Nonexistence of a Probability for the Klein-Gordon Equation David Bohm in his wonderful monograph Quantum Theory, in Section 4.6 discusses the difficulties one encounters in trying to develop a relativistic quantum mechanics. He starts from the relation 
\begin{equation}
\hbar^2 \omega^2 = m^2 c^4 + \hbar^2 k^2 c^4
\end{equation}
(which is equivalent to the classical relation $E^2=m^2 c^4 + p^2 c^2$), 
from which one derives (by proceeding as in Section 3.19) the second-order equation (Klein-Gordon equation):
\begin{equation}
\frac{\partial^2 \psi}{\partial t^2} = c^2 \Delta \psi - \frac{m^2 c^4}{\hbar^2} \psi.
\end{equation}
Then he tries to define a probability function $P$ involving $\psi$ and its partial drivatives $\frac{\partial \psi}{ \partial t}$, $\frac{\partial \psi}{\partial x_i}$:
\begin{equation}
P(x,t)= \hbar^2 \left| \frac{\partial \psi}{ \partial t} \right|^2 + \hbar^2 c^2 \lvert \nabla \psi \rvert^2 + m^2 c^4 \lvert \psi \rvert^2,
\end{equation}
which can be seen to have an integral $\int P(\mathbf{x},t) d\mathbf{x}$ which is conserved over time. Anyway, Bohm says that this function does not give rise to a physically acceptable probability, since if we choose e.g. $\psi= \exp i \left( \frac{Et-\mathbf{p} \cdot \mathbf{x} } {\hbar} \right)$, we get
\begin{equation}
P(x,t)=E^2+p^2c^2+m^2c^4=2E^2,
\end{equation}
so that $P$ behaves likes the (4,4)-component of a rank-2 tensor. From this he concludes that under a Lorentz transformation the integral $\int P(\mathbf{x},t) d\mathbf{x}$  transforms like an energy, that is like the fourth component of a four-vector, so it is not invariant (for a proof of the last statement see my post Tensors and the Klein-Gordon Equation). 
Bohm then states without proof that it is not possible to define any
(reasonable) probability density function, by using the solution $\psi$ of the wave equation above and its partial derivatives, which is invariant under Lorentz transformation. 
Does someone know some compelling reason why this is true?
 A: I have discovered that Kazemi, Hashamipour and Barati in their work Probability density of relativistic spinless particles suceeded in finding in the one-dimensional case a physically acceptable probability function for the Klein-Gordon equation. This probability function satisfies all the properties of a meaningful probability function, and in particular its integral is Lorentz invariant.
Anyhow, this probability function does not disprove Bohm's statement, since $P(x,t)$ does not depend on $\psi(x,t)$ and the values of the partial derivatives of $\psi$ computed in $(x,t)$, but it is a functional of $\psi$, that is it depends on the whole function $\psi$. So this probability function is not a counterexample to Bohm's statement.
Finally, I have found a very interesting work Uniqueness of conserved currents in quantum mechanics by Peter Holland, who shows that an essentially unique conserved four-vector current $\mathbf{J}$ exists for the Klein-Gordon equation, which has covariant components
\begin{equation}
J_{\mu} = \frac{i \hbar}{2m} \left( \psi^{*} \partial_{\mu} \psi - \psi \partial_{\mu} \psi^{*} \right).
\end{equation}
This current corresponds to the one usually defined for the Klein-Gordon equation. Its density component is $P=\frac{i \hbar}{2m} \left(\psi^{*}  \frac{\partial \psi}{\partial t} - \psi \frac{\partial \psi^{*}}{\partial t} \right) $, we see that $P$ does not satisfy the property $P \geq 0$ so that Bohm's statement follows. 
We must anyhow remark that Holland assumes in his proof that that $\mathbf{J}$ depends only on $\psi$ and its first derivatives, which is an assumption not explicitly made by Bohm, even though a perfectly plausible one (see the argument given by Holland in his work to justify the requirement that conserved currents should depend solely on the 'state variables').
