Physical interpretation of Klein-Gordon Equation conserved charge In the Klein-Gordon Equation the conserved charge is: $$\rho = \frac{i \hbar}{2m} (\psi^* \frac{\partial \psi}{\partial t} - \frac{\partial \psi^*}{\partial t} \psi) $$
rather than the conserved (probability) density in the Schrodinger equation:
$$\rho = \psi^* \psi.$$
If the physical interpretation from the Schrodinger equation is that $\rho$ is the conserved probability density of a particle, what is the analogous physical interpretation of $\rho$ from the Klein-Gordon Equation?
 A: The closure of the Klein Gordon equation's solutions under $\psi\mapsto\psi^\ast$, which multiplies $\rho$ as defined in your first equation by $-1$, precludes a straightforward probability interpretation. It does not, however, preclude a particle-minus-antiparticles count interpretation, which is equivalent to conserving a "charge" applicable to the relevant species (be it electric or otherwise).
There's another amusing aspect of this. Write $\psi=Re^{i\theta}$ so $\rho=\frac{-\hbar}{m}R^2\frac{\partial\theta}{\partial t}$, reminiscent of conserving the specific angular momentum $r^2\dot{\theta}$ in an orbit under a radial force. Indeed, plane-wave solutions $\propto\exp\operatorname{i}\omega t$ have $\ddot{\psi}/\psi\in\Bbb R^-$, in analogy with a force in the plane being antiparallel to a position construed as complex. (The overall sign makes the "force" attractive, preventing $|\psi|$ growing too large for unitarity.)
A: The answer is in the question: it is the charge density although a factor $e$ is missing.
