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.

  • $\begingroup$ 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. $\endgroup$ Oct 25, 2017 at 14:00
  • 2
    $\begingroup$ answer here: physics.stackexchange.com/a/216194/84967 $\endgroup$ Oct 25, 2017 at 14:20

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – MeMeansMe
    Oct 25, 2017 at 17:04
  • $\begingroup$ @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. $\endgroup$
    – knzhou
    Oct 25, 2017 at 17:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.