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In a pre-print I read that "The position of the particle is indeterminate as it could be anywhere along the wave packet. Hence compressing the wave packet to reduce the indeterminacy in position will change the wavelength and therefore the momentum, and thus make the momentum indeterminate, and the converse", as an explanation for HUP.

Is this explanation accurate for HUP?

Alex L
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  • Yes ,have a look http://hyperphysics.phy-astr.gsu.edu/hbase/Waves/wpack.html – anna v May 14 '19 at 06:06
  • @anna what is "wavelength change" in this explanation? Also, I thought HUP says in any case our knowledge is limited to it, but this guy says in order to compress the momentum we need to change the momentum and so the original position will change. Does HUP speak about disturbing the particle? – Alex L May 14 '19 at 06:11
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    This explanation sounds a bit naive. A particle does not have a position along the wave packet. A particle is the wave packet. The HUP applies to the detection of a particle when the wave function already has collapsed. Before that there is no particle, because it travels as a wave everywhere at once. – safesphere May 14 '19 at 06:15
  • @safesphere the link is about wavepackets from wave equation solutions. The inerpretation as a HUP needs quantum mechanics equations and probability distributions. It just shows that the uncertainty is inherent in wave packets . – anna v May 14 '19 at 06:29
  • @annav Agreed. The uncertainty principle indeed reflects the wave nature of matter. The explanation in question though could have been phrased much better. – safesphere May 14 '19 at 06:33
  • Griffith's Intro to Quantum Mechanics works out how to discover the minimum uncertainty wave packet. This might help illustrate quantitatively the issue presented her qualitatively. – R. Romero May 14 '19 at 15:05

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Well, the explanation is very qualitative but I think it works. What the explanation is trying to say is best conveyed by a picture:

enter image description here

As you can see, a broader wave packet in position space makes the corresponding packet narrower in momentum space. This is closely conected to a property of the Fourier transform (a very nice and easy to follow introduction of the Fourier transform, along with this particular property are discussed by 3Blue1Brown in this video and this other video).

You can also think of this argument in terms of the actual HUP $\Delta x \Delta p\geq\frac{\hbar}{2}$: if the distribution is narrow in $x$ then $\Delta x$ is small and thus $\Delta p$ must be large.

  • It says that the position is indeterminate due to the wavefunction spread, and if we try to compress wavefunction, we need to change the momentum. But HUP says it's fundamentally impossible to know position and momentum more accurate than what it says, it doesn't have anything to do with us disturbing the particle. – Alex L May 14 '19 at 06:06
  • And also I don't understand what it means by changing the "wavelength". – Alex L May 14 '19 at 06:07
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    +1 for the symmetry: 4-momentum is a Fourier conjugate of spacetime. – safesphere May 14 '19 at 06:09
  • @AliLavasani indeterminate means that the position is not a single specific point, the particle is not in any specific point in space. The wavefunction represents the possible positions the particle could be at if we measure where it is. What we do when we compress the wavefunction is decrease the standard deviation (if you don't know what this means, you should look for a course on probability and statistics) of the position. –  May 14 '19 at 06:21
  • The wavelength here refers to the idea of matter waves and the wave particle duality. I notice that you are unfamiliar with a lot of these topics, I would recommend you download an app called "Quantum" (the logo is a white "Q" in a red background), it is a good app with pictures that introduces quantum mechanics from a historical perspective. You can start to understand some of these topics by going through it. –  May 14 '19 at 06:24
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    The bit about a HUP for energy and time is at best misleading. Time is not an operator in QM. – Martin C. May 14 '19 at 06:42
  • Well, as you can see in this article, it is reasonable to introduce a time operator in QFT https://arxiv.org/ftp/quant-ph/papers/0211/0211047.pdf and after all particles in QM do not decay, but they do in QFT. Also, see this post https://physics.stackexchange.com/questions/220697/is-there-a-time-operator-in-quantum-mechanics –  May 14 '19 at 06:50
  • @SalvadorVillarreal The question that OP is asking is in the context of QM and not QFT. In QM, it is really misleading to put the so-called energy-time uncertainty relation in the same sentence as the true position-momentum uncertainty relation. There is a sense even in QM in which the relation that is referred to as energy-time uncertainty relation can be made sense of (see, Griffiths, for example) but it is certainly not an example of the standard Heisenberg uncertainty relations which fundamentally appeal to non-commuting nature of operators. –  May 14 '19 at 07:10
  • Also, while a huge fan of 3B1B, I am a bit irritated by the seeming attribution of a kind of triviality to the position-momentum uncertainty relations in the video. A priori, physically, the Fourier modes have nothing to do with momenta, so the uncertainty relation between the position-basis spread of a distribution/function and the Fourier-basis spread of the same doesn't tell us anything of physical relevance. The physical reason for the HUP is the commutator of the operators (as visible in any standard proof of the HUP) which then forces the Fourier-domain of positions to be momenta. –  May 14 '19 at 07:22
  • @Dvij I concede in the point of the energy and time. The question is indeed about QM and not QFT. But I don't think that 3B1B makes HUP trivial... What makes it work is precisely the fact that the momentum operator is proportional to the derivative with respect to position... Even if there is no a priori relation, it is a consequence of the equations that describe the system... I think it's similar to saying in classical mechanics that thrown objects follow a parabola, while there is nothing that requires that a priori, it is a consequence of the form of the equation (namely 2nd order ODE) –  May 14 '19 at 14:29