Let $g \in C^1(\mathbb R)$ with $g, xg, g' \in L^2(\mathbb R)$ and $\int_{\mathbb R} g^2 d \lambda = 1$.
Why does then hold true that $$\int_{\mathbb R} x^2 g^2 (x) dx \cdot \int_{\mathbb R} \mathcal E^2 |\mathcal F g |^2 (\mathcal E) d \mathcal E \geq \frac{\pi}{2}$$
where $\mathcal F$ is the Fourier transform?
Best Answer
First, a few quick things:
$1$. The uncertainty principle typically either has no $\pi$ in it or there is a $\pi$ in the denominator, not the numerator. I will provide it with no $\pi$, as it implies the version with it in the denominator. Also, if you’re working with the square of the $L^2$ norms, then the factor is $1/4$. At least, this is always how I've seen it stated.
$2$. There are a lot of scaling conventions for the Fourier transform. I'm going to use the one that makes it unitary, but I'll leave some extra steps in case you do not.
Onto the proof, observe that $$\xi\cdot \mathcal{F} f(\xi)=\mathcal{F}\left(\frac{1}{i}\frac{d}{dx}f\right)(\xi),$$ and $$\left[x,\frac{1}{i}\frac{d}{dx}\right] f=if.$$ Using these relations, Plancherel, and Cauchy-Schwarz, we see that
\begin{align*} \left\|xf\right\|_{L^2}\|\xi\cdot \mathcal{F}f\|_{L^2}&=\left\|xf\right\|_{L^2}\left\|\mathcal{F}\left(\frac{1}{i}\frac{d}{dx}f\right)\right\|_{L^2}=\left\|xf\right\|_{L^2}\left\|\left(\frac{1}{i}\frac{d}{dx}f\right)\right\|_{L^2}\\ &\geq \left|\left\langle\frac{1}{i}\frac{d}{dx}f, xf\right\rangle\right|\geq \left|\text{Im}\left\langle\frac{1}{i}\frac{d}{dx}f, xf\right\rangle\right|\\ &=\frac{1}{2}\left|\left\langle\left[x,\frac{1}{i}\frac{d}{dx}\right]f,f\right\rangle\right|\\ &=\frac{1}{2}\left\|f\right\|_{L^2}^2=\frac{1}{2}\left\|f\right\|_{L^2}\left\|\mathcal{F}f\right\|_{L^2}\\ &=\frac{1}{2}. \end{align*}
If one, instead, uses the semiclassical Fourier transform, then we pick up a factor of $h$ (or $\hbar$, if you like), which is consistent with the traditional version that you'll see.