= Solution
Treat $\phi$ and $\phi^*$ as independent fields, keep $A_\mu(x)$ as the prescribed background, and use the source's interaction with no added $e^2A^2|\phi|^2$ term. The momentum derivatives are
$$
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi)}
=\partial^\mu\phi^*-ieA^\mu\phi^*,\qquad
\frac{\partial\mathcal L}{\partial(\partial_\mu\phi^*)}
=\partial^\mu\phi+ieA^\mu\phi.
$$
Varying $\phi^*$ and $\phi$ gives respectively
$$
\boxed{(\Box+m^2)\phi+ie(\partial_\mu A^\mu)\phi
+2ieA^\mu\partial_\mu\phi=0,}
$$
$$
\boxed{(\Box+m^2)\phi^*-ie(\partial_\mu A^\mu)\phi^*
-2ieA^\mu\partial_\mu\phi^*=0.}
$$
No <Lorenz gauge condition> has been assumed. The background <divergence> term must be retained unless such a condition is separately imposed.
The <canonical momenta> conjugate to the two scalar fields are
$$
\pi=\dot\phi^*-ieA_0\phi^*,\qquad
\pi^*=\dot\phi+ieA_0\phi,
$$
so $\dot\phi=\pi^*-ieA_0\phi$ and $\dot\phi^*=\pi+ieA_0\phi^*$. Define the physical three-vector potential by $A_\mu=(A_0,-\mathbf A)$. The <Legendre transform> in the proper <canonical variables> is
$$
\boxed{\begin{aligned}
\mathcal H(\phi,\phi^*,\pi,\pi^*;x,t)
={}&(\pi+ieA_0\phi^*)(\pi^*-ieA_0\phi)
+\boldsymbol\nabla\phi^*\cdot\boldsymbol\nabla\phi+m^2|\phi|^2\\
&-ie\mathbf A\cdot(\phi\boldsymbol\nabla\phi^*-\phi^*\boldsymbol\nabla\phi).
\end{aligned}}
$$
Equivalently the first product is $\pi\pi^*+ieA_0(\phi^*\pi^*-\phi\pi)+e^2A_0^2|\phi|^2$. The last quadratic term arises from replacing velocities by <canonical momenta> even though the original <Lagrangian> has no quadratic interaction. It is not permission to modify the specified <Lagrangian>. An explicitly time-dependent background generally makes the scalar-field energy time dependent.
Back to article page
