Tensors and the Klein-Gordon Equation Consider the 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}
and define for each one of its solutions $\psi$ the quantity:
\begin{equation}
P(\mathbf{x},t)= \hbar^2  \frac{\partial \psi}{ \partial t} \frac{\partial \psi^{*}}{ \partial t} + \hbar^2 c^2  \nabla \psi \cdot \nabla \psi^{*} + m^2 c^4 \psi \psi^{*},
\end{equation}
Let us adopt the convention in which the generic point of the Minkowski space-time is $(x,y,z,ct)$. In Section (4.6) of his wonderful treatise Quantum Theory  Bohm states that under a Lorentz transformation
(i) $P(\mathbf{x},t)$ transforms as the (4,4)-coordinate a rank two tensor,
(ii) $\int P(\mathbf{x},t) d\mathbf{x}$ transforms as the fourth component of a four-vector.
Could someone give me a proof of these two statements, please?
NOTE (1). All that I know about the Klein-Gordon equation is that $\psi$ is invariant under Lorentz transformations, that is if $\psi(\mathbf{x},t)$ is a solution of the Klein-Gordon equation, then the new function $\phi(\mathbf{x'},t')$ obtained by replacing the equations of a Lorentz boost $(\mathbf{x},ct) \rightarrow (\mathbf{x'},ct')$ in $\psi$ is again a solution of the Klein-Gordon equation.
NOTE (2). Bohm justifies assertion (i) by considering the particular solution $\psi= \exp i \left( \frac{Et-\mathbf{p} \cdot \mathbf{x} } {\hbar} \right)$, for which we get
\begin{equation}
P=E^2+p^2c^2+m^2c^4=2E^2,
\end{equation}
so that $P$ transforms actually as the square of an energy.
 A: Actually, the $P(\mathbf{x},t)$ you mention is the 4,4-component of the stress-tensor $T_{ik}$ of the Klein-Gordon (K-G) field. In the following instead I will use the metric tensor $\eta_{ik}=diag(1,-1,-1,-1)$ and identify 
$P(\mathbf{x},t)$ with the 0,0-component of $T_{ik}$. Bohm aparently uses the other metric $diag(-1,-1,-1,1)$ convention. Moreover $c=1=\hbar$ is assumed.
One starts best off from the Lagrange density of the complex K-G: (Double appearing indices is summed over, i.e. Einstein summation convention):
$$ L = \partial_i \phi^\star \partial^i \phi - m^2 \phi^\star \phi$$
For a complex field $\phi$ and $\phi^\star$ are considered as independent variables. The definition of the stress tensor is given by: 
$$T_{ik}=\sum_{\varphi}\varphi_{,k}\frac{\partial L}{\partial \varphi^{,i}} -L\eta_{ik}$$
with $\varphi = (\phi, \phi^\star)$.
Upon inserting the expression for the Lagrange density of the K-G field in the stress tensor's definition we get: 
$$T_{ik} = \partial_i\phi^\star \partial_k\phi + \partial_k\phi^\star \partial_i\phi -L\eta_{ik}$$
The tensor property of $T_{ik}$ is rather obvious, as the partial derivatives $\partial_i$ respectively $\partial_k$ transform like covariant vectors and $\eta_{ik}$ is also a tensor. This is in particular true for the $(i,k)=(0,0)$-component:
$$T_{00} = 2\frac{\partial\phi^\star}{\partial t}\frac{\partial\phi}{\partial t} -L=\frac{\partial\phi^\star}{\partial t}\frac{\partial\phi}{\partial t}+\nabla\phi^\star\nabla\phi + m^2\phi^\star \phi\equiv P(\mathbf{x},t) $$
The 4-momentum vector $P_i$ yields: 
$$P_i = \int_{\partial\Omega} T_{ik} d\sigma^k$$
where $d\sigma^k$ is the vectorial hypersurface element which is parametrized by values $(u,v,w)$: 
$$d\sigma_i =\epsilon_{ijkm}\frac{\partial x^j}{\partial u}\frac{\partial x^k}{\partial v}\frac{\partial x^m}{\partial w}du dv dw$$
If now the hypersurface $t=const$ is chosen, as parameters $(u,v,w)=(x,y,z)\equiv(x^1,x^2,x^3)$ can be used:
$$d\sigma_i =\epsilon_{ijkm}\delta^j_1\delta^k_2\delta^m_3 dx^1 dx^2 dx^3= \epsilon_{i,1,2,3}d^3x = (d^3x,\mathbf{0})$$
so we get for this particular hypersurface $t=const$: 
$$P_i = \int_{t=const} T_{i0} d^3x $$
It can be shown that $P_i$ considered at another hypersurface $\partial\Omega$ which is Lorentz-transformed with respect to the original hypersurface $t=const$ has the same value if it is computed by the more general formula: 
$$P_i = \int_{\partial\Omega} T_{ik} d\sigma^k$$
But due to the covariant way of writing it is clear that  $P_i$ is a 4-vector  (but this would not be longer true in curved space-time)  and in particular 
$$P_0 = \int_{t=const} T_{00} d^3x  \equiv \int_{t=const}P(\mathbf{x},t) d^3x $$
the 0-component of the 4-vector $P_i$ (the momentum 4-vector), the energy of the K-G field. 
