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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 \begin{equation} \left(\partial_\mu \partial^\mu + \left(\frac{mc}{\hbar}\right)^2\right) \psi(x^\mu) = 0\qquad (1) \end{equation} 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.

MeMeansMe
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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.

knzhou
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