{c}
{title2=$\mathcal L'=-\tfrac12\partial_\mu A_\nu\partial^\mu A^\nu$}

With $D=\partial_\mu A^\mu$, expand the <Maxwell Lagrangian> and its <Feynman gauge> term to find
$$
-\frac14F_{\mu\nu}F^{\mu\nu}-\frac12D^2
=-\frac12\partial_\mu A_\nu\partial^\mu A^\nu+\partial_\mu K^\mu,
\qquad K^\mu=\frac12(A_\nu\partial^\nu A^\mu-A^\mu D).
$$
Commuting <partial derivatives> verifies the divergence identity. Both densities give $\Box A^\mu=0$ under the usual variational <boundary conditions>. Their <canonical momentum> components differ: the original density gives $-F^{0\nu}-\eta^{0\nu}D$, whereas the subtracted density gives $-\dot A^\nu$. Thus a free-photon <momentum> expansion using the latter requires a stated boundary-term convention.


 Feynman-gauge Maxwell kinetic density after a boundary-term subtraction

ID: feynman-gauge-maxwell-kinetic-density-after-a-boundary-term-subtraction