The Symmetry at the Heart of the Canonical Commutation Relation

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  • Опубликовано: 23 дек 2024

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  • @darrenalec6979
    @darrenalec6979 2 года назад +187

    The niche you’ve chosen is unique on RUclips, I believe. You delve into the math deeper than most dare, without apologizing too much. Which is fantastic for those of us who have exhausted the videos that just skim the surface, and who crave a more mathematically rigorous explanation of these concepts. Thank you.

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад +21

      Glad it's been helpful Darren!

    • @l1mbo69
      @l1mbo69 2 года назад +3

      See Sean Carrol's Biggest Ideas series as well

    • @user-sl6gn1ss8p
      @user-sl6gn1ss8p 2 года назад +3

      these are also really nice as refreshers

    • @adamstecklov3093
      @adamstecklov3093 2 года назад +2

      This exactly!

    • @corsaircaruso471
      @corsaircaruso471 23 часа назад

      I’m just starting to consume this content and I’m having a similar experience. So glad this is here

  • @sweetpotatoambassador
    @sweetpotatoambassador 2 года назад +11

    Elliot, you are becoming the new GOAT of youtube physics! Please keep the great vids coming bro :)

  • @jaw0449
    @jaw0449 2 года назад +9

    I'm in my 3rd year of PhD, and you've explained this better than anyone else I've heard/seen! Thank you! Great as a refresher especially

  • @kashu7691
    @kashu7691 2 года назад +8

    these videos are so valuable and are beginning to become real resources to use along with textbooks and lectures as i’m learning about all this. thank you!

  • @eigenchris
    @eigenchris 2 года назад +42

    Another great video. I'm curious: what software do you use to make the animations in your videos?

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад +15

      Thanks! For the last few videos I've done most of the animation in Keynote, and then brought it over to Final Cut to finish

    • @tanchienhao
      @tanchienhao 2 года назад +5

      Glad to see u here! Big fan of ur GR videos :) and awesome channel too Elliot!!

    • @finnjake6174
      @finnjake6174 2 года назад +2

      @@PhysicswithElliot Seriously you are reigniting my passion for physics.

    • @lexinwonderland5741
      @lexinwonderland5741 2 года назад

      hey, i watched your series on tensor calculus and Riemannian geometry! great series btw, I wish there was more actual math in youtube physics videos (like this channel, and your channel). keep up the great work!!

    • @JoeHynes284
      @JoeHynes284 2 года назад +1

      i always look for your comments on videos...

  • @LouisEdouardJacques
    @LouisEdouardJacques 2 года назад +8

    Thank you for making this video and making it available to us. I hope many students will find it. This simple origin of the canonical commutation relation is often glossed over, maybe because it removes a lot of it's mystical nature. You already discussed it a little in relation to vecort field with the poisson bracket but when you look at the ccr as shown in this video, you can easily then see the commutator as a derivative of operators and also the spatial derivative for the momentum in position space become obvious.

  • @khaledhosseini2030
    @khaledhosseini2030 2 года назад +1

    You don't really understand something unless you can explain it to your grandmother. This quote is attributed to Albert Einstein, but I must say that you have understood these things very well, that you explain them so beautifully, simply and fluently. Thank you very much.

  • @zainkarriem6344
    @zainkarriem6344 2 года назад +1

    Really nice videos! There is a deep connection between classical mechanics and quantum mechanics and something that cannot be easily explained. Your lectures are by far the most successful ones in making this connection, which in turn demystifies the structure of quantum mechanics as a machinery. Excellent indeed! Thanks for making these!

  • @hyprk5590
    @hyprk5590 2 года назад +5

    Wow… thank you for explanation of symmetry operators. I couldn’t understand that exact meaning when I learned about it in college. It’s very helpful for me. I’m interested in physics but it’s very hard for me to study alone.

