p}'}^{(\beta')}(x)]
-[\partial_0\Phi_{\bm p}^{(\beta)}(x)]^*\Phi_{{\bm p}'}^{(\beta')}(x)\Bigr).
\label{kg10}\end{aligned}$$ Its physical significance can be realized when considering the four-current density of charge, $j^\mu(x)$. In the absence of the external electromagnetic field, it is defined as [@Greiner1; @Greiner2] $$j^\mu(x)=\frac{{\mathrm{i}}e}{2m}\Bigl(\Phi^{(\beta)*}_{\bm p}(x)\bigl(\partial^\mu\Phi^{(\beta)}_{\bm p}(x)\bigr)-\bigl(\partial^\mu\Phi^{(\beta)*}_{\bm p}(x)\bigr)
\Phi^{(\beta)}_{\bm p}(x)\Bigr),
\label{kg11}$$ where $j^\mu(x)$ satisfies the continuity equation, $$\partial_\mu j^\mu(x)=0.
\label{kg12}$$ Taking into account , it follows from this equation that the quantity, $$Q=\int{\mathrm{d}}^3{\bm x}j^0(x),
\label{kg13}$$ is conserved. Since $j^0(x)$ can take either positive or negative values, it cannot be identified with the probability density. Instead, if we reinterpret the Klein-Gordon equation as satisfied by a quantum field $\hat{\Phi}(x)$, ${\hat{j}}^0(x)$ will describe the charge density of the field, whereas ${\hat{Q}}$ will be the field charge [@Greiner1; @Greiner2]. Going back to the definition of the Klein-Gordon inner product , we see therefore that it is related to the conservation of the field charge $Q$. In the presence of the electromagnetic field, the four-vector current density has to be redefined, $$\begin{aligned}
j^\mu(x)=\frac{{\mathrm{i}}e}{2m}\Bigl(\Phi^{(\beta)*}_{\bm p}(x)\bigl(\partial^\mu\Phi^{(\beta)}_{\bm p}(x)\bigr)-\bigl(\partial^\mu\Phi^{(\beta)*}_{\bm p}(x)\bigr)
\Phi^{(\beta)}_{\bm p}(x)\Bigr)-\frac{e^2}{mc}A^\mu(x)|\Phi_{\bm p}^{(\beta)}(x)|^2.
\label{kg14}\end{aligned}$$ Nevertheless, for as long as $A^0(x)=0$, which is the case discussed here, the definition of the Klein-Gordon inner product  does not change.

Keeping this in mind, we obtain for the inner product of the scalar 