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Use the supplied oscillator expansion, with the boundary-term convention for momentum stated in part (b). Use a real polarization vector basis, as in the supplied unconjugated polarization completeness relation. Raise its second field index to obtainThe mixed commutator contains only the annihilation-creation and creation-annihilation terms. Their signs are both positive after combining the minus sign in the momentum expansion with the negative Minkowski metric oscillator commutator. Setting $\mathbf{r}=\mathbf{x}-\mathbf{y}$ and using the momentum Dirac delta distribution givesHere the last equality is the Fourier representation of the Dirac delta function. Similarly,because the integrand is odd under $\mathbf{p}\mapsto -\mathbf{p}$. The momentum-momentum commutator is proportional to the same odd difference, now weighted by $|\mathbf{p}|/2$, and also vanishes. Therefore the equal-time canonical commutation relations areThese are identities of operator-valued distributions, understood after smearing. The extra indices in the TeX polarization relation are transcription defects; the PDF has the ordinary two-polarization completeness contraction used above. The direct momenta of part (b) also satisfy the canonical commutation relations after their boundary-generated shift, although they do not have the unmodified mode expansion printed here.

The oscillator-generated covariant photon Fock space has an indefinite Hermitian form, not a positive Hilbert space inner product. For a normalizable one-photon wave packet of polarization $\lambda$, its squared norm is proportional to $-{\eta }^{\lambda \lambda }\int d^{3}p|f(\mathbf{p}){|}^2$. Thus $\lambda =0$ gives a negative-norm photon state. The divergent ${\delta }^3(0)$ of an unsmeared momentum eigenstate is a separate normalization issue, avoided by the wave packet.

Choose contravariant polarization vectors ${ϵ}^{0\mu }=(1,0)$, ${ϵ}^{3\mu }=(0,\hat{\mathbf{p}})$, and two ${ϵ}^{r\mu }=(0,{\mathbf{e}}_r)$ with ${\mathbf{e}}_r\cdot \mathbf{p}=0$ for $r=1,2$. The third spatial polarization is longitudinal polarization; the first two are transverse polarization. The timelike photon polarization is distinct from the longitudinal one.

The Gupta-Bleuler quantization condition sets the divergence of the positive-frequency part of a quantum field to zero on physical states:Restore the factors $e^{-i|\mathbf{p}|t}$ to the annihilation terms. Since $p_{\mu }{ϵ}^{0\mu }=|\mathbf{p}|$ and $p_{\mu }{ϵ}^{3\mu }=-|\mathbf{p}|$, Fourier transform gives the equivalent conditionIts sign depends on the chosen sign of the longitudinal polarization vector; the covariant condition does not.

To see its content for a general Fock state, temporarily discretize momentum and decompose one unphysical oscillator sector as $|\Psi ⟩={\sum }_{n,m\ge 0}|n,m{⟩}_{0,3}\otimes |{\chi }_{nm}{⟩}_{\perp }$, where $|n,m⟩=(a^{0†}{)}^n(a^{3†}{)}^m|0⟩/\sqrt{n!m!}$. The temporal oscillator obeys $a^0|n,m⟩=-\sqrt{n}|n-1,m⟩$, while $a^3|n,m⟩=\sqrt{m}|n,m-1⟩$. The condition therefore becomesEquivalently, the allowed finite-particle states use the transverse creation operators and only $b^{†}=a^{0†}-a^{3†}$ in the unphysical sector. Indeed the constraint acts on a polynomial of the two unphysical creation operators as $-{\partial }_{a^{0†}}-{\partial }_{a^{3†}}$, whose kernel consists of polynomials in their difference. Also $[b,b^{†}]=-1+1=0$, and a state containing $b^{†}$ is orthogonal to every constrained state because $b$ annihilates every such state. This argument applies mode by mode and extends by smearing to continuum momentum.

For one photon, $c_0{a}^{0†}|0⟩+c_3{a}^{3†}|0⟩$ is constrained only when $c_3=-c_0$, producing a null state. The Gupta-Bleuler null-state quotient removes these null directions. The condition excludes negative-norm physical states; quotienting its null states leaves the two positive-norm transverse photon polarizations. The condition alone gives a positive semidefinite Hermitian form, not yet a positive definite Hilbert space.

Take ${\tau }^a={\sigma }^a/2$, $Y_{\varphi }=1/2$, metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, and a Higgs potential with ${\mu }_h^2>0$, $\lambda >0$. The gauge-scalar electroweak interaction isHere $B_{\mu \nu }={\partial }_{\mu }{B}_{\nu }-{\partial }_{\nu }{B}_{\mu }$. With the printed plus-sign convention for $D_{\mu }$, define $[D_{\mu },D_{\nu }]=ig{W}_{\mu \nu }^a{\tau }^a+i{g}^{\prime }{Y}_{\varphi }{B}_{\mu \nu }$; thus $W_{\mu \nu }^a={\partial }_{\mu }{W}_{\nu }^a-{\partial }_{\nu }{W}_{\mu }^a-g{ϵ}^{abc}{W}_{\mu }^b{W}_{\nu }^c$. The minus sign in the nonlinear gauge field strength follows from this convention.

