The Heisenberg Uncertainty Principle: Proof/Explanation!

In this video, I derive and discuss the Heisenberg Uncertainty Principle, perhaps one of the most famous relationships in Quantum Mechanics. I start by using the Generalized Uncertainty Principle (derived here: • The Generalized Uncert... ) to determine the commutator of the position and momentum operators whose expressions I found in the previous video (link: • Position and Momentum ... ).
Once I derive the Heisenberg Uncertainty Principle, I devote much of the video to explaining what it means - *that it's a statement on the fundamental nature of quantum mechanical particles which arises from certain mathematical principles (e.g. Fourier Transforms) describing those particles*. In the video, I state that the Heisenberg Uncertainty Principle 'is a consequence of mathematics', but keep in mind that this mathematics is used to model quantum mechanical particles. Thus, in the context of Quantum Mechanics, the Heisenberg Uncertainty Principle is a physical fact that can be derived mathematically (that's what I mean when I say 'consequence of mathematics').
Towards the end of the video, I emphasize that the Uncertainty Principle is NOT a statement about observer-induced limitations on measurements; that's the Observer Effect. The Heisenberg Uncertainty Principle is rather difficult to wrap one's head around, at least at first, so I encourage you to ask questions and comment down below!
Prerequisites: All the previous videos in this playlist (Playlist 1: th-cam.com/users/playlist?list..., and all the videos in this playlist: • Quantum Mechanics: Mat... .
Lecture Notes: drive.google.com/open?id=1S7B...
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EDIT: The deBroglie formula at 6:40 should have hbar instead of h.

[@ostensiblyquerulous]

Christ dude you’re production value and teaching skills are unmatched. Pretty much every topic you have covered/cover is super useful and helpful. Can’t say thank you enough - you are helping me succeed in my QM class.

[@FacultyofKhan]

I'm happy to hear that my videos have been useful for you! Thank you for the kind feedback!

[@duckymomo7935]

Faculty of Khan are you related to khan academy?

[@FacultyofKhan]

No

[@Ananta9817]

Bamboozled. I just checked your about section.
Thank you for uploading university level videos. They're all really interesting.

[@JoshuaAalampour]

Your videos are so underrated! Thank you so so so so so so so much!

[@steveymcginins7761]

Your videos are simply incredible

[@jamesmonteroso824]

this makes sense a lot thank you so mucccccccccccccccccchhhhh

[@sanjuansteve]

Since everything from gluons to solar systems are in orbit with something and moving as polarizable axial or helical apparent waves as they travel thru our expanding universe, it seems like the natural first assumption to explain particles behaving like polarizable axial or helical waves would be that they are orbiting undetectable dark matter particles, no? Wouldn't that explain the double slit, uncertainty, etc?
Does anyone know of any specific experiments, calculations, theories, etc that have disproved this possibility? Thx

[@Fearabbit]

Just the fact that you CAN measure dark matter by their mass. You would know that there is something there. But we know where all of the mass of particles comes from, there is no dark matter involved. Plus I'm not sure how them orbiting dark matter would explain any wave phenomena.

[@joshuakyanaalampour]

Thank you !

[@JRabba1995]

I think there's a mistake where you took the Fourier transnform of a delta function, the result should be a constant in momentum space. Great Channel, I enjoy all your videos.

[@FacultyofKhan]

Thank you for the kind words! The Fourier Transform of a delta function is actually a sinusoidal wave as I mentioned (see: mathworld.wolfram.com/FourierTransformDeltaFunction.html). The imaginary power on the e denotes the fact that it's sinusoidal (Euler's formula).

[@venkat4167]

You are wrong. You need to take magnitude i.e. absolute value of exp(-2pikx0) which is 1 throughout. The momentum however is still spread out.

[@freekiddionsmodestmenugta9759]

how can u say that fourier transform of postion space is momentum space ?

[@charansri503]

And this is the actual proof!

[@atharvas4399]

ist the de broglie formula at 6.40 wrong? lambda = h/p , why the 2 pi?

[@FacultyofKhan]

Yea I meant to write hbar instead of h; thanks for the correction!

[@jamesmonteroso824]

FINALLY !

[@simranak960]

Sir in 8:50, in the momentum space after Fourier transform of position space ,Sigma p supposed to be less and Sigma x supposed to be large isn't it?? If it's so then why the mistake in the video?

[@srishtikdutta8946]

Yes it's lit😁

[@maxunknowngamer2286]

Nice 👍

[@shubhammandot6547]

Theoretically speaking, if we can find the instantaneous velocity of the particle in a purely "particle wavefunction system" then we should be able to find atleast the instantaneous momentum of the particle with good accuracy, isn't that right?

[@FacultyofKhan]

I believe so, yes!

[@shubhammandot6547]

@@FacultyofKhan Then wouldn't that violate the uncertainty principle? I'm just curious to know how accurate this line of reasoning is.

[@FacultyofKhan]

@@shubhammandot6547 Why would it? A more certain velocity doesn't translate into a more certain position; it translates into a more certain momentum since momentum and velocity are closely related.

[@simranak960]

But in Quantum mechanics p is not equal to mv but p= h/lamda.

[@Ihsan_khan00]

Great, thanks

[@FacultyofKhan]

Glad you liked it!

[@1060michaelg]

I would totally get this if it weren't for the math part...

[@Pudin788]

thanks a lot

[@pondoknira117]

Jadi jika diambil limit mendekati nol bukan adalah salah

[@pondoknira117]

Artinya itu resolusi terendah yg bisa dicapai satu terhadap yang lain

[@pondoknira117]

Artinya penggunaan derivative dalam persamaan Schrodinger adalah cukup adanya

[@pondoknira117]

Ini turun kalau bukan dua
Delta x eq. p/m

[@pondoknira117]

Secara limit bukan nol /h

[@prasenjit7699]

Too hard