Spinors for Beginners 16: Lie Groups and Lie Algebras

Sdílet
Vložit
  • čas přidán 27. 07. 2024
  • Full spinors playlist: • Spinors for Beginners
    Leave me a tip: ko-fi.com/eigenchris
    Powerpoint slide files + Exercise answers: github.com/eigenchris/MathNot...
    Videos on momentum operator generating translations in quantum mechanics:
    Physics with Elliot: • The Symmetry at the He...
    Professor M Does Science's video: • The translation operat...
    Quantum Sense': • Ch 14: Where does the ...
    0:00 - Introduction
    2:45 - Groups & Lie Groups
    4:00 - Exponent of a so(3) Matrix
    7:40 - Calculating so(3) generators
    9:50 - Momentum generators translations
    10:58 - so(3) traceless proof
    13:35 - so(3) anti-symmetric proof
    15:11 - Warning about matrix exponentials
    16:43 - Lie Algebra Bracket
    20:24 - Structure coefficients
    21:48 - Lie Algebras as Tangent Spaces
    23:30 - Lie Algebra Property Proofs
    30:15 - Summary of so(3)
    31:05 - Overview of so+(1,3)
    32:50 - Spin-1 and Spin-1/2 representations
    34:40 - Math vs Physics conventions

Komentáře • 94

  • @eigenchris
    @eigenchris  Před 7 měsíci +15

    At 32:15, I messed up the 2nd last commutation relation on the bottom left. It should be [Ktz, Ktx] = - Jzx.

  • @rhke6789
    @rhke6789 Před 5 měsíci +8

    The only video I found that explains why Lie Algebra is important and has a use and how in the field of Quantum mechanics...that is also understandable. Thank you so much

  • @baruchspinoza4979
    @baruchspinoza4979 Před 7 měsíci +8

    Wow! A clear path through something that had appeared as a complex jungle. Thanks EigenChris.

  • @DavidAspden
    @DavidAspden Před 7 měsíci +2

    Probably one of your best yet. The ground you covered was vast.

  • @kikivoorburg
    @kikivoorburg Před 7 měsíci +6

    Very excited for this section! I’ve vaguely heard of Lie Algebras (and that they matter in physics) but never had quite a clear picture of what was going on. You have an incredible ability to make complex topics digestible and understandable, so I’m sure you’ll fill those gaps for me!
    Edit: I was not disappointed! It’ll take me a bit of time and a few rewatches to fully grasp it I’m sure, but wow have I already learnt so much! I finally get why “the Lie algebra is the tangent space of the Lie group” actually makes sense! Amazing work!!

  • @fizikchy
    @fizikchy Před měsícem

    Amazing video! You knit things very naturally. Thanks.

  • @francoisfjag4070
    @francoisfjag4070 Před 7 měsíci +2

    brilliant video. The best explanation I ever get on relationship between a Lie group and its corresponding algebra !!

  • @richardneifeld7797
    @richardneifeld7797 Před 7 měsíci

    Your videos keep getting better and better.

  • @TheJara123
    @TheJara123 Před 7 měsíci

    Block bustor man!! Your operational approach cuts through the topic...showing us way to learn!! So big thanks in order...!!

  • @jimlbeaver
    @jimlbeaver Před 7 měsíci

    I’ve been trying to understand this for a while. This was extremely helpful. Thank you.

  • @hugoreytinas5170
    @hugoreytinas5170 Před 7 měsíci

    I am so happy : i could not wait for this new video ! This is like a Christmas gift ! Thank you so much Eigenchris ! You are welcome anytime in south of France ! 😂

  • @sair9850
    @sair9850 Před 7 měsíci +3

    came from the joke videos, stayed for the spinors for Beginners 16: Lie Groups and Lie Algebras

  • @Impatient_Ape
    @Impatient_Ape Před 7 měsíci

    Bravo! Clap! Clap! Clap! This is a really really really good introduction to Lie algebras for motivated undergrad math majors. Lucid, accurate, and well visualized. For those in the comments... a typical math major in the USA might not see this material during their bachelor's degree; however, it is usually part of the graduate math curriculum.

