傅里叶变换

FouRier's transform, so it's a mathematical concept but comes from Physics.
For Nabla operator, we know Hamilton is a mathematician.

Symmetrical Relation

For a function \(x:\mathbb R\to\mathbb C\), the Fourier transform is
\( f(\omega)=\int_{-\infty}^{\infty} x(t)\,e^{-i\omega t}\,dt \)
and the inverse Fourier transform is
\( x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty} f(\omega)\,e^{i\omega t}\,d\omega \)

where:

• t is the original variable (time or space),
• \(\omega\) is the frequency variable,
• i is the imaginary unit, (\(i^2\)=-1),

The transform is valid whenever the integrals converge — absolutely, or in the finite-energy sense

\(\int_{-\infty}^{\infty}|x(t)|^2\,dt<\infty,\)

in which case the second integral is understood as a limit

\(f(\omega)=\lim_{R\to\infty}\int_{-R}^{R} x(t)\,e^{-i\omega t}\,dt\)

That is the definition.

Fundamental

\(\psi(x, t) = Ae^{ikx}e^{-i\omega t}\)

Space and Time pairs

\(\phi(k) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \psi(x, 0) dx\)
\(\psi(x, 0) = \int_{-\infty}^{\infty} \phi(k) dk\)
\(f(\omega) = \int_{-\infty}^{\infty} \psi(0, t) dt\)
\(\psi(0, t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(\omega) d\omega\)

De Broglie relation

\(E=\hbar \omega = \frac{h}{2\pi} \frac{2\pi} {t} = hv\)
\(x = \frac{2\pi}{k}\)
\(p=\hbar k = \frac{h}{2\pi} \frac{2\pi}{x} = \frac{h}{\lambda}\)

where \(x\) is the space coordinate but in mathematical induction, it could be like this.

Heisenberg uncertainty relation

$f(p)=\int h(x) e^{-ipx/\hbar} dx = f(\hbar k) = \int h(x) e^{-i\hbar k x/\hbar} dx = \int h(x) e^{-ik x} dx $

$h(x)=\frac{1}{2\pi\hbar}\int f(p)e^{ipx/\hbar}dp = \frac{1}{2\pi\hbar}\int f(\hbar k)e^{i\hbar kx/\hbar}d\hbar k = \frac{1}{2\pi}\int f(\hbar k)e^{ikx}dk $

\((\Delta x)^2 = \int x^2 |h(x)|^2dx\)
\((\Delta \hbar k)^2 = \int (\hbar k)^2 |f(\hbar k)|^2d\hbar k\)

\(\Delta x \Delta p \geq \frac{h}{2\pi \times 2}\)

\([\Delta x][\Delta p] = \text{m}\cdot\frac{\text{kg}\cdot\text{m}}{\text{s}} = \frac{\text{kg}\cdot\text{m}^2}{\text{s}} = \text{J}\cdot\text{s} = [\hbar]\)

\(\Delta p = \frac{h}{2\pi} \times \Delta \omega\)

\(\Delta x \Delta \omega \geq \frac{1}{2}\)

for monochromatic LASER, \(\Delta \omega \to 0\), \(\Delta x \to \infty\)

| \((x,p)\) | \(\hbar/2\) |
| \((x,k)\) | \(1/2\) |
| \((x,\omega)\) | \(1/2\) |
| \((x,\nu)\) | \(1/(4\pi)\) |
| \((t,E)\) | \(\hbar/2\) |
| \((t,\omega)\) | \(1/2\) |
| \((t,\nu)\) | \(1/(4\pi)\) |