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?

  • 3
    $\begingroup$ 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
    Commented Dec 23, 2018 at 21:39

2 Answers 2


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.


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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