TITLE: Heisenberg uncertainty principle in $d$ dimensions
QUESTION [8 upvotes]: Suppose $f(x)$ is a $d$-dimensional real function and $$\int_{R^{d}}|f(x)|^2 \,\mathrm d x=1$$  Show that
  $$ \left( \int_{R^{d}}|x|^2|f(x)|^2 \,\mathrm d x \right) \left( \int_{R^{d}}|\xi|^2|\hat f(\xi)|^2 \,\mathrm d\xi \right) \geq \frac{d^2}{16\pi^2}$$

I derived that $$1=\int_{R^{d}}x \left(\frac{d}{dx}\right)|f(x)|^2 \,\mathrm dx$$
but I lost my way. I need your help.

REPLY [12 votes]: Consider the equation
$$
\sum_{i=1}^n\frac12x_i\frac{\mathrm{d}}{\mathrm{d}x_i}|f|^2=\mathrm{Re}\left(\nabla f\cdot\overline{xf}\right)\tag{1}
$$
Integrating $(1)$ over $\mathbb{R}^n$ and then integrating by parts on the left side:
$$
\begin{align}
\frac n2\|f\|_2^2
&=\mathrm{Re}\left(-\int_{\mathbb{R}^n}\nabla f\cdot\overline{xf}\,\mathrm{d}x\right)\\
&\le\left|\int_{\mathbb{R}^n}\nabla f\cdot\overline{xf}\,\mathrm{d}x\right|\\[6pt]
&\le\|\nabla f\|_2\|xf\|_2\\[9pt]
&=2\pi\|\xi\hat{f}\|_2\|xf\|_2\tag{2}
\end{align}
$$
Thus,
$$
\|\xi\hat{f}\|_2\|xf\|_2\ge\frac{n}{4\pi}\|\hat{f}\|_2\|f\|_2\tag{3}
$$
The last inequality says that the $L^2$ support radius for $f$ and $\hat{f}$ cannot have a product less than $\frac{n}{4\pi}$.  This inequality is sharp as can be seen using the function $f(x) = e^{-\pi x\cdot x}$, whose Fourier Transform is itself, and whose $L^2$ support radius is $\sqrt{\frac{n}{4\pi}}$.