# 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 $$0$$?

• To get time independent Schrödinger you never use $\imath\psi_{t}=E\psi$ rather you use separation of variables. Commented Jan 11, 2019 at 22:16
• @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? Commented Jan 12, 2019 at 1:17
• @Alberto Navarro In the Schrödinger equation, separation of variables and finding stationary solutions ($i\partial_t\psi=E\psi$) are equivalent. Commented May 27, 2019 at 21:26

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.

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

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.