# Heisenberg uncertainty principle in $d$ dimensions.

Suppose $f(x)$ is a $d$-dimensional real function and $\int_{R^{d}}|f(x)|^2dx=1$. Show that

$$(\int_{R^{d}}|x|^2|f(x)|^2dx)(\int_{R^{d}}|\xi|^2|\hat f(\xi)|^2d\xi)\geq\frac{d^2}{16\pi^2}$$

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

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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 $\sqrt{\frac{n}{4\pi}}$.
Why is it $$\|\nabla f \|_2 = 2 \pi \|\xi \hat f \|_2 \quad ?$$ What definition of Fourier transform are you using? – Brainstorming Jan 18 '15 at 19:21
I am using $$\hat{f}(\xi)=\int_{\mathbb{R}^n}f(x)\,e^{-2\pi ix\cdot\xi}\,\mathrm{d}x$$ It is the one for which $e^{-\pi x\cdot x}$ is its own Fourier Transform. – robjohn Jan 18 '15 at 19:24