In David Tong's notes in QFT he states that the degrees of freedom decouple in momentum space for the Klein-Gordon eq. He writes that this can be seen by using the Fourier transform (see picture below). I've tried to reproduce this below, but I cannot show the last equality. Does anyone have an idea on what to do?

\begin{align} 0&=\left(\partial_\mu\partial^\mu+m^2\right)\phi(\vec{x},t)\\ &=(\partial^2_t-\partial_i\partial^i+m^2)\phi(\vec{x},t)\\ &=(\partial^2_t+m^2)\phi(\vec{x},t)-\int_{-\infty}^{\infty}\frac{d^3p}{(2\pi)^3} \partial_i\partial^ie^{i\vec{p}\cdot\vec{x}}\phi(\vec{p},t)\\ &=(\partial^2_t+m^2)\phi(\vec{x},t)-\int_{-\infty}^{\infty}\frac{d^3p}{(2\pi)^3} (ip^i)\partial_ie^{i\vec{p}\cdot\vec{x}}\phi(\vec{p},t)\\ &=\int_{-\infty}^{\infty}\frac{d^3p}{(2\pi)^3}\left(\partial^2_t+p^2+m^2\right) e^{i\vec{p}\cdot\vec{x}}\phi(\vec{p},t)\\ &\stackrel{??!}{=}\left(\partial^2_t+p^2+m^2\right) \phi(\vec{p},t)\\ \end{align}

From David Tongs notes on QFT page 22, http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf

  • $\begingroup$ I realise that Peskin and Schroder do the same on page 20 of their QFT book. They do not add anything to the explanation though. $\endgroup$ – Holger R. H. Jan 20 at 21:01
  • $\begingroup$ Try to show that $\phi(p)$ satisfies (2.6) iff $\phi(x)$ satisfies (2.4). (also, your $p^2$ should be $\vec{p}^2$). $\endgroup$ – Oбжорoв Jan 21 at 13:23

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