Non-Relativistic Limit of Klein-Gordon Probability Density

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.
  • 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.


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.