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22KiB, 544x300, uncertainty principle.jpg

Anonymous Tue 27 Oct 2015 16:10:43 No.7619583 View Reply Report Delete

Quoted By: >>7619603 >>7619660 >>7619669 >>7619809 >>7620250 >>7620319

Does anyone know a physical reasoning behind the uncertainty principle? I know the mathematical reasoning is from the commutation of the position and momentum operators. But, why physically can we not localize position and momentum?

Anonymous

Anonymous Tue 27 Oct 2015 16:18:30 No.7619603 Report Delete

Quoted By:

>>7619583

I would say that the commutation relation between the operators is the mathematical representation of the observable fact. There's no real explanation for why the uncertainty principle exists in the first place.

Anonymous

Anonymous Tue 27 Oct 2015 16:36:39 No.7619660 Report Delete

Quoted By:

>>7619583
>I know the mathematical reasoning is from the commutation of the position and momentum operators

No, the mathematical and physical reason is you can't be localized in position/time space and frequency (spacial or time) space from Fourier analysis.

When you link momentum with the de Broglie wavelength, you immediately get the uncertainty relation.

Anonymous

Anonymous Tue 27 Oct 2015 16:40:17 No.7619669 Report Delete

Quoted By:

>>7619583
The why boils down to the mathematics, really. The universe has dimensions of space, time, and probability through which particles can move.

Anonymous

Anonymous Tue 27 Oct 2015 16:47:01 No.7619693 Report Delete

Quoted By:

https://www.youtube.com/watch?v=wBbMUphqU_0

Pretty good explanation tbh

Anonymous

Anonymous Tue 27 Oct 2015 17:17:59 No.7619744 Report Delete

Quoted By: >>7619750 >>7619752

The best reason I have heard is that for something like an electron, in order to see it, you must hit it with a photon, and on that scale the momentum of a photon makes a huge difference and this you know where the electron is but now because of momentum, you dont know its speed

Anonymous

Anonymous Tue 27 Oct 2015 17:21:34 No.7619750 Report Delete

Quoted By:

>>7619744
To see a electron accurately you use a high energy photon, this allows you to see the photons location due to it's high frequency, however the energy is absorbed by the electron and the momentum changes but you can't know by how much. To find the electrons momentum you use a low frequency photon, this measures momentum but due to low frequency -->large displacement

Anonymous

Anonymous Tue 27 Oct 2015 17:22:29 No.7619752 Report Delete

Quoted By: >>7620209

>>7619744
This is not a good way to think about this. What you are saying is true, but that is not the underlying phenomenon of the uncertainty principle. The uncertainty / delocalization is there whether you hit it with a photon or not.

Anonymous

Anonymous Tue 27 Oct 2015 17:30:48 No.7619774 Report Delete

Quoted By:

I wouldn't say that there is reasoning behind it. Remember that the mathematics is something we've developed to math physical observations, not the other way around. That the uncertainty principle that applies to Fourier analysis happens to work out in this case is pleasant for us, but it doesn't have to be that way.

Anonymous

Anonymous Tue 27 Oct 2015 17:52:29 No.7619809 Report Delete

Quoted By: >>7620334

>>7619583
In order to get better precision on position, you need to use higher frequencies of light (or whatever you are using to bombard with). Higher frequency light has higher energy so if you were to bombard an electron, you will know it's position better but that extra energy will significantly change whatever the electron's momentum was before you bombarded it leading to uncertainty of momentum.

The opposite is also true. You can use a very low frequency photon which has low energy, When you bombard an electron, you won't be changing its momentum much because the photon has low energy. However, now you have uncertainty in the electron's position.

Anonymous

Anonymous Tue 27 Oct 2015 18:00:20 No.7619822 Report Delete

Quoted By:

daily reminder to popsci NEETs that the uncertainty principle is not the observer effect

Anonymous

Anonymous Tue 27 Oct 2015 21:35:18 No.7620209 Report Delete

Quoted By: >>7620334

>>7619752
Well I guess not - I dont study quantum mechanics - but for any layman, how else should one explain it in an understandable manner? As I said, I dont study quantum, but I have had to use it on occasion, for example I had to show on an exam that you cannot balance a pencil on its tip indefinitely because of the uncertainty principle, and well, honestly I just like to think of it as sooner or later some electron hits the pencil and it has to go down lol but that stuff is way above my head. I only deal with classical mechanics

Anonymous

Anonymous Tue 27 Oct 2015 21:50:08 No.7620250 Report Delete

Quoted By:

>>7619583
Why does the picture use chi and rho instead of x and p?

Anonymous

Anonymous Tue 27 Oct 2015 22:19:51 No.7620319 Report Delete

Quoted By:

>>7619583
This is just a property of all waves, and so is naturally in QM because QM is wave mechanics.

Here is a "physical" description: the position of a wave is where it is, this is intuitive. You also know that the wavelength of a wave is related to its momentum.

So lets imagine a wave with one bump traveling along. This wave is easy to define where the wave is, it is at the bump. It'd be especially easy if the bump were infinitely thin and we could say exactly where it is, but it wouldn't be a wave then. With the solitary bump traveling along though, we have a hard time saying anything about the wavelength, since it isn't well defined seeing as it is only a bump.

Now let's imagine the opposite scenario, many many bumps, if it were a wave on a string it has crests and troughs all the way. In this case we can't say anything about where the wave is besides on the string, while the wavelength is now very well defined.

Anonymous

Anonymous Tue 27 Oct 2015 22:28:18 No.7620334 Report Delete

Quoted By: >>7620402

>>7620209

I imagine there isn't really a good way to intuitively explain this. We're used to thinking and looking at things at a macrolevel, where there is very limited mechanical uncertainty at all.

From my limited knowledge, I gather that describing phenomena as waves/particles is just a way to simplify some quantity in time/space and make it possible to analyze and describe more succinctly. Scientists are not measuring something that makes sense as only a wave or a particle, but as both under different conditions.

>>7619809

Isn't that the observer effect?

Anonymous

Anonymous Tue 27 Oct 2015 22:57:51 No.7620402 Report Delete

Quoted By: >>7620405

>>7620334
The intuitive way that I personally make sense of it (which maybe is wrong), is to take the Fourier transform of the delta function, which is constant with respect to frequency. Hence, the energy of the dirac function is equally distributed across ALL frequencies. Since the dirac function can be thought of as a probability distribution with zero variance, it corresponds to a particle who's position is known exactly.

On the other hand, the Fourier transform of a perfect sine or cosine wave is a set of pulses (and zero everywhere else), so you have the other extreme of an extremely unlocalized particle, with no spread in the frequency.

Anonymous

Anonymous Tue 27 Oct 2015 23:01:09 No.7620405 Report Delete

Quoted By:

>>7620402
> energy of the dirac function
meaning the signal's energy, NOT physical energy.

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