Quantum field theory, Lecture 4

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  • Опубликовано: 8 ноя 2024

Комментарии • 44

  • @MewNewPhysics
    @MewNewPhysics 4 месяца назад +1

    Best lectures on Quantum Field Theory on the internet for mathematically minded people. Love from Pakistan

  • @robertgernert8481
    @robertgernert8481 7 лет назад +19

    Thanks for that bit with the circulant matrix! I've always wondered how the Fourier transform comes into play in all of these physical theories.

    • @UnforsakenXII
      @UnforsakenXII 5 лет назад +4

      Indeed. Fourier transforms always just seemed like the magic that worked.

  • @solaireofastora4091
    @solaireofastora4091 3 года назад +3

    I like the mention of numerical methods to provide an intuition for using Fourier transforms. It’s not necessarily obvious to people that you can represent derivatives as matrices when you discretise a system, nor is it obvious how being able to solve a differential equation using Fourier transforms is related to circulant matrices in the discretised system.
    Edit: I just realised the lecturer has a peculiar likeness to Ricky Gervais

  • @Josh-p5b7u
    @Josh-p5b7u 9 месяцев назад +1

    Maybe one comment on why the Fourier transform:
    because the analogy of 'diagonalizing the matrix' in the function space case is the 'spectral theorem', which states that you can conjugate the operator (self-adjoint, let's say, for simplicity) by a unitary operator to make it as a multiplication operator under the spectral measure. In a lot lot of specific cases (and indeed, here), this unitary operator is the Fourier transform.

  • @galSyehoshua
    @galSyehoshua Год назад +1

    Thanks for your videos. I am confused with the last description of the Hamiltonian, I hope you can help me:
    I understand the motivation of change of coordinates and desire for ladder operation. I also see the resembles in the last equitation. What is mysterious to me and required assistance is the computation of the d^3x and d^3p' integrals that left us with this simple form. It seems to be more than working out the algebra.

  • @aditinawani-e3n
    @aditinawani-e3n 2 месяца назад +1

    Sir can you recommend a resource ot read about circulant matrices in this qft context ? I cant find any at all, only you have mentioned this .
    Thank you sir

  • @anubhabsur8376
    @anubhabsur8376 4 года назад +3

    Thanks a ton for these wonderful lectures.
    I would be highly obliged if you could provide some material on the diagonalization of a circulant matrix and a bit more insight on the emergence of the fourier transform here.

    • @tobiasjosborne
      @tobiasjosborne  4 года назад +1

      Many thanks for your comment. The wikipedia article is actually pretty good:
      en.wikipedia.org/wiki/Circulant_matrix
      Sincerely,
      Tobias Osborne

  • @Joao-uj9km
    @Joao-uj9km 6 месяцев назад

    Thanks a lot for this high-quality, public set of lectures :)
    One tiny question: at 1:12:40 I don't get why that integral is infinite. The delta seems well-behaved to me.

  • @prasadpawar7027
    @prasadpawar7027 3 года назад +3

    1:02:08 There should be different exponential terms with the two commutators and I think we assumed that the two annihilation operators commute with each other and the same with creation operators.

    • @tobiasjosborne
      @tobiasjosborne  3 года назад +3

      many thanks for the comment and the correction.
      Sincerely,
      Tobias Osborne

  • @rodrigoromero537
    @rodrigoromero537 4 года назад +2

    Thank you so much Dr. Tobias for uploading this course, I found it more physically reasonable than many others I've attended or watched online. Nevertheless , there are specific points in the lecture where I believe the "analogy" argument is not enough. For instance, at the point where you state that the commutation relation between the "momentum" and "position" operators in some way defines them in a unique way. In non-relativistic quantum mechanics the momentum operator is uniquelly defined by the Stone's theorem as the generator of translations in the physical space, and then, the position operator by the canonical commutation relation. I assume that there must be an analogous procedure in this case, but unless some more information is provided, these quantum momentum and position fields appears to me as fallen from the sky. Am I right or there is something I'm missing?

  • @ms-uj3qe
    @ms-uj3qe 6 лет назад +6

    Sir, thanks for the Lectures! Very usefull. I have a question, in the time mark 1:13:00 you discuss the measurement axioms of QM. You claim that the students may be confused about that, and say "that's what happens when you present the axioms of QM in the wrong way". Could you give me a reference that states the axioms in the "right" way? And maybe a reference for common books that do it wrong (if there's some in mind). The point of my question is that, as a undergrad, I want to make sure I don't get misled by wrong information on basic stuff. Thanks!

    • @tobiasjosborne
      @tobiasjosborne  6 лет назад +8

      The "right way" (this is my opinion, not shared by many of my colleagues!) is to present QM as a probabilistic theory and avoid measurement collapse. This sort of approach is sometimes called operational QM and can be found in, e.g., Foundations of Quantum Mechanics I by Ludwig, G.. I dislike anything to do with projective measurements and measurement collapse (i.e., almost all QM texts). However, I cannot prove measurement collapse is wrong; I just dislike it and have strong opinions. Sincrely,
      Tobias

  • @fawzyhegab
    @fawzyhegab 6 лет назад +3

    You mentioned that giving the right quantization for the Hamiltonian H=qp whould imply the Riemann Hypothesis. Could you give me a reference for this result? And what does it mean here to give "the right quantization" here? Is not the right quantization the one which give results that agree with ones we get experimentally? What does this have to do with Riemann Hypothesis?

