Ch 10: What's the commutator and the uncertainty principle? | Maths of Quantum Mechanics
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- Опубліковано 23 січ 2023
- Hello!
This is the tenth chapter in my series "Maths of Quantum Mechanics." In this episode, we'll define the commutator, and we'll derive how commuting observables share a simultaneous eigenbasis. We'll then dive into how non-commutation necessarily leads to uncertainty relations in quantum mechanics.
If you have any questions or comments, shoot me an email at:
quantumsensechannel@gmail.com
Thanks!
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Music:
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♪ Astral 1 by Patricia Taxxon
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Hi everyone, a point of clarification:
At 13:53 when I say “constant times the identity” I mean that we have “i * hbar * Identity”, where the identity operator isn’t shown (since no one ever shows it, but it’s there!). Since we have operators on the left, we need an operator on the right, so that’s why the identity is there. Sorry if this caused some confusion!
-QuantumSense
Professor Basil J. Hiley: an anti-commutator or Baker bracket ( known as the Jordan product). The Lie bracket becomes the Poisson bracket as we go to the classical limit, while the Jordan product becomes the normal inner product. In symbols (AB + BA)/2 -> AB. The Jordan product is the most neglected product in the whole discussion of the foundations of quantum mechanics."
Please do a video on this anti-commutator. thanks
I've been fascinated at how there is this deeper dynamic hiding in the noncommutative transition frequencies that Heisenberg discovered - as Alain Connes focuses on in that 2015 lecture to physicists. It's such a simple yet radical view that Connes realizes can be understood through music training as he also started at an early age.
Yakir Aharonov's research group has stated that the negative-mass particles of weak measurements "imply gravitational repulsion" (2022) - and this is actually the secret of levitation in deep meditation. My own teacher, Chunyi Lin, of did 28 days of cave nonstop meditation in full lotus padmasana - at Mt. Qingcheng - in the mid-1990s. He said when he finished that 28 days meditation with no sleep - he then left the cave and was in full lotus meditation outside the cave when he levitated up, spiraling up nine feet, in full lotus padmasana while he was next to a pine tree. hahaha. It's too bad that Aharonov, et. al., do not adapt the noncommutativity truth.
Actually there is a new Ph.D. student who created a youtube channel and he points out how the origin of Heisenberg Uncertainty is in fact the nonlocal superposition that is noncommutative!
So I mentioned to him about your [Basil J. Hiley] research - maybe he'll switch over to that for his Ph.D. focus. haha.
All the best,
drew hempel
Certainty is dual to uncertainty -- the Heisenberg certainty/uncertainty principle!
Commutators = two paths (duality).
"Always two there are" -- Yoda.
Synergy is dual to energy -- energy is actually dual.
Potential energy is dual to kinetic energy.
Concreteness in speeches, expository clarity, very clear voice for non-native english speakers, excellent video editing and superb quality images. This channel deserves a big round of applause. Thanks, thanks, thanks
Yes. Well-prepared lectures.
I just have to say how big of a fan I am of all the hard work you are putting into this. Beautifully explained, well-illustrated, and logically-sequenced. Thanks a million.
Hello, thanks so much for watching, and thank you for the kind words! I’m glad you’re enjoying the series!
-QuantumSense
@@quantumsensechannel Enjoy doesn't even describe it! Can't wait for another episode and I just keep repeating for myself, "wow, this is so good".
That recap on the proof made u better than 80% of physic/maths explainer out there. Its like they dont see how lengthy their explaination, even though intuitive, are.
dude, you are legit lifesaver. I couldn't attend my university lectures on quantum mechanics because I fell ill and this has literally saved my life. Thank you so much, I am literally going to cry. Best channel ever!
