This series is like the imaginary numbers explained one: absolutely phenomenal. Thank you for making it!

@condor6222

Жыл бұрын

whoa, which imaginary numbers one?

@swaree

Жыл бұрын

@@condor6222 imaginary numbers are real by welch labs, a classic series on youtube

I think it’s worth going into the difference between hermitian (symmetric) and self-adjoint operators. In physics literature these terms are often used synonymously, which they are for bounded operators. But in fact most quantum mechanical operators are unbounded (momentum, position, laplacian, etc…) and an unbounded operator A has real spectrum, and thus the necessary condition for modeling a physical quantity, if and only if A is self-adjoint, which is a much more restrictive property than symmetry, and much harder to check in practice Edit: typo

@quantumsensechannel

Жыл бұрын

Hello! Thanks for watching. And that’s really interesting. To be completely honest, I wasn’t even aware that the two definitions differed, I just assumed they were used interchangeably (as everyone does in physics, apparently). I’ll read into the differences, thanks for pointing that out! -QuantumSense

@HilbertXVI

Жыл бұрын

For some more details, an unbounded self adjoint operator can't be defined on the entire Hilbert space (by the Hellinger Toeplitz theorem) so its domain in general is just a dense subspace. The condition for self adjointness of an operator A then requires that the adjoint of A (which is again a densely defined unbounded operator) have the same domain as A in addition to being symmetric.

@KCOSumedhaMandal

7 ай бұрын

cool dp

Linear and abstract algebra courses saved me a lot of time in quantum mechanics when I took the courses. This is awesome to see being put out on youtube! Wonderful!

The bra and the ket being hermitian adjoints of each other makes perfectly sense, seeing the similarities of the hermitian adjoint operator and the vector transpose

Hey, I know you’ve got a lot of comments saying how brilliant and amazing these videos are, so I’ll keep it short. I’m starting uni next year, and I’ve been trying to learn as much as I can about QM before the course. I’ve read many books on the subject, but these videos took my clarity to the next level. Best QM series I’ve found by far. Thanks for the content, keep it up :) - Tom

never loved physics, always was more into mathematics however if this was how I was taught physics, I would have certainly loved it. The modeling part is just so fun and cool nice series for sure

Wow, this video is just wow! I learned so much thanks to your series so far, it's amazing!!!

how do you get these out so fast?

@quantumsensechannel

Жыл бұрын

Hello! Thanks for watching. The series is finished in advance, and I’m just releasing episodes at a regular schedule. I’ve been working on the series for a little over 1.5 years in my free time. Hopefully that cleared it up! -QuantumSense

@zokalyx

Жыл бұрын

@@quantumsensechannel and you're a hero for doing it! These are the best introduction to QM any student can have. Plus it's a fresh new perspective for people who already know the math (like me).

@physicsstudies2611

5 ай бұрын

​​@@quantumsensechannelI am curious that how u did complex conjugate of inner product equal to transpose of it? I Also tried to look from other sources but they are bit confusing, sometimes it's written as transpose complex conjugate i.e. deger but sometimes they simply do tranjugat by saying complex conjugate only!! Please clear my doubt. It's at 4:38

@ammanullahrajpoot3160

19 күн бұрын

​@@physicsstudies2611 I think it's simple because dagger 🗡️ of the inner product means first you have to take transpose (i e, flip the position) and then you have to take complex conjugate of this inner product!

4:01 This is kinda interesting, because complex numbers are basically 2d rotation operators. If e^ix gives you a rotation by x degrees, then its complex conjugate e^-ix gives you a rotation by -x degrees. If you imagine/draw it, you see that when you take two random vectors and transform them using these rotations and then project them onto one another, the projection will have the same length as shown at the beginning of the video.

This video together with the 7th one is a gem for students

Really very thank you sir. Your doing great. So many students will inspire by your vedios and may become scientists for the future exploations

From next month I have my exam on quantum mechanics and iam going to write what you teach here ...ohh goodness for you

Thank you! This is a much better way of teaching.

진짜 영상 말도 안되게 잘 만드셨네요!!! wow

This was good. Well explained

6:54 I love how intuitive math becomes, once you understand the basic principles involved.

Hi! I'm following closely this series and I love it. I'm finishing my BSc in Maths so the math here is pretty clear for me, but my Physics knowledge is non-existent. Are you planning on also doing a series on the Quantum Mechanics of Quantum Mechanics after this perhaps? haha Cheers and thanks for the great work!

I hope more episodes will come out soon because I have my QM exam in 3 days :DD Now for real, thank you so much for this series, you give a refreshing look on topics I learned this semester.

