# Show: Lorentz-invariance of solution of Klein-Gordon equation [closed]

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.

EDIT:

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.