  • @comrade_marshal
    @comrade_marshal Год назад

    All we need is a mathematically rigorous treatment of physics with a nice physical explanation of things. Most people just explain the theory (very well though), but math takes a back seat. I like your way of balancing both the essential components of physics. I know that your community would be a niche, but whatever be it , you are giving a "complete" explanation as far as RUclips videos are concerned

  • @Aloka145
    @Aloka145 2 года назад +1

    amazing to go back to the quantum mechanics fundamentals, exactly what I needed before going deeper in quantum field theory and quantum loop gravity

  • @summerQuanta
    @summerQuanta 2 года назад +1

    Actually you have assumed that the position operator is continuous and that the translation operator can be expanded. This approach is very nice as it shows how our intuition from classical mechanics should lead to the commutation relation. I would add that the commutation relation is somehow not just a consequence though, but at least as fundamental and things go both ways. By defining it the way we do, we can then derive the eigenspectrums of the position and momentum operators to be continuous as we physically expect (that is the relation ^x|x>=x|x> and ^p|p>=p|p>) . The commutation relation is actually somehow even more central as it is what defines the algebra of those observables and also naturally leads to uncertainty principles. It is also a concept easy to generalize to other pair of observables too as you mentioned in your previous video I think.

  • @dahleno2014
    @dahleno2014 2 года назад +4

    I’m loving these videos. They’re an excellent supplement to the physics courses I’ve been taking at university c

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад

      Love to hear it Dahlen!

    • @dahleno2014
      @dahleno2014 2 года назад

      @@PhysicswithElliot They really help with a conceptual understanding of things. It’s one thing to be able to do the math, it’s another thing to interpret things, which you do an excellent job of communicating. Keep up the good work!

  • @gafus4309
    @gafus4309 2 года назад +5

    You make amazing videos. Please never stop ✌🏻

  • @MRF77
    @MRF77 2 года назад +16

    This is another excellent video Elliot! I hope you will continue to produce this kinda gems and take the Grant's (from 3blue1brown) place for Physics. All the best!

  • @chrisr9320
    @chrisr9320 2 года назад +3

    I applaud your approach, you go way deeper into calculations than I would dare in a youtube video :)
    In this video, I would say there is a potential snag in that at one point you simply define the translation generator to be -i*p/hbar and this gives the result. Maybe proving or at least motivating this connection would be helpful?

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад

      Thanks Chris! Make sure you watch the previous video about how momentum is the classical generator of translations for some of the motivation

    • @chrisr9320
      @chrisr9320 2 года назад +1

      @@PhysicswithElliot Will check that out!

  • @ccw1612
    @ccw1612 2 года назад +4

    Thank you! Your video are very good for me. I am a Mathematics student who want to learn more Physics. Will you make some video about Gauge theory in future?

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад +2

      Thanks! Yes I will surely start discussing gauge theories before too long

  • @12345anvesh
    @12345anvesh 2 года назад +3

    Wow. The video was great. Thanks a lot for your efforts :)

  • @kennethmui88
    @kennethmui88 2 года назад +2

    At 8:34, I am confused as to why you put the Unitary Operator in between the bracket, Why did you do "U inverse, x, U"

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад

      Under the transformation, |psi> turned into U|psi> while

  • @kristim1
    @kristim1 Год назад +1

    We need more videos like this bruv

  • @MrOvipare
    @MrOvipare 2 года назад +2

    how can I NOT click on a title like this?

  • @diatribeeverything
    @diatribeeverything 2 года назад +1

    Amazing content! Keep it coming.

  • @dirichlettt
    @dirichlettt 2 года назад +1

    Thank you for these videos and the notes! They help so much in understanding these topics.

  • @TheCrunchyGum
    @TheCrunchyGum Год назад

    I understand the point of the commutation relations now... thank you

  • @justinburzachiello1897
    @justinburzachiello1897 2 года назад +5

    These videos are great. Conviniently, I recently bought "A Students Guide to Lagrangians and Hamiltonians" by Patrick Hamill to learn more about this very subject.

  • @enriquemacias9957
    @enriquemacias9957 Год назад

    Excellent video. Thank you for sharing these ideas.

  • @ליאורפ-ה8כ
    @ליאורפ-ה8כ 2 года назад

    Thanks for the video!
    Just a small question: what needs to be preserved is the probability ||^2 so the phase of isn't important, right? which means you can also have operators that satisfy U^daggerU = -1 (or any other phase), and not just the U^daggerU=1 operators, no?