Thanks for another great video. I just wanted to point out a common misconception that you repeated in your summary. Namely that one cannot measure position and momentum at the same time. The simple follow-up question is “so,how do we measure momentum?”, because the detector is always located somewhere and seems like it must de facto measure both. Indeed, the most common way to measure momentum is via a time-of-flight experiment. Here, the quantum particle is trapped in a region of small size and at t=0, the trap is released. Then the particle is allowed to expand in free space until detected. We measure this time of flight along with the particle being detected at a specific position. Then we can determine the momentum (from distance travelled and time travelled and the mass). Uncertainty only enters when the experiment is repeated, because you get a different answer and ultimately determine the momentum probability distribution of the original quantum state. Cindy Regal’s group has recently used this method to image the harmonic oscillator Fock states in momentum space with time-of-flight of atoms. Other applications include the momentum microscope used in XFELs. So, it actually is very common to measure position and momentum at the same time. Proper interpretation of the uncertainty principle requires care.
can you please elaborate? what exactly is the proper interpretation that you mentioned in the last line? Thanks
@@kabeerkumar4334 The proper interpretation is not to say uncertainty does not allow you to measure position and momentum at the same time--a common mistake. Instead, you should recognize that the uncertainty principle is about fluctuations, and to find those, you need many repeated measurements. Also understanding properly the meaning, in the experiment mentioned, one can only get the initial momentum uncertainty of the quantum state, since that is all that is being measured.
@@quantum4everyone okay, I'll find out more about it. Thanks for your response.
This video was absolutely ILLUMINATING, just as the one on observables as linear operators was (where I finally understood how central the property of an operator’s linearity is). Textbooks and other introductory material could well take your exposition as a model to emulate - perhaps even rewriting them accordingly. If only all physics teachers had as clear and meaningful a way to explain things perhaps Feynman’s famous (apocryphal?) quip might be proven wrong, or at least exaggerated, that “Anyone who thinks they understand QM clearly doesn’t understand it”. - Though this of course begs the question: What does it mean “to understand” something? (There are different types of “understanding”! There’s an excellent Mindscape podcast by Sean Carroll on this).
watch Professor Basil J. Hiley - How does the Classical World Emerge from the Implicate Order? on the Pari Center channel - he exposes the error of Richard Feynman. Hiley: "it [kinetic energy] blows up but was richard feynman worried? not at all what he said is you know how how do i where is this what about how can i get rid of this energy and then he said let me change the rest mass of the particle to the rest mass one plus delta where delta is a small chain for a short time say epsilon that's my epsilon so the major thing is why and he doesn't tell us he just says do it and nobody seems to say why but we now have a story about mass re-normalization blah blah blah but hold it this is a squared term and what we find look at the work of morris and myself what we find is that the quantum potential never appears until you go to order h squared so what he's missing here is the quantum potential and if you put the quantum potential in you get the right result and now the interesting thing is de broglie remember chris [Dewdney] was saying it's the de broglie bohm theory that that he was talking about"
A sign of understanding of a subject and intelligence is the clarity and simplicity of how one presents ideas. Quantum Sense's video presentation here is brilliant!
this is so powerful!! It's like saying momentum and energy are same thing. While momentum and position are different thing!
Hey creator it's been a long time you didn't uploaded any such videos like this...your content is appreciable ...pls keep posting..
The series is finished. The length of the serie was announced at the start.
Hey, as I said before, your videos are really mind opening. It would be awsome, for me, if I may suggest, if at the end of this series, will be included one or more videos showing typical applications in problems of some of the explained topics!
Why must the eigenvectors of operator B span the eigenspace of A at 9:05 ? Am I missing something from a past video?
Hello! Thank you for watching.
This comes from Chapter 7, where we proved that the eigenvectors of an observable must span the entire eigenspace. Thus, the eigenvectors of B have to span this eigenspace in some way. Let me know if there’s still some confusion!
-QuantumSense
This really connects the formal math and the intuition in a beautiful way. Well structured, excellent execution.
This series is invaluable! ^^ Keep up the fantastic work! ❤❤
That was a really nice explanation, keep these coming!
