Time-independent Klein-Gordon PDE Given the KG PDE:
$$\psi_{tt} - \psi_{xx} + m^2 \psi = 0.$$
Wikipedia describes the time-independent variant of this as just setting $\psi_{tt}=0$.
My question is this:
For the Schrödinger equation, the time independence is achieved by setting $i\psi_t = E\psi$, is it legittimate to consider setting $\psi_{tt}=E^2 \psi$ in the KG equation rather than $0$? Why is it preferencial to set it to $0$?
 A: The "time-independent" Schrodinger equation is called so because it doesn't contain time derivatives. The physical solutions, however, do contain explicit time dependence, as the energy eigenstates evolve as
$$i\partial_t\psi=H\psi=E\psi,$$
or
$$\psi(x,t)=\psi(x,0)e^{-iEt}.$$
This is physically irrelevant when only dealing with one energy level, but it very important when superimposing states from multiple energy levels. In this case, we would write
$$\psi(x,t)=\sum_{n=0}^{\infty}\psi_{n}(x)e^{-iE_nt}.$$
(Note later that the spectrum is discreet [for bound states] due to the physical requirement that $\psi$ is square normalizable.) Another way to write this is to introduce a Fourier transformed wavefunction in the frequency domain given by
$$\psi(x,t)=\int_{-\infty}^{\infty}\frac{\mathrm{d}\omega}{2\pi}\widetilde{\psi}(x,\omega)\,e^{-i\omega t}.$$
The above equation tells us that the Fourier components of $\psi$ can be written as
$$\widetilde{\psi}(x,\omega)=2\pi\sum_{n=0}^{\infty}\psi_n(x)\,\delta(\omega-E_n).$$
In fact, we could have started with the Fourier transformed wavefunction in the first place, and the Schrodinger equation ends up to be
$$H\widetilde{\psi}(x,\omega)=\omega\,\widetilde{\psi}(x,\omega).$$
That is, the time independent Schrodinger equation is just the normal Schrodinger equation in frequency space.
We can apply the same logic to the Klein-Gordin equation. We have
$$\partial_t^2\psi(x,t)\Longrightarrow -\omega^2\widetilde{\psi}(x,\omega).$$
Thus, the Klein-Gordon equation when acting in frequency space is given by
$$\left(-\omega^2-\partial_x^2+m^2\right)\widetilde{\psi}(x,\omega)=0.$$
This is the appropriate generalization of the time-independent Schrodinger equation.
The reason that wikipedia set $\partial^2_{t}\psi=0$ is because "time-independent" can be taken to mean that the function simple doesn't depend on time, whereas in the Schrodinger equation, "time-independent" should really be rephrased as "frequency space." Often the two usages don't overlap (after all, the Klein-Gordon equation isn't an evolution equation for a wavefunction).

As a little bonus, you can go further and Fourier expand your field in both frequency and momentum space to get
$$\psi(x,t)=\int\frac{\mathrm{d}\omega}{2\pi}\frac{\mathrm{d}k}{2\pi}\widetilde{\psi}(k,\omega)\,e^{i(kx-\omega t)}.$$
In these variables, the Klein-Gordon equation takes the form
$$\left(m^2-\omega^2+k^2\right)\widetilde{\psi}=0.$$
This implies that $\widetilde{\psi}$ must take the form
$$\widetilde{\psi}(k,\omega)=2\pi\,C(k,\omega)\,\delta(m^2-\omega^2+k^2).$$
Now, we have $\omega^2-k^2-m^2=(\omega-\omega_k)(\omega+\omega_k)$, where $\omega_k=\sqrt{m^2+k^2}$, and so
$$\delta(\omega^2-k^2-m^2)=\frac{1}{2\omega_k}\left[\delta(\omega-\omega_k)+\delta(\omega+\omega_k)\right],$$
and thus, we have
$$\psi(x,t)=\int\mathrm{d}\omega\int\frac{\mathrm{d}k}{2\pi}\,\frac{1}{2\omega_k}\,C(\omega,k)\,e^{i(kx-\omega t)}\left[\delta(\omega-\omega_k)+\delta(\omega+\omega_k)\right].$$
Evaluating the delta functions and letting $C(\omega_k,k)=A_k$ and $C(\omega_k,-k)=B_k$, we have
$$\psi(x,t)=\int\frac{\mathrm{d}k}{2\pi}\frac{1}{2\omega_k}\left[A_ke^{i(kx-\omega_kt)}+B_ke^{-i(kx-\omega_kt)}\right].$$
This is the most general solution to the Klein-Gordon equation, and pops up all over the place in QFT textbooks.
A: In the article referred to "time-independence" simply means $\psi({\bf r},t) = \psi({\bf r})$, which implies that $\psi_{tt}=0$. 
A: A quantum mechanical wave-function is not a measurable quantity. What you can measure are observables of the form $\langle\psi| A|\psi^*\rangle$. If $\psi(x,t)$ is of the form $\psi(x,t)=e^{-i \omega t} \phi(x)$ then all observables are time-independent. This means despite $\psi$ depending on $t$, the physical situation is time independent. The equation, which describes this time-independent physical situation is obtained by using the ansatz from above to modify the wave-equation. This equation is usually called the "time-independent wave equation". Just defining the time-independent wave equation by setting $\partial_t\, \psi(x,t)=0$ in the original equation is possible, but useless.
