(J_c\cdot k_2)(n\cdot k_1)/s$, nonzero in the explicit configuration there for $n=(1,0,0,0)$. Even with other states physical, the pure-gauge vector $k$ is the relevant null combination; separate timelike and spatial longitudinal basis vectors are not themselves pure gauge, and their separate amplitudes need not be zero.

These auxiliary components do not define separately observable positive probabilities. Physical states are classes in BRST cohomology, and the S-matrix preserves that physical state space in this anomaly-free vectorlike gauge spectrum. There is no physical transition from a physical state to a nonphysical state. The incorrect stronger statement would be that every covariant component called “unphysical” has a zero matrix element.

For loop Feynman diagrams or unitarity cuts, summing covariant gauge indices with $-g_{\mu \nu }$ propagates those auxiliary components. In a Non-Abelian gauge theory, their cancellation requires the interacting Faddeev-Popov ghost fields as well as the gauge boson diagrams. Ghost contributions carry their Grassmann field loop signs and restore the Slavnov-Taylor identities and physical unitarity. One may instead use a formulation with only physical propagating polarization vectors, but one cannot simply discard the ghost diagrams while retaining the Feynman gauge rules from part a.

Use natural units and the Minkowski metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$. Varying the Proca action with respect to $A_{\nu }$ gives $\partial \mathcal{L}/\partial A_{\nu }=m^2{A}^{\nu }$ and $\partial \mathcal{L}/\partial ({\partial }_{\mu }{A}_{\nu })=-F^{\mu \nu }$. The Euler-Lagrange equation is thereforeTaking a divergence annihilates the antisymmetric field strength, since commuting derivatives are symmetric in their indices. For $m\ne 0$ this gives the Lorenz constraint in Proca theory ${\partial }_{\nu }{A}^{\nu }=0$, and hence $(\square +m^2)A^{\nu }=0$. The constraint follows from dynamics; it is not an additional gauge fixing condition for this massive theory.

For four-momentum $p^{\mu }=(E,0,0,p)$, longitudinal polarization of a massive vector boson lies in the time–$z$ plane, has $p\cdot \epsilon =0$ and is normalized by $\epsilon \cdot \epsilon =-1$. Takingsatisfies both requirements because $E^2-p^2=m^2$. It can also be obtained by boosting the rest-frame unit spatial polarization along the direction of motion. Together with two transverse polarizations it gives the three physical spin states of a massive spin one particle.

In the Higgs boson rest frame, choose $p^{\mu }=(E,0,0,p^{\ast })$ and $p^{\prime \mu }=(E,0,0,-p^{\ast })$, with $E=m_H/2$ and $p^{\ast 2}=m_H^2/4-m_W^2$. Consistent normalized longitudinal polarization vectors areTheir Lorentz contraction is $(p^{\ast 2}+E^2)/m_W^2=(2p^{\ast 2}+m_W^2)/m_W^2$. The Higgs boson coupling to W bosons consequently givesUsing the two-body Lorentz-invariant phase space, $p^{\ast }=m_H\sqrt{1-4x}/2$ and $v^{-2}=\sqrt{2}{G}_F$ yields the longitudinal partial decay widthThe charged W bosons are distinguishable, so there is no identical-particle factor of $1/2$. This expression includes the entire angular phase space but only the specified longitudinal final polarizations. The Higgs decay to two W bosons summed over all polarizations instead has the factor $1-4x+12x^2$; the additional transverse contribution is $8x^2$. At $m_H\gg m_W$ the longitudinal channel dominates and the leading width grows as $G_F{m}_H^3$.

Use a mostly-plus target metric tensor and the standard open-string mode expansion convention ${\alpha }_0^{\mu }=\sqrt{2{\alpha }^{\prime }}{p}^{\mu }$. The integer string oscillator level is $N={\sum }_{n\ge 1}{\alpha }_{-n}\cdot {\alpha }_n$, whose eigenvalues are nonnegative integers. Then $L_0=N+{\alpha }^{\prime }{p}^2$, so the string mass-shell condition givesThe ground state is a tachyon, and the level-one state is massless. The physical string state conditions in old covariant string quantization areOnly the positive-mode Virasoro constraints annihilate a ket; their negative-mode partners annihilate the corresponding bra. Imposing both signs on every ket would contradict the nonzero Virasoro central extension.

The central charge comes from normal ordering the quadratic oscillator expressions. Classically their Poisson brackets have no central term. Quantum mechanically, moving annihilation operators through creation operators produces double contractions. From the oscillator commutator one obtains $[L_m,{\alpha }_n^{\mu }]=-n{\alpha }_{m+n}^{\mu }$, which gives the noncentral part $(m-n)L_{m+n}$.

