Feynman-gauge Maxwell kinetic density after a boundary-term subtraction (${\mathcal{L}}^{\prime }=-\frac{1}{2}{\partial }_{\mu }{A}_{\nu }{\partial }^{\mu }{A}^{\nu }$)

With $D={\partial }_{\mu }{A}^{\mu }$, expand the Maxwell Lagrangian and its Feynman gauge term to findCommuting partial derivatives verifies the divergence identity. Both densities give $\square A^{\mu }=0$ 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.

In mostly-minus signature, adding the boundary term ${\partial }_{\mu }{K}^{\mu }$ with $K^{\mu }=\frac{1}{2}(A_{\nu }{\partial }^{\nu }{A}^{\mu }-A^{\mu }{\partial }_{\nu }{A}^{\nu })$ to the subtracted Feynman gauge density changes ${\tilde{\pi }}^{\mu }=-{\dot{A}}^{\mu }$ to the displayed canonical momenta. For fields with vanishing spatial boundary terms, $\int K^0{d}^{3}x=\int A_0{\partial }_i{A}_i{d}^{3}x$, whose functional derivatives are these momentum shifts. They depend only on coordinates, so $[A_{\mu },{\pi }^{\nu }]=i{\delta }_{\mu }{}^{\nu }{\delta }^3$ is unchanged. The only potentially nonzero mixed momentum bracket is $[{\pi }^0(x),{\pi }^i(y)]=i({\partial }_{y^i}+{\partial }_{x^i}){\delta }^3(x-y)=0$. Thus both density representatives define canonical commutation relations, although their momenta must not be equated componentwise.