= Solution
Use $g_{\mu\nu}=\operatorname{diag}(1,-1,-1,-1)$ throughout. Vary the <Maxwell action> with respect to the <electromagnetic four-potential>. Antisymmetry of the <electromagnetic field tensor> gives $\delta\mathcal L=-F^{\mu\nu}\partial_\mu\delta A_\nu$. After integration by parts, the <Euler-Lagrange field equation> is
$$
\boxed{\partial_\mu F^{\mu\nu}=0},\qquad
\Box A^\nu-\partial^\nu(\partial_\mu A^\mu)=0.
$$
A <gauge transformation> $A_\mu\mapsto A_\mu+\partial_\mu\chi$ leaves $F_{\mu\nu}$ unchanged because <partial derivatives> commute. It therefore leaves the <Maxwell action> and observable electromagnetic fields unchanged: different potentials represent the same physical field. This redundancy matters for <canonical quantization>: before <gauge fixing>, the momentum conjugate to $A_0$ vanishes and the four potential components cannot be treated as four independent physical oscillators.
Choose the <Lorenz gauge> $D=\partial_\mu A^\mu=0$, attainable by solving $\Box\chi=-D$ with suitable boundary data. The <Euler-Lagrange field equation> then reduces to
$$
\boxed{\Box A^\mu=0}.
$$
<Residual Lorenz gauge symmetry> consists of transformations with $\Box\chi=0$. The gauge condition and this residual redundancy ultimately leave two <transverse polarizations> of a free <photon>.
For the specified kinetic density $\mathcal L_C=-\frac12\partial_\mu A_\nu\partial^\mu A^\nu$, one has $\partial\mathcal L_C/\partial(\partial_\mu A_\nu)=-\partial^\mu A^\nu$. Its <Euler-Lagrange field equation> is therefore $-\Box A^\nu=0$. This density is the <Feynman-gauge Maxwell kinetic density after a boundary-term subtraction>, since
$$
-\frac14F_{\mu\nu}F^{\mu\nu}-\frac12D^2
=\mathcal L_C+\partial_\mu K^\mu,\qquad
K^\mu=\frac12(A_\nu\partial^\nu A^\mu-A^\mu D).
$$
The divergence follows by expanding both sides and commuting the second derivatives. Thus the wave equation follows from a <Feynman gauge> action; the additional physical-state condition will implement the <Lorenz gauge>. One must not impose $D=0$ inside the original action before varying it.
Take the covariant components $A_\nu$ as canonical coordinates. The <canonical momentum> computed from $\mathcal L_C$ is
$$
\boxed{\Pi^\nu=\frac{\partial\mathcal L_C}{\partial(\partial_0A_\nu)}=-\partial_0A^\nu}.
$$
The equal-time canonical Poisson brackets are $\{A_\mu(t,\mathbf x),\Pi^\nu(t,\mathbf y)\}=\delta_\mu{}^\nu\delta^3(\mathbf x-\mathbf y)$, with coordinate-coordinate and momentum-momentum brackets zero. Applying <canonical quantization> gives
$$
\boxed{[A_\mu(t,\mathbf x),\Pi^\nu(t,\mathbf y)]=i\delta_\mu{}^\nu\delta^3(\mathbf x-\mathbf y)},\qquad
[A_\mu,A_\nu]=[\Pi^\mu,\Pi^\nu]=0.
$$
In particular $[A_\mu,\dot A_\nu]=-ig_{\mu\nu}\delta^3$. The negative temporal sign is the source of the indefinite state-space norm in covariant photon quantization.
The spatial <Fourier transform> of the wave equation gives an oscillator of frequency $\omega=|\mathbf k|$ for each component. The <Hermitian adjoint> condition on the <electromagnetic four-potential> then yields the <mode expansion of a free field>
$$
\boxed{A_\mu(x)=\int\frac{d^3k}{(2\pi)^3 2\omega}\,[a_\mu(\mathbf k)e^{-ik\cdot x}+a_\mu^\dagger(\mathbf k)e^{ik\cdot x}]},\qquad k^\mu=(\omega,\mathbf k).
$$
These are massless <on shell> modes. Consistently with the preceding equal-time <canonical commutation relation>,
$$
[a_\mu(\mathbf k),a_\nu^\dagger(\mathbf l)]=-g_{\mu\nu}(2\pi)^3 2\omega\delta^3(\mathbf k-\mathbf l),\qquad
[a_\mu,a_\nu]=[a_\mu^\dagger,a_\nu^\dagger]=0.
$$
The normalization is the same as for the scalar modes, with $-g_{\mu\nu}$ replacing the species identity matrix.