The coefficient is fixed by the free-boson Virasoro central term. On a formal zero-momentum vacuum, for $m>0$,The two contraction pairings giveHere ${\eta }_{\mu \nu }{\eta }^{\mu \nu }=D$; the time coordinate contributes one, just as each spatial coordinate does. Comparing with the Virasoro algebra provesThis counts covariant target coordinates. It is distinct from the $D-2$ physical transverse polarizations. Including the central charge of reparameterization ghosts gives $c_{\mathrm{total}}=D-26$; cancellation of the worldsheet Weyl anomaly gives the usual critical dimension of the bosonic string $D=26$.

A general level-one state isThe zero-mode constraint requires $k^2=0$. Since $[L_1,{\alpha }_{-1}^{\mu }]={\alpha }_0^{\mu }$, the remaining nontrivial positive-mode constraint is $k\cdot ϵ=0$; all $L_m$ with $m\ge 2$ act trivially at this level. Thus the polarization vector is transverse to a nonzero null momentum.

There is also a null string state $L_{-1}|0;k⟩=\sqrt{2{\alpha }^{\prime }}k\cdot {\alpha }_{-1}|0;k⟩$. It is physical when $k^2=0$ and orthogonal to every physical polarization. The null-state quotient of a string therefore identifiesChoose $k=(E,E,0,\dots ,0)$. Transversality sets ${ϵ}^1={ϵ}^0$, and the null-state identification removes this common component. The remaining $D-2$ spatial components have positive norm. Hence the physical level-one states form a massless vector with $D-2$ polarizations, or 24 at the critical dimension. The removed component is gauge redundancy. States violating the constraints, including independent timelike negative-norm polarizations, are unphysical; physical null states are removed by the quotient rather than counted as additional particles.

Vacuum Maxwell's equations are invariant under spatial rotations. Rotate the preceding coordinate system so that ${\mathbf{e}}_z$ becomes the propagation direction $\hat{\mathbf{k}}$, and rotate the transverse electric direction to a unit polarization vector $\hat{\mathbf{n}}$. Dot and cross products transform covariantly under this rotation, so transversality and the field relation are unchanged. With $\omega =c|\mathbf{k}|$,These complex fields encode real fields through their real parts. The displayed form describes linear electromagnetic polarization; a general transverse complex amplitude can describe elliptical or circular electromagnetic polarization as well.

Denote the momentum in the printed mode expansion by ${\tilde{\pi }}^{\nu }$, as it equals $-{\dot{A}}^{\nu }$ for the boundary-subtracted density discussed in part (b). Take a real complete basis of polarization vectors. Its completeness relation with one index raised isHere raising or lowering the polarization labels uses the same diagonal matrix $\eta$, so this is just the completeness relation in the PDF. Write $\mathbf{r}=\mathbf{x}-\mathbf{y}$. In the mixed field-momentum commutator, the $a{a}^{†}$ term gives the product of the minus sign in $\tilde{\pi }$'s creation part and the minus sign in the oscillator commutator. The $a^{†}a$ term gives the same sign. The two mode normalization factors multiply to $1/2$, and the momentum delta function removes one integral. ConsequentlyFor the field-field commutator, the polarization sum gives insteadbecause the integrand is odd under $\mathbf{p}\mapsto -\mathbf{p}$. Similarly,These are the requested canonical commutation relations, derived by photon oscillator completeness and canonical brackets.

They also hold for the literal momenta of part (b). The shifts areSince the fields commute at equal times, their mixed brackets with the momenta do not change. The only potentially new momentum bracket isAll other momentum brackets vanish separately. Thus both canonical conventions have $[A_{\mu },{\pi }^{\nu }]=i{\delta }_{\mu }{}^{\nu }{\delta }^3$ and vanishing coordinate-coordinate and momentum-momentum brackets, even though the momentum operators differ by spatial derivatives. This resolves the density/expansion mismatch without changing the printed oscillator algebra.

The oscillator commutator givesThe temporal mode $\lambda =0$ has negative norm. More precisely, the nonzero wave packet $|f,0⟩=\int d^{3}pf(\mathbf{p})a_{\mathbf{p}}^{0†}|0⟩/(2\pi {)}^3$ has norm $-\int d^{3}p|f(\mathbf{p}){|}^2/(2\pi {)}^3<0$. The difficulty is an indefinite Hermitian form, not the divergent normalization of an unsmeared momentum eigenstate. The covariant photon Fock space is therefore not a positive physical Hilbert space.

