Klein-Gordon equation in the non-relativistic and semiclassical limit in a Wigner approach 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.
 A: I would go to the nonrelativistic limit in a different way. Substitute $\Psi = e^{-mc^2/\hbar} \phi$ and neglect the second time derivative of $\phi$. The result is the Schrödinger equation, so it is sufficient to study the Schrödinger equation in the semiclassical limit. 
A: Assuming you are considering the plane-wave solution of the K-G equation, take it to be in 1+1 for computational simplicity, 
$$
\phi \propto \exp \left (-it\sqrt{k^2 + M^2c^2/\hbar^2} +ikx\right ), 
$$
the (real) Wigner transform of its density matrix  would be proportional to the  unavoidable
$$
\int dy e^{-iyp} e^{-ik(x-y\hbar/2)+ik(x+y\hbar/2  ) }         \propto  \delta (p-\hbar k).$$
The $k\mapsto -k$ case is, equivalently, also a solution. c does not enter because the time dependence washes out (energy-momentum conservation), while $\hbar$ may be absorbed into the units of the initial (conserved) wavenumber k. As un-normalizable plane waves, issues of normalization are moot. Perhaps you wish to provide more clearly and explicitly your normalization considerations.
In consequence, as you might expect for a free wave, the Wigner function is nothing but conservation of momentum, in any limit. You never had to take such.
It is independent of time, of course, and a plain flat line parallel to the x-axis, as in the non-relativistic case. 
