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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.

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.