14.2. Fourier Transform#

Contents: definition; properties; inverse transform; Plancherel’s theorem; uncertainty relation

14.2.1. Definition#

…

\[\mathscr{F}\left\{ g(t) \right\}(f) := \int_{t = -\infty}^{+\infty} g(t) \, e^{-i 2 \pi f t} \, dt .\]

14.2.2. Properties#

Linearity

Dirac delta.

(14.3)#\[\mathscr{L}\left\{ \delta(t) \right\} = \int_{t=-\infty}^{+\infty} \delta(t) \, e^{-i 2 \pi f t} \, dt = 1 \]

Time delay.

Derivative.

Integral.

Initial value.

Final value.

14.2.3. Inverse Fourier Transform#

Under the assumptions …todo, the inverse Fourier transform reads

\[\mathscr{F}^{-1}\left\{ G(f) \right\}(t) := \int_{f = -\infty}^{+\infty} G(f) \, e^{i 2 \pi f t} \, df .\]
Proof using Dirac’s delta expression.
\[\begin{split}\begin{aligned} \mathscr{F}^{-1}\left\{ G(f) \right\}(t) := \int_{f = -\infty}^{+\infty} G(f) \, e^{i 2 \pi f t} \, df & = \int_{f = -\infty}^{+\infty} \int_{\tau=-\infty}^{+\infty} g(\tau) e^{-i 2 \pi f \tau} \, e^{i 2 \pi f t} \, df = \\ & = \int_{f = -\infty}^{+\infty} \int_{\tau=-\infty}^{+\infty} g(\tau) e^{-i 2 \pi f \tau} \, e^{i 2 \pi f t} \, df = \\ & = \int_{f = -\infty}^{+\infty} \int_{\tau=-\infty}^{+\infty} g(\tau) e^{i 2 \pi f (t-\tau)} \, df = \\ & = \int_{\tau=-\infty}^{+\infty} g(\tau) \delta(t-\tau) \, d\tau = g(t) \ . \end{aligned}\end{split}\]
Proof using dominated convergence theorem and Fubini’s lemma.

Proof. By the dominated convergence theorem, it follows that

\[\begin{split}\begin{aligned} \int_{\mathbb{R}} e^{i 2 \pi x \xi} F(\xi) \, d \xi & = \lim_{\varepsilon \rightarrow 0} \int_{\mathbb{R}} \underbrace{e^{-\pi \varepsilon^2 \xi^2 + i 2 \pi x \xi}}_{G(\xi;x,\varepsilon)} F(\xi) \, d \xi = \\ & = \lim_{\varepsilon \rightarrow 0} \int_{\mathbb{R}} g(y;x,\varepsilon) f(y) \, dy = \\ & = \lim_{\varepsilon \rightarrow 0} \int_{\mathbb{R}} \varphi_{\varepsilon}(x-y) \, f(y) \, dy = \\ & = \int_{\mathbb{R}} \delta(x-y) \, f(y) \, dy = f(x) \end{aligned}\end{split}\]

Lemma 1. The Fourier transform of function \(\varphi(t):= e^{-\pi|t|^2}\) reads

\[\begin{split}\begin{aligned} \mathscr{F}\{ \varphi(t) \}(\omega) & = \int_{t=-\infty}^{+\infty} \varphi(t) e^{-i \omega t} \, dt = \\ & = \int_{t=-\infty}^{+\infty} e^{-\pi|t|^2} e^{-i \omega t} \, dt = \\ & = \int_{t=-\infty}^{+\infty} e^{-\pi \left( t^2 + i \frac{\omega}{\pi} t - \frac{\omega^2}{4 \pi^2} \right)} \, dt \, e^{- \frac{\omega^2}{4 \pi^2}} = \\ & = \int_{t=-\infty}^{+\infty} e^{-\pi \left( t + i \frac{\omega}{2 \pi} \right)^2} \, dt \, e^{- \frac{\omega^2}{4 \pi}} = \\ & = e^{- \frac{\omega^2}{4 \pi}} \ , \end{aligned}\end{split}\]

having evaluated the integral \(\int_{-\infty}^{+\infty} e^{-\alpha x^2}\) with \(\alpha = \pi\). todo justify the result for complex exponential. Use Bromwich contour integrals

Lemma 2. Fourier transform of \(f(\alpha t)\), \(\alpha > 0\)

\[\mathscr{F}\{ f(\alpha t) \}(\omega) = \int_{\mathbb{R}} f(\alpha t) e^{-j\omega t} \, dt = \int_{\tau \in \mathbb{R}} f(\tau) e^{-j \frac{\omega}{\alpha} \tau} \, d\tau \frac{1}{\alpha} = \frac{1}{\alpha} F\left(\frac{\omega}{\alpha} \right) \]

