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