  • @AkamiChannel
    @AkamiChannel Před 7 měsíci

    Can't wait for the next video! Thanks!

  • @kousarshaheen4801
    @kousarshaheen4801 Před 7 měsíci

    Thanks for your videos, it helps me alot in my thesis on particle physics, kindly make sure when we reach particle physics

  • @n810h
    @n810h Před 7 měsíci

    This was really good. Great job!

  • @dariushimani8956
    @dariushimani8956 Před 7 měsíci +2

    Banger video

  • @Anonymous-kj6cu
    @Anonymous-kj6cu Před 7 měsíci +33

    I love your voice. Sounds like a robot.

    • @alexleibovici4834
      @alexleibovici4834 Před 7 měsíci +1

      Is it his voice?

    • @AdrianBoyko
      @AdrianBoyko Před 7 měsíci +1

      Vocal fry

    • @ryansamuel8835
      @ryansamuel8835 Před 7 měsíci +3

      I like how he doesn’t get all emotional when he speaks. Like he has no emotions and is just presenting information

    • @MiroslawHorbal
      @MiroslawHorbal Před 6 měsíci +2

      It sounds like a robot who is also slightly annoyed that he has to explain things to you.
      But jokes aside. I love these videos and his teaching style.

    • @AdrianBoyko
      @AdrianBoyko Před 6 měsíci +1

      @@MiroslawHorbal Marvin

  • @pieterkok7486
    @pieterkok7486 Před 7 měsíci +8

    To find the generator from the group element, you apply the "standard" derivative rule to matrices. However, there are many types of derivatives one can define for matrices (for example Lie brackets, as in the Liouville/Von Neumann equation). Do we need additional justification (other than that it works in this case) for this choice of derivative?
    Great series, by the way! 😀

    • @eigenchris
      @eigenchris  Před 7 měsíci +3

      I haven't thought about that too much. I'm just used to using the standard derivative. Maybe there are cases where you need another kind if derivative, but I'm not familiar with that.

    • @gijsb4708
      @gijsb4708 Před 7 měsíci +5

      Whenever a differentiable manifold (such as a Lie group) is represented as elements of a vector space (here the space of matrices) then the standard derivative gives the right answer when calculating tangent vectors (which are in this case the generators).

  • @thevikifalcon7670
    @thevikifalcon7670 Před 7 měsíci +2

    I have never seen a good explanation for convergences between geometric algebra and lie theory (lie algebras + lie groups). Oftentimes, you can use both to represent the same things- spinors in this case, but more generally rotation-like transformations on some manifold. In this series, you should consider explaining why this convergence exists. For instance, is there a way to directly convert the lie theory way of thinking about things into the geometric algebra way and vice versa? Thanks and keep up the great work!

    • @eigenchris
      @eigenchris  Před 7 měsíci +8

      This might become more obvious in the next video. In the spin-1/2 representation, the generators are Clifford Bivectors, or equivalently, quaternions. And the SU(2) matrices are the equivalent of the Spin(3) group.

  • @bantix9902
    @bantix9902 Před 7 měsíci

    the goat back at it again!!!!

  • @richardneifeld7797
    @richardneifeld7797 Před 7 měsíci

    Excellent! Also, I find that mathematician's normally pro ide more comprehensible video lectures than physicists. As of they better understand the underlying principles. I put you in the mathematician lecturer class. Clear and insightful.

  • @sgurdmeal662
    @sgurdmeal662 Před 7 měsíci

    This made my day

  • @breezy1118
    @breezy1118 Před 17 hodinami

    Great video! However, the summation indices at 8:38 should start at n=1 after taking the derivative.