Lemma 3. \(\frac{1}{\varepsilon} \varphi\left(\frac{t}{\varepsilon} \right) \rightarrow \delta(x)\) for \(\varepsilon \rightarrow 0\)

\[\mathscr{F}\left\{\frac{1}{\varepsilon}\varphi\left(\frac{t}{\varepsilon} \right) \right\}(\omega) = \frac{1}{\varepsilon} \varepsilon e^{-\frac{\omega^2}{4 \pi \varepsilon^2}} = e^{-\frac{\omega^2}{4 \pi \varepsilon^2}}\]

  1. Fourier transform

\[G(f) = \int_{t=-\infty}^{\infty} e^{-i \omega t} g(t) \, dt\]
\[g(t) = e^{i \alpha t} \psi(t)\]
\[\mathscr{F}\{ g(t) \}(\omega) = \int_{t=-\infty}^{+\infty} g(t) e^{-i \omega t} \, dt = \int_{t=-\infty}^{+\infty} \psi(t) e^{i \alpha t} e^{-i \omega t} \, dt = \int_{t=-\infty}^{+\infty} \psi(t) e^{-i (\omega-\alpha) t} \, dt = \mathscr{F}\{ \psi(t) \}(\omega-\alpha) \ .\]
\[\psi(t) = \phi(\alpha t)\]
\[ \mathscr{F}\{ \psi(t) \} = \int_{t=-\infty}^{+\infty} \psi(t) e^{-i \omega t} \, dt = \int_{t=-\infty}^{+\infty} \phi(\alpha t) e^{-i \omega t} \, dt = \int_{\tau = -\infty}^{+\infty} \phi(\tau) e^{-i \frac{\omega}{\alpha} \tau} \, \frac{d \tau}{\alpha} = \frac{1}{\alpha} \mathscr{F}\{ \phi(t) \}\left( \frac{\omega}{\alpha} \right) \ . \]

  1. Fubini’s theorem

\[\varphi(t):= e^{-\pi t^2}\]
\[ \mathscr{F}\{ \varphi(t) \} = \int_{t=-\infty}^{+\infty} \varphi(t) e^{-i \omega t} \, dt = \int_{t=-\infty}^{+\infty} e^{-\pi t^2} e^{-i \omega t} \, dt \]
\[0 = \oint_{\gamma} e^{-\alpha |z|^2} \, dz = \int_{\dots} \dots\]
\[z = R e^{i \theta}, \quad dz = i R e^{i \theta} d \theta\]
\[\int_{C/4} e^{-\alpha |z|^2} \, dz = \int_{\theta=0}^{\frac{\pi}{2}} e^{-\alpha R^2} i R e^{i\theta} d \theta = i R e^{-\alpha R^2 } \frac{e^{-i \theta}}{i}|_{\theta= 0}^{\frac{\pi}{2}}\]
\[\begin{split}\begin{aligned} \int_{t=0}^{+\infty} e^{-\pi t^2} e^{-i \omega t} \, dt & = \int_{t=0}^{+\infty} e^{-\left( \pi t^2 + i \omega t - \frac{\omega^2}{4 \pi} \right)} \, dt \, e^{-\frac{\omega^2}{4 \pi}} = \\ & = \int_{t=0}^{+\infty} e^{-\pi \left( t + i \frac{\omega}{2 \pi} \right)^2} \, dt \, e^{-\frac{\omega^2}{4 \pi}} \\ \end{aligned}\end{split}\]

  1. \(\varphi_{\varepsilon}(t) = \frac{1}{\varepsilon^n} \varphi\left( \frac{t}{\varepsilon} \right)\), \(t \in \mathbb{R}^n\), is an approximation of Dirac’s delta for \(\varepsilon \rightarrow 0\), so that

    \[\begin{split}\begin{aligned} & \lim_{\varepsilon \rightarrow 0} \int_{t = -\infty}^{+\infty} \varphi_{\varepsilon}(t- \tau) f(t) \, dt = f(\tau) \\ & \lim_{\varepsilon \rightarrow 0} \int_{t = -\infty}^{+\infty} \varphi_{\varepsilon}(t) \, dt = 1 \\ \end{aligned}\end{split}\]

    As the Fourier transform \(\mathscr{F}\left\{\varphi_{\varepsilon}(t)\right\}(\omega) \rightarrow 1\) for \(\varepsilon \rightarrow 0\), then \(\varphi_{\varepsilon}(t) \rightarrow \delta(t)\).

