Heisenberg Uncertainty Principle
Buildup
There exists a special relationship between $f$ and its Fourier transform $\hat{f}$. If for some constant $\Omega$, $\hat{f} (\omega) = 0\ for\ | \omega | \ge \Omega$ holds, then it is not possible for $f$ to exhibit the same property. In other words, it indicates that both $f$ and $\hat{f}$ cannot be concentrated in a narrow location simultaneously, mathematically speaking, this means $f$ and $\hat{f}$ cannot both have a finite narrow support at the same time, and statistically, it implies that the variances of $f$ and $\hat{f}$ cannot both be small at the same time.
This fact can be seen in the following property of the Fourier transform.
$$ \mathcal{F} \left[ f(\delta x) \right] (\xi) = \dfrac{1}{\delta} \hat{f} \left( \dfrac{\xi}{\delta} \right),\quad \delta > 0 $$
In the equation above, if $\delta$ increases, it means that $f$ is being compressed closer to the origin, while at the same time, $\hat{f}$ is spreading out. If $\delta$ decreases, the opposite occurs.
Before introducing the theorem, let’s introduce a new notation. Let’s define the dispersion of function $f$ at $a$ as follows.
$$ \Delta_{a} f: =\frac{\displaystyle \int(x-a)^{2}|f(x)|^{2} d x}{\displaystyle \int|f(x)|^{2} d x} $$
This represents how spread out the values of $f$ are around $a$. A large value of $\Delta _{a} f$ means that $f$ hardly takes any large values near $a$, and conversely, a small value of $\Delta _{a} f$ means that $f$ takes a lot of large values near $a$. Therefore, the statement that $f$ and $\hat{f}$ cannot both be concentrated in a narrow interval means that there is a limit to both $\Delta_{a}f$ and $\Delta_{\alpha}\hat{f}$ values being small for all $a, \alpha \in \mathbb{R}$.
In a nutshell, as the value of $f$ becomes certain, the value of $\hat{f}$ becomes uncertain, and vice versa. If you are familiar with statistics, you can understand $\Delta_{a}f$ as the variance of a random variable with a mean of $a$ and probability density function of $|f(x)|^{2}$.
This content is expressed in the following theorem.
Theorem
If $f^{\prime}$ is piecewise continuous, and $f(x), xf(x), f^{\prime}(x) \in L^{2}$ holds, then the following inequality is valid.
$$ \left( \Delta_{a} f \right) ( \Delta_{\alpha} \hat{f} ) \ge \dfrac{1}{4},\quad \forall a,\alpha \in \mathbb{R} $$
Proof
$a=\alpha=0$
Let’s first consider the case of $a=\alpha=0$. By integration by parts, we obtain the following equation.
$$ \int_{A}^{B} \overline{xf(x)} f^{\prime}(x) dx = x|f(x)|^{2} \bigg|_{A} ^{B}-\int_{A}^{B}\left(|f(x)|^{2} + x f(x) \overline{f^{\prime}(x)}\right) dx $$
This can be simplified as follows.
$$ \begin{align*} \int_{A}^{B} |f(x)|^{2} dx &= x|f(x)|^{2} \bigg|_{A} ^{B} - \int_{A}^{B} xf(x) \overline{f^{\prime}(x)}dx - \int_{A}^{B} \overline{xf(x)} f^{\prime}(x) dx \\ &= x|f(x)|^{2} \bigg|_{A} ^{B} - 2 \text{Re} \int_{A}^{B} xf(x) \overline{f^{\prime}(x)}dx \end{align*} $$
Since we assumed $xf(x) \in L^{2}$, $\lim \limits_{x \to \pm \infty}xf^{2}(x) \le \lim \limits_{x \to \pm \infty} x^{2}f^{2}(x) = 0$ holds. Thus, taking the limit of $B\to \infty, A \to -\infty$ in the equation above results in the following.
$$ \int_{-\infty}^{\infty} |f(x)|^{2} dx = - 2 \text{Re} \int_{-\infty}^{\infty} xf(x) \overline{f^{\prime}(x)}dx $$
Then, by the Cauchy-Schwarz inequality, the following equation holds.
$$ \begin{align} \int_{-\infty}^{\infty} |f(x)|^{2} dx &\le 2 \text{Re} \left( \int_{-\infty}^{\infty} x^{2} |f(x)|^{2} dx \right)^{\frac{1}{2}} \left( \int_{-\infty}^{\infty} |f^{\prime}(x)|^{2} dx\right)^{\frac{1}{2}} \nonumber \\ \implies \left( \int_{-\infty}^{\infty} |f(x)|^{2} dx \right)^{2} &\le 4 \left( \int_{-\infty}^{\infty} x^{2} |f(x)|^{2} dx \right) \left( \int_{-\infty}^{\infty} |f^{\prime}(x)|^{2} dx\right) \label{eq1} \end{align} $$
Then, by Plancherel’s theorem $\| \hat{f} \|^{2} = 2\pi \| f \| ^{2}$ is true, and the Fourier transform of the derivative is $\mathcal{F} \left[ f^{\prime} \right] (\xi) = i \xi \mathcal{F} f (\xi)$, thus the following holds.
