Feynman gauge

With $D={\partial }_{\mu }{A}^{\mu }$, expand the Maxwell Lagrangian and its Feynman gauge term to findCommuting partial derivatives verifies the divergence identity. Both densities give $\square A^{\mu }=0$ under the usual variational boundary conditions. Their canonical momentum components differ: the original density gives $-F^{0\nu }-{\eta }^{0\nu }D$, whereas the subtracted density gives $-{\dot{A}}^{\nu }$. Thus a free-photon momentum expansion using the latter requires a stated boundary-term convention.

In mostly-minus signature, adding the boundary term ${\partial }_{\mu }{K}^{\mu }$ with $K^{\mu }=\frac{1}{2}(A_{\nu }{\partial }^{\nu }{A}^{\mu }-A^{\mu }{\partial }_{\nu }{A}^{\nu })$ to the subtracted Feynman gauge density changes ${\tilde{\pi }}^{\mu }=-{\dot{A}}^{\mu }$ to the displayed canonical momenta. For fields with vanishing spatial boundary terms, $\int K^0{d}^{3}x=\int A_0{\partial }_i{A}_i{d}^{3}x$, whose functional derivatives are these momentum shifts. They depend only on coordinates, so $[A_{\mu },{\pi }^{\nu }]=i{\delta }_{\mu }{}^{\nu }{\delta }^3$ is unchanged. The only potentially nonzero mixed momentum bracket is $[{\pi }^0(x),{\pi }^i(y)]=i({\partial }_{y^i}+{\partial }_{x^i}){\delta }^3(x-y)=0$. Thus both density representatives define canonical commutation relations, although their momenta must not be equated componentwise.

Gupta-Bleuler quantization keeps Lorentz covariance by first quantizing all four photon polarizations in an indefinite inner-product space. The condition $({\partial }_{\mu }{A}^{\mu }{)}^{(+)}|\mathrm{phys}⟩=0$, followed by quotienting null states, leaves the two transverse physical polarizations.

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.

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 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.

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.

For a temporal photon wave packet $|f,0⟩=\int d^{3}pf(\mathbf{p})a_{\mathbf{p}}^{0†}|0⟩/(2\pi {)}^3$, the oscillator commutator gives $⟨f,0|f,0⟩=-\int d^{3}p|f(\mathbf{p}){|}^2/(2\pi {)}^3<0$ when $f\ne 0$. This is the negative direction of the indefinite Hermitian form, not the divergent normalization of an unsmeared momentum eigenstate. It is excluded by the physical condition in Gupta-Bleuler quantization.