# Uncertainty Principle Intuition

So, as your usual physics undergrad, I read Griffiths's derivation of the general uncertainty principle. I understood it but there was no physical intuition given behind it in the book. It was basically a bunch of manipulations and inequalities used to get just the right result.

Other than "It's because of the statistical interpretation." is there a more intuitive reasoning behind uncertainty. Like commutators, behavior of particles, etc.

Thank you for the replies and as always, much obliged.

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I am not aware of an intuitive interpretation of the general case of two non-commuting observables, except the one that you don't seem to find very intuitive at all :) – Danu Aug 22 '14 at 14:29
"possible duplicate of..." Not really... That is a very specific question regarding whether we can make measurements arbitrarily close in time if not at once. I am asking about the intuition behind the general uncertainty principle. Use of mathematics is welcome. Fourier transform reasoning is something i'd like to hear as well. Or about non commuting observable's. – SilverSlash Aug 22 '14 at 14:31
I know that the question may not seem very similar, but I think my answer there answers your question, at least for the case of $x$ and $p$, where the Fourier transform is enlightening. Perhaps someone can generalize the answer I presented there, but I'm not sure. – Danu Aug 22 '14 at 14:34
Related: physics.stackexchange.com/questions/118960/… (I now retracted my duplicate-vote). – Danu Aug 22 '14 at 14:38

## 1 Answer

Heisenberg's Uncertainty Principle is, in essence, a consequence of two basic facts: de Broglie's relation between a particle's momentum and its associated wavelength, and a mathematical fact known as the bandwidth theorem, which states an equivalent uncertainty relation between a wave's position and its wavelength.

The fundamental physical leap here is of course the first one, the statement that particles have a 'wavelength' of some meaningful sort. This leap takes you directly outside of classical physics and there is nothing within the classical sphere that is really enough to motivate it. De Broglie's relation remains a cornerstone of quantum theory, though it is usually stated in a more sophisticated form as the canonical commutation relation $[\hat x, \hat p]=i\hbar$.

The bandwidth theorem, on the other hand, is a very intuitive statement about waves, and it essentially says that if you want to measure a wave's wavelength $\lambda$ to some uncertainty $\Delta \lambda$, you will need to measure a number of periods $n$ that is inversely proportional to $\Delta \lambda$. Thus, on one end, to have zero uncertainty in the wavelength you need to measure an infinite number of periods, and on the other end, a wave pulse that's so short it only fits one peak and no troughs can hardly be said to have a defined wavelength.

The precise statement is cleanest in terms of the wavevector $k=2\pi/\lambda$, and the theorem states that the size $\Delta x$ of the region occupied by some waveform, and the width $\Delta k$ of the region of wavevectors that appreciably contribute to the waveform's plane-wave decomposition, are related via $$\Delta x\,\Delta k\geq 2\pi.$$ Since de Broglie's relation reads $p=\hbar k$, the extension to the Heisenberg principle is obvious.

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This really helped me. Thanks a lot! – SilverSlash Aug 22 '14 at 16:39
Great answer. My own intuition improved by reading it. Does this bandwidth theorem to which you refer go by a different name in mathematical circles? – joshphysics Aug 23 '14 at 23:03
@joshphysics I'm not sure. The name 'bandwidth theorem' is not really that widespread, I believe, but I don't really know many other names for it. That's probably a question for a Fourier analyst, but it's likely that it's known more as a fundamental fact than by a specific name. – Emilio Pisanty Aug 23 '14 at 23:16