Also at 7:10, this is the key fact used to prove the spectral theorem for self adjoint operators in finite dimensions!
Wow! This really connects the formal math and the intuition in a beautiful way. Well structured, excellent execution.
I took a semester of introductory QM and none of it made sense until watching this series. I don't comment much, but this video on particular is really phenomenal :) thank you so much for making this, it's the first engaging, coherent explanation of QM that I've seen
So cool! Amazing video
Phenomenal quality of content sir
Excellent presentation,vow !
Well summarized. Easy to understand
Wow you're dropping these videos so fast for how high-quality they are. Great work.
Hello! Thank you for watching and for the kind words!
To be frank, though, the series is made in advance, and I am releasing them at a steady schedule. It is the product of around 1.5 years of work in my free time.
-QuantumSense
@@quantumsensechannel Wow. Well thanks for that 1.5 years. What I really like about this series is that you often take concepts that are presented as postulates or without proof or intuition in textbooks and other videos / courses and you present an intuition and reasoning for them. It's great to get a deeper understanding.
8:46 after thinking about this a while I have one question... Since the parallel and orthogonal eigenvectors share the same eigen value that would mean that the parallel and orthogonal eigenvectors should lie in the same degenerate eigenspace i.e. that would imply that the orthogonal eigenvector should lie in the degenerate eigenspace to which it should be orthogonal to which is contradiction i.e. the orthogonal vector is a 0 vector...
Is this why you state the eigenvectors of B must lie inside or orthogonal to the degenerate eigenspace? if so it wasn't clear
very helpful series
amazing
if A and B are two observables that commute then we know for sure that all eigenvectors of A are also eigenvectors of B. But if they do not commute then by using the commutation relation, is there a way to know the no of eigenvectors common to both A and B?
Hello! Thank you for watching.
In the case of momentum and position, yes! It’s actually quite easy to do so: assume position and momentum share an eigenvector (which by definition must not be the zero vector). Now apply that eigenvector to both sides of the commutation relation. This should show that the eigenvector must actually be the zero vector. Hence, you have a contradiction, and therefore they cannot share any eigenvectors.
-QuantumSense
And one more thing: I would be careful in saying that if A and B commute, that all eigenvectors of A are eigenvectors of B. This may not be true in the degenerate case. As we showed, we may only have a specific set of eigenvectors in the degenerate eigenspace of A that are eigenvectors of B. So although everything in the degenerate eigenspace of A is an eigenvector of A, it may be that only certain vectors in the eigenspace of A are eigenvectors of B. Part of the challenge of finding the simultaneous eigenbasis is finding these special degenerate vectors that are eigenvectors of both.
-QuantumSense
@@quantumsensechannel Oh right! thanks for the correction.......so then is it accrurate to say that if A and B commute then there exists a set of eigenvectors common to A and B that spans the entire hilbert space?
In case where the eigenvectors of B lying in the degenerate space of A have the same corresponding eigenvalues then I think all eigenvector of A are also eigenvectors of B right?
A simple counter example is with angular momentum. We say that one can only have simultaneous eigenvectors of L squared and one component of L. But, the s-wave state has an eigenvalue of zero for Lx, Ly, and Lz and also of L squared. So, just because operators do not commute does not mean, in some cases, they cannot have simultaneous eigenvectors.
Yes, this is a good point! I tried to make it clear in the video that noncommutation only implies the nonexistence of a simultaneous eigenbasis, but you can very much still have some shared eigenvectors, as you have shown!
-QuantumSense
At 4:36, in nondegenerated eigenvalue case, μ has to be 1, ie, B|α> is just the same vactor α, with the same direction and same length, right ? If μ ≠ 1, then B|α> and α are two different vectors with the same eigenvalue λ respect to opperator A. Please tell me where I get losted, thanks!
I had exactly the same thought!
3:45 Why do we assume that B_hat only rotates the vector and not scale the vector?