14.2.4. Plancherel’s theorem#

…assumptions…todo

(14.4)#\[\int_{f=-\infty}^{+\infty} |G(f)|^2 \, df = \int_{t=-\infty}^{+\infty} |g(t)|^2 \, dt \]

and

(14.5)#\[\int_{f=-\infty}^{+\infty} A^*(f) \, G(f)\, df = \int_{t=-\infty}^{+\infty} a^*(t) \, g(t) \, dt \ .\]
Proof of Plancherel’s thm for the magnitude
\[\begin{split}\begin{aligned} \int_{f=-\infty}^{+\infty} |G(f)|^2 \, df & = \int_{f=-\infty}^{+\infty} G(f)^* G(f) \, df = \\ & = \int_{f=-\infty}^{+\infty} \left( \int_{t_1=-\infty}^{+\infty} g(t_1) e^{-i 2 \pi f t_1} dt_1 \right)^* \left( \int_{t_2=-\infty}^{+\infty} g(t_2) e^{-i 2 \pi f t_2} dt_2 \right) \, df = \\ & = \int_{t_1, t_2=-\infty}^{+\infty} g^*(t_1) \, g(t_2) \, \int_{f=-\infty}^{+\infty} e^{i 2 \pi f ( t_1 - t_2 )} \, df \, dt_1 \, dt_2 = && (1) \\ & = \int_{t_1, t_2=-\infty}^{+\infty} g^*(t_1) \, g(t_2) \, \delta( t_1 - t_2 ) \, dt_1 \, dt_2 = && (2) \\ & = \int_{t_1=-\infty}^{+\infty} g^*(t_1) \, g(t_1) \, dt_1 = && (3) \\ & = \int_{t_1=-\infty}^{+\infty} |g(t_1)|^2 \, dt_1 \ . \end{aligned}\end{split}\]

having used (1) the approximation (11.3) of Dirac’s delta, and (2) property (11.2) of Dirac’s delta, and (3) the expression of the absolute value of complex functions \(g^*(t_1) g(t_1) = |g(t_1)|^2\).

Proof of Plancherel’s thm for the product of functions
\[\begin{split}\begin{aligned} \int_{f=-\infty}^{+\infty} A^*(f) \, G(f) \, df & = \int_{f=-\infty}^{+\infty} G(f)^* G(f) \, df = \\ & = \int_{f=-\infty}^{+\infty} \left( \int_{t_1=-\infty}^{+\infty} a(t_1) e^{-i 2 \pi f t_1} dt_1 \right)^* \left( \int_{t_2=-\infty}^{+\infty} g(t_2) e^{-i 2 \pi f t_2} dt_2 \right) \, df = \\ & = \int_{t_1, t_2=-\infty}^{+\infty} a^*(t_1) \, g(t_2) \, \int_{f=-\infty}^{+\infty} e^{i 2 \pi f ( t_1 - t_2 )} \, df \, dt_1 \, dt_2 = && (1) \\ & = \int_{t_1, t_2=-\infty}^{+\infty} a^*(t_1) \, g(t_2) \, \delta( t_1 - t_2 ) \, dt_1 \, dt_2 = && (2) \\ & = \int_{t_1=-\infty}^{+\infty} a^*(t_1) \, g(t_1) \, dt_1 \ . \end{aligned}\end{split}\]

having used (1) the approximation (11.3) of Dirac’s delta, and (2) property (11.2) of Dirac’s delta.

14.2.5. Uncertainty relation#

An uncertainty relation holds linking standard deviations of a probability density function in time domain and a probability density function built with its Fourier transform. From this very same relation, Heisenberg uncertainty relation between position and momentum in Quantum Mechanics seamlessly follows.

Given a function \(g(t)\) whose square of the absolute value is normalized to one, and thus it can be used as a probability density function in time domain,

\[\int_{t = -\infty}^{+\infty} |g(t)|^2 \, dt = \int_{t = -\infty}^{+\infty} g^*(t) \, g(t) \, dt = 1 \ .\]

for Plancherel’s theorem, the square of the magnitude of Fourier transform \(G(f)\) is unitary as well,

\[\int_{f = -\infty}^{+\infty} |G(f)|^2 \, df = \int_{f = -\infty}^{+\infty} G^*(f) \, G(f) \, df = 1 \ ,\]

and thus it can be interpreted as a probability density function in frequency domain. The following uncertainty relation holds

\[\sigma_{t,g}^2 \sigma_{f,G}^2 \ge \left( \frac{1}{2 \pi} \frac{1}{2} \right)^2 \ , \]

or in terms of pulsation \(\omega = 2 \pi \, f\),

\[\sigma_{t,g}^2 \sigma_{\omega,G}^2 \ge \left( \frac{1}{2} \right)^2 \ , \]

