# Decoupling of degrees of freedom in Klein-Gordon equation

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} • 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. – Holger R. H. Jan 20 '19 at 21:01
• Try to show that $\phi(p)$ satisfies (2.6) iff $\phi(x)$ satisfies (2.4). (also, your $p^2$ should be $\vec{p}^2$). – Oбжорoв Jan 21 '19 at 13:23

## 1 Answer

I also struggled with this point, but below is the solution I ended up with, which uses another Fourier transform.

Starting from the line where you got stuck, define: $$F(\vec{p},t) = \left(\partial_t^2 + \vec{p}^2 + m^2\right)\phi(\vec{p},t)$$ so we can write that line as: $$\int_{-\infty}^\infty \frac{d^3\vec{p}}{(2\pi)^3} F(\vec{p},t) e^{i\vec{p}\cdot\vec{x}} = 0$$ this expression says a function of $$\vec{x}$$ is zero for all values of $$\vec{x}$$. Since it is a function we can apply an inverse fourier transform to both sides of the equation, making it a function of a new momentum variable $$\vec{q}$$: \begin{aligned} \int_{-\infty}^\infty d\vec{x} \, e^{-i\vec{q}\cdot\vec{x}} \int_{-\infty}^\infty \frac{d^3\vec{p}}{(2\pi)^3} F(\vec{p},t) e^{i\vec{p}\cdot\vec{x}} &= 0 \\ \Rightarrow \int_{-\infty}^\infty d\vec{p} \, F(\vec{p},t) \left(\int_{-\infty}^\infty d\vec{x}^3 \frac{e^{i(\vec{p}-\vec{q})\cdot\vec{x}}}{(2\pi)^3} \right) &= 0 \end{aligned} Recognising that the term in parenthesis is a definition of the delta function, we can write: \begin{aligned} &\Rightarrow \int_{-\infty}^\infty d\vec{p} \, F(\vec{p},t) \delta^3(\vec{p}-\vec{q}) = 0 \\ &\Rightarrow F(\vec{q},t) = 0 \qquad \forall\vec{q} \\ &\Rightarrow \left(\partial_t^2 + \vec{p}^2 + m^2\right)\phi(\vec{p},t) = 0 \qquad \forall\vec{p} \end{aligned} where we have relabelled $$\vec{q}\to\vec{p}$$ in the last line.