Time-independent Klein-Gordon PDE

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

$$ipartial_tpsi=Hpsi=Epsi,$$

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

$$psi(x,t)=int_{-infty}^{infty}frac{mathrm{d}omega}{2pi}widetilde{psi}(x,omega),e^{-iomega t}.$$

The above equation tells us that the Fourier components of $psi$ can be written as

$$widetilde{psi}(x,omega)=2pisum_{n=0}^{infty}psi_n(x),delta(omega-E_n).$$

In fact, we could have started with the Fourier transformed wavefunction in the first place, and the Schrodinger equation ends up to be

$$Hwidetilde{psi}(x,omega)=omega,widetilde{psi}(x,omega).$$

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

$$partial_t^2psi(x,t)Longrightarrow -omega^2widetilde{psi}(x,omega).$$

Thus, the Klein-Gordon equation when acting in frequency space is given by

$$left(-omega^2-partial_x^2+m^2right)widetilde{psi}(x,omega)=0.$$

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).

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As a little bonus, you can go further and Fourier expand your field in both frequency and momentum space to get

$$psi(x,t)=intfrac{mathrm{d}omega}{2pi}frac{mathrm{d}k}{2pi}widetilde{psi}(k,omega),e^{i(kx-omega t)}.$$

In these variables, the Klein-Gordon equation takes the form

$$left(m^2-omega^2+k^2right)widetilde{psi}=0.$$

This implies that $widetilde{psi}$ must take the form

$$widetilde{psi}(k,omega)=2pi,C(k,omega),delta(m^2-omega^2+k^2).$$

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

$$delta(omega^2-k^2-m^2)=frac{1}{2omega_k}left[delta(omega-omega_k)+delta(omega+omega_k)right],$$

and thus, we have

$$psi(x,t)=intmathrm{d}omegaintfrac{mathrm{d}k}{2pi},frac{1}{2omega_k},C(omega,k),e^{i(kx-omega t)}left[delta(omega-omega_k)+delta(omega+omega_k)right].$$

Evaluating the delta functions and letting $C(omega_k,k)=A_k$ and $C(omega_k,-k)=B_k$, we have

$$psi(x,t)=intfrac{mathrm{d}k}{2pi}frac{1}{2omega_k}left[A_ke^{i(kx-omega_kt)}+B_ke^{-i(kx-omega_kt)}right].$$

This is the most general solution to the Klein-Gordon equation, and pops up all over the place in QFT textbooks.