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Degenerate matter: How reality deals with uncertainty
- sdflhasjd 4y ago> When the position probability distribution is made narrower, the momentum probability distribution becomes wider ... Note that this is not a limitation of measurement I see this a lot, but what's the reason we know it's not the case?
- russdill 4y agoIt's just describing quantum mechanics. If it's not the case, quantum mechanics is wrong.
- azeemba 4y agoBecause it is part of the theory of quantum mechanics itself. It is not just an empirical fact, it is a prediction of the theory itself. Similar behavior shows up in Fourier Transforms [1]. When an audio sample is converted into frequency-space, it shows similar "narrowing/widening" relationship with the original audio sample. This again isn't a limitation of measurement, its just the relationship between the two bases [1]: https://en.wikipedia.org/wiki/Fourier_transform https://en.wikipedia.org/wiki/Fourier_transform
- mannerheim 4y agoNot just similar, position and momentum are Fourier transforms of each other in QM (but that's in the article).
- shonenknifefan1 4y agoThis is the Heisenberg uncertainty principle, which is the foundation of quantum mechanics. https://en.wikipedia.org/wiki/Uncertainty_principle?wprov=sfla1 https://en.wikipedia.org/wiki/Uncertainty_principle?wprov=sf...
- illini1 4y agoIt has to do with the wave-particle duality of matter. As you more finely attempt to measure the momentum (a more fundamental measurement compared to kinetic energy) and its location, you run into the wave-like nature of matter. It becomes more of a distribution rather than a point measurement. I won’t attempt to show the mathematics behind this, I’ll leave that for a physicist or mathematician.
- incompatible 4y agoThere was an article posted here recently that argued quite well for the wave-only view, avoiding the wave-particle duality by accepting that particles don't exist: https://news.ycombinator.com/item?id=33937900 https://news.ycombinator.com/item?id=33937900
- platz 4y agoThey are conjugate variables (i.e. Fourier transform duals) which arise in quantum mechanics simply due to the matter wave nature of all quantum objects https://en.wikipedia.org/wiki/Uncertainty_principle#cite_ref-Rozema_11-0 https://en.wikipedia.org/wiki/Uncertainty_principle#cite_ref... https://en.wikipedia.org/wiki/Uncertainty_principle#Introduction https://en.wikipedia.org/wiki/Uncertainty_principle#Introduc... https://en.wikipedia.org/wiki/Conjugate_variables https://en.wikipedia.org/wiki/Conjugate_variables
- aviplane 4y agoI think this is fundamentally a question of interpretation of quantum mechanics. We (society/physicists) do some set of experiments, and come up with a model to explain all of the resulting measurements. Quantum mechanics is a model that explains almost all of the phenomena that we observe. However, it is fundamentally only a description of our observations. To answer the question directly, in quantum mechanics, a state (of some particle, or collection of particles, as in the blog post) is represented as some vector, and we represent "operators" (which can represent position, momentum, spin, etc...) as matrices acting on the state vectors. When we do a measurement, the state is projected onto an eigenvector of the operator. If we do multiple measurements we have to project multiple times, along different bases. This process is not necessarily commutative. If I measure position and then momentum I will get a fundamentally different result than if I measure momentum and then position. If we make measurements of two observables that have the same basis, then the two matrices will commute, and there is no such limitation. However, with non-commuting observables this is a fundamental limitation, no matter how good your measurement is, you will always be projecting the initial state in different ways depending on how you measure the different observables.
- krackers 4y agoAs for why we use eigenfunctions to represent the observables, I found [1] to have a satisfying explanation that was satisfying to me. [1] https://physics.stackexchange.com/questions/583835/why-do-we-use-eigenvalues-to-represent-observed-values-in-quantum-mechanics https://physics.stackexchange.com/questions/583835/why-do-we...
- l33t233372 4y agoHere is an excellent video from an excellent channel explaining the concepts involved: https://m.youtube.com/watch?v=MBnnXbOM5S4 https://m.youtube.com/watch?v=MBnnXbOM5S4
- sabellito 4y agoThis is the video that got me to really understand the uncertainty principle 4 years ago.
- ziofill 4y agoQuantum physicist here. This is the way I like to think about it. A quantum system corresponds to some vector (i.e. the state of the system) and there are an infinitude of possible bases to write down the vector. Position and momentum are two bases of the same vector space that don’t share any element (not only that, they are particularly incompatible in the sense that you need ALL the element of one basis to decompose a single element of the other), so if you have a very localized system (e.g. that roughly corresponds to only one basis element in the position basis) you’ll need a lot of elements of the momentum basis to describe the same vector, and vice versa. The key to the puzzle is that there’s no escaping the uncertainty principle because position and momentum are two incompatible points of view of the same vector space (and of the same vector if you think about the state). Update: I think part of what’s confusing about this is that classically we can have pretty much any combination of position and momentum, but quantum mechanically it’s the opposite: once you know the state in the position basis, its momentum representation is also determined and vice versa.
- dataflow 4y agoQuestion: Let's say we added a single particle X (or, well, multiple) to our otherwise quantum world whose position and momentum we could pin down with arbitrary accuracy. What would happen if we used that particle to measure another particle Y? Normally Y has some fundamental uncertainty around its position and momentum, but would having access to X let us bypass that limit? If not, what would happen to the total uncertainty in the system after the measurement? (If that would allow us to break the limit, then it would seem like this really is a measurement limitation in some sense, so I'm assuming the answer is that that's not the case, but then I'm wondering what would happen afterward.)
