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?

  • 1
    $\begingroup$ 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

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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.