= Solution
Vary the <Maxwell Lagrangian> with respect to $A_\nu$. Since $\delta F_{\mu\nu}=\partial_\mu\delta A_\nu-\partial_\nu\delta A_\mu$ and $F$ is antisymmetric,
$$
\delta\mathcal L=-F^{\mu\nu}\partial_\mu\delta A_\nu.
$$
After <integration by parts>, the <Euler-Lagrange field equation> is
$$
\boxed{\partial_\mu F^{\mu\nu}=0,\qquad
\Box A^\nu-\partial^\nu(\partial_\mu A^\mu)=0.}
$$
The <gauge transformation> $A_\mu\mapsto A_\mu+\partial_\mu\chi$ leaves $F_{\mu\nu}$ unchanged because <partial derivatives> commute. This <gauge invariance> means that potentials related in this way represent the same electromagnetic field; all four potential components are not independent physical <photon> degrees of freedom. In particular, the ungauged kinetic operator is singular and its temporal <canonical momentum> vanishes.
Choose the <Lorenz gauge> $\partial_\mu A^\mu=0$. The field equation then becomes $\boxed{\Box A_\mu=0}$. Residual <gauge transformations> with $\Box\chi=0$ preserve this condition. The correctly named <Lorenz gauge condition> concerns the divergence of the potential; <Lorentz covariance> is the separate transformation property of the equations.
For the alternative quadratic <Lagrangian density>, direct differentiation gives
$$
\frac{\partial\mathcal L'}{\partial(\partial_\rho A_\nu)}=-\partial^\rho A^\nu,
$$
so its <Euler-Lagrange field equation> is also $\Box A^\nu=0$. This density is the <Feynman-gauge Maxwell kinetic density after a boundary-term subtraction>. Indeed, with $D=\partial_\mu A^\mu$,
$$
-\frac14F_{\mu\nu}F^{\mu\nu}-\frac12D^2
=-\frac12\partial_\mu A_\nu\partial^\mu A^\nu+\partial_\mu K^\mu,\qquad
K^\mu=\frac12(A_\nu\partial^\nu A^\mu-A^\mu D).
$$
The divergence identity follows by expanding $\partial_\mu K^\mu$ and commuting its second derivatives. The quadratic density alone describes four wave equations; the gauge constraint must still be incorporated in the physical states. It is not an unconstrained four-positive-polarization replacement for <Maxwell equations>.
Using this specified boundary-term convention, the <canonical momentum> conjugate to $A_\nu$ is
$$
\boxed{\Pi^\nu=-\partial_0A^\nu.}
$$
The <canonical commutation relations> are therefore
$$
\boxed{[A_\mu(t,\mathbf x),\Pi^\nu(t,\mathbf y)]
=i\delta_\mu{}^\nu\delta^3(\mathbf x-\mathbf y),}
$$
with field-field and momentum-momentum brackets zero. Equivalently $[A_\mu,\dot A_\nu]=-ig_{\mu\nu}\delta^3$. The temporal component has the opposite sign to the spatial components; replacing the metric by a positive Euclidean one would destroy these brackets.
The <wave equation> gives massless <frequencies> $\omega_{\mathbf k}=|\mathbf k|$. Hermiticity pairs its positive- and negative-frequency solutions, so the <Heisenberg picture> field has the expansion
$$
\boxed{A_\mu(x)=\int\frac{d^3k}{(2\pi)^3 2\omega_{\mathbf k}}
\left[a_\mu(k)e^{-ik\cdot x}+a_\mu^\dagger(k)e^{ik\cdot x}\right],\qquad
k^0=\omega_{\mathbf k},\quad k^2=0.}
$$
This is a real field expansion for each component, but its oscillator metric is indefinite. The assumed brackets, with the dagger's vector index restored, are
$$
[a_\mu(k),a_\nu^\dagger(k')]=-g_{\mu\nu}(2\pi)^3 2\omega_{\mathbf k}\delta^3(\mathbf k-\mathbf k').
$$
They verify the <photon oscillator completeness and canonical brackets>. Inserting $\Pi^\nu=-\dot A^\nu$ gives at equal times
$$
[A_\mu(t,\mathbf x),\Pi^\nu(t,\mathbf y)]
=\frac{i\delta_\mu{}^\nu}{2}\int\frac{d^3k}{(2\pi)^3}
\left[e^{i\mathbf k\cdot(\mathbf x-\mathbf y)}+e^{-i\mathbf k\cdot(\mathbf x-\mathbf y)}\right]
=i\delta_\mu{}^\nu\delta^3(\mathbf x-\mathbf y).
$$
The other brackets cancel between the two <frequency> contributions. Thus both the $2\omega$ normalization and the sign $-g_{\mu\nu}$ are needed.
