Assume $\psi$ is a solution of the Klein-Gordon equation (KGE). Let $\Lambda$ be a Lorentz transformation.

Show: $\phi = \psi(\Lambda^{-1} \cdot )$ is also a solution of the KGE.

I try to compute $$(\partial^\mu \partial_\mu \phi)(x) = (\partial^\mu\partial_\mu(\psi(\Lambda^{-1}\cdot))(x) = (\partial^\mu (\partial_\mu\psi)(\Lambda^{-1}\cdot)\underbrace{(\partial_\rho (\Lambda^{-1})_\nu^\rho x^\nu)}_{=\text{tr}\Lambda^{-1}})(x)$$

Is this correct so far? I am convinced this is wrong, but I don't know where! Can someone help me herefore.


I just figured out that my mistake was in the chain rule of derivatives: $$(\partial_\mu \psi(\Lambda^{-1} \cdot ) (x) = (\partial_\nu \psi)(\Lambda^{-1}x )(\partial_\mu (\Lambda^{-1}x)^\nu) = (\partial_\nu \psi)(\Lambda^{-1}x ) (\partial_\mu( (\Lambda^{-1})_\rho^\nu x^\rho)) =(\Lambda^{-1})^\nu_\mu (\partial_\nu \psi)(\Lambda^{-1}x ) $$

Considering this, it is easy to verify that $$(\partial^\mu\partial_\mu \phi)(x) = (\partial^\mu\partial_\mu \psi)(\Lambda^{-1} x)$$ and therefore obviously $\phi$ a solutionof the KGE.



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