Hi Brandon, do you have any textbook or lecture notes recommendation for "quantum mechanics for mathematicans?" None of the ones I looked for seem to line up with your content? They are either super diffeq heavy or representation theory heavy
Hello, thank you for watching.
I haven’t found a textbook that really follows the way I’m approaching quantum mechanics. The closest I’ve read would probably be “Quantum Mechanics: A Modern Development” by Ballentine. This is my favorite quantum textbook, and he does a great job going into every single nook and cranny of quantum mechanics. Note that he does dip into representation theory, as he derives the foundation of quantum mechanics from the generators of Galilean group representations. This is a graduate level quantum textbook and he tackles some deep mathematical aspects, but his exposition and story telling is phenomenal, and it was one of the books that inspired my series.
If anyone else knows a good textbook, feel free to pitch in!
-QuantumSense
@@quantumsensechannel hi, thanks for your response! I'll will surely check it out later tonight. I've been working through the beginning of Hall's Quantum Theory for Mathematicans and it's pretty good so far, too, but nothing like the stuff you're putting out.
Thanks for the super entertaining and educational videos!
@@brianhu6277 Brian Hall's book is probably best suited for a first year math grad student, and I've found that there's no easier book if you want to learn QM the rigorous way. The math involved (functional analysis/representation theory) is undeniably hard, but if you slowly work through it it's very rewarding.
If you want mathematical references, for representation theory I'd suggest Hall's own "Lie Groups, Lie Algebras, and Representations: An Elementary Introduction". For functional analysis Kreyszig's "Introductory functional analysis with applications" is a gentle book that introduces everything you'd need.
I like Ballantine, but the best quantum book with mathematical rigor and reference to the proper Hilbert space approaches is the two-volume series by Galindo and Pascual. Pity it was originally written in Spanish and then translated into english or more people would know about it. It is a fantastic book. And this is from a Ballantine lover. But, Ballantine has some issues with some of the things he discusses. His book is great for discussing special cases (like bound states in the continuum). So do yourself a favor and get both!
Could you do the proof with non commuting observables?
Would be very enlightening for me.
Regards Hans
study Basil J. Hiley for the answer on noncommutative nonlocality. He was the collaborator of David Bohm. The secret is in the 2nd derivative of H-bar. thanks
Very beautiful.
can you have 0 uncertainty in one quantity and infinite uncertainty in another? how do you calculate their product in that case?
Around 8:50 You say that: «we can take the eigenvectors of B to lie either inside the eigenspace of A, or orrhogonal to it - there is no in between». I think that for this to be considered a proof, this statement needs further justificatio. Hopefully you could clarify
I also had the same doubt but I've been able to come up with a justification myself which I don't know is correct or not but here it goes regardless:
since the orthogonal component and the parallel component of the eigenvector share the same eigenvalue they must lie in the same degenerate eigenspace but the orthogonal component lieing in the plane to which it's orthogonal to is a contradiction! which implies his quote «we can take the eigenvectors of B to lie either inside the eigenspace of A, or orrhogonal to it - there is no in between»
Hopefully this made some sense!
Looking at the explanation of B scaling by mu on both the parallel and perpendicular components of the beta eigenvector at around 8:00, at first I thought "Any linear combination of beta should yield the same result if B is linear, so is this sophisticated argument required?" But then I realised if I extend to a matrix like a shear matrix, which only has eigenvectors in a few directions, and decompose them as linear combination of vectors in some set of arbitrary directions, those vectors would also have to be eigenvectors by my argument. What am I missing? Why can you not decompose eigenvectors into linear combinations and get more eigenvectors? I suppose I am misunderstanding how B would work on each vector in the linear combination?
Hello! Thank you for watching.