The <Gupta-Bleuler formalism> imposes the <Lorenz gauge condition> weakly on states:
$$
(\partial_\mu A^\mu)^{(+)}|\mathrm{phys}\rangle=0,
\qquad k^\mu a_\mu(k)|\mathrm{phys}\rangle=0\quad\text{for every }k.
$$
Here $(+)$ is the annihilation, positive-frequency part. Setting the full divergence to zero as an operator would conflict with the four-component <canonical commutation relations>. The subsidiary condition instead selects a physical pre-space inside the <covariant photon Fock space>, followed by a quotient of its null states.
Apply it to $|k,\epsilon\rangle=\epsilon^\nu a_\nu^\dagger(k)|0\rangle$. Commuting $k^\mu a_\mu$ through the creator gives a multiple of $-k_\nu\epsilon^\nu|0\rangle$. Hence the allowed <photon polarization vectors> obey
$$
\boxed{k\cdot\epsilon=0.}
$$
Take $k=(\omega,0,0,\omega)$. The complete solution is
$$
\epsilon=(c,\alpha,\beta,c)
=c(1,0,0,1)+\alpha(0,1,0,0)+\beta(0,0,1,0),
$$
with complex <coefficients>. The last two vectors are transverse linear <photon polarizations>, and $(0,1,\pm i,0)/\sqrt2$ gives the two circular <photon polarizations>. The first vector is $k/\omega$, the allowed temporal-longitudinal combination. A purely spatial longitudinal vector $(0,0,0,1)$ by itself does not satisfy the subsidiary condition.
The oscillator brackets give, apart from the positive continuum normalization factor,
$$
\langle k,\epsilon|k,\epsilon\rangle=-\epsilon^{*\mu}g_{\mu\nu}\epsilon^\nu
=\boxed{|\alpha|^2+|\beta|^2}.
$$
<Momentum> eigenstates are distributionally normalized; a box or a <wave packet> makes the norm statement literal. The allowed longitudinal combination has zero norm because $k^2=0$, and it is orthogonal to every constrained state since $k\cdot\epsilon=0$. It also gives a pure-gauge field wave: the field strength of $A_\mu\propto k_\mu e^{-ik\cdot x}$ vanishes. Thus $\epsilon$ and $\epsilon+ck$ represent the same physical <photon polarization>. The <Gupta-Bleuler null-state quotient> removes this null direction and leaves \b[two positive-norm transverse <photon> <photon polarizations>]. The negative-norm temporal oscillator of the unrestricted space is not an extra physical state. This is the <transverse one-photon physical quotient>.
Finally verify the <two-photon fermion Ward identity> directly for <electron-positron annihilation into two photons>. Define $D(r)=\not r-m$ and the reduced <Dirac propagator>
$$
G(r)=D(r)^{-1}=\frac{\not r+m}{r^2-m^2}.
$$
The equality follows from the <Clifford algebra>, since $(\not r-m)(\not r+m)=(r^2-m^2)I$. The incoming <Dirac spinors> obey $D(p)u(p)=0$ and $\overline v(q)(\not q+m)=0$. On replacing one inserted <photon polarization vector> by $k$, the <spinor> factor in the two diagrams is
$$
\overline v(q)\left[\not\epsilon'G(p-k)\not k+\not kG(p-k')\not\epsilon'\right]u(p).
$$
At the electron end, $\not k=D(p)-D(p-k)$, so
$$
G(p-k)\not k\,u(p)=-u(p).
$$
For the other diagram, <four-momentum conservation> gives $p-k'=k-q$. Hence
$$
\not k=D(p-k')+(\not q+m),\qquad
\overline v(q)\not kG(p-k')=\overline v(q).
$$
The two contributions therefore cancel exactly:
$$
T\big|_{\epsilon\to k}
=-e^2\left[-\overline v(q)\not\epsilon' u(p)+\overline v(q)\not\epsilon' u(p)\right]
=\boxed{0}.
$$
Individual diagrams need not vanish; their sum implements the <Ward identity>. Linearity then gives invariance under $\epsilon\mapsto\epsilon+ck$, as required by <gauge invariance>. In particular, the null longitudinal <photon polarization> proportional to $k$ has zero physical production amplitude. \b[Physical scattering produces the transverse <photon> classes], not an independent longitudinal massless <photon>.
Back to article page