  • @enterprisesoftwarearchitect
    @enterprisesoftwarearchitect Před 7 měsíci +1

    It’s a semester course in 30 minutes!!! ❤️❤️❤️

  • @lazzatv
    @lazzatv Před 7 měsíci

    I'd love to see a well made derivation of rank equal spin quantum number relations hinted at 1:24, since I only saw some details about it during a lecture.
    If I'm not mistaken, from commutation relations you get |spin angular momentum vector|^2=hbar^2*(s+1)*s, the |spin angular momentum vector| is then expressed with the sum of the single components. Single components are expressed with independent spin-projection operators eigenvalues and communtation relationship when not independent, so I think the tensor field dimension can be expressed as 2*rank+1. Spin-projection operators apparently have a rank dependentent representations as shown later in the video which determine their rank dependent eigenvalues.
    I always thought it was a cool explanation of the fact that rank of the particle tensor field implies a certain value for the spin quantum number s and then the non-relativistic spin angular momentum can actually be derived, since it just becomes a first order term for the wavefunction angular momentum derived when the SO(3) operator for small rotation is applied to the tensor field.

  • @kikivoorburg
    @kikivoorburg Před 7 měsíci +1

    Upon rewatching I've noticed how similar the structure coefficients f_ij^k are to the connection coefficients Γ_ij^k from differential geometry / GR! Both define the "multiplication" of basis vectors in terms of the basis vectors:
    [g_iᵢ, g_j] = Σ f_ij^k g_k
    ∂_i ∂_j = Σ Γ_ij^k ∂_k
    The placement of the indices on the coefficients is even identical (upper/Lower indices ftw, understanding co/contra-variance is so clarifying). Always more to learn in these videos!!
    Now that I think about it, doesn't the fact we can "multiply" basis vectors in differential geometry mean that it too is an algebra?
    Edit: in case it wasn't super clear, due to the definition of ∂ / ∂x^i = ∂_i = e_i, we have ∂_i ∂_j = ∂ e_i / ∂ x^j, which is the formula that Wikipedia uses with the Christoffel symbols.

    • @eigenchris
      @eigenchris  Před 7 měsíci

      Yeah, I'm not sure how deep or shallow the connection is, other than "two things come together to give a third thing". The Lie bracket can be thought of as a kind of derivative... the Jacobi Identity I show briefly can be thought of as a kind of product rule if you take one of the terms and put it on the other side of the = sign. Would be interesting if there was a deeper connection between the two, but I'm not aware of anything at this point in time.

    • @kikivoorburg
      @kikivoorburg Před 7 měsíci

      @@eigenchris Even shallow connections are interesting, though it would indeed be exciting if there was more going on. Thanks for the reply!

  • @AMADEOSAM
    @AMADEOSAM Před 7 měsíci

    Great explanation …

  • @saber9468
    @saber9468 Před 6 měsíci

    Hello Chris, thank you so much for your video. I was interested in knowing if you had a similar video covering lie algebra derivatives? I really like your teaching style and was hoping for a more visual representation of lie algebra derivatives i.e derivative of point w.r.t to rotation etc...

  • @juleskurianmathew1983
    @juleskurianmathew1983 Před 6 měsíci

    After finishing spinor series, kindly do a quantum mechanics/ quantum chromodynamics/ quantum electrodynamics/ quantum field theory series. I love your tensor series.

    • @eigenchris
      @eigenchris  Před 6 měsíci +1

      Hi. While I will cover a little bit of QFT in the last part of this series, I won't be doing a full QFT series. It's just too big a topic and I don't understand it very well.

  • @sebastiandierks7919
    @sebastiandierks7919 Před 7 měsíci

    Really great video! Can't wait for the next episode. At 9:28, the group element R_zx should have the minus sign in front of the other sin function if I am not mistaken.

    • @eigenchris
      @eigenchris  Před 7 měsíci +1

      I think it's correct. If you put in theta=90 degrees, and plug in the z-vector [0,0,1]^T, you end up getting the x-vector [1,0,0]^T. So it rotates z into x, as expected.

