How is the Fourier uncertainty obtained $\sigma_x\sigma_f\ge1/4\pi$ for function $f$ and its Fourier transform $F$ ?
where the $\sigma_x=\bar{x^2}-\bar{x}^2$, $\sigma_f=\bar{f^2}-\bar{f}^2$ are the variances. The moments give $$\bar{x^2}=\frac{-1}{4\pi^2}\frac{F^{(2)}(0)}{F(0)}, \bar{x}=\frac{-1}{2\pi i}\frac{F^{(1)}(0)}{F(0)}$$
$$\sigma_x=\frac{-1}{4\pi^2}\frac{F^{(2)}(0)}{F(0)}+\frac{1}{4\pi^2}[\frac{F^{(1)}(0)}{F(0)}]^2$$
We have a function $\psi (x)$ with $\tilde{\psi}(f):=\int_{\Bbb R}\psi (x)\exp 2\pi ifx dx$ and without loss of generality $0=\overline{x}=\int_{\Bbb R}\psi^\ast x\psi dx,\,0=\overline{f}=\int_{\Bbb R}\tilde{\psi}^\ast f\tilde{\psi}df$, so $\sigma_x^2=\langle \psi|x^2|\psi\rangle,\,\sigma_f^2=\langle\psi|f^2|\psi\rangle$ where $\langle g|A|h\rangle:=\int_{\Bbb R}g^\ast Ah$ is an inner product on the vector space of functions $\Bbb R\to\Bbb C$, with $A$ a linear operator from that space to itself. Integrating by parts,$$f\tilde{\psi}(f)=\frac{1}{2\pi i}\int_{\Bbb R}\psi (x)\partial_x\exp 2\pi ifx dx=\frac{i}{2\pi}\int_{\Bbb R}\psi'(x)\exp 2\pi ifx dx=\frac{i}{2\pi}\widetilde{\psi'},$$so $f=\frac{i}{2\pi}\partial_x$ in the sense of operators, giving the commutator $[x,\,f]=-\frac{i}{2\pi}$. Next we have use for an inequality on $\Bbb C$, $|z|^2\ge\Im^2 z=\big(\tfrac{z-z^\ast}{2i}\big)^2$. Hence$$|\langle g|AB|g\rangle|^2\ge\bigg(\frac{\langle g|AB-B^\dagger A^\dagger|g\rangle}{2i}\bigg)^2.$$ If $A,\,B$ are Hermitian, the Cauchy-Schwarz inequality gives$$\langle g|A^2|g\rangle\langle g|B^2|g\rangle\ge|\langle g|AB|g\rangle|^2\ge\bigg(\frac{\langle g|[A,\,B]|g\rangle}{2i}\bigg)^2.$$The choice $A=x,\,B=f$ gives $\sigma_x^2\sigma_f^2\ge\big(\tfrac{1}{4\pi}\big)^2,\,\sigma_x\sigma_f\ge\tfrac{1}{4\pi}$. See here for a more thorough treatment that discusses some unbounded-operator issues I've glossed over.