Degenerate matter: How reality deals with uncertainty

https://azeemba.com/posts/degenerate-matter.html

by azeemba • 4 years ago

123 51 4 years ago

sdflhasjd• 4 years 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• 4 years ago

It's just describing quantum mechanics. If it's not the case, quantum mechanics is wrong.

azeemba2• 4 years ago

Because 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

mannerheim2• 4 years ago

Not just similar, position and momentum are Fourier transforms of each other in QM (but that's in the article).

shonenknifefan1• 4 years ago

This is the Heisenberg uncertainty principle, which is the foundation of quantum mechanics.

https://en.wikipedia.org/wiki/Uncertainty_principle?wprov=sf...

illini1• 4 years ago

It 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• 4 years ago

There 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

platz• 4 years ago

They 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...

https://en.wikipedia.org/wiki/Uncertainty_principle#Introduc...

https://en.wikipedia.org/wiki/Conjugate_variables

aviplane• 4 years ago

I 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• 4 years ago

As 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...

l33t233372• 4 years ago

Here is an excellent video from an excellent channel explaining the concepts involved: https://m.youtube.com/watch?v=MBnnXbOM5S4

sabellito• 4 years ago

This is the video that got me to really understand the uncertainty principle 4 years ago.

ziofill2• 4 years ago

Quantum 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.

dataflow3• 4 years ago

Question: 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.)

elorant• 4 years ago

Is determined by what though? The measurement itself or some "exotic" relationship/factor we still haven't managed to figure out?

a9h74j• 4 years ago

One key word there, AFAIK, being representation.

To lower the abstraction level a bit: In tangible 1-D Schrodinger first-year QM terms: 1) All of the information about the state is in the wave function. 2) Per Fourier type limits, you cannot even represent in a wave function known position and known momentum past a limit on the multiplcative product of calculable "spreads" or uncertainties. (Smaller spread in position as implied in a wavefunction forces larger spread in momentum as calculated from the same.)

This can clarify that the limit is not because "measurement introduces disturbance". Rather, representation forces spread(s).

xeonmc2• 4 years 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.

frisco• 4 years ago

Imagine 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• 4 years ago

I tried to FFT an audio sample with a single value and the FFT app barfed.

jacquesm2• 4 years ago

That's a beautiful way to describe this.

marcosdumay3• 4 years ago

Are 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.

whatshisface2• 4 years ago

Because you can observe matter diffraction in slit experiments, and any theory where that didn't happen would be at odds with our observations.

chadcmulligan• 4 years ago

If particles are waves then the uncertainty principle follows from the Fourier transform - https://www.quora.com/How-does-the-uncertainty-principle-rel...

civilized• 4 years ago

In 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

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• 4 years ago

I'm pretty sure you can find a lot of degenerate matter on reddit/twitter xd

mahathu• 4 years ago

Why should I listen to a Software Engineering Manager on this?

tromp• 4 years 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• 4 years ago

It 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.

simne2• 4 years ago

Yes, 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.

danbruc2• 4 years ago

In 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.

snarkconjecture2• 4 years ago

You 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).

danbruc2• 4 years ago

Is this actually true? I am not talking about the time derivative for all times, only for the initial conditions.

justaj2• 4 years ago

Unable to view videos because my IP is blocked. Please don't block IP addresses: https://news.ycombinator.com/item?id=34178168

azeemba2• 4 years ago

I checked and there don't seem to be any IP restrictions or geo-restrictions.

Do you mind sharing what browser/platform you are using?

justaj2• 4 years ago

I checked back again and it works now. Could have been a HN hug of death?

simne2• 4 years ago

I wonder, when will appear analog of Uncertainty Principle for coding and Open Source :)

Degenerate matter: How reality deals with uncertainty

Related Stories