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?

  • 3
    $\begingroup$ There is some uncertainty about this, is there some principle involved? Perhaps Cauchy Schwartz has something to say about it? $\endgroup$
    – copper.hat
    Commented Dec 10, 2019 at 14:35
  • $\begingroup$ Use how the Fourier transform relates to multiplication, the Cauchy-Schwarz inequality, relate everything to a commutator that you can compute directly, then use Plancheral. $\endgroup$
    – cmk
    Commented Dec 10, 2019 at 15:20
  • $\begingroup$ Totally agree with @cmk , with the exception that the name is Plancherel. $\endgroup$
    – Maksim
    Commented Dec 12, 2019 at 14:21
  • $\begingroup$ @cmk How does this work with Plancherel? $\endgroup$
    – StMan
    Commented Dec 12, 2019 at 18:15
  • $\begingroup$ And @Maksim too; I edited the question. $\endgroup$
    – StMan
    Commented Dec 12, 2019 at 18:16

1 Answer 1


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.


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