# Fourier Transforming the Klein Gordon Equation

Starting with the Klein Gordon in position space, \begin{align*} \left(\frac{\partial^2}{\partial t^2} - \nabla^2+m^2\right)\phi(\mathbf{x},t) = 0 \end{align*} And using the Fourier Transform: $\displaystyle\phi(\mathbf{x},t) = \int \frac{d^3p}{(2\pi)^3}e^{i \mathbf{p} \cdot\mathbf{x}}\phi(\mathbf{p},t)$: \begin{align*} \int \frac{d^3p}{(2\pi)^3}\left(\frac{\partial^2}{\partial t^2} - \nabla^2+m^2\right)e^{i \mathbf{p} \cdot\mathbf{x}}\phi(\mathbf{p},t)&=0 \\ \int \frac{d^3p}{(2\pi)^3}e^{i \mathbf{p} \cdot\mathbf{x}}\left(\frac{\partial^2}{\partial t^2} +|\mathbf{p}|^2+m^2\right)\phi(\mathbf{p},t)&=0 \end{align*} Now I don't understand why we are able to get rid of the integral, to be left with \begin{align*} \left(\frac{\partial^2}{\partial t^2} +|\mathbf{p}|^2+m^2\right)\phi(\mathbf{p},t)=0 \end{align*}

• 4 good answers in 4 minutes! You hit the jackpot... Feb 25, 2014 at 12:58
• And, strangely, none of the answerers bothered to upvote the question. Feb 25, 2014 at 14:00

## 4 Answers

The functions $e^{i \bf p \cdot \bf x}$ as functions of $\bf x$ are linearly independent for different $\bf p$'s, hence every coefficient in the linear superposition (that is, in the integral) must be zero.

The reason you can get rid of the integral and the exponential is due to the uniqueness of the Fourier transform. Explicitly we have,

\begin{align} \int \frac{ \,d^3p }{ (2\pi)^3 } e ^{ i {\mathbf{p}} \cdot {\mathbf{x}} } \left( \partial _t ^2 + {\mathbf{p}} ^2 + m ^2 \right) \phi ( {\mathbf{p}} , t ) & = 0 \\ \int d ^3 x \frac{ \,d^3p }{ (2\pi)^3 } e ^{ i ( {\mathbf{p}} - {\mathbf{p}} ' ) \cdot {\mathbf{x}} } \left( \partial _t ^2 + {\mathbf{p}} ^2 + m ^2 \right) \phi ( {\mathbf{p}} , t ) & = 0 \\ \left( \partial _t ^2 + {\mathbf{p}} ^{ \prime 2} + m ^2 \right) \phi ( {\mathbf{p}'} , t ) & = 0 \end{align} where we have used,

$$\int d ^3 x e ^{- i ( {\mathbf{p}} - {\mathbf{p}} ' ) \cdot x } = \delta ( {\mathbf{p}} - {\mathbf{p}} ' )$$ and $$\int \frac{ d ^3 p }{ (2\pi)^3 } \delta ( {\mathbf{p}} - {\mathbf{p}} ' ) f ( {\mathbf{p}} ) = f ( {\mathbf{p}} ' )$$

• How did you get from the first line to the second line? why can you just take $p \rightarrow p-p'$ and integrate over $d^3x$? Feb 18 at 20:36

You can see $\left(\frac{\partial^2}{\partial t^2} +|\mathbf{p}|^2+m^2\right)\phi(\mathbf{p},t)$ as the Fourier transform of your function. And which is your function? 0, and what is 0-s Fourier transform, well zero.

Since $e^{i\textbf{p}\cdot \textbf{x}}$ is a plane wave, it will integrate to infinity when the integral is taken over all of momentum space. If the integral is to evaluate to zero then the remaining term $\left(\partial_t^2 + |\textbf{p}^2| + m^2\right)\phi\left(\textbf{p}, t\right)$ has to be identically zero.