# 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

• 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.
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.