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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?

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    $\begingroup$ For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein. $\endgroup$ – Qmechanic Dec 23 '18 at 21:39
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The trick is to make the approach for the relativistic Klein-Gordon wave function $$ \Phi(\vec{r},t) = \Psi(\vec{r},t) e^{-imc^2t/\hbar} $$ (The physical reasoning behind this approach is: The fast oscillating exponential is the solution for the particle at rest. And compared to that, $\Psi$ will give only slow variations.)

From that you find its derivatives $$ \begin{eqnarray} \frac{\partial\Phi}{\partial t} &=& \left( \frac{\partial\Psi}{\partial t} - \frac{imc^2}{\hbar}\Psi \right) e^{-imc^2t/\hbar} \\ \vec{\nabla}\Phi &=& \vec{\nabla}\Psi e^{-imc^2t/\hbar} \end{eqnarray} $$

Plug these into the definitions of the Klein-Gordon probability density and current ($P$ and $\vec{j}$) and you get $$ \begin{eqnarray} P &=& \Psi^* \Psi + \frac{i\hbar}{2mc^2} \left( \Psi^*\frac{\partial\Psi}{\partial t} - \Psi \frac{\partial\Psi^*}{\partial t} \right) \\ \vec{j} &=& \dfrac{\hbar}{2mi}\left(\Psi^*\vec{\nabla}\Psi-\Psi\vec{\nabla}\Psi^*\right) \end{eqnarray} $$

Now you can do the limit $c \to \infty$ and get the 'known' expressions of the non-relativistic Schrödinger probability density and current.

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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.

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