# Klein Gordon eqn, decoupling degrees of freedom

1. Oct 24, 2011

### Onamor

Having some trouble following my notes in QFT. Any help greatly appreciated.

We have the Klein Gordon equation for a real scalar field $\phi\left(\overline{x},t\right)$; $\partial_{\mu}\partial^{\mu}\phi + m^{2}\phi = 0$.

To exhibit the coordinates in which the degrees of freedom decouple from each other, we take the Fourier transform, $\phi\left(\overline{x},t\right)= \int\frac{d^{3}p}{\left(2\pi\right)^{3}}e^{i \overline{p} .\overline{x}}\phi\left(\overline{p},t\right)$.

Then $\phi\left(\overline{p},t\right)$ satisfies $\left(\frac{\partial^{2}}{\partial t^{2}}+\left(\overline{p}^{2} + m^{2}\right)\right)\phi\left(\overline{p},t\right) = 0$.

If you do it by brute force you get $\int\frac{d^{3}p}{\left(2\pi\right)^{3}}e^{ i \overline{p} .\overline{x}}\frac{\partial^{2}}{\partial t^{2}}\phi\left(\overline{p},t\right) - \int\frac{d^{3}p}{\left(2\pi\right)^{3}}\partial^{2}_{i}e^{ i \overline{p} . \overline{x}}\phi\left(\overline{p},t\right) + m^{2}\int\frac{d^{3}p}{\left(2\pi\right)^{3}}e^{ i \overline{p} . \overline{x}}\phi\left(\overline{p},t\right) = 0$

then $\int\frac{d^{3}p}{\left(2\pi\right)^{3}}e^{ i \overline{p} .\overline{x}}\frac{\partial^{2}}{\partial t^{2}}\phi\left(\overline{p},t\right) + \int\frac{d^{3}p}{\left(2\pi\right)^{3}} \overline{p}^{2} e^{ i \overline{p} . \overline{x}}\phi\left(\overline{p},t\right) + m^{2}\int\frac{d^{3}p}{\left(2\pi\right)^{3}}e^{ i \overline{p} . \overline{x}}\phi\left(\overline{p},t\right) = 0$

Now I dont see how to get rid of the intergrals. I can see its similar to a delta function, but you cant just take the $\phi\left(\overline{p},t\right)$ out of the integrals because the measure is $p$.

Thanks for helping me with this, please let me know if I haven't been clear.

Last edited: Oct 24, 2011
2. Oct 25, 2011

### mahnamahna

Your last line can be rewritten as
$\int\frac{d^{3}p}{\left(2\pi\right)^{3}}[\frac{\partial^{2}}{\partial t^{2}}\phi\left(\overline{p},t\right) + \overline{p}^{2} \phi\left(\overline{p},t\right) + m^{2}\phi\left(\overline{p},t\right)]e^{ i \overline{p} .\overline{x}} = 0$
Each term of e^{ip.x} is linearly independent (as they form a basis) and thus each term must be identically 0, giving the desired relation. Identically you can say the fourier transform of $\frac{\partial^{2}}{\partial t^{2}}\phi\left(\overline{p},t\right) + \overline{p}^{2} \phi\left(\overline{p},t\right) + m^{2}\phi\left(\overline{p},t\right)$ is 0. So it must be 0 as well.

3. Oct 25, 2011

### Onamor

Ah, thats why. Thanks very much, much appreciated.