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад +1

      The state is only defined up to rescaling, so changing the normalization isn't physically meaningful. But there is a second class of symmetry operators which are called anti-unitary---the classic example is time reversal symmetry

  • @finaltheorygames1781
    @finaltheorygames1781 Год назад

    This is excellent. I took quantum, but I never saw this derivation.

  • @deepakjanardhanan7394
    @deepakjanardhanan7394 2 года назад +2

    Hurray new upload....

  • @StaticYouTubeGaming
    @StaticYouTubeGaming 2 года назад +1

    Great video! When you are talking about ‘generators’ is that related to generators within algebra like for a cyclic group? Or are they unrelated and just share the name?

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад +1

      They're different because here we're talking about the infinitesimal generator of a continuous symmetry, so that there is a continuous parameter like \lambda or t that you can smoothly dial down to zero. Cyclic groups are discrete, so each transformation is either turned on or off.
      The idea is similar though in that you're writing a group element as a power of the generator. The difference is roughly whether you can take a continuous power or only discrete powers

    • @alphalunamare
      @alphalunamare 2 года назад

      Interpretation of the english word 'generator' in some way implies a motivation twixt one state and another. Its actual meaning depends on the context within which it is being used. Its use in Group Theory is entirely independent of its usage in Quantum Mechanics (I think).

  • @meroramo8179
    @meroramo8179 Год назад

    what a freakin gold mine

  • @tommy51s5
    @tommy51s5 2 года назад +1

    Is p the generator of spatial translation symmetry or just spatial translation? Wouldn't only conserved momentum yield translation symmetry?

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад

      Well it's a matter of terminology; sometimes we call any unitary operator in quantum mechanics a symmetry because it preserves the probabilities. But yes, the statement of translation invariance is that the translation operator U(\lambda) leaves the Hamiltonian invariant, U(\lambda)^{-1} H U(\lambda) = H, which requires that [p, H] = 0. That's the quantum version of the classical condition {p, H} from the last video.

  • @Imran52Feb
    @Imran52Feb Год назад

    Knowledge of Linear Algebra is necessary to understand the Relativistic QM . Is there any video for that ?

  • @evangelion045
    @evangelion045 2 года назад

    Is there an intuitive way to justify the choice of the rescaling factor used in the video?

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад +1

      In i p/hbar? The hbar is there so that p has the usual units for momentum, and the i is there so that p comes out real in an appropriate sense (called Hermitian). It's analogous to writing a complex phase e^{i \theta} = 1 + i \theta + ..., where \theta is a real angle.

  • @urnext6874
    @urnext6874 2 года назад

    This is very helpful
    Thank you

  • @vahidsf4808
    @vahidsf4808 2 месяца назад

    It's perfect to understand

  • @narfwhals7843
    @narfwhals7843 2 года назад +1

    You introduce the i/hbar factor to get the right units and to ensure that the result is real. Do we get a different commutation relation if we don't enforce that requirement? Could we have a version of quantum mechanics with complex momenta and [x,p]=1?

    • @zray2937
      @zray2937 2 года назад +1

      Real momentum has a direct interpretation in experiments. How would you relate the results from experiments with the eigenvalues of an anti-Hermitian operator?

    • @narfwhals7843
      @narfwhals7843 2 года назад +1

      @@zray2937 I haven't the faintest idea. Perhaps a subset of a theory from which observable reality emerges.
      But I'm not actually necessarily looking for a theory that describes reality. Just wondering if it makes sense mathematically. if it could describe _some_ reality.

    • @RizkyMaulanaNugraha
      @RizkyMaulanaNugraha 2 года назад +2

      The i/hbar factor came straight out from the generator. We just conveniently extract it because we want the generator to be real, but it’s always there. Let’s suppose we don’t do that, that means to make it generic, any operator might not include i, but it’s adjoint will have i, due to the requirement that it needs to be hermitian. so it will still include i at the end.

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад +1

      A different choice would just amount to rescaling the definition of p, but we want it to coincide with the usual definition of momentum in the classical limit

    • @patrickyoukidean6938
      @patrickyoukidean6938 2 года назад

      @@PhysicswithElliot I think the point is about 'i', h bar is OK since we need to have the right unit for momentum. However, you did not tell us why do we need an 'i' here. Why it would ensure the generator to be real. It would be great if you could explain it. Otherwise it's like a magic trick to introduce the 'i' to the commutation relation. Anyway this is still a nice video.

