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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?

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    $\begingroup$ What is the Noether charge you find for U(1), when you switch photons off? $\endgroup$ Oct 12, 2021 at 15:54
  • $\begingroup$ 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? $\endgroup$
    – Alex Gower
    Oct 12, 2021 at 16:01
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    $\begingroup$ 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. $\endgroup$ Oct 12, 2021 at 16:07
  • $\begingroup$ Related, on the interpretation of $\rho$ as a probability: physics.stackexchange.com/q/340023/226902 and physics.stackexchange.com/q/622975/226902 $\endgroup$
    – Quillo
    Oct 12, 2021 at 16:49

2 Answers 2

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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.)

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

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