= Solution
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 $-\eta^{\lambda\lambda}\int d^3p\,|f(\mathbf p)|^2$. 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> $\epsilon^{0\mu}=(1,\mathbf0)$, $\epsilon^{3\mu}=(0,\widehat{\mathbf p})$, and two $\epsilon^{r\mu}=(0,\mathbf e_r)$ 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:
$$
(\partial_\mu A^\mu)^{(+)}|\Psi\rangle=0.
$$
Restore the factors $e^{-i|\mathbf p|t}$ to the annihilation terms. Since $p_\mu\epsilon^{0\mu}=|\mathbf p|$ and $p_\mu\epsilon^{3\mu}=-|\mathbf p|$, <Fourier transform> gives the equivalent condition
$$
\boxed{(a_{\mathbf p}^0-a_{\mathbf p}^3)|\Psi\rangle=0\quad\text{for every nonzero momentum, distributionally}.}
$$
Its 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 $|\Psi\rangle=\sum_{n,m\geq0}|n,m\rangle_{0,3}\otimes|\chi_{nm}\rangle_\perp$, where $|n,m\rangle=(a^{0\dagger})^n(a^{3\dagger})^m|0\rangle/\sqrt{n!m!}$. The temporal oscillator obeys $a^0|n,m\rangle=-\sqrt n|n-1,m\rangle$, while $a^3|n,m\rangle=\sqrt m|n,m-1\rangle$. The condition therefore becomes
$$
\sqrt{n+1}\,|\chi_{n+1,m}\rangle+\sqrt{m+1}\,|\chi_{n,m+1}\rangle=0.
$$
Equivalently, the allowed finite-particle states use the transverse <creation operators> and only $b^\dagger=a^{0\dagger}-a^{3\dagger}$ in the unphysical sector. Indeed the constraint acts on a <polynomial> of the two unphysical <creation operators> as $-\partial_{a^{0\dagger}}-\partial_{a^{3\dagger}}$, whose kernel consists of <polynomials> in their difference. Also $[b,b^\dagger]=-1+1=0$, and a state containing $b^\dagger$ 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, $c_0a^{0\dagger}|0\rangle+c_3a^{3\dagger}|0\rangle$ is constrained only when $c_3=-c_0$, producing a null state. The <Gupta-Bleuler null-state quotient> removes these null directions. \b[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>.
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