# 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. – Alberto Navarro Jan 11 '19 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? – Jepsilon Jan 12 '19 at 1:17
• @Alberto Navarro In the Schrödinger equation, separation of variables and finding stationary solutions ($i\partial_t\psi=E\psi$) are equivalent. – Bob Knighton May 27 '19 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.