3 ms·

The infinite sum of a harmonic series is an impulse train at its fundamental frequency. That is, if you push a swing at all overtone frequencies, the only push

by xeonmc 4y ago

The infinite sum of a harmonic series is an impulse train at its fundamental frequency. That is, if you push a swing at all overtone frequencies, the only push that lines up is the initial push, while all the other ones don't line up and cancel to zero for the rest of the period. And if your resonance length is all of free space, your fundamental frequency is zero and you get a single delta impulse. In other words, the more harmonic components there are that interferes with each other, the more localized your spatial distribution is. The fewer interfering components there are, the more spread out it is. At the classical limit, we assume there are practically infinitely many interfering components, with all of free space as the resonance length. Ergo we have perfect localization At the quantum scale, interacting components are few and the length scale is highly constrained, ergo we get very spread-out (due to having few overtones) and quantized (due to small resonance length) position wavefunctions.

snarkconjecture 4y ago
I think that's not quite it. The relevant theorem is that variance(position-space function f) * variance(Fourier transform of f) >= some constant. A wavefunction is a wavefunction, it doesn't have more frequency components at the classical scale. It's just that what looks like a large position spread at the quantum scale is pretty tiny compared to classical length scales, and what looks like a large momentum spread (variance of the Fourier transform) is pretty tiny compared to classical-scale momenta. You shouldn't think of a wavefunction as being perfectly localized, with infinitely spread out frequency components, because that would mean the particle's momentum is infinitely uncertain. Instead, think of a Gaussian function, whose Fourier transform is a Gaussian. The widths of those two Gaussians are inversely proportional to each other. Also, you should think about continuous Fourier transforms, not discrete Fourier series. Periodic wavefunctions are only the norm in situations like crystals where the environment itself is periodic.
- xeonmc 4y ago
  Isn't continuous Fourier Transform simply the limit of Fourier series as the fundamental frequency approaches zero though? That's what I meant when I said "all of free space", it's the same idea with the emergence of "continuous" band structures in bulk materials -- although their sizes aren't strictly infinite, for all practical purposes it's long enough that the energy levels are sufficiently close-together to be considered a continuum. And yes, I stand corrected. My analogy with overtones bandwidth vs spatial localization instead pertains to the position-momentum tradeoff, it is indeed incorrect to overgeneralize it to what the classical limit means. The classical limit, as you correctly pointed out, is more about the tradeoff between the two being practically negligible in the length/impulse scale being dealt with in the classical regime -- effectively a "rounding error", so to speak.

by xeonmc 4y ago

snarkconjecture 4y ago

I think that's not quite it. The relevant theorem is that variance(position-space function f) * variance(Fourier transform of f) >= some constant. A wavefunction is a wavefunction, it doesn't have more frequency components at the classical scale. It's just that what looks like a large position spread at the quantum scale is pretty tiny compared to classical length scales, and what looks like a large momentum spread (variance of the Fourier transform) is pretty tiny compared to classical-scale momenta. You shouldn't think of a wavefunction as being perfectly localized, with infinitely spread out frequency components, because that would mean the particle's momentum is infinitely uncertain. Instead, think of a Gaussian function, whose Fourier transform is a Gaussian. The widths of those two Gaussians are inversely proportional to each other. Also, you should think about continuous Fourier transforms, not discrete Fourier series. Periodic wavefunctions are only the norm in situations like crystals where the environment itself is periodic.

xeonmc 4y ago
Isn't continuous Fourier Transform simply the limit of Fourier series as the fundamental frequency approaches zero though? That's what I meant when I said "all of free space", it's the same idea with the emergence of "continuous" band structures in bulk materials -- although their sizes aren't strictly infinite, for all practical purposes it's long enough that the energy levels are sufficiently close-together to be considered a continuum. And yes, I stand corrected. My analogy with overtones bandwidth vs spatial localization instead pertains to the position-momentum tradeoff, it is indeed incorrect to overgeneralize it to what the classical limit means. The classical limit, as you correctly pointed out, is more about the tradeoff between the two being practically negligible in the length/impulse scale being dealt with in the classical regime -- effectively a "rounding error", so to speak.