Heisenberg uncertainty principle from the integral of norm of schwartz function?

Question

If we only have a schwartz function $f$: $\mathbb{R} \to \mathbb{C}$, that $\int_{-\infty}^{\infty} |f(x)|^2 \,dx = 1$,

• how can we show that $\int_{-\infty}^{\infty} x^2 |f(x)|^2 \,dx \cdot \int_{-\infty}^{\infty} k^2 |\hat{f}(k)|^2 \,dk \geq 1/4$?
• how can we show that the above inequality becomes equality when $f(x) = C_1 \exp\left[-\dfrac{x^2}{\sigma^2}\right]$ for some $\sigma$ (So the integrals become some variances about $0$-mean Gaussian functions)?

The Wikipedia gives a different lowerbound on the integral. Why is that? Thanks in advance.

Answer 1

Answer to Question 1:

Integration by parts yields:

\begin{equation} \begin{split} 1 &= \int_{-\infty}^{\infty}|f(x)|^{2}dx = x|f(x)|^{2}\Big|_{-\infty}^{\infty} - \int_{-\infty}^{\infty}x\frac{d}{dx}|f(x)|^{2} \\ &= -\int_{-\infty}^{\infty}x\frac{d}{dx}|f(x)|^{2}\text{, since }f \in \mathcal{S}(\mathbb{R}) \\ &= -\int_{-\infty}^{\infty}x(f'(x)\overline{f(x)} + f(x)\overline{f'(x)})dx\text{, by how }|f(x)|^{2} = f(x)\overline{f(x)} \\ \Rightarrow 1 &\leq 2\int_{-\infty}^{\infty}|x||f'(x)||f(x)|dx \\ &\leq 2\left(\int_{-\infty}^{\infty}x^{2}|f(x)|^{2}dx\right)^{1/2}\left(\int_{-\infty}^{\infty}|f'(x)|^{2}dx\right)^{1/2}\text{, by Cauchy-Schwarz} \\ &= 2\left(\int_{-\infty}^{\infty}x^{2}|f(x)|^{2}dx\right)^{1/2}\left(\int_{-\infty}^{\infty}k^{2}|\hat{f}(k)|^{2}dk\right)^{1/2}; \end{split} \end{equation}

The last equality follows from Plancherel's theorem and the Fourier transform of a function's derivative:

$$\int_{-\infty}^{\infty}|f'(x)|^{2}dx = \int_{-\infty}^{\infty}|\hat{f'}(k)|^{2}dk = \int_{-\infty}^{\infty}k^{2}|\hat{f}(k)|^{2}dk$$

The result follows.

Answer to Question 2:

For equality to hold in the Cauchy-Schwarz inequality $\langle f, g \rangle_{L^{2}(\mathbb{R})} \leq ||f||_{L^{2}(\mathbb{R})}||g||_{L^{2}(\mathbb{R})}$, we must have $g(x) = Cf(x)$ a.e., for some $C \in \mathbb{C}$. Thus, for equality to hold in our above result, we must have $f'(x) = Cxf(x)$, for some $C \in \mathbb{C}$. Then:

$$\frac{f'(x)}{f(x)} = Cx \Rightarrow \int \frac{f'(x)}{f(x)} dx = C\int x dx \Rightarrow \text{ln}[f(x)] = \frac{C}{2}x^{2} + D\text{, for some }D\in \mathbb{C} $$ $$\Rightarrow f(x) = C_{1}e^{Bx^{2}}\text{, for some }C_{1}, B \in \mathbb{C}$$

In order to satisfy our assumption that $f \in \mathcal{S}(\mathbb{R})$, we must have $B = -C_{2}$, for some $C_{2} > 0$. Take $\sigma = \frac{1}{\sqrt{C_{2}}}$. Furthermore, we must have $\int_{-\infty}^{\infty}|f(x)|^{2}dx = 1$, so:

$$1 = \int_{-\infty}^{\infty}|f(x)|^{2}dx = |C_{1}|^{2}\int_{-\infty}^{\infty}e^{-2C_{2}x^{2}}dx = \frac{|C_{1}|^{2}}{\sqrt{4C_{2}}}\int_{-\infty}^{\infty}e^{-\frac{x^{2}}{2}}dx = |C_{1}|^{2}\sqrt{\frac{\pi}{2C_{2}}}$$ $$\Rightarrow |C_{1}|^{2} = \sqrt{\frac{2 C_{2}}{\pi}}$$