# Solving Klein-Gordon-Foch equation with initial conditions

Suppose wave function $$\psi=\psi(\vec{r},t)$$ satisfies Klein-Gordon-Foch equation $$\square\psi=-m^2\psi.$$ Also suppose initial conditions $$\psi_0=\psi(\vec{r},0)$$ and $$\psi_{t0}=(\partial_t\psi)(\vec{r},0)$$

Let us try to find function $$\psi$$ using all of the conditions above. We know plane wave solutions of the KGF equation $$\psi=Ae^{i\vec{p}\vec{r}-iEt}$$ where momenta satisfies $$E^2=p^2+m^2.$$

Decomposing the function $$\psi_0$$ into initial waves using Fourier transform

$$F(\psi_0)=\frac{1}{(\sqrt{2\pi})^3}\int d^3r\,\psi_0e^{-i\vec{p}\vec{r}}$$

The Fourier transform gives us amplitude of every wave, so $$\psi$$ could be defined for positive or negative energy as

$$\psi=F^{-1}(F(\psi_0)e^{i\vec{p}\vec{r}-iEt})=\frac{1}{(\sqrt{2\pi})^3}\int d^3p\,e^{i\vec{p}\vec{r}-iEt}F(\psi_0)$$

Substituting $$F(\psi_0)$$

$$\psi=\frac{1}{(2\pi)^3}\int d^3p\,d^3r'\,e^{i\vec{p}(\vec{r}-\vec{r}')-iEt} \,\psi_0(\vec{r}')$$

I am not sure if this solution is correct and if it is not, please tell me where I have made mistake. And I am not sure about the fact, that $$\psi_{t0}$$ is not used in the solution.

Where was the mistake made and why don't I have a time derivative dependent solution?

• You can check if your function is a solution simply by replacing it in the differential equation and checking if it holds
– user319197
Commented Feb 20, 2023 at 19:17

The equation has positive and negative frequencies solutions precisely because it is second order. This two dimensional phase space allows you to set the initial conditions with $$\psi_0$$ and $$\psi_{t0}$$. You were a bit too fast when selecting only one mode.
From the wave decomposition, solutions of the KGE are uniquely written as: $$\psi(x,t)= \int\frac{d^3k}{(2\pi)^3}\left(a(k)e^{i(kx-\omega(k)t)}+ b(k)e^{i(kx+\omega(k)t)}\right)$$ With $$a,b$$ arbitrary complex functions is $$\psi$$ complex or $$a=b$$ when $$\psi$$ is real. I’ve chosen: $$\omega=\sqrt{k^2+m^2}$$ You can solve for $$a,b$$ by taking the Fourier transform of the initial conditions. I will rather use the notation: $$\tilde f(k)= \int d^3x f(x) e^{-ikx}\\ f(x)= \int\frac{d^3k}{(2\pi)^3} \tilde f(k) e^{ikx}$$
This gives you: \begin{align} \tilde \psi_0 &= a+b\\ \tilde \psi_{t0} &= -i\omega a+i\omega b \end{align} which you can solve to: \begin{align} a &= \frac{1}{2}\tilde\psi_0-\frac{1}{2i\omega}\tilde\psi_{t0}\\ b &= \frac{1}{2}\tilde\psi_0+\frac{1}{2i\omega}\tilde\psi_{t0} \end{align} Note the dependence on the full initial conditions. To make it more explicit, you can rewrite is as a sum of spatial convolutions: $$\psi=K*\psi_0+K_t*\psi_{t0}$$ with the kernels deduced from the previous formula: \begin{align} K(x,t) &= \int\frac{d^3k}{(2\pi)^3}\cos(\omega(k)t)e^{ikx} \\ K_t(x,t) &= \int\frac{d^3k}{(2\pi)^3}\frac{\sin(\omega(k)t)}{\omega(k)} e^{ikx} \end{align} Note that in the final formula, the sign choice of $$\omega$$ is irrelevant as it should be. Also note that $$K$$ gives the even in time contribution while $$K_t$$ gives the odd contribution as expected (think of $$\cos,\sin$$ when you only have one mode). I’m not sure that there exists a closed form, but if you are interested in large space/time asymptotics, the saddle point method will give you the answers directly.