# Is the Uncertainty Principle a logical consequence from the Wave-Particle duality?

I always thought of the Uncertainty Principle as a logical consequence that follows from the Wave-Particle duality, or more precise, from the fact that all particles behave as waves as long as they do not interact with other particles.

From basic wave properties, it's clear that you cannot know both the position and the speed/impulse of a wave exactly. A strongly localized wave dissipates fast, so it has a "blurry" impulse. Moreover, a wave with an exact impulse has no precise location. So, if a fundamental "particle" behaves like a wave, then the Uncertainty Principle (for the position/impulse variable pair) is a logical consequence.

Is that correct so far?

So I wonder, why do many physicists describe the Uncertainty Principle like it was some axiom that is built into this world, and that has no reason or explanation for why it exists? There are even physicists who describe it like Nature playing masquerade. "The more we look at the position, the more Nature conceals the impulse."

No, that's not what the Uncertainty Principle is about.

The fundamental fact is that everything behaves like waves. This is the axiom, the fact without a known reason behind it. The Uncertainty Principle is just a consequence of that fact, one of the first consequences discovered by mankind.

EDIT @ACuriousMind: Well that makes sense, but I have the impression that this just shifts the discussion over to the fact that there are non-commutating operators. It's clear that if two operators don't commutate, then there's the uncertainty that is quantitatively described by the formula you wrote. But I'm still convinced that the fact that there are non-commutating operators has a qualitative reason. After all, if classical mechanics is formulated using the Hilbert Spaces tools, your inequality becomes trivial because $[A, B] = 0$ for all operators in classical mechanics.

While, in settings like the double-slit, it is true that you may think about the quantum objects as being represented by a probability wave, this breaks down whenever one considers finite-dimensional Hilbert spaces, as they occur e.g. in the setting of quantum information and its qubits. There's no continuous set of generalized position operators - not ever a position operator at all - and hence no "wavefunction". Nevertheless, the relation $$\sigma_A(\psi)\sigma_B(\psi) \geq \frac{1}{2}\lvert\langle \psi \vert [A,B] \vert \psi \rangle\rvert$$ holds for all operators $A,B$ and all states $\psi$.