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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 $0$?

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    $\begingroup$ To get time independent Schrödinger you never use $\imath\psi_{t}=E\psi$ rather you use separation of variables. $\endgroup$ – Alberto Navarro Jan 11 at 22:16
  • $\begingroup$ @Alberto Navarro I see. Would it still be possible to extract an energy value from the time independent equation? Or would it strictly require time due to it being relativistic? $\endgroup$ – Jepsilon Jan 12 at 1:17
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    $\begingroup$ @Alberto Navarro In the Schrödinger equation, separation of variables and finding stationary solutions ($i\partial_t\psi=E\psi$) are equivalent. $\endgroup$ – Bob Knighton May 27 at 21:26
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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.

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In the article referred to "time-independence" simply means $\psi({\bf r},t) = \psi({\bf r})$, which implies that $\psi_{tt}=0$.

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