The problem is to show that $$\psi=\psi_0e^{iP_\mu x^\mu}$$ is a solution to Klein Gordon equation $$(\partial _\mu \partial ^ \mu +m^2)\psi=0$$ if and only if the 4-momentum $p _\mu$ satisfies the relativistic dispersion relation. Tensor calculus is a new subject for me so I do not really know how to solve it but I assumed that $p$ is a momentum operator such that $$\partial _\mu \psi=i p_\mu \psi \mathrm{\,\ and \,\ } \partial^\mu \psi=ip^ \mu \psi$$ and then $$\partial _\mu \partial^ \mu \psi=ip^ \mu ip_\mu \psi=-p^2\psi=-m^2\psi\implies p^2=m^2$$ which leads to the answer. How do I solve it explicitly using differentials and not momentum operators?

  • $\begingroup$ In order to prove the "if and only if" statement you must demonstrate both the directions of the sentence. In the answer you shown the backward one (using the definition of $p_\mu$). Now try to prove the other direction (hint: put directly the expression of $\psi$ into the KG equation and do the derivatives) $\endgroup$ – ndrearu Feb 10 '18 at 21:09
  • $\begingroup$ Please take a minute to read our guidelines for homework and exercise questions as well as check-my-work questions. We intend our questions to be potentially useful to a broader set of users than just the one asking, and we prefer conceptual questions over those just asking for a specific computation. $\endgroup$ – Emilio Pisanty Feb 28 '18 at 13:44

You can show your result $\partial_\mu \psi = i p_\mu \psi$ by applying $\partial_\mu$ to $\psi$: $$ \partial_\mu \psi = \partial_\mu \left(\psi_0 e^{i p_\nu x^\nu}\right) = \psi \partial_\mu \left(i p_\nu x^\nu\right) = i p_\nu \psi \left(\partial_\mu x^\nu\right) = i p_\nu \delta_\mu^\nu \psi = i p_\mu \psi $$

  • $\begingroup$ Thank you but why in step $$\psi \partial _\mu (ip_\mu x^\mu)=ip_\mu \psi(\partial_\mu x^\mu)$$ the momentum $p_ \mu$ is factored out in front of the differential? Isn't the differentiation carried over it as well as position? $\endgroup$ – Convaly Feb 10 '18 at 21:31

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