    • @sebastiandierks7919
      @sebastiandierks7919 Před 7 měsíci

      ​@@eigenchris sorry, my bad, you're right of course. I always assumed that the signs in front of the sin functions were the same for all three rotation matrices so that if you deleted the column and row with ones and zeros you'd get the 2-dimensional rotation matrix. It's wrong though, turns out the rotation about the y-axis has different signs. I guess as a physicist you rarely rotate about the y-axis, you mostly use a coordinate system where you rotate about x and z, so my mistake never got caught.

    • @eigenchris
      @eigenchris  Před 7 měsíci

      @@sebastiandierks7919 You can imagine taking the xy rotation matrix and moving the cos/sin 2x2 block down 1 row and to the right 1 row. If you do this twice, while "wrapping around" from bottom to top and from right to left, you get the zx rotation matrix. And doing it a 3rd time returns you to the xy rotation matrix.

  • @user-se9pc1ut6z
    @user-se9pc1ut6z Před 6 měsíci

    24:36 one slide summary of Lie group and Lie algebra!

  • @fjaresj
    @fjaresj Před 5 měsíci

    Long round of applause! ❤

  • @tobiaspower381
    @tobiaspower381 Před 7 měsíci

    I cannot say thank you enough. :)

  • @pavgogo
    @pavgogo Před 7 měsíci

    Excellent.
    If aba(inverse) leads to commutator ab-ba, what about bab(inverse)? also is it also ab-ba or ba-ab?

  • @cmilkau
    @cmilkau Před 7 měsíci

    Actually, we only proved the properties of an algebra, but not the Jacobi identity to make it a Lie algebra (antisymmetry is obvious). You can prove that this identity always holds when the multiplication (Lie bracket) [a,b] looks like a commutator a·b - b·a and "·" is associative and distributes over +.

  • @ericbischoff9444
    @ericbischoff9444 Před 7 měsíci +1

    Again, great video!
    I have a question. At 27:27 you prove that the commutator of tangent vectors is a tangent vector. Am I correct to consider this does not prove the general case of a multiplication in a Lie algebra, using the "more abstract" definition of a Lie algebra based on the Jacobi identity? If I understood well, commutators are the most common example of a Lie bracket, but not necessarily the only ones. Or am I missing something?

    • @sebastiandierks7919
      @sebastiandierks7919 Před 7 měsíci +1

      You are correct, it doesn't prove the general case, although if I remember my Lie Groups and Lie Algebras course correctly, the proof of the general case is very similar. You also take the conjugation of two paths through the manifold to show that for every Lie group an object called the Lie bracket can be constructed that satisfies alternating and antisymmetry properties as well as the Jacobi identity. (I'm a physicist, not a mathematician though, although the course at university was a pure maths course from a maths professor; it's just been 10 years ago so yeah...)

    • @ericbischoff9444
      @ericbischoff9444 Před 7 měsíci

      Thank you@@sebastiandierks7919 .

  • @HelloWorld-lv4we
    @HelloWorld-lv4we Před 3 měsíci

    I need a physics expert!
    I am a layperson who enjoys physics, and I have a question about spinors. My understanding is that spinors are equivalent to space-time indices (indices on a 4D manifold). Do spinors ever exist on odd-dimensional manifolds? In other words, do spinors occur on even-dimensional manifolds only? I ask this because I’ve heard that bosons are “non-spinorial”, so I’m mainly curious to know whether the force fields are odd-dimensional. If I’m conceptualizing these things wrong, please tell me.

  • @siguc
    @siguc Před 4 měsíci

    Thank you for this amazing series!
    Speaking of the proof at 14:21, isn't it obvious that dR^T/dTheta=(dR/dTheta)^T? After all, it doesn't matter whether we first differentiate every cell of a matrix and then transpose it, or, alternatively, transpose the matrix first and then differentiate every of its cells.
    If we accept the above, then we can differentiate dR/dTheta->MR (as you explained earlier) and then substitute R(0)=I, which completes the proof.

    • @eigenchris
      @eigenchris  Před 4 měsíci +1

      What's obvious for one person might not be obvious for someone else. I didn't want to skip too many steps and risk confusing people.

