Momentum operator, energy operator, and a differential equation

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  • Опубліковано 19 гру 2024

КОМЕНТАРІ • 52

  • @ricardocesargomes7274
    @ricardocesargomes7274 7 років тому +49

    The best Lectures of Quantum Physics!

    • @shadmanyakub1494
      @shadmanyakub1494 7 років тому +1

      Ricardo Césa what is the teachers name?

    • @lexhuismans3604
      @lexhuismans3604 7 років тому +2

      It is in the description: Barton Zwiebach

    • @michaelcordova1803
      @michaelcordova1803 2 роки тому

      Barton Zwiebach es peruano.

    • @demr04
      @demr04 Рік тому +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.

  • @VijayKumar-pk1sx
    @VijayKumar-pk1sx 3 роки тому +5

    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 роки тому +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. :-)

  • @richardhall9815
    @richardhall9815 5 років тому +8

    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.

  • @waibenglam756
    @waibenglam756 6 років тому +11

    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 роки тому

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

  • @TrungKiênNguyễn-b8u
    @TrungKiênNguyễn-b8u Рік тому

    18:31 energy operator - 2nd derivative

  • @RajeevSingh-ki8bc
    @RajeevSingh-ki8bc 6 років тому +7

    I want more lectures from that professor

  • @vinodkancherla4504
    @vinodkancherla4504 4 роки тому +3

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

  • @TrungKiênNguyễn-b8u
    @TrungKiênNguyễn-b8u Рік тому

    6:22 momentum operator

  • @AirborneLRRP
    @AirborneLRRP 7 років тому +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.

    • @demr04
      @demr04 Рік тому

      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 7 місяців тому

      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.

  • @cafe-tomate
    @cafe-tomate 2 роки тому +1

    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)

    • @AdenKhalil
      @AdenKhalil 3 місяці тому

      The integral was the superposition of many exponentials(wave function), so a wave function also.

  • @romeovalentin5524
    @romeovalentin5524 6 років тому

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

    • @mfoucault1984
      @mfoucault1984 6 років тому +1

      somebody told him just a moment later..

  • @anishgade4050
    @anishgade4050 3 роки тому +1

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

    • @schmetterling4477
      @schmetterling4477 2 роки тому +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 2 роки тому

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

  • @juliogodel
    @juliogodel 7 років тому +2

    Just marvellous.

  • @ismailbaris7181
    @ismailbaris7181 4 роки тому

    Amazing. Thank you very much.

  • @kaushaljain5999
    @kaushaljain5999 4 роки тому

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

    • @philos22
      @philos22 4 роки тому +1

      Look at subtitles

  • @not_amanullah
    @not_amanullah 4 місяці тому

    Thanks ❤️🤍

  • @surendrakverma555
    @surendrakverma555 2 роки тому

    Very good 🙏🙏🙏🙏🙏🙏

  • @ashokpillay6343
    @ashokpillay6343 5 років тому

    Well explained!!!

  • @Ajaysingh-fl1vf
    @Ajaysingh-fl1vf 3 роки тому

    just wow sir

  • @theconstellation__
    @theconstellation__ 4 роки тому

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

    • @schmetterling4477
      @schmetterling4477 2 роки тому

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

    • @rahilshaik1603
      @rahilshaik1603 Рік тому

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

  • @rezokobaidze8501
    @rezokobaidze8501 3 роки тому

    thanks a lot

  • @iwonakozlowska6134
    @iwonakozlowska6134 4 роки тому +2

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

    • @demr04
      @demr04 Рік тому

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

  • @not_amanullah
    @not_amanullah 4 місяці тому

    This is helpful ❤️🤍

  • @joshramer7
    @joshramer7 6 років тому +1

    Truly, the Harrison Ford of physics!

  • @wagsman9999
    @wagsman9999 3 роки тому

    Awesome

  • @oscaraguilar6906
    @oscaraguilar6906 Рік тому

    guapisimo

  • @vinodkancherla4504
    @vinodkancherla4504 4 роки тому

    Love you, Professor!

  • @sumitbhoi0074
    @sumitbhoi0074 2 роки тому

    Better than Wikipedia lol

  • @juanfa98
    @juanfa98 5 років тому

    Jefe

  • @zackarykutler5348
    @zackarykutler5348 4 роки тому

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

  • @faustogadani3075
    @faustogadani3075 4 роки тому

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