00:00 : Introduction and prerequisites 00:40 : Motivation for defining the adjoint operator associated with an operator 02:20 : Definition of the Hermitian adjoint operator associated with an operator 02:37 : 3 properties of the Hermitian adjoint operator 03:20 : Hermitian adjoints examples (of a complex number, of a ket and a bra, of an operator) 05:26 : Physics: Hermitian adjoint operator of an observable operator: reminder about observables 05:51 : Physics: Hermitian adjoint operator of an observable operator: reminder about 3 properties of observables 06:27 : Physics: a useful way to express an observable operator as a sum of projectors on its eigenspaces 07:55 : Physics: using this expression to prove that observable operators are self adjoints, or Hermitian operators 09:20 : Physics: reflection and summary about the derivation that observables operators must be Hermitian operators

When you represent an operator as a matrix, how do you determine the values in the columns and the rows?

I’ve heard from some colleagues at the Math department that “Hermitian” operators are not necessarily the same as “self-adjoint” operators. Although it’s probably a small technicality, this got me very curious and I was wondering if you knew anything about this. Any help would be appreciated! Congrats on the amazing series of videos :)

you are the best...thank you so much..

Great video

I appreciate you thanks

Can you please make one video about the spectrum of quantum harmonic oscillator??

This vid should be include in any space probe launched that potentially could reach other intelligent life within our galaxy. Or beyond the Milky Way.

Can you suggest the books on Quantum mechanics?

I am barely following the math, but I am enjoying the beauty of it, the way I do with Blue-Brown... So, this reminds me of the Hadamard gate... am I intuiting this correctly?

@quantumsensechannel

Жыл бұрын

Hello! Thank you for watching, and for the kind words. And yes! The Hadamard gate is indeed a Hermitian matrix, although the “H” designation is merely a coincidence. One thing though: the more fundamental property of quantum logic gates is that they are *unitary operators* , which we’ll cover in their own episodes. We’ll find that unitary operators conserve probability in quantum mechanics, which is why they are used to mix and manipulate qubits! So the Hadamard gate is a unitary operator that also happens to be Hermitian. -QuantumSense

I'm not sure whether this is explained later on, but the idea of eigenstates being orthogonal is something that still isn't intuitive to me. I feel like everything else you said makes sense as long as you assume this, but I still don't see why it has to be true. One of the major roadblocks for me is: Sure, you can make the eigenstates of one observable operator (e.g. energy) orthonormal. This then forces the inner product to work exactly how you have said above. But then take another observable operator (e.g. position). What forces the eigenstates of that operator to be orthonormal? You've already nailed down exactly what the inner product has to do from how it acts on energy eigenstates, so what is forcing the inner product of two (distinct) position eigenstates to be zero? I'm really enjoying the videos, and apart from this one thing, everything flows so intuitively well for me, more so than for any other quantum physics course I've seen before. Thank you for the wonderful explanations!

Proof for property 3 (spoiler alert): apply operators rightmost first and leftmost last. = =

Is there a connection between the Hermitian adjoint dagger operator and the transpose operator? Both of them obey similar rules: (Aᵀ)ᵀ = A (A + B)ᵀ = Aᵀ + Bᵀ (AB)ᵀ = BᵀAᵀ Also, when the Hermitian adjoint is applied to a ket, it returns its respective bra linear functional (and vice-versa), similar to how the transpose operation when applied to a vector returns its row matrix (and vice-versa), which causes, respectively, a similarity between the equivalence between the inner product and a linear functional (⟨Φ|Ψ⟩ = ⟨Φ||Ψ⟩ = |Φ⟩† |Ψ⟩) and the equivalence between the dot product and row matrix and vector multiplication (u⋅v = uᵀv). Edit: Also, is the Hermit adjoint in general just a map from a vector space to its respective dual space?

After this, because this is a premade series, will you make kind of a Q&A so we can ask abt thing we were confused or not so sure abt?

@quantumsensechannel

Жыл бұрын

Hello! Thank you for watching. I don’t plan to make a Q&A, but I’m always open to answering any questions you have. I do my best to answer all questions in my videos, so don’t be afraid to leave a comment, even if it is only tangentially related to the video. -QuantumSense

@khiemgom

Жыл бұрын

@@quantumsensechannel @quantumsensechannel if the square of the dot product/bra ket already give you the probability, why do we need the Observable Operators and they are transformation so what effect will they have on the quantum state?

