quantum mechanics - Uncertainty and Classical waves

My professor, introducing Heisenberg uncertainty principle, started from the Fourier transform and the classical uncertainty for waves.

He told about the localized impulsive wave $\delta(x)$ which has defined position but total uncertainty of impulse (its Fourier transform is composed of every possible momentum). On the other hand, a wave of defined impulse is a monochromatic wave, which spreads over the entire position axis and doesn't have a proper localization.

I'm perfectly comfortable those considerations, but then, out of noting, he writes

$$\Delta x \: \Delta k \geq 1/2$$

From this it's easy to derive the Heisenberg principle, but I can't understand where the previous formula comes from.

Does it come from Fourier transform properties, from the properties of optical waves, or from something else?

Answer

The Heisenberg Uncertainty Principle has two distinct aspects:

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  One is the identification of matter as a wave and, in particular, the relationship between a particle's momentum $p$ and its wavelength $\lambda$ through de Broglie's relationship $p=h/\lambda$. This is the crucial bit of physical input.

  
  


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  The second one is purely mathematical, and it's the relationship $\Delta x\, \Delta k\geq 1/2$. This is a general fact about waves and their Fourier transforms, and in a signal-processing context it's known as the bandwidth theorem.

  
  

In general, the bandwidth theorem is a bit hard to state precisely - or rather, there are multiple valid slightly different ways to state it, depending on exactly how you define the terms that appear in it and the classes of functions you're considering. However, in all its incarnations it is simply a fundamental fact of the theory of Fourier transforms.

As an example, if you have a complex-valued function $f(x)$ normalized to $\int_{-\infty}^\infty |f(x)|^2\:\mathrm dx=1$ and you define the position uncertainty as $$ \Delta x=\sqrt{\int_{-\infty}^\infty x^2 \: |f(x)|^2\:\mathrm dx - \left(\int_{-\infty}^\infty x \: |f(x)|^2\:\mathrm dx\right)^2} $$ the Fourier transform as $$ \tilde f(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-ikx}f(x)\:\mathrm dx, $$ and the wavevector uncertainty as $$ \Delta k=\sqrt{\int_{-\infty}^\infty k \: |\tilde f(k)|^2\:\mathrm dk - \left(\int_{-\infty}^\infty k \: |\tilde f(k)|^2\:\mathrm dx\right)^2} ,$$ then the uncertainty relation $$\Delta x\:\Delta k\geq \frac12$$ holds at least for all continuously differentiable $f$ such that $f'$, $\hat xf$ and $\hat k\tilde f$ are in $L_2$ (example proof). The uncertainty principle does hold for broader classes of functions, at least in a moral sense, but as I said there are multiple valid variants and it's a pain to list them all. However, for any suitable class of (generalized) functions, and definitions of the uncertainties, as long as the left-hand side's uncertainty product makes sense then it will have some sort of lower bound of order unity.