- marcosdumay 4y ago> Let's say we added a single particle X (or, well, multiple) to our otherwise quantum world whose position and momentum we could pin down with arbitrary accuracy. That particle X does not exist. We can't add it to the world.
- 4y ago
- frisco 4y agoImagine you want to calculate what frequencies are present in a sound. At the width of the time window you are looking at goes to zero, you just have a single point. There's no frequency content of a single point! It's just a number. As you look at a longer time window, you can see what frequencies are present over the window, but you lose some sense of time "resolution" over which that is meaningful. This is fundamental and inherent in the Fourier transform. Put differently, you can't have a signal that is simultaneously both "band-limited" and "time-limited". Other conjugate variables, like position and momentum, work the same way.
- spiritplumber 4y agoI tried to FFT an audio sample with a single value and the FFT app barfed.
- jacquesm 4y agoThat's a beautiful way to describe this.
- marcosdumay 4y agoAre you asking why we say that the momentum is physically spread through a distribution instead of a precise value that we just happen to not know? The hypothesis that it has a precise value is called "hidden variable theory" if you want to search for it. We know that several variants of it are false because of how particles interfere with each other and how they can become correlated. The interference math is quite simple (the fact that particles have destructive interference means they aren't just a lot of stuff with unknown properties), but the math on correlation is complicated. There is still an hypothesis called "superdeterminism" that say that all the information about the entire past and future of particles always existed with infinite precision. That one we can't rule out.
- whatshisface 4y agoBecause you can observe matter diffraction in slit experiments, and any theory where that didn't happen would be at odds with our observations.
- chadcmulligan 4y agoIf particles are waves then the uncertainty principle follows from the Fourier transform - https://www.quora.com/How-does-the-uncertainty-principle-relate-to-Fourier-transforms https://www.quora.com/How-does-the-uncertainty-principle-rel...
- civilized 4y agoIn quantum mechanics, the impossibility of localizing in both position and momentum simultaneously is a corollary of the more fundamental principle of wave-particle duality: 1. Everything has both particle and wave characteristics. 2. Position comes from looking at the thing from the particle perspective, while momentum comes from looking at the thing from the wave perspective. 3. It's mathematically impossible for a thing to be a "perfect" point particle and also a "perfect" wave (i.e. a sine function). Wikipedia covers this well: https://en.wikipedia.org/wiki/Uncertainty_principle https://en.wikipedia.org/wiki/Uncertainty_principle > the uncertainty principle is inherent in the properties of all wave-like systems... it arises in quantum mechanics simply due to the matter wave nature of all quantum objects. Thus, the uncertainty principle actually states a fundamental property of quantum systems and is not a statement about the observational success of current technology.
- TDiblik 4y agoI'm pretty sure you can find a lot of degenerate matter on reddit/twitter xd
- mahathu 4y agoWhy should I listen to a Software Engineering Manager on this?
- tromp 4y ago> computer chips are a common example of something that only works because of quantum mechanics Does that mean that quantum effects play an essential role in the operation of a transistor?
- mikrl 4y agoIt has been a while since I studied solid state physics but yes, exactly correct. The fun effects of semiconductors rely on the electronic properties of certain solids (or their electrons, equivalently their absence ie ‘holes’) which take place at a small scale, hence the use of quantum mechanics. Some of these effects can manifest at larger scales, and can be modelled classically too.
- simne 4y agoYes, and not exact. While every mosfet transistor have mode could considered genuine quantum (tunnel effect), it is not used in computers, it used in microwave generators/detectors. What used in computers, are just switch mode (electric field creating/moving/changing conducting zone), which is mostly classic physics and created as fruit of garage engineering, without understanding of quantum mechanics. And yes, this is very abstract border, as I know, 100% quantum effects considered when working with single atom or single electron (something <= 1nm), but for example, 100nm transistors, usually considered as classic physics. BTW it is very interest, what will show image sensor with ~1nm sized cells. Now their cells are much larger than wavelength, but already created superlenses, could focus on nm-sized things.
- danbruc 4y agoIn classical mechanics momentum is the derivative of position with respect to time. In quantum mechanics momentum is the gradient, i.e. the combination of the partial derivatives with respect to the three directions of space, of the wave function [1], i.e. the probability amplitude of the particle as a function of position. In consequence, classically one can prepare a particle with any position and any momentum independently, i.e. one fixes the position and the time derivative of the position. Quantum mechanically fixing the position distribution also fixes the momentum distribution as the later is the derivative of the former with respect to space and therefore can not be specified independently. Similarly to the classical case one can also fix the time derivative of the wave function but that corresponds to the energy, not the momentum. [1] In position space.
- snarkconjecture 4y agoYou can't specify the time derivative of the wavefunction independently, because the Schrodinger equation is first-order (unlike classical dynamical laws which are second-order).
- danbruc 4y agoIs this actually true? I am not talking about the time derivative for all times, only for the initial conditions.
- justaj 4y agoUnable to view videos because my IP is blocked. Please don't block IP addresses: https://news.ycombinator.com/item?id=34178168 https://news.ycombinator.com/item?id=34178168
- simne 4y agoI wonder, when will appear analog of Uncertainty Principle for coding and Open Source :)