# 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?

• 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. – Qmechanic Dec 23 '18 at 21:39

## 2 Answers

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.