Momentum operator, energy operator, and a differential equation

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  • Опубликовано: 30 июл 2017
  • MIT 8.04 Quantum Physics I, Spring 2016
    View the complete course: ocw.mit.edu/8-04S16
    Instructor: Barton Zwiebach
    License: Creative Commons BY-NC-SA
    More information at ocw.mit.edu/terms
    More courses at ocw.mit.edu

Комментарии • 49

  • @ricardocesargomes7274
    @ricardocesargomes7274 7 лет назад +45

    The best Lectures of Quantum Physics!

    • @shadmanyakub1494
      @shadmanyakub1494 6 лет назад +1

      Ricardo Césa what is the teachers name?

    • @lexhuismans3604
      @lexhuismans3604 6 лет назад +2

      It is in the description: Barton Zwiebach

    • @michaelcordova1803
      @michaelcordova1803 2 года назад

      Barton Zwiebach es peruano.

    • @rajinfootonchuriquen
      @rajinfootonchuriquen Год назад +1

      @@michaelcordova1803 al buscarlo pensé que iba a ser de esos latinos criados en EEUU, pero es nacido y criado en el Perú. Que bueno que hay gente que independiente de donde venga logré grandes cosas.

  • @richardhall9815
    @richardhall9815 4 года назад +7

    It's cool that they actually show the students in the front at the beginning of the video. I've been wondering for a while whether he was actually talking just in an empty room in front of a camera or if it was an actual lecture with students there the whole time.

  • @VijayKumar-pk1sx
    @VijayKumar-pk1sx 2 года назад +3

    Professor is so cool...he explains everything with patience and...best part is that he carries proof...MIT stands apart from the rest of universities around the world...I miss studying in this university ...buut any ways seems to have a substitute..love you MIT but you may not ....but i love you anyways:-)

    • @schmetterling4477
      @schmetterling4477 2 года назад +1

      He teaches nothing that you can't get exactly the same way at any other university in a Western country. Physics undergrad education is pretty much the same all over the developed world. If there is a working restroom in the building, then this is what you get in QM101. Your problem is that you don't even have a working restroom. :-)

  • @juliogodel
    @juliogodel 7 лет назад +2

    Just marvellous.

  • @RajeevSingh-ki8bc
    @RajeevSingh-ki8bc 6 лет назад +7

    I want more lectures from that professor

  • @ismailbaris7181
    @ismailbaris7181 3 года назад

    Amazing. Thank you very much.

  • @waibenglam756
    @waibenglam756 6 лет назад +9

    Excellent. Wished his lectures were available when I was I was an undergrad. p/s Professor Zweich does look a bit like Harrison Ford without his glasses on!

    • @rahuldhungel
      @rahuldhungel 3 года назад

      Did he ever take that Voigt-Kampff test himself☺️😉?

  • @ashokpillay6343
    @ashokpillay6343 4 года назад

    Well explained!!!

  • @vinodkancherla4504
    @vinodkancherla4504 4 года назад +2

    Walter Lewin, Barton Zwiebach, Leonard Susskind, R.Shankar can motivate students!!

  • @rezokobaidze8501
    @rezokobaidze8501 3 года назад

    thanks a lot

  • @Ajaysingh-fl1vf
    @Ajaysingh-fl1vf 3 года назад

    just wow sir

  • @user-kb4wz1yq1g
    @user-kb4wz1yq1g 9 месяцев назад

    6:22 momentum operator

  • @wagsman9999
    @wagsman9999 2 года назад

    Awesome

  • @oscaraguilar6906
    @oscaraguilar6906 Год назад

    guapisimo

  • @AirborneLRRP
    @AirborneLRRP 6 лет назад +2

    My only qualm is that Ehat operator is actually defined as ihbar d/dt. In this case, with no potential energy it in fact is equal to the energy operator. With potential total energy would not just equal kinetic, so the way we defined the energy would not be an eigenstate of the energy operator, which defeats the our purpose of the operator.