    • @siguc
      @siguc Před 4 měsíci

      @@eigenchris Agree. Most of your logic seems very clear. It’s also very helpful that you keep reiterating the same conclusions across the course, especially when you approach those conclusions from different angles.

  • @cmilkau
    @cmilkau Před 7 měsíci

    All hermitian or antisymmetric matrices have an EVD; once you have that exponentiation is super simple

  • @user-po9mm4cb4d
    @user-po9mm4cb4d Před 7 měsíci

    Is there going to be a QFT series after this, similar to hoe you had a relativoty series aftef tensor calculus? Love your content btw.❤

    • @eigenchris
      @eigenchris  Před 7 měsíci

      I'm afraid not. QFT is too big and confusinf. I don't understand it, and don't have the energy to make a series long enough to cover it. I'll talk a bit about QFT at the end of this series.

  • @Daniel-nm4kt
    @Daniel-nm4kt Před 6 měsíci

    Can you make a series on the basics of linear algebra?

    • @eigenchris
      @eigenchris  Před 6 měsíci +1

      I don't plan on doing any more long series after this. Have yoh watched 3blue1brown's "Essensce of Linear Algebra"?

  • @Cat-yz1tk
    @Cat-yz1tk Před 7 měsíci

    in the part where you prove that the lie algebra is closed under commutators you multiplied two vectors in the tangent space as is required for the commutator a*b-b*a this makes sense where the tangent vectors are matricies but how does that work in a general lie group? how do tangent vectors multiply with each other?

  • @himanshuchaudhary4197
    @himanshuchaudhary4197 Před 7 měsíci

    Dear Eigenchris, Could you please create a video or provide information on deriving the Einstein field equation from the Einstein-Hilbert Action? Alternatively, if you could share some notes or relevant links, that would be greatly appreciated.

    • @eigenchris
      @eigenchris  Před 7 měsíci

      Hi. Sorry but I'm done with making GR videos. I'm focusing on the spinor videos now.

  • @ritahall8148
    @ritahall8148 Před 7 měsíci +1

    Typo at 32:19?, on penultimate line, should have bracket of tz and tx boosts equal to the negative zx rotation

    • @eigenchris
      @eigenchris  Před 7 měsíci +1

      Yup. My bad...

    • @Dc4nt
      @Dc4nt Před 7 měsíci

      Similar typo at 10:58 ?
      Last line, right hand side should have g_zx

    • @Dc4nt
      @Dc4nt Před 7 měsíci

      Just spotted you use the same slide again at 16:44 , same typo.

  • @tomgraupner171
    @tomgraupner171 Před 7 měsíci

    I understand that the "physics part" is coming along next, but please allow one question beforehand: Does the [Ktz, Ktx] = - Jzx means that a boost in the x-y plane automatically includes a rotation around the z axes?

    • @eigenchris
      @eigenchris  Před 7 měsíci +1

      I'm not sure what the physical meaning of 2 boosts commuting to a rotation is. In the Lie Group, the basic rotations and boosts are independent from each other. You can always do a boost with no rotation and vice-versa.

    • @tomgraupner171
      @tomgraupner171 Před 7 měsíci +1

      @@eigenchris I assume my question was silly. I found this: en.wikipedia.org/wiki/Wigner_rotation
      So, it looks like a second boost in a different direction (e.g. -> y), while performing the first boost (e.g. -> x). That leads to the rotation, as the product/sequence of these 2 boosts must no longer be a "single boost".
      Does that make any sense?
      Thanks a lot for all your work!

  • @curtishorn1267
    @curtishorn1267 Před 7 měsíci

    This is the way.

  • @eduardoGentile720
    @eduardoGentile720 Před 7 měsíci

    One question, why should a Det=0 Matrix to the zeroth power be the identity if it's non invertible?

    • @eigenchris
      @eigenchris  Před 7 měsíci

      Why does the invertibility matter?

