I would like to analyse the semiclassical and non-relativistic limit of the Klein-Gordon equation, \begin{equation} \frac{1}{c^2} \partial_t^2 \phi - \Delta \phi + \frac{M^2 c^2}{\hbar^2} \phi =0. \label{KGEscalar} \end{equation}
Therefore I used a Wigner transformation and passed to the limit $\epsilon = \frac{1}{c^2}=\hbar \to 0$. The evolution equation for the Wigner density $\omega^{\epsilon}$ in the limit which I obtained in both cases is \begin{equation} \frac{\partial}{\partial t} \omega^{0} =0. \end{equation} This means that the "quasi probability" to detect a particle at a specific location in the phase space is constant, i.e. it is zero when we demand any normalization conditions.
I mean this is not surprisingly when we consider the Klein-Gordon equation above, if we let $\epsilon \to 0$ this of course would lead to problems. But this seems really strange to me from a physical point of view. In the nonrelativistic limit we have that the rest energy become unbounded and therefore can cause problems. But what happens in the semiclassical limit, i.e. $\hbar \to 0$? At the moment I'm not sure how to intepret this result.
To be clear, I defined the Wigner transformation for $L^2$ functions $f, g$ as
\begin{equation} \omega^{\epsilon} (f, g) (x, \xi) := \int _{\mathbb{R}} f(x-\frac{v} {2} \epsilon) g(x-\frac{v} {2} \epsilon) e^{i v \xi } d v. \end{equation}
Then choose $f=g= \phi$. The evolution equation for $\omega^{\epsilon}_q (\phi, \phi) $ can easily be obtained by a Fourier transformation $\mathscr{F} (x \mapsto \zeta) $and reads \begin{equation} \partial_t \omega^{\epsilon}_q - i \zeta \int _{-\frac{1}{2} }^{\frac{1}{2} } \nabla \lambda_q (\xi - s \epsilon \zeta) ds=0. \end{equation} Where $\lambda_q $ are the eigenvalues of the associated symbol of the differential operator of the Klein Gordon equation. The constants $c$ and $\hbar$ enter in the symbol.