$$ \int_{-\infty}^{\infty} |f^{\prime}(x)|^{2} dx = \dfrac{1}{2\pi} \int_{-\infty}^{\infty} |\mathcal{F}[f^{\prime}] (\xi)|^{2} d\xi = \dfrac{1}{2\pi} \int_{-\infty}^{\infty} \xi^{2} | \hat{f} (\xi)|^{2} d\xi $$
Substituting this in for $\eqref{eq1}$ results in the following.
$$ \left( \int_{-\infty}^{\infty} |f(x)|^{2} dx \right)^{2} \le 4 \left( \int_{-\infty}^{\infty} x^{2} |f(x)|^{2} dx \right) \dfrac{1}{2\pi} \int_{-\infty}^{\infty} \xi^{2} | \hat{f} (\xi)|^{2} d\xi $$
Furthermore, expressing Plancherel’s theorem in integral form gives $\int |f(x)|^{2}dx = \frac{1}{2\pi} \int |\hat{f}(\xi)|^{2} d\xi$, thus substituting this into the left side of the equation results in the following.
$$ \left( \int_{-\infty}^{\infty} |f(x)|^{2} dx \right) \left( \int_{-\infty}^{\infty} |\hat{f}(\xi)|^{2} d\xi \right) \le 4 \left( \int_{-\infty}^{\infty} x^{2} |f(x)|^{2} dx \right) \left( \int_{-\infty}^{\infty} \xi^{2} | \hat{f} (\xi)|^{2} d\xi \right) $$
Thus, we obtain the following result.
$$ \dfrac{1}{4} \le \dfrac{ \left(\displaystyle \int_{-\infty}^{\infty} x^{2} |f(x)|^{2} dx \right) }{ \left(\displaystyle \int_{-\infty}^{\infty} |f(x)|^{2} dx \right) } \dfrac{\left(\displaystyle \int _{-\infty}^{\infty} \xi^{2} | \hat{f} (\xi)|^{2} d\xi \right)}{\left(\displaystyle \int_{-\infty}^{\infty} |\hat{f}(\xi)|^{2} d\xi \right)} = \left( \Delta_{0} f \right) ( \Delta_{0} \hat{f} ) $$
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Generalization
Let’s assume $F(x)=e^{-i \alpha x}f(x+a)$. Then, the following equation holds.
$$ \begin{align*} \Delta_{0} F = \dfrac{\displaystyle \int x^{2} |F(x)|^{2} dx}{\displaystyle \int |F(x)|^{2}dx} &= \dfrac{ \displaystyle \int x^{2} |f(x+a)|^{2} dx }{ \displaystyle \int |f(x+a)|^{2}dx} \\ &= \dfrac{ \displaystyle \int (x-a)^{2} |f(x)|^{2} dx }{ \displaystyle \int |f(x)|^{2}dx} & (\text{change of variable } x +a=x) \\ &= \Delta_{a}f \end{align*} $$
Now, if we find $\hat{F}$, it results in the following.
$$ \begin{align*} \hat{F} (\xi) = \int F(x) e^{-i \xi x}dx &= \int f(x+a)e^{-i(\alpha + \xi)x}dx \\ &= \int f(x)e^{-i(\alpha + \xi)x} e^{i(\alpha + \xi)a} dx & (\text{change of variable } x +a=x) \\ &= e^{i(\alpha + \xi)a} \int f(x)e^{-i(\alpha + \xi)x} dx \\ &= e^{i(\alpha + \xi)a} \hat{f}(\alpha + \xi) \end{align*} $$
Now, finding $\Delta_{0}\hat{F}$ results in the following.
$$ \begin{align*} \Delta_{0}\hat{F} = \dfrac{\displaystyle \int \xi^{2} |\hat{F}(\xi)|^{2} d\xi}{\displaystyle \int |\hat{F}(\xi)|^{2}d\xi} &= \dfrac{\displaystyle \int \xi^{2} | \hat{f}(\alpha + \xi)|^{2} d\xi}{\displaystyle \int | \hat{f}(\alpha + \xi)|^{2}d\xi} \\ &= \dfrac{\displaystyle \int (\xi-\alpha)^{2} | \hat{f}(\xi)|^{2} d\xi}{\displaystyle \int | \hat{f}(\xi)|^{2}d\xi} & (\text{change of variable } \xi +a=\xi) \\ &= \Delta_{\alpha}\hat{f} \end{align*} $$
Therefore, using the result from the case of $a=\alpha=0$, we obtain the following equation.
$$ \dfrac{1}{4} \le (\Delta_{0}F)(\Delta_{\alpha}\hat{F}) = (\Delta_{a}f) (\Delta_{\alpha}\hat{f}) $$
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