@quantumsensechannel

Жыл бұрын

@@khiemgom this is a good question. To calculate probabilities, we use the amplitude squared of the inner product with the eigenstate (eg | |^2), as you say, and we get these eigenstates from the corresponding operator. You seem to be asking the following: if we only ever use these eigenstates for probabilities, then why do we need to even bother with the operator? In other words, *what do the operators actually do* in QM. First note that we sometimes want to calculate more than just probabilities. For example, I mentioned last chapter that the expectation value of a quantity is calculated using the following inner product , where A is the observable of interest. So here we need the abstract operator to calculate what we want. Now more fundamentally, what do these operators do? In the following chapters, we’ll derive that every observable generates some change in the state of our particle. So, we’ll find that momentum generates position changes (which is why it acts as a wavefunction derivative), position generates momentum changes, and energy generates time changes. So fundamentally, observable operators generate some change in the position, momentum, time, etc. of our particle. We’ll use this approach when deriving the schrodinger equation. If this is confusing, don’t worry, I have an episode dedicated to generators and what they are, so we’ll learn about where this idea comes from. Hopefully that somewhat answered your question! My claims will make more sense in later chapters, but take my word for it for now!

Like everyone else I think your videos are amazing! I have a question for you though. At 6:55 in the video you go from E (the linear operator) acting on the eigenbasis |E i> becomes E i (the eigenvalue) multiplied by |E i> (eigenbasis). My question is why is the eigenbasis still in this second line? Doesn't the operator acting on the eigenbasis just give you the eigenvalue?

@quantumsensechannel

Жыл бұрын

Hello, thank you for watching! Remember that eigenvectors of an operator are scaled by their eigenvalue when acted upon by said operator. So the fundamental eigenvector definition is M_hat v = lambda* v, where v is a vector. This is the same as what we have in the video. Let me know if this didn’t clear it up! -QuantumSense

kzread.info/dash/bejne/lpVl1KppeNScYrQ.html.... so are both the bra and the ket vectors? I thought bras were linear mappings and kets were vectors... but in this case they are both orthogonal vectors to each other, which is similar to another video that I saw.. I suppose in the context of taking an inner product that one of the vectors has to transform into a row 'co-vector'.

What do you mean by intuition? That is not well defined: my first degree was in Anthropology, where intuition becomes an existential problem with a high probability of pie fights.

would anyone recommend any books for understanding the math for quantum mechanics

@grates3652

4 ай бұрын

If youre into Particle Physics this could be helpful (although its not as complete; im sure i found the pdf for free on the internet somewhere, should be easy to find): J.J. Sakurai - Modern Quantum Mechanics, Prentice Hall Another recommendation: Ramamurti Shankar - Principles of quantum mechanics, Springer

So sad that I didn't see these videos two years ago, or 9 months ago. I feel myself wasting my time before.

2:20 “in front” :P

6:36

I’m in QM II right now and you aren’t posting fast enough to keep up with the material. Please post faster 😅 I’m just joking ❤❤ kindof haha I appreciate your channel so much

7:23 If I'm not mistaking, you assumed we were alowed to multiply a vector and a bra, and therefore a vector and a linear form. Since the linear form gives us a scalar we can indeed consider that a linear form times a vector u is a linear map which, when applied on a vector x gives us the (linear form applied on x) times u and we are back to a scalar times a vector which is what we are used to. However, even if I ended up understanding it, I think this was not obvious at all. The fact that you've never talk about this kind of multiplication before makes this manipulation a bit confusing. Maybe you should have explain this at least once. Oustide of that, I admire how you manage to make QP intuitive while explaining the whole mathematical theory behind it, those videos are just awesome!!

@quantumsensechannel

Жыл бұрын

Hello! Thank you for watching. In a previous chapter, I covered the bra, which is a member of the dual space of linear forms, as you say. In that episode we go in depth into what this means, and how it relates to the inner product via the Riesz representation theorem. So hopefully anyone confused by it could find further information in that episode. Let me know if this wasn’t what you were quite getting at! -QuantumSense

@theanonimus7549

Жыл бұрын

@@quantumsensechannel To be honest, I watched all the previous videos today, but from what I remember the multiplication between a bra and a vector is never directly explained. But anyway I was just struggling to understand what type of mathematical object were on the sum, since there are many things hinding behind bra-ket manipulations, but I think I do understand now. Thank you for answering so quickly, sorry if my comments were confused and once again, those videos are extraordinary

Daggers?! So we ARE wizards! ...Or maybe thieves :P

👍👍👍👍👍👍👍👍👍👍👍

HOME, huh? good taste.

bras and daggers - starting to get interesting. A hermit's dagger kets the bra.

Calling operators "Hermitian operators" when they are equal to there own hermitian, is like calling ce^x a "derivative function" cuz it's equal to it's own derivative.

Read the Emperors New Mind By Roger Penrose for homework please

POV : ur lost 😭😂😹🙏🏻