    • @eduardoGentile720
      @eduardoGentile720 Před 7 měsíci

      @@eigenchris We say that a number to the zeroth power is 1 because by the proprieties of exponents by multiplying the same base with different exponents the exponents should add up, and since the negative power is equal to the inverse of a of a number, a number to the zeroth power is the number divided by itself, witch is 1 (when it's not 0). If you define it the other way around, so a number to the zeroth power is 1 and then you apply the rule of exponents to obtain that a number to the -1 is its inverse then you should be allowed to exponentiate 0^0 or even a 0 Matrix to the 0, I'm asking this just because this doesn't make much sense to me

  • @BakedAlaska187
    @BakedAlaska187 Před 7 měsíci

    All of the tangent vectors pictured are actually matrices, right?

  • @phitsf5475
    @phitsf5475 Před 6 měsíci +1

    No offence, but I think your script combined with Taylor Swift doing the reading would make for a badass combination.

  • @tedsheridan8725
    @tedsheridan8725 Před 7 měsíci +1

    Great video! It really helped demystify Lie groups/algebras for me. (BTW, I love your content, but as a constructive criticism from someone who also makes CZcams videos, I'd encourage you to work on your narration. It's clear, but very monotone and over-pronounced. I'd really try to adopt a more conversational style as opposed to reading off a powerpoint presentation. I think you'd find it much more engaging and pleasant to listen to.)

    • @ericbischoff9444
      @ericbischoff9444 Před 7 měsíci

      I'm of the contrary opinion. I find Chris' voice appeasing and it lets me concentrate on the contents. I would certainly not be able to do the same with a more "conversational" or even "journalistic" tone, like there is in so many youtube videos.

    • @eigenchris
      @eigenchris  Před 7 měsíci +2

      My voice seems a bit divisive among some people. Some like it, others really don't. I'm probably just going to keep doing it in the current style since that's why I'm used to.

    • @onebronx
      @onebronx Před 7 měsíci

      @tedsheridan8725 Just increase video playback speed

    • @tedsheridan8725
      @tedsheridan8725 Před 7 měsíci +1

      ​@@ericbischoff9444 I agree with you on the 'journalistic' tone - it's so overused. It's quite distracting to listen to a narrator pretend to be an TV news anchor.

  • @omegapirat8623
    @omegapirat8623 Před 7 měsíci +1

    Can you extend this session to (quantized) spinor fields, spinor structures, spinor bundles, etc on a manifold?
    I'd like to understand modern physics in terms of a proper mathematical model of spacetime following this amazing lecture series czcams.com/video/V49i_LM8B0E/video.html
    Unfortunately, in this lecture, there is not so much said about spinor matter since it mainly focuses on differential geometry (and tensor matter).
    Spinor matter is the last import ingredient that I don't fully understand to get a complete picture of modern standard physics (standard model and general relativity).

    • @eigenchris
      @eigenchris  Před 7 měsíci

      I'm not sure I'm going to get that far in this series. What exactly do you want to see covered? By "spinor field", do you just mean the Dirac Equation in QFT?

    • @omegapirat8623
      @omegapirat8623 Před 7 měsíci

      @@eigenchris
      I am talking about spinors in curved spacetime (maybe as a smooth section of a spinor bundle?). Or to be more precise how to define it on a lorentzian manifold. In contrast to tensor fields, you need a metric for spinor fields. Therefore it won't work for general manifolds I guess.
      That's something that I didn't master so far and I don't find good literature related to that topic either.
      Tensor fields and spinor fields are key aspects in understanding matter and as they are constructed on a 4D lorentzian manifold in physics one should understand this to get a complete picture of modern physics.

  • @funkfusiontale
    @funkfusiontale Před 6 měsíci +1

    Δ++ (uwu)

  • @ltobias93
    @ltobias93 Před 7 měsíci

    If I have found a Genie, I would ask for your videos minus all the proves

  • @eccotom1
    @eccotom1 Před 7 měsíci

    God his voice is so hot

  • @miguelaphan58
    @miguelaphan58 Před 3 měsíci

    .. a very dicult subject

  • @stevelam5898
    @stevelam5898 Před 6 měsíci

    Why differentiate to get the Lie group from the algebra though?