  • @chick3n71
    @chick3n71 Год назад

    If x is to be conserved in time dx/dt=0, then also [x, H]=0?

  • @mustafaathar1070
    @mustafaathar1070 Год назад

    Dude you are just awesome

  • @marlovsk1
    @marlovsk1 2 года назад

    Thanks!

  • @satyanarayantripathy
    @satyanarayantripathy Год назад +1

    What is the physical significance of commutators?

    • @channeldoesnotexist
      @channeldoesnotexist 6 месяцев назад

      A nonzero commutator physically indicates that one cannot simultaneously measure two conjugate variables. In other words, it means Heisenberg's uncertainty principle is true. Operators that do commute indicates that those physical quantities can be measured simultaneously.

  • @dlrmfemilianolako8
    @dlrmfemilianolako8 2 года назад +3

    Thank you so much for your videos . You are the best . 😉😉😉😉😉
    I want to be a theoretical physicist in future . Please can you tell me what books should I read to understand quantum mechanics . I have " Quantum Mechanics Concepts and Applications by Nouredine Zettili " Is this book , good ?

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад +1

      Thanks Emiliano! I don't know the book, but the main thing is to find one that's at the right level for you based on what you've already learned. So if that one feels good then go for it!

    • @dlrmfemilianolako8
      @dlrmfemilianolako8 2 года назад +1

      @@PhysicswithElliot
      In terms of math I think I went to a considerable level to understand things at the micro and macro level (when I was in the 11th grade I worked with national and international Mathematics and Physics Olympiads) ( Last year I ranked as First in my country for Physics )
      Being a fan of Elon Musk ,I want to not be influenced by school because education today ( at high school)does not measure intelligence but memory ( Of course you can say that I need a degree , but my principles are two : 1 - education is not the learning of many facts but the training of the mind to think
      2 - when something is important enough , you will do it even if the odds are not in you favour
      In Short ( one person must have a master degree but still be an idiot if he has paid money to get that degree )
      In my country, believe it or not, we get cosine and sine theorem in the last year of high school ( is shameful for education today )
      Since no Physics professor in my city has reached this level, then I started working on my own.
      So I want to thank you professor for every videos that you have posted on youtube ,,, have helped me a lot .
      Now I am learning theories of these topics ( Quantum Meachanics and Astronomy ) and certainly associated with relevant problems,
      can you help me with a list of books( 2 or 3) that you think I should read on both of the topics I mentioned above to get the best education ( certainly with your videos )

  • @arezaajouneghani3082
    @arezaajouneghani3082 Год назад

    Brilliant!!!!!!!!THANKS

  • @likaspokas5481
    @likaspokas5481 2 года назад

    I still don't get what it means by "Momentum is the generator of spatial translations" classically before quantization and probabilities.

  • @amandobhal4264
    @amandobhal4264 2 года назад

    As always great video. Just a correction at 7:19, It's not adjoint of U. It's U dagger. adjoint is different.

  • @mauromichi2139
    @mauromichi2139 2 года назад +1

    greats concepts mate! a little bit too fast in my opinion... not only for understanding the maths (if you dont have the background you will never get) but to digest what you've just said... it's hard to follow.

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад

      Definitely a ton in there to digest! I suggest watching again ;)

  • @xephyr417
    @xephyr417 2 года назад +1

    You said "we introduce this factor of i/h as a matter of convention" but then it ended up being integral to the solution :|

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад +1

      The hbar is there to get the units right and the i makes p real in an appropriate sense (called a Hermitian operator). You could pick a different overall scaling, but then p wouldn't correspond to what we usually call momentum in the classical limit