    • @rajinfootonchuriquen
      @rajinfootonchuriquen Год назад

      Given the relation (operator -> observable). If you have a decomposition of the observable, its operator is also a decomposition of other operators. The potential occurs if you find a scalar field that satifies a gradient equation, so in a big sense, defining the energy operator as a composition of potential operator is more general that thinking of a composition of potential operator plus a potential operator.

    • @ingeniouswild
      @ingeniouswild 2 месяца назад

      Yeah I reacted to this part as well, he says a couple of times that the p^2 is the "energy operator" but conceptually I think it helps to think of it in reverse - p^2 is just, to start with, the "p^2 operator" and the *physics* of the particle in this situation that you already have (E=p^2/m) tells you that this operator operating on the wavefunction should give the same result as the energy operator operating on the wavefunction that was previously defined.

  • @surendrakverma555
    @surendrakverma555 2 года назад

    Very good 🙏🙏🙏🙏🙏🙏

  • @anishgade4050
    @anishgade4050 3 года назад +1

    Deriving Schrodinger's Equation by my own hand is the most exciting thing in the world

    • @schmetterling4477
      @schmetterling4477 2 года назад +1

      Schroedinger's equation is unphysical, so you can't "derive" it from rational first principles. Not sure what you think you are doing, but it is certainly not a "derivation".

    • @ladyalexandra2980
      @ladyalexandra2980 Год назад

      Congratulations, not even Schrödunger could do it. He dreamt of it or so.

  • @user-kb4wz1yq1g
    @user-kb4wz1yq1g 9 месяцев назад

    18:31 energy operator - 2nd derivative

  • @romeovalentin5524
    @romeovalentin5524 6 лет назад

    At 14:10, it should be E*psi=p^2/(2*m) I think

    • @mfoucault1984
      @mfoucault1984 5 лет назад +1

      somebody told him just a moment later..

  • @iwonakozlowska6134
    @iwonakozlowska6134 4 года назад +2

    It's like a "heat equation" with an imaginary coefficient.

    • @rajinfootonchuriquen
      @rajinfootonchuriquen Год назад

      Yeah. It should be called the Schrodinger's diffusion equation, but maybe "wave" is more catchy

  • @cafe-tomate
    @cafe-tomate 2 года назад

    He is talking about the exponential but the wavefunction Ψ is the the integral of these exponentials times a function Φ
    Note that Ψ is an eigenstate of the p operator (p hat) only if Φ peaks narrowly around a certain value (the eigenvalue being p of course)

  • @vinodkancherla4504
    @vinodkancherla4504 4 года назад

    Love you, Professor!

  • @kaushaljain5999
    @kaushaljain5999 4 года назад

    12:15 to 12:24 What did you say at this time?

    • @philos22
      @philos22 3 года назад +1

      Look at subtitles

  • @theconstellation__
    @theconstellation__ 3 года назад

    But E=p^2/2*m is for non relativistic particles
    What about photons

    • @schmetterling4477
      @schmetterling4477 2 года назад

      Those need the real theory. This is just a toy theory for those who will never do serious physics.

    • @rahilshaik1603
      @rahilshaik1603 Год назад

      you just use the relativistic formula, E^2 = p^2c^2 m^2c^4

  • @joshramer7
    @joshramer7 5 лет назад +1

    Truly, the Harrison Ford of physics!

  • @juanfa98
    @juanfa98 4 года назад

    Jefe

  • @sumitbhoi0074
    @sumitbhoi0074 Год назад

    Better than Wikipedia lol

  • @faustogadani3075
    @faustogadani3075 4 года назад

    the pronunciation of De Broglie name is uncorrect. But the prof is ok

  • @zackarykutler5348
    @zackarykutler5348 3 года назад

    I give my girlfriend my *De Br OG Ley Wavelengths* sometimes.