You are correct that you can decompose beta however you like, but your decomposition may not have the properties of the parallel/perpendicular ones we derived. B acting on an arbitrary decomposition could move it all over the place, it doesn’t have to scale it by lambda. But B acting on an in-eigenspace/perpendicular component must scale it by lambda, because we derived that those components must remain in the eigenspace/perpendicular to it. It’s this special property that constrains the components to only move in the “scale” direction. So components that don’t have these properties could move anywhere in the space. Hopefully this cleared it up a bit!
-QuantumSense
@@quantumsensechannel Yes, thank you. I may work through an example to see exactly how the operator diverges from what I expect, but overall I think I get the idea.
Love the video series btw. I have been patiently waiting for new installments. Great introductory exposition of quantum mechanics.
8:14 Why do they both have to be scale by the same amount?
here before you become famous!
8:06 Why must B times the parallel component be a scalar multiple of the parallel component? I thought that it could be pointed in any direction in the plane.
Also, why does B apply the same scaling factor to both components?
Hello, thank you for watching.
Our only condition was that the parallel component must stay on the plane. But when we add the condition that it is a component of an eigenvector of B, then that constrains the parallel component to only be scaled by B. Think about it, if the parallel component moved anywhere else on the plane, then it would change the direction of the beta eigenvector, and so it wouldn’t be an eigenvector anymore!
And both components are scaled by lambda because beta is scaled by lambda. If the components had any other scaling relation, then beta wouldn’t be scaled by lambda when we add the parallel and perpendicular components together.
This is a really subtle argument, so let me know if there’s still any mystery. I myself had to stare at this argument for a while to make sense of it myself.
-QuantumSense
Got it. So here’s my attempt at clarifying the argument:
B has an eigenvector beta with the associated eigenvalue mu.
We want to decompose B*beta into the plane component and the perpendicular component. Since vectors work like they do, the decomposition is unique.
Since beta is an eigenvector of B, the plane component of (B*beta) is equal to the plane component of (mu*beta), which is equal to mu*(the plane component of beta). Similarly, the perpendicular component of (B*beta) is equal to mu*(the perpendicular component of beta). This gives a decomposition of (B*beta). This part of the argument doesn’t depend on the plane being related to A.
Since A and B commute, we have that B*(the plane component of beta) is a vector in the plane. Similarly, B*(the perpendicular component of beta) is a vector perpendicular to the plane. By linearity, this gives a decomposition of (B*beta) into a plane and a perpendicular component. This part of the argument doesn’t depend on beta being an eigenvector.
So, with the two parts of the arguments,this gives two decompositions of (B*beta). Since decompositions are unique, we can equate them. This shows that (mu*(the plane component of beta))= B*(the plane component of beta). And similarly for the perpendicular component.
Therefore, the two components of each eigenvector of B are also themselves eigenvectors of B with the same eigenvalue.
@@quantumsensechannel Second part of the argument:
For each eigenvector of B, we can decompose it into a parallel eigenvector and a perpendicular eigenvector. The span of these two component contains the span of the original eigenvector. These two components are also eigenvectors of A.
Pick an eigenbasis of B. (I think it always exists?) Decompose it into twice as many eigenvectors. These vectors span the space. So a subset of these vectors form a basis (I think?). This gives a set of vectors that are an eigenbasis for both A and B. And then we cane somehow (I think?) show that all eigenvectors of A are eigenvectors of B and vice versa.
I don’t know if this argument works for infinite dimensional vector spaces.
If you make a PDF I will buy it, and probably I will not the only one.
00:00 : introduction commutators and uncertainty relationsa
00:45 : Motivation for commutation of operators
01:35: Definition of the commutator
02:35 : Motivation and importance of the commutator operator of operators in terms of eigenvectors
04:20 : Motivation and importance of the commutator operator of operators in terms of eigenvectors: nondegenerate case
04:57 : Motivation and importance of the commutator operator of operators in terms of eigenvectors: degenerate case
09:40: Motivation and importance of the commutator operator of observables in terms of eigenvectors: degenerate case recap
10:40: Consequences for observable operators: simultaneous diagonalization
11:40: Importance in physics: commutating vs noncommutating observables
13:40: Importance in physics: noncommutating observables
14:40 : Importance in physics: Heisenberg uncertainty principle
15:36 : Importance in physics: Generalized uncertainty principle and other uncertainty principles
16:30 Recap about commutation of observables
17:00 : Next video
I've been taught that in case of degenerate, you can't argue that B only scales |alpha>, but all we can say is that B transforms |alpha> from eigen space to eigen space. no more than that.