Heisenberg uncertainty relation in quantum mechanics

Space and momentum representation of the state function \(\Psi\) are related by the transformation,

\[\langle x | \Psi \rangle := \psi(x,t) = \int_{p=-\infty}^{+\infty} \psi_p(p,t) \, e^{i \frac{p}{\hbar} x} \, dp \ ,\]

as it’s shown in the section Quantum Mechanics:From position to momentum representation. The wave number reads \(k = \frac{p}{\hbar}\). Starting from the uncertainty relation between the space coordinate \(x\) and the wave number \(k\),

\[\sigma_{x} \sigma_{k} \ge \frac{1}{2} \ , \]

Heisenberg uncertainty principle for position and momentum (for the same Cartesian coordinates) reads

\[\sigma_{x} \sigma_{p} \ge \frac{\hbar}{2} \ . \]
Proof of the uncertainty relation

Assuming zero average \(\overline{t} = 0\), \(\overline{f} = 0\) (see below for proof without this assumption)

\[\begin{split}\begin{aligned} \sigma_{t,g}^2 \sigma_{f,G}^2 & = \int_{t=-\infty}^{+\infty} \left| t \right|^2 \, g^*(t) \, g(t) \, dt \, \int_{f=-\infty}^{+\infty} \left| f \right|^2 \, G^*(f) \, G(f) \, df = \\ & = \int_{t=-\infty}^{+\infty} \left| t \, g(t) \right|^2 \, dt \, \int_{f=-\infty}^{+\infty} \left| f \, G(f)\right|^2 \, df = && (1) \\ & = \int_{t=-\infty}^{+\infty} \left| t \, g(t) \right|^2 \, dt \, \int_{t=-\infty}^{+\infty} \left| -\frac{i}{2 \pi} \, \dot{g}(t) \right|^2 \, dt \ge && (2) \\ & = \left| \int_{t=-\infty}^{+\infty} - t \, g^*(t) \frac{i}{2 \pi} \, \dot{g}(t) \, dt \right|^2 \ge && (3) \\ & = \left( \frac{1}{2\pi} \right)^2 \left( \frac{1}{2} \right)^2 \end{aligned}\end{split}\]

having used in

(1)
\[\mathscr{F}\{ \dot{g}(t) \}(f) = \int_{t=-\infty}^{+\infty} \dot{g}(t) e^{-i 2 \pi f t} \, dt = \dots = i 2 \pi \, f \, G(f) \ , \]
\[f \, G(f) = -i \, \frac{\mathscr{F}\{ \dot{g}(t) \}}{2 \pi}\]

and thus Plancherel’s theorem

\[\begin{split}\begin{aligned} \int_{f=-\infty}^{+\infty} | f \, G(f) |^2 \, df & = \int_{f=-\infty}^{+\infty} \left| -\frac{i}{2 \pi} \, \mathscr{F}\{ \dot{g}(t) \} \right|^2 \, df = \\ & = \int_{t=-\infty}^{+\infty} \left| -\frac{i}{2 \pi} \, \dot{g}(t) \right|^2 \, dt \end{aligned}\end{split}\]

in (2) Cauchy-Schwartz inequality,

(3)
\[\begin{split}\begin{aligned} a & := \int_{t=-\infty}^{+\infty} t g^*(t) \dot{g}(t) \, dt = \\ & = \underbrace{\left[ t \, g^*(t) \, g(t) \right]|_{-\infty}^{+\infty}}_{=0} - \int_{t=-\infty}^{+\infty} \dfrac{d}{dt} \left( t g^*(t) \right) g(t) \, dt = \\ & = - \int_{t=-\infty}^{+\infty} g^*(t) \, g(t) \, dt - \int_{t=-\infty}^{+\infty} t \dot{g}^*(t) g(t) \, dt = \\ & = - 1 - a^* \ . \end{aligned}\end{split}\]
\[-1 = a + a^* = 2 \, \text{re}\{a\} \ ,\]

and thus

\[|a|^2 \ge \text{re}\{a\}^2 = \frac{1}{4} \ .\]

If \(\overline{t} \ne 0\), or \(\overline{f} \ne 0\),

\[\begin{split}\begin{aligned} \sigma_{t,g}^2 \sigma_{f,G}^2 & = \int_{t=-\infty}^{+\infty} \left| t - \overline{t} \right|^2 \, g^*(t) \, g(t) \, dt \, \int_{f=-\infty}^{+\infty} \left| f - \overline{f} \right|^2 \, G^*(f) \, G(f) \, df = \\ & = \int_{t=-\infty}^{+\infty} \left| ( t - \overline{t} ) \, g(t) \right|^2 \, dt \, \int_{f=-\infty}^{+\infty} \left| ( f - \overline{f} ) \, G(f)\right|^2 \, df = && (1) \\ & = \dots \\ \end{aligned}\end{split}\]