I personally find the proof of the uncertainty principle quite amusing. Here it is, if any of you are interested. DHB Let us consider a real, continuously differentiable, $L^2$ function $f(t)$, which further satisfies $f(t) \sqrt{t} \rightarrow 0$ as $|t| \rightarrow \infty$. For convenience, we will also assume here that $f(-t) = f(t)$, as this insures that the Fourier transform $\hat{f}(x)$ of $f(t)$ is purely real, although in general this condition is not necessary. Define \begin{eqnarray*} E(f) &=& \int_{-\infty}^\infty f^2(t) \, dt \\ V(f) &=& \int_{-\infty}^\infty t^2 f^2(t) \, dt \\ \hat{f}(x) &=& \int_{-\infty}^\infty f(t) e^{-i t x} \, dt \\ Q(f) &=& \frac{V(f)}{E(f)} \cdot \frac{V(\hat{f})}{E(\hat{f})} \end{eqnarray*} \vspace{1ex}\noindent {\bf Theorem} With the above assumptions and definitions, $Q(f) \geq 1/4$, with equality if and only if $f(t) = a e^{-(b t)^2/2}$ for real constants $a$ and $b$. \vspace{1ex}\noindent {\bf Proof.} By applying Schwarz' inequality to the functions $t f(t)$ and $f'(t)$, we can write \begin{eqnarray*} \left| \int_{-\infty}^\infty t f(t) f'(t) \, dt \right|^2 &\leq& \left[\int_{-\infty}^\infty t^2 f^2(t) \, dt \right] \left[\int_{-\infty}^\infty (f'(t))^2 dt \right] \end{eqnarray*} \noindent Furthermore, \begin{eqnarray*} \int_{-\infty}^\infty t f(t) f'(t) \, dt &=& \frac{1}{2} \int_{-\infty}^\infty t \frac{d f^2(t)}{dt} \, dt \; = \; -\frac{1}{2} E(f) \end{eqnarray*} \noindent by applying integration by parts. Let $g(t) = f'(t)$. By noting that $\hat{g}(x) = - i x \hat{f}(x)$, and by applying Parseval's identity \cite[pg. 65]{papoulis1977} to $f(t)$ and $g(t)$, we obtain, respectively, \begin{eqnarray*} \int_{-\infty}^\infty |f(t)|^2 \, dt &=& \frac{1}{2 \pi} \int_{-\infty}^\infty |\hat{f}(x)|^2 \, dx \\ \int_{-\infty}^\infty |f'(t)|^2 \, dt &=& \frac{1}{2 \pi} \int_{-\infty}^\infty x^2 |\hat{f}(x)|^2 \, dx \end{eqnarray*} \noindent Combining these results, we obtain our desired inequality. Recall that equality in the Schwarz inequality occurs only when the two functions ($t f(t)$ and $f'(t)$ in this case) are linear scalings of each other, i.e. $f'(t) = c t f(t)$ for some $c$. By solving this elementary differential equation, we conclude that the minimum value $1/4$ is achieved if and only if $f(t) = a e^{-(b t)^2/2}$ for constants $a$ and $b$. By the way, it is worth noting that $Q(f)$ is unaffected by a linear scaling of either the function argument $t$ or the function value $f(t)$.