Non-Relativistic Limit of Klein-Gordon Probability Density In the lecture notes accompanying an introductory course in relativistic quantum mechanics, the Klein-Gordon probability density and current are defined as:
$$
\begin{eqnarray}
P & = & \dfrac{i\hbar}{2mc^2}\left(\Phi^*\dfrac{\partial\Phi}{\partial t}-\Phi\dfrac{\partial\Phi^*}{\partial t}\right) \\
\vec{j} &=& \dfrac{\hbar}{2mi}\left(\Phi^*\vec{\nabla}\Phi-\Phi\vec{\nabla}\Phi^*\right) 
\end{eqnarray}
$$
together with the statement that:

One can show that in the non-relativistic limit, the known expressions for the probability density and current are recovered.

The 'known' expressions are:
$$
\begin{eqnarray}
\rho &=& \Psi^*\Psi \\
\vec{j} &=& \dfrac{\hbar}{2mi}\left(\Psi^*\vec{\nabla}\Psi-\Psi\vec{\nabla}\Psi^*\right)
\end{eqnarray}
$$
When taking a 'non-relativistic limit', I am used to taking the "limit" $c \to \infty$, which does give the right result for $\vec{j}$, but for the density produces $P=0$. How should one then take said limit to recover the non-relativistic equations?
 A: The trick is to make the approach for the relativistic Klein-Gordon wave function
$$ \Phi(\vec{r},t) = \Psi(\vec{r},t) e^{-imc^2t/\hbar} \tag{1}$$
The physical reasoning behind this approach is:

*

*The fast oscillating exponential (with the extremely high frequency
$\frac{mc^2}{\hbar}$) is the solution for the particle at rest
(i.e. with energy $E=mc^2$).

*Compared to that high-frequency oscillation, $\Psi$ is assumed to give
only slow timely variations. Or more precisely:
$$\frac{\hbar}{mc^2}\frac{\partial\Psi}{\partial t}\ll\Psi \tag{2}$$
which just means that $\Psi$ will change only by a relatively small amount
during a time-interval $\Delta t=\frac{\hbar}{mc^2}$.

From (1) you find its derivatives
$$
\begin{eqnarray}
\frac{\partial\Phi}{\partial t}  &=& \left( \frac{\partial\Psi}{\partial t} - \frac{imc^2}{\hbar}\Psi \right) e^{-imc^2t/\hbar} \\
\vec{\nabla}\Phi                 &=& \vec{\nabla}\Psi e^{-imc^2t/\hbar}
\end{eqnarray}
\tag{3}$$
Plug (3) into the definitions of the Klein-Gordon probability density and current ($P$ and $\vec{j}$) and you get
$$
\begin{eqnarray}
 P       &=& \Psi^* \Psi + \frac{i\hbar}{2mc^2} \left( \Psi^*\frac{\partial\Psi}{\partial t} - \Psi \frac{\partial\Psi^*}{\partial t} \right) \\
 \vec{j} &=& \dfrac{\hbar}{2mi}\left(\Psi^*\vec{\nabla}\Psi-\Psi\vec{\nabla}\Psi^*\right)
\end{eqnarray}
\tag{4}$$
In (4) the expression for $\vec{j}$ is already the known non-relativistic
Schrödinger probability current.
But the expression for $P$ differs from the expected non-relativistic
probability density $\Psi^*\Psi$.
Now you can do the non-relativistic limit on the first equation of (4)

*

*either by the simple heuristics to use $c \to \infty$,

*or by using the above condition (2) about slow timely variations.

With both methods you get
$$P \approx \Psi^*\Psi \tag{5}$$
and thus recover the non-relativistic Schrödinger probability density.
A: You can substitute $\Phi = e^{-mc^2t/\hbar} \Psi$ and then neglect the second order time derivative of $\Psi$. Drop the constant $mc^2$ and you will have recovered the Schrödinger equation. 
