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




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


(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


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


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


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


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


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


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


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