How to Fourier Invert a Plane Wave?

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SUMMARY

The discussion focuses on solving the Fourier transform of the Klein-Gordon equation, specifically the expression for \(\varphi(x)\) involving the terms \(a(\bold{k})\) and \(a^{\dagger}(\bold{k})\). The confusion arises from the integration process, particularly in the transition from the double integral involving \(d^3x\) and \(\tilde{dk}\) to the simplified result. The key takeaway is the distinction between the dummy variable \(k'\) and the original wave vector \(k\), which clarifies the integration steps outlined in Srednicki's text.

PREREQUISITES
  • Understanding of the Klein-Gordon equation
  • Familiarity with Fourier transforms in three dimensions
  • Knowledge of quantum field theory concepts, particularly creation and annihilation operators
  • Basic proficiency in mathematical integration techniques
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  • Study the Fourier transform properties in quantum field theory
  • Review Srednicki's Quantum Field Theory, particularly Chapter 1 on the Klein-Gordon equation
  • Learn about the role of dummy variables in multiple integrals
  • Explore advanced integration techniques for handling complex wave functions
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Students and researchers in theoretical physics, particularly those focused on quantum field theory and the mathematical foundations of particle physics.

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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?
 
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dummy!

Hi what! :smile:

It's a different k …

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

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