# Heisenberg uncertainty principle: a proof for a beginner

Scrolling in a book on real analysis I found, as a last exercise, the request to prove the Heisenberg uncertainty principle. The exercise states

Let $$f\in \mathcal L_{1}^{2}(\mathbb R)$$, such that $$\|f\|_{2}=\|\hat f\|_{2}=1$$. Prove that $$\left(\int_{\mathbb R}|x|^{2}|f(x)|^{2}\,\mathrm dx\right)\cdot\left(\int_{\mathbb R}|\xi|^{2}|\hat f(\xi)|^{2}\,\mathrm d\xi\right)\geq \frac{1}{(4\pi)^{2}}$$ Hint: suppose that $$f\in C_{c}^{\infty}$$ and use the following identities: $$\int_{\mathbb R}x\overline{f}(x)f'(x) = -{1\over 2} \|f\|_{2}^{s}$$ Plancherel's identity and Cauchy-Schwarz inequality.

The book states that $$\mathcal L^2_s(\mathbb R^d) := \{f:\mathbb R^d\rightarrow\mathbb C \text{ measurable} \text{ : }(1+|x|^2)^{s\over 2}f\in L^2(\mathbb R^d)\}$$ which calles it weighted $$L^2$$ spaces.

I was very curious on how to solve this problem because I don't have nearly as much knowledge as it's needed to solve this! So I'm here to ask you if you could give me the solution.

I can't give you my working because, as just stated, I don't know much about real analysis and I was just very curious to see the solution! Probably I'll understand it if I see one but searching on the internet didn't gave me any help.

• When you copy text from a book, you ought to identify what book it is. – Nate Eldredge Sep 21 '18 at 16:19
• @NateEldredge I'm sorry but probably I used the word book in the wrong way. What I was reading are "notes": in practise a book on real analysis wrote by my professor that he gave to us some time ago – Local Mathmatician Sep 21 '18 at 17:19

You have by the given identities

$$\frac{1}{4} = \left(-\frac{1}{2}\|f\|_2^ 2 \right)^2 = \left(\int_\mathbb{R}{x \overline{f}(x) f'(x) dx} \right)^2 \le \|x \mapsto x \overline{f}(x) \|_2^2 \cdot \|f'\|_2^2 = \|x \mapsto x \overline{f}(x) \|_2^2 \cdot \|\widehat{f'}\|_2^2$$

The first inequality is Cauchy Schwarz and the last equality is Plancherel Theorem. Using $$\|\widehat{f'}\|_2 = 2\pi \| x \mapsto x\hat{f}(x)\|_2$$ we get

$$\frac{1}{(4 \pi)^2} \leq \|x \mapsto x \overline{f}(x) \|_2^ 2 \| x \mapsto x\hat{f}(x)\|_2^ 2 = \left( \int_\mathbb{R} {|x|}^2{|f(x)|}^2dx \right) \cdot \left( \int_\mathbb{R} {|x|}^2{|\hat{f}(x)|}^2dx \right)$$

• Just a square typo on the pie. – mathreadler Sep 21 '18 at 13:44
• Oh sry. And thanks for the info. – Christian Sep 21 '18 at 13:45
• Thanks for the solution! The only obscure thing to me now is the notation $$\|x\rightarrow x\overline{f}(x)\|_2$$ what's that all about? – Local Mathmatician Sep 21 '18 at 13:47
• It's he $L^2$ norm of the mapping $g(x) = x \overline{f}(x)$. It's just short hand syntax to not explicity define the new map every time. – Christian Sep 21 '18 at 13:50
• @DavideMorgante one has to choose if one wants to be exactly precise in your language, or if one wants to avoid awkward notation. If you look in papers, you might see people write $\| x\bar f\|$ – Calvin Khor Sep 21 '18 at 13:56