# 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?

• You will need to provide more context if you want us to explain your professor's derivation. In general, the Heisenberg principle was a result of the cauchy schwarz inequality if I remember correctly. – Sanya Jun 28 '16 at 8:11
• The Heisenberg principle is straight from the inequality in the question. The problem is, I don't know where the latter comes from. I have just the consideration written in the question. – Drebin J. Jun 28 '16 at 8:18
• The inequality from the question IS the heisenberg principle; that was not my point – Sanya Jun 28 '16 at 9:50
• Ok, i didn't got that. How is that possible however? He said it applies to classical waves. I know the De Broglie hypothesis for which every particle is a wave and vice-versa, but he stressed the term "clasical". I assume it derives from some theorem on Fourier transform, but I'm not sure about it and, if that's the case, I don't know which one. (Hence the question). – Drebin J. Jun 28 '16 at 10:08
• I think his argument is pretty bad handwaiving: a delta peak has indefinite $\Delta k$ while 0 $\Delta x$ and for a monochromatic wave vice versa, thus there needs to be some inequality - at least that's what I'd guess at. But it's hard to reconstruct his argument. The Heisenberg Inequality in Quantum Mechanical framework is derivable and I'm positive you'll find a good derivation. – Sanya Jun 28 '16 at 10:25

## 1 Answer

The Heisenberg Uncertainty Principle has two distinct aspects:

• 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.

• 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.