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?