At around argument of 9:00, I took it as " since the orthogonal remains orthogonal, and the parallel remains parallel, B should only allows scale transform of each."
But I think I have counter example.
Let's say that |beta> that is eigen vector of B and is composed of four vectors, two vectors |beta p1>, |beta p2> live inside the eigen space of A and two vectors |beta o1>, |beta o2>live orthogonal to the eigen space of A.
that is to say B|beta>=lamda|beta>, |beta>= |beta p1>+|beta p2> + |beta o1>+|beta o2>.
since we proved that (by the argument that B is Hermitian) B transforms the orthogonals to be stay orthogonal, and transforms others lives in side eigen space of A to stay inside A, This video suggested that all conditions will only allow |beta p1>, |beta p2> to be scaled but no change of direction.
However the thing is if four vectors |beta p1>, |beta p2> , |beta o1>, |beta o2> composes |beta>, we can think of the case where |beta p1>, |beta p2> ,/ |beta o1>, |beta o2> , each two pairs changes direction (while orthogonals stay orthgonal and |beta p1>, |beta p2> stays inside A's eigen space) but still does not changes direction of |beta>.
just think of the matrix
|1 2 0 0|
|2 1 0 0|
|0 0 1 2|
|0 0 2 1|
this doesn't change the eigen vector (1,1,1,1)'s direction but changes it's composing vector (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1).
lets say (1,0,0,0), (0,1,0,0) |beta p1>, |beta p2>,
(0,0,1,0), (0,0,0,1) to be |beta o1>, |beta o2>.
and this doesn't taking |beta p1>, |beta p2> out of eigen space of A, and |beta o1>, |beta o2> out of orthogonal space.
I'm still struggling because this can be a counter example of argument around 9:00 of this vedio..
I really need someone to point out what went wrong inside my counter example. pls help me
My bad,
Since |beta p> can't never leave eigen space of lamda of A, there exist at least one linear combination of |beta p> which it self is eigen vector of B operator.
Then since we are free to choose the eigenvector |alpha i> that lives inside lamda space and having same direction with |beta p> thus become eigenvector of both A and B.
and since they should comply completeness, we can well find |beta p_i> which are both eigenvector of B and A and also span all the eigen space of lamda of A
Got it 👍
The Heisenberg Uncertainty principle isn't at all weird, when you accept, that particles have wave properties. It's a typical wave property and you get a good understanding of it by studying what happens when waves go through a single slit.
Of course you need the mathematics anyway to understand the consequences of the behavior in situations like the hydrogen atom and more complicated arangements of atoms.
14:00 where does this come from?
If beta is an eigenvector of B then shouldn't it be an eigenstate, as B is an observable ? And if so, then you said that beta could be describe as a superposition of 2 other eigenvectors of B ( one parallel and one perpandicular from the eigenspace ) but every eigenvectors of an obsevable are supoosed to be perpandicular, but beta is an eigenstate that is a superposition of 2 other eigenstates of B. But eigenstates can not be superpositions of other eigenstates as their are suposed to be perpandicular. So what am I missing ? Because it feels like it goes against the fact that beta as a eigenvector B should have only one possible value.
Hello! Thank you for watching.