The minima have ${\varphi }^{†}\varphi =v^2/2$, with $v^2={\mu }_h^2/\lambda$. A gauge transformation rotates the vacuum to $(0,v{)}^T/\sqrt{2}$. In unitary gauge, the three angular Goldstone bosons are removed, leaving $\varphi =(0,v+H{)}^T/\sqrt{2}$ and one real Higgs boson. The generator $Q=T^3+Y$ annihilates the vacuum, because its lower component has $T^3=-1/2$ and hypercharge $1/2$. This identifies the unbroken electromagnetic gauge group.

The quadratic terms from the Higgs field kinetic term areDefine the Weinberg angle and physical fields byThe neutral combinations areThe neutral gauge-boson mass matrix is $(v^2/4)\left(\begin{array}{cc}{g}^2& -g{g}^{\prime }\\ -g{g}^{\prime }& g^{\prime 2}\end{array}\right)$. It has eigenvalues zero and $(g^2+g^{\prime 2})v^2/4$, with the zero eigenvector giving $A_{\mu }$. Expanding the Higgs potential about its minimum gives $V=\text{constant}+\lambda v^2{H}^2+\lambda v{H}^3+\lambda H^4/4$. Therefore the tree-level electroweak gauge-boson masses and scalar mass areAlso $m_W=m_Z\cos {\theta }_W$ and $e=g\sin {\theta }_W=g^{\prime }\cos {\theta }_W$. The three removed Goldstone bosons supply the longitudinal polarization vector degrees of freedom of the massive electroweak gauge bosons.

Replacing $v^2$ by $(v+H{)}^2$ in the neutral mass term gives all the Higgs boson couplings to Z bosons in unitary gauge:There are exactly the cubic $HZZ$ and quartic $HHZZ$ elementary tree-level Feynman diagrams. Differentiating with respect to the identical fields gives Feynman rules $2i{m}_Z^2{g}_{\mu \nu }/v$ and $2i{m}_Z^2{g}_{\mu \nu }/v^2$, respectively. There is no elementary $ZHH$ vertex for the real neutral radial field.

The bosonic annihilation operator $a^{\lambda }(p)$ removes a vector particle of four-momentum $p$ and polarization vector ${\epsilon }^{\mu ,\lambda }(p)$. The bosonic creation operator $c^{\lambda †}(p)$ creates its antiparticle with the same four-momentum and polarization label. The polarization vector describes the corresponding spin-one mode expansion of a free field.

Because charge conjugation acts through a unitary operator, it leaves the numerical polarization vector coefficients unchanged. Applying the specified charge conjugation to the operators givesComparison with the Hermitian conjugation of the original expansion therefore givesThe factor ${\eta }_C$ is the intrinsic charge-conjugation phase.

A Feynman diagram specifies a term in the perturbative scattering amplitude, not a measured microscopic trajectory. Assign external spinors or polarization vectors, an interaction factor to each Feynman vertex and a propagator to each internal line; impose four-momentum conservation and integrate over independent loop momenta. Sum amplitudes for indistinguishable alternatives, then square, sum final spins and average unpolarized initial spins. A decay rate uses $d\Gamma =|\mathcal{M}{|}^{2}d{\Phi }_n/(2M)$ in natural units; a cross section also divides by incident flux. interference means probabilities of individual diagrams must not simply be added.

The electromagnetic fermion vertex is $f\bar{f}\gamma$, with coupling proportional to $e{Q}_f{\gamma }^{\mu }$; it preserves fermion flavour. There is no elementary three-photon or four-photon vertex in quantum electrodynamics. The weak charged current has $W\ell \nu$ and $W{u}_i{d}_j$ vertices, with a left-chiral projector and a Cabibbo-Kobayashi-Maskawa matrix element at the quark vertex. The weak neutral current has $Zf\bar{f}$ vertices with vector and axial couplings; it includes neutrinos and is flavour diagonal at tree level. Electroweak gauge interactions also include $WW\gamma$, $WWZ$ and quartic $WWWW$, $WWZZ$, $WWZ\gamma$ and $WW\gamma \gamma$ vertices. Higgs-fermion and Higgs-vector interactions complete the mass-related scalar sector.^[ref]

Quantum chromodynamics has $q\bar{q}g$ vertices proportional to $g_s{\gamma }^{\mu }{T}^a$, plus three-gluon and four-gluon Feynman vertices because gluons carry color charge. A quark-gluon vertex preserves quark flavour. Internal particles need not be on their mass shells, but electric charge and four-momentum are conserved at each vertex. The figures below show representative leading flavour topologies; binding and hadronization are implicit, and additional strong-interaction contributions can modify hadronic amplitudes.

