How to derive the conserved current of the Klein Gordon equation? Similarly to the probability current in non-relativistic quantum mechanics, there is a conserved current for the Klein Gordon equation, however a different one. I'm trying to calculate that.
The KG equation reads
$$(\partial^2 + m^2)\psi. $$
From this I want to show that when $\psi$ satisfies the KG equation, then the following is also satisfied:
$$\psi^* \partial^2 \psi - \psi \partial^2 \psi^* = 0.$$
Which would imply the current conservation law
$$\partial_{\mu}[\psi^* \partial^{\mu} \psi - \psi \partial^{\mu}\psi^*] = 0.$$
However I can't seem to find a way to get rid of the mass term by simply slapping the complex conjugated on the left side or something like that.
What am I missing?
 A: Consider the KG equation for a complex scalar field $\phi(x) \in \mathbb{C}$
$$ \tag{1} ( \square + m^2 ) \phi(x) = 0,$$
and its complex conjugate
$$ \tag{2} ( \square + m^2 ) \phi^*(x) = 0.$$
Multiplying at the left (1) by $\phi^*(x)$ and (2) by $\phi(x)$ you have respectively
$$ \tag{3} \phi^*(x) (\square + m^2 ) \phi(x) = 0$$
and
$$ \tag{4} \phi(x) ( \square + m^2 ) \phi^*(x) = 0.$$
Subtracting now (4) from (3) you have
$$ \tag{5} \phi^*(x) (\square + m^2 ) \phi(x) - \phi(x) ( \square + m^2 ) \phi^*(x) = 0 $$
$$ \tag{5-1} \Longrightarrow (\phi^* \square \phi - \phi \square \phi^*) + m^2 (\phi^*\phi - \phi \phi^*) = 0$$
where in the last step we used the fact that $m \in \mathbb{R}$ and hence $\phi(x) m = m \phi(x)$ (and omitted the $x$ argument for lazyness).
Because also $\phi(x), \phi^*(x) \in \mathbb{C}$, i.e. they are just complex number (as opposite to operator fields in QFT), you have $\phi(x)\phi^*(x)=\phi^*(x)\phi(x)$ and hence the conclusion:
$$ \phi^* \square \phi - \phi \square \phi^* = 0
\Longleftrightarrow \partial_\mu ( \phi^*(x) \partial^\mu \phi(x) - \phi(x) \partial^\mu \phi^*(x) ) = 0,  $$
i.e.
$$ \partial_\mu J^\mu(x) = 0, \qquad J^\mu(x) \equiv \phi^*(x)\partial^\mu\phi(x) - \phi(x)\partial^\mu \phi^*(x). $$

Note: I'm using the notation
$$ \square \equiv \partial^2 \equiv \partial_\mu \partial^\mu, $$
and my $\phi(x)$ is your $\psi(x)$.
