Theorem (Heisenberg Uncertainty Principle).
Let $f \in \mathcal{S}(\mathbb{R})$. Then:
Equality is achieved by $f(x) \propto \exp(-x^2/2)$.
We claim that (H) is equivalent to
\[\int_{\mathbb{R}} x^2 |f(x)|^2\,dx \;+\; \int_{\mathbb{R}} \xi^2 |\hat f(\xi)|^2\,d\xi \ge \frac{\|f\|_{L^2}^2}{2}. \tag{1}\]Assume first that $|f|_{L^2}=1$. If we set
\[A := \int_{\mathbb{R}} x^2 |f(x)|^2\,dx, \qquad B := \int_{\mathbb{R}} \xi^2 |\hat f(\xi)|^2\,d\xi,\]then (1) says $A+B \ge \frac{1}{2}$. Among all $A,B>0$ with $A+B=C$, the product $AB$ is maximized at $A=B=C/2$; in particular the best universal lower bound on $AB$ given a lower bound on $A+B$ is obtained in the symmetric case. Thus from $A+B \ge \frac{1}{2}$ we obtain
\(AB \ge \left(\frac{1}{4}\right)^2 = \frac{1}{16},\) i.e. $(H)$ in the normalized case. For general $f$, write $f = g * | f |{L^2}$ with $| g |{L^2} = 1$ and apply the normalized inequality to $g$, which yields $(H)$.
Let
\[H = -\frac{1}{2}\frac{d^2}{dx^2} + \frac{1}{2}x^2\]be the harmonic oscillator. For any $f \in \mathcal{S}(\mathbb{R})$,
\[\begin{aligned} 2\langle Hf,f\rangle &= \int_{\mathbb{R}} \left(-f''(x)\right)\overline{f(x)}\,dx + \int_{\mathbb{R}} x^2|f(x)|^2\,dx \\ &= \int_{\mathbb{R}} |f'(x)|^2\,dx + \int_{\mathbb{R}} x^2|f(x)|^2\,dx \qquad \text{(IBP)}\\ &= \int_{\mathbb{R}} |\widehat{f'}(\xi)|^2\,d\xi + \int_{\mathbb{R}} x^2|f(x)|^2\,dx \qquad \text{(Plancherel)}\\ &= \int_{\mathbb{R}} \xi^2|\hat f(\xi)|^2\,d\xi + \int_{\mathbb{R}} x^2|f(x)|^2\,dx. \end{aligned}\]Therefore (1) is equivalent to the spectral gap estimate
\[\langle Hf,f\rangle \ge \frac{1}{2}\|f\|_{L^2}^2. \tag{2}\]We compute
\[\begin{aligned} H &= -\frac{1}{2}\frac{d^2}{dx^2} +\frac{1}{2}x^2 \\ &= \frac{1}{2}\left(x+\frac{d}{dx}\right)\left(x-\frac{d}{dx}\right) +\frac{1}{2}\left(\frac{d}{dx}x-x\frac{d}{dx}\right). \end{aligned}\]Since
\[\left(x\frac{d}{dx}-\frac{d}{dx}x\right)f(x) = xf'(x) - (xf(x))' = xf'(x) - (xf'(x)+f(x)) = -f(x),\]we have $\frac{d}{dx}x - x\frac{d}{dx} = \mathrm{Id}$, and hence
\[H = \frac{1}{2}\left(x+\frac{d}{dx}\right)\left(x-\frac{d}{dx}\right) +\frac{1}{2}\mathrm{Id}.\]Define
\[a=\frac{1}{\sqrt{2}}\left(x+\frac{d}{dx}\right), \qquad a^*=\frac{1}{\sqrt{2}}\left(x-\frac{d}{dx}\right).\]Then
\[H = a^*a + \frac{1}{2}\mathrm{Id}.\]Consequently, for any $f$,
\[\begin{aligned} \langle Hf,f\rangle &= \langle a^*af,f\rangle + \frac{1}{2}\langle f,f\rangle \\ &= \|af\|_{L^2}^2 + \frac{1}{2}\|f\|_{L^2}^2 \\ &\ge \frac{1}{2}\|f\|_{L^2}^2, \end{aligned}\]which is exactly (2), and hence proves (1) and (H).
Equality holds if and only if $af=0$, i.e.
\[\left(x+\frac{d}{dx}\right)f(x)=0.\]Equivalently,
\[f'(x) = -x f(x),\]so
\[f(x)=C\exp\left(-\frac{x^2}{2}\right)\]for some constant $C>0$.
This completes the proof.