How to Fourier Invert a Plane Wave?

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 2K views
waht
Messages
1,502
Reaction score
4

Homework Statement



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

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

[tex]\tilde{dk} = \frac{d^{3}k}{(2 \pi)^{3} 2 \omega}[/tex]

The way it's done in Srednicki p.26 has me confused when taking the Fourier transform of [itex]\varphi[/itex]

[tex]\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} )[/tex]

Homework Equations



[tex]kx = \bold{k} \cdot \bold{x} - \omega t[/tex]


The Attempt at a Solution




[tex]\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][/tex]


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


so the problem is how do these integrals with respect to dx and dk disappear?
 
Physics news on Phys.org
dummy!

Hi what! :smile:

It's a different k …

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