# Fourier transforming the Klein-Gordon equation

I'm aware of the fact that there are similar questions on this forum but I could not find an answer that fits my problem.

Many textbooks state that a general solution to the Klein-Gordon equation $$\left(\partial_\mu \partial^\mu + \left(\frac{mc}{\hbar}\right)^2\right) \psi(x^\mu) = 0\qquad (1)$$ is given by $$\psi(x^\mu) = \int \frac{d^4k}{\sqrt{2\pi}^4}\delta\left(k_\mu k^\mu-\left(\frac{mc}{\hbar}\right)^2\right) A(k^\mu) \text{e}^{-ik_\mu x^\mu},$$ where $k^\mu$ is the Lorentz invariant wave four-vector, $\delta(.)$ is the $\delta$-distribution and $A(k^\mu)$ is some arbitrary complex function.

I assume that this result is obtained by applying a Fourier transformation to equation $(1)$, but I cannot find out where the $\delta$-function in the integral comes from. The solution cannot be that difficult (since I've not found an answer yet), so I hope someone is willing to show me how one gets the expression for $\psi(x^\mu)$ by a Fourier transformation of the Klein-Gordon equation.

• All solutions of the K-G equation fulfill a dispersion equation which is $(\frac{mc}{\not h})^2 = k_\mu k^\mu = k_0^2 - \vec{k}^2$, therefore space-components and time-components of the $k$-4-vector are not independent. – Frederic Thomas Oct 25 '17 at 14:00
• answer here: physics.stackexchange.com/a/216194/84967 – AccidentalFourierTransform Oct 25 '17 at 14:20

Taking the Fourier transform of both sides of your starting equation gives $$(-k^2 + m^2) \tilde{\psi}(k) = 0$$ where I set some constants to one. So for every $k$, at least one of these factors must be zero. If the first factor is not zero, then $\tilde{\psi}(k)$ is, so I might as well write $$\tilde{\psi}(k) = \delta(-k^2 + m^2) A(k)$$ to enforce this; this yields a valid $\tilde{\psi}(k)$ given any $A(k)$. Applying an inverse Fourier transform gives your second equation.
• I think this solves my problem. Could you eloborate a little bit on why you chose the $\delta$-distribution, though? I want to make sure that I got it right. – MeMeansMe Oct 25 '17 at 17:04
• @MeMeansMe It’s arbitrary, I could have taken anything that vanishes for $k^2 \neq m^2$. The delta is nice because it lets you explicitly get rid of one of the $k$ integrals if desired. – knzhou Oct 25 '17 at 17:17