With outgoing physical photon polarization vectors, the Feynman rule givesThe overall phase does not affect the decay width. Replacing either polarization vector by its own four-momentum gives zero by antisymmetry. Thus the Ward identity permits replacement of each physical photon polarization sum by $-{\eta }^{\mu \nu }$ inside this amplitude. The printed sum is not an equality of free-standing physical two-polarization tensors: for a photon moving along $z$, its physical $00$ component is zero whereas $-{\eta }_{00}=-1$. Gauge-dependent longitudinal terms cancel only after contraction with the transverse amplitude.

The contraction of the two Levi-Civita symbols givessince $q_1^2=q_2^2=0$ and $2q_1\cdot q_2=m_a^2$. No initial spin average is needed for the spin-zero axion. The result is independent of direction in its rest frame. Including the identical-particle factor explained in solution therefore givesOmitting $1/2!$ would overestimate the physical rate by two. The normalization agrees with the standard axion-photon convention.^[ref]

With real complete polarization vectors, the covariant photon Fock space oscillator commutators and ${\sum }_{\lambda ,{\lambda }^{\prime }}{ϵ}_{\mu }^{\lambda }{ϵ}^{\nu {\lambda }^{\prime }}{\eta }_{\lambda {\lambda }^{\prime }}={\delta }_{\mu }{}^{\nu }$ imply the equal-time canonical commutation relations $[A_{\mu }(\mathbf{x}),-{\dot{A}}^{\nu }(\mathbf{y})]=i{\delta }_{\mu }{}^{\nu }{\delta }^3(\mathbf{x}-\mathbf{y})$. The two mixed creation-annihilation terms add; the two coordinate-coordinate or momentum-momentum terms cancel after their polarization sums. The sign of the time-component oscillator is essential to the identity. Boundary-induced canonical transformation in Feynman gauge explains the momenta for the unsubtracted density.

Two orthonormal transverse polarization vectors span the plane perpendicular to a nonzero momentum. Summing their dyadic products gives the spatial transverse projector. It has trace and rank two and acts as the identity on physical photon polarizations.

The sum over the two physical transverse photon polarization vectors is not the full tensor $-{\eta }_{\mu \nu }$. It differs by gauge-dependent terms proportional to the photon four-momentum. When the amplitude satisfies the Ward identity, those terms vanish on contraction and one may use $-{\eta }_{\mu \nu }$ as a replacement inside the squared amplitude. This replacement must not be mistaken for a free-standing equality of physical polarization tensors.

For $m>0$, the three physical polarization vectors are transverse to $k$ and have norm $-1$ in metric $(+---)$. In the rest frame their sum is $\mathrm{diag}(0,1,1,1)$; Lorentz covariance then gives the displayed expression for $k^2=m^2$. This includes the longitudinal mode. It cannot be specialized to $m=0$ by setting its denominator to zero.

For a resting W boson with real spatial polarization vector along $z$, the fermion spin sum gives $\sum |\mathcal{M}{|}^2=g^2{M}_W^2{\sin }^2\theta /2$ for massless charged lepton and antineutrino momenta. The two-body Lorentz-invariant phase space gives the displayed angular decay rate. Since $\int {\sin }^2\theta d\Omega =8\pi /3$, the integrated width is $g^2{M}_W/(48\pi )$, equal to the unpolarized width. The massive-vector spin average changes the angular distribution, not the total width. No initial spin average is taken for a specified pure polarization.

A Slavnov-Taylor identity is a BRST symmetry constraint relating correlation functions or interaction vertices in a gauge-fixed Non-Abelian gauge theory. Unlike a simple statement that every open gauge index is transverse, the general identities include Faddeev-Popov ghost fields. After putting external legs on shell and making all the other external states physical, they imply decoupling when one massless vector polarization vector is replaced by its four-momentum. They also control cancellations of auxiliary covariant components needed for physical unitarity.

Choose a temporal polarization vector, two transverse polarization vectors and a spatial longitudinal polarization vector. The Gupta-Bleuler quantization condition is $(a^0-a^3)|\chi ⟩=0$. Applied to ${\sum }_{\lambda }{c}_{\lambda }{a}^{\lambda †}|0⟩$, it gives $c_3=-c_0$. The remaining temporal-longitudinal excitation $a^{0†}-a^{3†}$ has zero norm and is orthogonal to every constrained state. Its field wavefunction is proportional to the null four-momentum, hence is pure gauge. Quotienting this direction leaves the two positive transverse photon states. Use wave packets or a box normalization to avoid treating continuum momentum eigenstates as normalizable vectors.

The interaction $\lambda \Phi F_{\mu \nu }{F}^{\mu \nu }/4$ gives the displayed on-shell scattering amplitude, up to an overall phase, for a scalar field decaying to two photons. Substituting either photon four-momentum for its polarization vector makes the expression vanish, verifying its Ward identity. In the parent rest frame, summing the two physical transverse photon polarizations gives $\sum |\mathcal{A}{|}^2=|\lambda {|}^2{M}^4/2$. Including the identical-particle factor in a final-state phase-space integral gives $\Gamma =|\lambda {|}^2{M}^3/(64\pi )$.