I recently saw a video by 3Blue1Brown, in which he explained that the uncertainty principle isn't a quantum phenomenon, but a result of basic signal properties. The basic premise was: the more precise you single in on a time value in the Fourier transform, the less precise you get in the frequency domain. Thus, uncertainty is not a quantum property but a property of all signals.

Now, I asked a professor of quantum physics what his take on this was, and if that means that quantum uncertainty is a result of the wave properties of particles.

His answer was: "It's the other way around, actually. All of signal processing comes from quantum physics."

And this seems like a copout to me. Why do we say that particles have wave properties, and not that waves have quantum properties? Is it fair to assume that quantum physics is more fundamental than wave properties?

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    $\begingroup$ Nope, that's totally backwards. Both quantum (wave) mechanics and signal processing come from Fourier analysis. $\endgroup$ – knzhou Mar 16 '18 at 10:29

I was quite impressed with the 3Blue1Brown video you referred to and its nuanced presentation of the uncertainty principle and its epistemological status, which it got right to a tee. In short, the Heisenberg Uncertainty Principle as embodied within quantum mechanics has two distinct components:

  • A mathematical property of waves (which I know as the Bandwidth Theorem, though that name is not that widely used) which provides a lower bound on the product of the spatial extent $\Delta x$ of the wave and the width $\Delta k$ of its support on (spatial) frequency space, $$\Delta x \,\Delta k \gtrsim 2\pi.$$ This is a purely mathematical theorem (given sufficiently rigorous definitions of $\Delta x$ and $\Delta k$) and it can be squarely inscribed within signal processing without any reference at all to quantum mechanics.
  • A core physical principle, in the form of de Broglie's relation $p=h/\lambda = \hbar k$ between the mechanical momentum $p$ of a particle and its wavelength $\lambda$ (and through that to its spatial frequency, a.k.a. wavevector, $k=2\pi/\lambda$). This is how you get to uncertainty products of the form $\Delta x \,\Delta p$; no amount of mathematics will get you from frequencies to mechanical momentum, because the core identification is a physical principle.

In short, your initial assertion that

uncertainty is not a quantum property but a property of all signals

isn't right or wrong: it depends on what quantity's uncertainty you're talking about. Uncertainty in frequency is a property of all signals; uncertainty in momentum is a distinctly quantum property. On the other hand, your second assertion,

quantum uncertainty is a result of the wave properties of particles

is correct in a quite fundamental sense, thought it is still lacking half of the picture.

And, on the opposite side, your professor's response,

It's the other way around, actually. All of signal processing comes from quantum physics.

strikes me as completely wrong. Maybe he was thinking of some historical perspective? (In which case, it's still unlikely to be right, I should think ─ Fourier analysis was well established by the time Heisenberg came along.) Conceptually, signal processing stands on its own two feet, and it is quantum physics that uses some of the same mathematics together with its own physics sauce to produce descriptions of the physical world.

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  • $\begingroup$ One could, I think, take the notion that you express in your second bullet point a little further by introducing the idea of commuting versus non-commuting observables to get a more generalized uncertainty principle in quantum mechanics. But perhaps that would just cloud the immediate issue. $\endgroup$ – dmckee --- ex-moderator kitten Mar 16 '18 at 14:33
  • $\begingroup$ I know I'm missing the "quantized momentum" half of the picture, but that is simply responsible for the quantized boundary in the uncertainty principle, right? Maybe what my professor meant was, that the most fundamental basis we know is quantum physics, and thus all signals we know are just a result of quantum behavior. Which I think is still a very weak claim, since it simply assumes that quantum physics is the most fundamental physics. $\endgroup$ – BigBadWolf Mar 17 '18 at 11:25
  • $\begingroup$ I can't speak to your professor's frame of mind and I think it's pretty pointless to speculate about what he meant. Signal processing is math, not physics, so QM cannot be "more fundamental". As to what you mean by "that is simply responsible for the quantized boundary in the uncertainty principle", I'm completely in the dark - that's just word salad from what I can see, really. $\endgroup$ – Emilio Pisanty Mar 17 '18 at 12:51

Well, one could think of that professor's statement as referring to how that quantum mechanics is needed to describe the basic physics of electrons and atoms in the components that make up the electronic devices used to do signal processing - however the behavior of many components, especially in analogue circuits, at high signal intensities (i.e. currents carrying much higher charge flow rates than a few electrons per second and with energy levels much greater than a single photon of comparable frequency to the signal(s) involved) can be usefully approximated very well with classical mechanics - namely Maxwell's equations. Components like transistors, however, that are often used in digital signal processing do require a more thoroughly quantum understanding to fully make sense of. But I don't think it's necessarily connected to what the video is describing.

Regarding the question of waves versus particles in quantum mechanics and particles versus fields in quantum field theory - all this is explained by the mathematical notion of the Hilbert Space which is used to describe the system(s) in question. All quantum states are members of a Hilbert Space which is associated with the system and it is possible to represent the same space in different (mathematical jargon: isomorphic) ways, in particular, as amplitude functions (wave functions) giving a quantum amplitude for each possibile value of some particular measurable quantity, and which of these you take as representation determines which is "fundamental".

In particular Hilbert Spaces are a special kind of vector space from linear algebra, and in linear algebra every vector space has one or more basis sets which are a "just big enough" subset of the vectors that can be combined together to generate all other vectors in the space - and there is one basis set for each kind of measurable quantity and associated with a certain mathematical operator defined on the space. If you use the position operator, your wave functions will be represented as amplitudes for particle positions, and this corresponds to the particle view. If you use the momentum operator, your wave functions will be represented as giving amplitudes for particle momenta, and this corresponds to the wave view. For quantum fields, if you use the number operator, that is the particle view, and if you use the field operator, that is the field view. Each one is another way of describing the exact same mathematical object by looking at it through a different lense, so they are entirely equivalent and neither one is superior to the other in just the minimum mathematical setup of quantum mechanics.

This is also what is going on in the 3Blue1Brown video - the mathematical space of Fourier-transformable signals and waves is essentially the same (not sure if exactly but one may include the other as a suitably broad subset) as the Hilbert Space used to describe a single particle in 1-dimensional space, and thus both by virtue of having the same mathematical description will have analogous mathematical properties - and it is also this that can be considered as a reason why one would think to talk of the quantum spaces as "particles versus waves". The Fourier transform itself corresponds to a change of basis vectors (isomorphism of equivalent representations) from time to frequency in the signal case and position to momentum in the quantum case. Any time that two such quantities can be considered as different bases of a vector space the uncertainty relation will exist between them - it is no different than the tradeoff of x- vs y-component of a vector in 2D space as you rotate that vector, or perhaps better, hold the vector steady and rotate the 2D coordinate axes about the origin (such a change of axes is effectively a change of basis).

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