<Gupta-Bleuler quantization> starts with the <Fock vacuum> annihilated by every $a_\mu$ and imposes only the annihilation, or <positive-frequency part of a quantum field>, of the <Lorenz gauge> condition:
$$
\boxed{(\partial_\mu A^\mu)^{(+)}|\mathrm{phys}\rangle=0}
\quad\Longleftrightarrow\quad
k^\mu a_\mu(\mathbf k)|\mathrm{phys}\rangle=0\quad\text{for every }\mathbf k.
$$
Imposing the entire Hermitian field divergence as a strong operator identity would be incompatible with the four-component <canonical commutation relation>. The weaker condition suffices: between two physical states its positive-frequency part vanishes on the ket, and its negative-frequency part vanishes on the bra. Hence $\langle\mathrm{phys}'|\partial_\mu A^\mu|\mathrm{phys}\rangle=0$.
For a one-photon state $|\mathbf k,\epsilon\rangle=\epsilon^\nu a_\nu^\dagger(\mathbf k)|0\rangle$, commuting the constraint through the creator gives
$$
l^\mu a_\mu(\mathbf l)|\mathbf k,\epsilon\rangle
=-(k\cdot\epsilon)(2\pi)^3 2\omega\delta^3(\mathbf l-\mathbf k)|0\rangle.
$$
Thus the allowed <photon polarization vectors> satisfy $\boxed{k\cdot\epsilon=0}$. Their state inner product, with the common momentum-normalization factor suppressed, is $\langle\epsilon'|\epsilon\rangle=-\epsilon'^*\cdot\epsilon$. Momentum eigenstates are delta-normalized; the same statements hold with finite norms for normalizable wave packets.
Rotate the spatial axes so that $k=\omega(1,0,0,1)$. The constraint is $\epsilon^0=\epsilon^3$, so the complete allowed family is
$$
\boxed{\epsilon=\alpha(1,0,0,1)+\beta(0,1,0,0)+\gamma(0,0,1,0)},\qquad \alpha,\beta,\gamma\in\mathbb C.
$$
The first vector is the allowed longitudinal gauge polarization, proportional to the null <four-momentum> $k$. Its state has zero norm because $k^2=0$; it is also orthogonal to every constrained state because $k\cdot\epsilon=0$. A purely spatial vector $(0,0,0,1)$ alone is not an allowed physical polarization: its time component must accompany it to satisfy the constraint. The other two vectors are <transverse polarizations>, perpendicular to $\mathbf k$, with positive unit norms. For the displayed family,
$$
-\epsilon^*\cdot\epsilon=|\beta|^2+|\gamma|^2.
$$
The timelike negative-norm excitation has been excluded by the constraint, and the remaining form is positive semidefinite. The <Gupta-Bleuler null-state quotient> identifies $\epsilon$ and $\epsilon+\lambda k$, removing the null direction. It leaves exactly \b[two transverse physical photon polarizations] with a positive <inner product>; their <bosonic Fock space> is the physical photon state space.
Finally use <Feynman slash notation>, $\not a=\gamma^\mu a_\mu$, and impose <four-momentum conservation> $p+q=k+k'$. The external <Dirac spinors> obey
$$
(\not p-m)u(p)=0,\qquad \bar v(q)(\not q+m)=0.
$$
Set $r=p-k$ and $r'=p-k'$. Replacing $\epsilon$ by $k$ in the first term of the <electron-positron annihilation into two photons> amplitude, the rightmost contracted vertex obeys
$$
\not k\,u=(\not p-m)u-(\not r-m)u=-(\not r-m)u.
$$
Since the <Clifford algebra> implies $(\not r+m)(\not r-m)=(r^2-m^2)I_4$, that term becomes
$$
\bar v\not\epsilon'\frac{\not r+m}{r^2-m^2}\not k\,u
=-\bar v\not\epsilon'u.
$$
For the second term, <four-momentum conservation> gives $k=q+r'$. The contracted vertex is now adjacent to the outgoing positron spinor on the left, so
$$
\bar v\not k=\bar v(\not q+\not r')=\bar v(\not r'-m).
$$
Consequently
$$
\bar v\not k\frac{\not r'+m}{r'^2-m^2}\not\epsilon'u
=+\bar v\not\epsilon'u.
$$
The two terms cancel, proving the <two-photon fermion Ward identity>
$$
\boxed{\bar v(q)M\big|_{\epsilon=k}u(p)=0,\qquad T\big|_{\epsilon=k}=0}.
$$
Exchanging the two photon labels proves the same result for $\epsilon'=k'$. By linearity the amplitude is invariant under $\epsilon\mapsto\epsilon+\lambda k$ on either external photon leg. Therefore the null longitudinal gauge polarization has zero production amplitude: \b[physical scattering produces only the two transverse photon polarizations]. This cancellation also shows why both photon orderings in the <tree-level Feynman diagrams> are essential; either term alone fails the <Ward identity>.
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