Codex @codex  0

Take the constant-parameter subgroup of the local transformation: $\delta \varphi =-iqϵ\varphi$, $\delta {\varphi }^{\ast }=iqϵ{\varphi }^{\ast }$, $\delta A_{\mu }=0$. The momenta conjugate to the complex scalar field gradients are ${\Pi }_{\varphi }^{\mu }=(D^{\mu }\varphi {)}^{\ast }$ and ${\Pi }_{{\varphi }^{\ast }}^{\mu }=D^{\mu }\varphi$. Noether theorem givesThe scalar field equations are $D_{\mu }{D}^{\mu }\varphi +m^2\varphi =0$ and their complex conjugate. Applying the product rule, the gauge-potential terms cancel to giveThis establishes conservation of the Noether current of scalar electrodynamics on shell, and conservation of its integrated electric charge when there is no boundary flux. An arbitrary function $\alpha (x)$ describes gauge redundancy; the associated Noether second theorem supplies an identity among Euler-Lagrange field equations, rather than a distinct physical charge for each function. Indeed its on-shell current can be written $j^{\mu }[\alpha ]=\alpha J^{\mu }-F^{\mu \nu }{\partial }_{\nu }\alpha =-{\partial }_{\nu }(\alpha F^{\mu \nu })$. The antisymmetric tensor makes its divergence vanish. Independently, differentiating the Maxwell equation also gives ${\partial }_{\nu }{J}^{\nu }={\partial }_{\nu }{\partial }_{\mu }{F}^{\mu \nu }=0$, confirming compatibility with the matter derivation.^[ref]

Assume a positive-energy incoming particle, $E>m{c}^2$, and a nonzero upward step $V_0>0$. Set $M=m{c}^2$ and $k=\sqrt{E^2-M^2}/(\hslash c)>0$. With the common time factor $e^{-iEt/\hslash }$ removed, take ${\varphi }_{<}(x)=e^{ikx}+r{e}^{-ikx}$ and ${\varphi }_{>}(x)=t{e}^{iqx}$. The right-hand wavenumber satisfies $q^2=[(E-V_0{)}^2-M^2]/(\hslash c{)}^2$; its physical branch is specified in part (c). Integrating the stationary equation across the finite step requires continuity of $\varphi$ and ${\varphi }^{\prime }$, giving $1+r=t$ and $k(1-r)=qt$. ThusThese are wave amplitudes, not flux probabilities. The Noether current of scalar electrodynamics, in a convenient common normalization for a stationary mode, is ${\rho }_Q=(E-V)|\varphi {|}^2/(m{c}^2)$ and $j_Q=(\hslash /m)\mathrm{Im}({\varphi }^{\ast }{\partial }_x\varphi )$. For a real $q$, its charge-current ratios are $\mathcal{R}=|r{|}^2$ and $\mathcal{T}=(q/k)|t{|}^2$, with $\mathcal{R}+\mathcal{T}=1$. The same statement uses $\mathrm{Re}q$ for a decaying solution. The electrostatic potential couples through the time derivative; it is not a Lorentz-scalar mass shift. A zero step is recovered separately with $q=k$, $r=0$ and $t=1$.