And your confusion stems from the fact that we found that beta is composed of two other eigenvectors with the same eigenvalue, thus beta is degenerate in that case. It’s perfectly valid to have a degenerate eigenvectors in QM that aren’t perpendicular: they all correspond to the same measurement outcome, so it’s ok if they can be written as a superposition of each other. The idea that eigenvectors HAVE to be orthogonal only holds for nondegenerate eigenvectors.
That being said, you can always choose a set of orthogonal vectors in your degenerate eigenspace, so we usually assume that we’ve already done that. But they don’t have to be! Let me know if this cleared it up.
-QuantumSense
@@quantumsensechannel Ok thank you for responding, I misunderstood what degenerate means so thanks for clearing things up.
Why no in between? But there are eigenvectors of B that can be composed of other degenerate eigenvectors with a common μ eigenvalue
Another question: This video shows that special things happen when AB=BA. But then why is the commutator AB-BA important? Surely, we only care about whether this value is zero or nonzero, right?
Hello! Thank you for watching.
In terms of understanding whether observables share an eigenbasis, you’re correct that we only care about whether the commutator is zero or nonzero.
But don’t forget the first part of the video where we motivated the commutator: its original purpose is as a mathematical tool. Being able to flip two operators using their commutator is very, very powerful. I really mean it when I say that it often simplifies expressions. This is especially apparent in quantum field theory, where commutation relations essentially define the mathematics of your operator, and you use them to move things around in calculations. So we often care about the commutator itself when doing explicit calculations/derivations.
Hopefully this somewhat answered your question!
-QuantumSense
With the right manipulations you can calculate the energy spectra and eigenfunctions just from the commutator. This is routinely done for the harmonic oscillator and angular momentum, but it can be done for the Coulomb problem and much more too. Go look at Pauli’s original solution of hydrogen, which all follows just from the commutator of position with momentum.
Tiny point, but there is an error in this video. 12:47 says contradiction but should say contrapositive
There is a more proper title for this series: Master Quantum Machanics in One Day
At 13:54 you say "it actually equals a constant times the identity". This should say "times the imaginary unit"(as is shown on screen).
Hello, thank you for watching.
I did mean the identity here. It’s suppressed (as it always is, conventionally) but the “accurate” statement is i*hbar*Identity. This is just semantics though, but we technically need an operator on both sides of a commutation relation, so there is a hidden Identity operator lurking on the right hand side.
-QuantumSense
@@quantumsensechannel oh I see what you mean.
That's a confusing way to say it then, though. Showing two constants but not the identity but saying one constant times the identity. But maybe that's just me.
@@narfwhals7843 It got me, too.
Sorry about the confusion, I completely see why it doesn’t sound quite right. I’ve added a pinned comment addressing this. Thanks for letting me know!
-QuantumSense
10:39
Make a patreon!
Another fascinating way to see uncertainty principle to inevitably pop out comes from the simple consequence of quantization. See ua-cam.com/video/mpY1WUJ9FMk/v-deo.html (around minute 36). I hope there are subtitles, because it is in Italian (almost Sardinian 😄😄)
Quem disse que e intuição 😑 mais que insulto 🤣🤣🤣🤣🤣🤣🤣
pretty frightening stuff....the extraordinarily bizarre, arcane, and abstruse lengths that must be employed to describe quantum mechanics make me believe that quantum computers will never achieve any useful coherence or function if they are based on this kind of reasoning
Hello, thank you for watching.
And do not fret! I’ll tell you something that one of my professors once told me:
“Quantum mechanics is not complicated. Classical mechanics is complicated. Classical mechanics involves symplectic manifolds in 2^N phase space, wherein equations of motion are reduced to coupled differential equations. Quantum mechanics is just linear algebra; you take vectors, move them around, and see what you get. What makes quantum mechanics seem difficult is that it is non intuitive.”
Granted, we can make the linear algebra feel complicated with partial differential equations and what not - but at its core, it really just is a bunch of vectors moving around in space, and they are very easy to model in quantum computing.
So have no fear of the math; remember that math exists to help understand, not to obscure.
-QuantumSense