= Solution
The oscillator <commutator> gives
$$
\langle0|a_{\mathbf p}^{\lambda}a_{\mathbf q}^{\lambda'\dagger}|0\rangle
=-\eta^{\lambda\lambda'}(2\pi)^3\delta^{(3)}(\mathbf p-\mathbf q).
$$
The temporal mode $\lambda=0$ has negative norm. More precisely, the nonzero <wave packet> $|f,0\rangle=\int d^3p\,f(\mathbf p)a_{\mathbf p}^{0\dagger}|0\rangle/(2\pi)^3$ has norm $-\int d^3p\,|f(\mathbf p)|^2/(2\pi)^3<0$. 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:
$$
\boxed{(\partial_\mu A^\mu)^{(+)}|\mathrm{phys}\rangle=0.}
$$
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 $\langle\Phi|\partial\cdot A|\Psi\rangle=0$ between physical states: the positive-frequency part annihilates the ket, and its adjoint annihilates the bra.
For a nonzero <momentum> choose upper-index <polarization vectors>
$$
\epsilon^{\mu0}=(1,\mathbf0),\quad
\epsilon^{\mu1}=(0,\mathbf e_1),\quad
\epsilon^{\mu2}=(0,\mathbf e_2),\quad
\epsilon^{\mu3}=(0,\widehat{\mathbf p}),
$$
where $\mathbf e_1,\mathbf e_2$ 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 $-i\omega(a^0_{\mathbf p}-a^3_{\mathbf p})$. The physical condition is therefore
$$
(a^0_{\mathbf p}-a^3_{\mathbf p})|\mathrm{phys}\rangle=0.
$$
For a one-mode state $|\chi\rangle=\sum_\lambda c_\lambda a^{\lambda\dagger}|0\rangle$, the condition reads $-c_0-c_3=0$. Thus
$$
|\chi\rangle=c_1a^{1\dagger}|0\rangle+c_2a^{2\dagger}|0\rangle
+c_0(a^{0\dagger}-a^{3\dagger})|0\rangle,
\qquad
\langle\chi|\chi\rangle=|c_1|^2+|c_2|^2.
$$
The last direction has zero norm and is orthogonal to every constrained state. Its field wavefunction is proportional to $\epsilon^{\mu0}+\epsilon^{\mu3}=p^\mu/\omega$, hence it is a pure <gauge symmetry> direction. A purely longitudinal creator $a^{3\dagger}$ alone is not physical, despite having positive norm.
For the multiparticle sketch, put $C=a^0-a^3$ and $C^\dagger=a^{0\dagger}-a^{3\dagger}$. The <commutator> $[C,C^\dagger]=0$ allows polynomials in $C^\dagger$ times transverse creator states to obey the constraint. Conversely, on creator polynomials the constraint is $-\partial_{a^{0\dagger}}-\partial_{a^{3\dagger}}$, so these are all the constrained polynomials. Every term containing $C^\dagger$ is orthogonal to every constrained bra, since $\langle\Phi|C^\dagger=\langle C\Phi|=0$. Quotienting those null vectors and completing the remaining positive space is the <Gupta-Bleuler null-state quotient>. It gives
$$
\boxed{\text{two physical transverse photon polarizations, or helicities }+1,-1.}
$$
The one-particle instance is the <transverse one-photon physical quotient>.
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