Gupta-Bleuler quantization imposes only the annihilation, or positive-frequency part of a quantum field, of the divergence:A strong operator identity $\partial \cdot A=0$ is incompatible with the four unconstrained canonical commutation relations; in the literal momentum convention it would set ${\pi }^0=0$. The weaker condition instead implies $⟨\Phi |\partial \cdot A|\Psi ⟩=0$ between physical states: the positive-frequency part annihilates the ket, and its adjoint annihilates the bra.

For a nonzero momentum choose upper-index polarization vectorswhere ${\mathbf{e}}_1,{\mathbf{e}}_2$ are an orthonormal pair perpendicular to $\mathbf{p}$. Labels 1 and 2 have transverse polarization, label 3 has spatial longitudinal polarization, and label 0 has timelike photon polarization. With $p^0=|\mathbf{p}|=\omega$, the divergence of the annihilation part is proportional to $-i\omega (a_{\mathbf{p}}^0-a_{\mathbf{p}}^3)$. The physical condition is thereforeFor a one-mode state $|\chi ⟩={\sum }_{\lambda }{c}_{\lambda }{a}^{\lambda †}|0⟩$, the condition reads $-c_0-c_3=0$. ThusThe last direction has zero norm and is orthogonal to every constrained state. Its field wavefunction is proportional to ${ϵ}^{\mu 0}+{ϵ}^{\mu 3}=p^{\mu }/\omega$, hence it is a pure gauge symmetry direction. A purely longitudinal creator $a^{3†}$ alone is not physical, despite having positive norm.

For the multiparticle sketch, put $C=a^0-a^3$ and $C^{†}=a^{0†}-a^{3†}$. The commutator $[C,C^{†}]=0$ allows polynomials in $C^{†}$ times transverse creator states to obey the constraint. Conversely, on creator polynomials the constraint is $-{\partial }_{a^{0†}}-{\partial }_{a^{3†}}$, so these are all the constrained polynomials. Every term containing $C^{†}$ is orthogonal to every constrained bra, since $⟨\Phi |C^{†}=⟨C\Phi |=0$. Quotienting those null vectors and completing the remaining positive space is the Gupta-Bleuler null-state quotient. It givesThe one-particle instance is the transverse one-photon physical quotient.

There are two closely related objects to distinguish. Inverting the quadratic Proca action gives the usual covariant Proca propagator. After integration by parts, its kernel isThe matrix inverse with the Feynman i-epsilon prescription givesMultiplication by $K^{ab}$ gives $i{\delta }_b^a$ in the distributional limit. The numerator agrees with the polarization sum for a massive vector boson at the poles, but is not a transverse linear projection at arbitrary four-momentum.

For the literal canonical time-ordered product of $A_a$ and $A_b$, the nondynamical component $A_0$ produces the Proca time-ordering contact term. In the chosen time coordinate the full answer isTo see the local term directly, the three physical polarization vectors give ${\sum }_{\lambda }{ϵ}_{\lambda 0}^2=|\mathbf{q}{|}^2/m^2$. The canonical $00$ two-point correlation function therefore has Fourier transformBy contrast, ${\eta }_{00}+q_0^2/m^2=|\mathbf{q}{|}^2/m^2-(q^2+m^2)/m^2$, so the covariant expression contains an additional $i/m^2$. The mixed and spatial components have no additional contact term. ThusIf $T$ is used to mean covariant time ordering, commonly denoted $T^{\ast }$, the conventional answer is instead just ${\Delta }^{\mathrm{cov}}$. Both conventions have the same propagating poles and agree away from coincidence; explicitly separating them respects the printed definition as an ordinary time-ordered product.

Put $q^a=(E,0,0,k)$ with $E=\sqrt{k^2+m^2}$. An orthonormal real basis of polarization vectors isEach polarization vector obeys $q_a{ϵ}_{\lambda }^a=0$. The Minkowski metric gives ${ϵ}_{\lambda }\cdot {ϵ}_{{\lambda }^{\prime }}={\delta }_{\lambda {\lambda }^{\prime }}$, since $E^2-k^2=m^2$. The first two are transverse to the spatial momentum; the longitudinal polarization of a massive vector boson is spatially longitudinal but still orthogonal to the full four-momentum. The polarization sum for a massive vector boson isIn the rest frame all three polarization vectors are spatial unit vectors. Their three independent positive-norm states are the physical spin states of a massive spin-one particle.

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.