# How to fourier invert a plane wave

1. Mar 28, 2009

### waht

1. The problem statement, all variables and given/known data

Ok so got a solution to the Klein-Gordon equation and need to solve for a(k)

$$\varphi(x) = \int \tilde{dk} \left[ a(\bold{k}) e^{ikx} + a^{\dagger}(\bold{k} ) e^{-ikx} \right]$$

$$\tilde{dk} = \frac{d^{3}k}{(2 \pi)^{3} 2 \omega}$$

The way it's done in Srednicki p.26 has me confused when taking the fourier transform of $\varphi$

$$\int d^3x e^{-ikx} \varphi(x) = \frac{1}{2\omega} a(\bold{k}) + \frac{1}{2\omega} e^{2i\omega t} a^{\dagger}(\bold{-k} )$$

2. Relevant equations

$$kx = \bold{k} \cdot \bold{x} - \omega t$$

3. The attempt at a solution

$$\int d^3x e^{-ikx} \varphi(x) =\int d^3x e^{-ikx} \int \tilde{dk} \left[ a(\bold{k}) e^{ikx} + a^{\dagger}(\bold{k} ) e^{-ikx} \right]$$

$$= \int d^3x \int \tilde{dk} a(\bold{k}) + \int d^3x \int \tilde{dk} e^{-2kx} a^{\dagger}(\bold{k} )$$

so the problem is how do these integrals with respect to dx and dk disappear?

2. Mar 29, 2009

### tiny-tim

dummy!

Hi waht!

It's a different k …

you have ∫∫ eikx eik'x … for a dummy k'