# 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?

• What is the Noether charge you find for U(1), when you switch photons off? Oct 12, 2021 at 15:54
• I'm sorry, I have only just started my QFT course so this is a bit beyond me, is there a way of 'dumbing it down' or should I just wait? Oct 12, 2021 at 16:01
• Yes, wait until you "gauge" the phase transformation by coupling it to photons. Your charge is the electric charge effecting $\psi\to e^{i\theta} \psi$. See WP. Oct 12, 2021 at 16:07
• Related, on the interpretation of $\rho$ as a probability: physics.stackexchange.com/q/340023/226902 and physics.stackexchange.com/q/622975/226902 Oct 12, 2021 at 16:49

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.)
The answer is in the question: it is the charge density although a factor $$e$$ is missing.