Solving Klein-Gordon-Foch equation with initial conditions

Question

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?

Answer 1

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.

Hope this helps.