    • @tobiasjosborne
      @tobiasjosborne  6 лет назад +5

      This is called the Berry conjecture: you can find out more, e.g., here: mathworld.wolfram.com/BerryConjecture.html
      Sincerely,
      Tobias Osborne

  • @shivammahajan303
    @shivammahajan303 Год назад

    17:00 so that's where they get the connection between QFT and the zeta function that the pop-sci articles talked about.

  • @PeeterJoot
    @PeeterJoot 6 лет назад +5

    37:25 Looks like you have an extra (1/2) riding along with each of the m^2's

    • @tobiasjosborne
      @tobiasjosborne  5 лет назад +2

      Many thanks for the correction! this is a mistake.
      Sincerely,
      Tobias Osborne

    • @kooroshsadri8888
      @kooroshsadri8888 3 года назад +1

      I came to say this to. The mistake takes place at 30:20 or so.

  • @karabomothupi9759
    @karabomothupi9759 4 года назад +1

    Beautiful

  • @a.s.4309
    @a.s.4309 6 лет назад +1

    38:58 "And instead of writing q-subscript-x, which I hate."
    I feel ya man.

  • @kostrahb
    @kostrahb 3 года назад +2

    52:00 How about a little funny thing - an inverted hat/caret (ˇ)? :D
    It's graphically distinguishing, easy to type and moreover, if you apply inversion on hat twice you get the same symbol, just like with coordinates after applying Fourier transform twice. (Yeah, the hat looks better and the coordinates after applying FT twice are inverted, but still...)

    • @tobiasjosborne
      @tobiasjosborne  3 года назад +5

      That is a great idea, I love it! I will use this in future for the FT
      Many thanks!
      Sincerely,
      Tobias

  • @peterwildemann6375
    @peterwildemann6375 7 лет назад +2

    The underbar always stands for the spatial part of the four-vector, right?

    • @tobiasjosborne
      @tobiasjosborne  7 лет назад +4

      Yes, that is the convention I tried to adhere to. Best wishes!
      Sincerely,
      Tobias

  • @robertgernert8481
    @robertgernert8481 7 лет назад +1

    Concerning the infinity: does it make sense to say that we are only interested in differences of energy between two states and as the infinity is the same, it cancels?

    • @tobiasjosborne
      @tobiasjosborne  7 лет назад +2

      Dear Romeo,
      Many thanks for your comment! Yes: this is exactly the reason we can neglect the infinite ground state energy shift (at least in the absence of gravity).
      Sincerely,
      Tobias

  • @Carbenicillin
    @Carbenicillin 6 лет назад +1

    Thanks a lot. In the first lecture you choose the metric as ( + - - -). But in the momentum density , it seems ipx=-i{\omega}·t+ip·x. Is this a typo or you just changed the metric?

    • @tobiasjosborne
      @tobiasjosborne  6 лет назад +2

      Thankyou for your comment! Indeed, our metric is meant to be (+,-,-,-). I couldn't find the place you are referring to, but if I wrote ipx=-i{\omega}·t+ip·x then that would be a mistake. Sincerely,
      Tobias

  • @zonglangfrancis9794
    @zonglangfrancis9794 4 года назад +1

    Hi, can I have access to the questions you used in this course and their solutions ?

    • @tobiasjosborne
      @tobiasjosborne  4 года назад +3

      The problems and solutions will not be distributed.
      Sincerely,
      Tobias Osborne

  • @frederickburke9944
    @frederickburke9944 4 года назад

    42:15 in the discrete case, we have q on the RHS and q' on the LHS. But the continuous case has phi on both sides. That would make K a delta function. I'm thinking something else was meant. The phi on the LHS is not supposed to be the same operator as the phi on the RHS, right?

  • @sergeyliflandsky3231
    @sergeyliflandsky3231 4 года назад +1

    It is not clear to me what the circulant matrix has to do with KG Hamiltonian. Sure the DFT diagonalizes the circulant matrix
    but how do you infer that the guess by analogy should diagonalize the KG Hamiltonian, what is circulant about it? For all I can see from the lecture this educated guess is as good as a guess that came out of the blue.

    • @tobiasjosborne
      @tobiasjosborne  4 года назад +4

      Many thanks for your comment. When you discretise the KG hamiltonian, e.g., by finite differences for the derivatives, you get a matrix representation which is circulant. This could be helpful if you know about the DFT. Otherwise, translation invariance is already enough to suggest the fourier transform.
      Sincerely,
      Tobias Osborne

    • @sergeyliflandsky3231
      @sergeyliflandsky3231 4 года назад +1

      @@tobiasjosborne Thank you for your answer!

  • @bryanjcampbell1
    @bryanjcampbell1 6 лет назад

    Why do the spatial derivatives not act on the creation/annihilation operators when substituting phi back into the KG Hamiltonian? It seems that the only effect of the spatial derivatives is to bring down factors of +- ip.

    • @tobiasjosborne
      @tobiasjosborne  6 лет назад +1

      That is correct: in momentum space derivative operators act as multiplication by p.

  • @rv706
    @rv706 4 года назад

    "Analogious"

  • @toddsecor288
    @toddsecor288 7 лет назад +2

    I wish I could watch these lectures but chalk boards drive me crazy.

  • @rq4740
    @rq4740 4 года назад

    no mods no mods no mods