  • @drdca8263
    @drdca8263 2 года назад

    Let me see if I can summarize the steps here, in order to test my understanding:
    We want U(\lambda) to be a one-parameter unitary family of operators (or, more specifically, a continuous group homomorphism from (\br,+) to the unitary group) such that, for any state |\psi> , U(\lambda) |\psi> is such that the expected value of the position in this state, is the same as \lambda plus the expected value of the position in the state |\psi>
    i.e. () = + \lambda =
    and to first order, U(\lambda) = 1 + \lambda * (-i /\hbar) * U'(0) , where we name U'(0) to be called \hat{p} ,
    and where the factor of (-i /\hbar) makes it so that \hat{p} is hermitian, and the minus sign part of it is there to... make it have the sign we would expect for momentum?
    and so then,
    well, there's the \hat{x} + (i /\hbar) \lambda \hat{p} \hat{x} - (i/ \hbar) \lambda \hat{x} \hat{p} + \lambda^2 \hbar^{-1} \hat{p}^2
    ok, yeah, and, uh, this being the same (up to first order in \lambda) as \hat{x} + \lambda, implies that (i/ \hbar) \lambda [\hat{p},\hat{x}] = \lambda
    and so [\hat{p},\hat{x}] = - i \hbar
    and so [\hat{x},\hat{p}] = i\hbar
    but I glossed over the "why can we go from it looking at it through expectations, to it being like this" step...
    Uh, something about, quadratic forms and bilinear forms being in correspondence under some conditions, maybe, gives us that?
    Unsure.
    Can we phrase all these "to first order in \lambda" as just, taking the derivative wrt \lambda? Seems like should be able to.
    If U(\lambda)^\dagger \hat{x} U(\lambda) = \hat{x} + \lambda
    and then we take the derivative of this with respect to \lambda,
    then we get U'(\lambda)^\dagger \hat{x} U(\lambda) + U(\lambda)^\dagger \hat{x} U'(\lambda) = 1
    then, evaluating this at \lambda = 0, the U(0) = 1
    and so we get the U'(0)^\dagger \hat{x} + \hat{x} U'(0) = 1
    Which is, basically the same thing, yeah,
    just need to get that U'(0)^\dagger = - U'(0)
    And, that comes from, uh, the generators of the unitary group being, err..
    the skew-hermitian operators?
    Yeah. That's exactly what is needed.
    Ok, cool!
    That gives that [\hat{x},U'(0)] = 1
    (and the i \hbar comes from just, pulling that out of the U'(0) )
    And so, if you have some operator A,
    and there's an analogy of translation but for the values of this operator,
    then, a similar thing happens, yeah. Ok.
    What if, like, we try to do this with "angle"?
    Like, suppose the operator A gives us the direction something is facing,
    and we are trying to get the generator of the rotation?
    Well, a problem with that is that the angle doesn't really, uh,
    if it has real number values that we are adding (say, they are from 0 to 2 \pi)
    then, at the point where they wrap around, they don't add correctly across that.
    Can still like, do stuff for little intervals of angles I guess,
    and definitely we can define the rotations,
    ok yeah, we can define the U(\lambda) which rotates things around by \lambda radians,
    and we can talk about the generator of that,
    but, I guess there's just no appropriate "angle" operator, for which it would have the corresponding commutation relation?
    But I expect that the operators that do exist and are kind of like one, would be in some sense close to having the corresponding commutation relation? idk.

    • @PhysicswithElliot
      @PhysicswithElliot  2 года назад +1

      Yes you don't need to restrict to infinitesimal transformations; taking the derivative of the finite transformation e^{i \lambda p/\hbar} x e^{-i \lambda p} = x + \lambda at \lambda = 0 will give the same result

    • @alphalunamare
      @alphalunamare 2 года назад

      A perfect example of the dual usage of the term 'generator' in a quantum thesis.

    • @shamsurrehman3566
      @shamsurrehman3566 2 года назад

      Sir can you help me on writing synopsis for MPhil degry?

  • @cacorami95
    @cacorami95 2 года назад

    🤯

  • @sensorer
    @sensorer 7 месяцев назад

    Hate to be a hater, but I find the explanation in this video to be lacking and circular. Okay, so we're going to pull out this factor (1/ih_bar) and this here is a quantum generator(because it just is), and now that we've pulled out of our hat that it's the quantum generator, it should be momentum. For me, Dirac's approach of getting the quantum Poisson bracket from the classical one is much more satisfying and then using the canonical commutation relations you got from that you find the translation operator and all the other stuff you want.

  • @DeepLyricist
    @DeepLyricist 2 года назад +3

    You should add some cuts in your video showing your face. I feel I understand the material better when someone explains it to me with a serious face while wearing glasses.