# Fourier transform and wave packet

Given a generic integrable function $$f \in L^1(\mathbb{R})$$, we define its Fourier transform in the following way:

$$\mathcal{F}_f(y)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} f(x)e^{-ixy}\text{d}x$$

We can then show that (if $$\mathcal{F}_f \in L^1(\mathbb{R})$$):

$$f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \mathcal{F}_f(y)e^{ixy}\text{d}y$$

That makes sense for me.

Studying wave packets, given by the superposition of infinitely many plane waves, I found the following formula for a generic component of the electric field associated to the wave packet:

$$E(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \mathcal{F}_E(k)e^{i(kx-\omega t)}\text{d}k$$, where $$\omega=\omega(k)$$ is a function of $$k$$ and $$\mathcal{F}_E(k)$$ is the Fourier transform of $$E$$ at time $$t=0$$, namely $$\mathcal{F}_E(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} E(x,0)e^{-ikx}\text{d}x$$

I don't know how to justify this formula for $$E(x,t)$$. Namely, I don't understand how from the well known Fourier transform:

$$E(x,0)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \mathcal{F}_E(k)e^{ikx}\text{d}k$$

we derive the following expression:

$$E(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \mathcal{F}_E(k)e^{i(kx-\omega(k) t)}\text{d}k$$

To see why, one must recognize the time evolution of a single frequency mode. The electromagnetic wave must satisfy the wave equation $$\frac{\partial^2 E}{\partial t^2}=c^2\nabla^2 E$$, so for a given signed angular frequency $$\omega$$, an oscillation in complex exponential form must be given as $$\exp(i(kx-\omega t))$$. Since the wave equation is linear, the time evolution of an initial state $$E=\frac{1}{\sqrt{2\pi}}\int \mathcal{F}_E(k)\exp(ikx)dk$$ which is is superposition of the $$\exp(ikx)$$ will be a superposition of the functions satisfying $$f(x,0)=\exp(ikx),\quad \frac{\partial^2 f}{\partial t^2}= c^2 \frac{\partial^2 f}{\partial x^2}$$ namely $$\exp(i(kx-\omega t))$$.
• Thank you so much! Got it! You are saying that at $t=0$ we have that $E(x,0)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \mathcal{F}_E(k)e^{ikx}\text{d}k$, and THEN for a generic $t$ we have that $E(x,t)$ is given by the superposition of the plane waves evoluted in time according to D'Alembert wave equation. Am I right? Aug 21, 2021 at 21:27