Position, velocity, and momentum from the wavefunction

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  • Опубликовано: 25 ноя 2024

Комментарии • 66

  • @sphericalchicken
    @sphericalchicken  11 лет назад +7

    That's the one. Griffiths is a well-written text, and about the right level for a first course.

  • @ceeerson1
    @ceeerson1 4 года назад +6

    Hey Thanks a lot for putting these videos on RUclips. I'm just a guy trying to continue learning and I am using these for self studying. I'm grateful.

  • @sphericalchicken
    @sphericalchicken  11 лет назад +9

    The constant A is not necessarily positive, and in general is a complex number. The normalization of the wavefunction requires that the infinite integral of the squared absolute magnitude of the wavefunction is 1, and that "squared absolute magnitude" means that positive, negative, real, and complex A can all work. We usually assume A is positive and real since the phase of A (taken as a complex number) doesn't affect any observable properties of the system.

  • @speeshers
    @speeshers 5 лет назад +7

    loved the background bell sounds lol. they were kinda soothing

    • @jolez_4869
      @jolez_4869 4 года назад +2

      They are the bells of doom ringing to anyone trying to grasp QM.

  • @rio7858
    @rio7858 4 года назад +3

    This is great teaching
    Please continue more

  • @brendansmith670
    @brendansmith670 8 лет назад +19

    Did you forget to bring down the negative at 12:30?

    • @saybrook_7283
      @saybrook_7283 8 лет назад +1

      Yes.

    • @Harry-ub2fv
      @Harry-ub2fv 4 года назад +2

      Yeah, bro. Thanks, you commented I spent 15 mins to check whether I'm wrong or not.

    • @hershyfishman2929
      @hershyfishman2929 3 года назад +2

      He finally gets back to the negative sign (without comment) at 16:20

  • @davidbakaj5104
    @davidbakaj5104 8 лет назад +1

    = 4/3 because when you take the d^2/dx^2 of (1-x^2) you will get 2 so you bring it outside the derivative you will be intergrating (1-x^2) so you will get 4/3

    • @DocGould
      @DocGould 7 лет назад +7

      You forgot your Constant (h^2 A^2)/m

  • @tokkia1384
    @tokkia1384 5 лет назад +1

    Love the bells in the background .

  • @xXxBladeStormxXx
    @xXxBladeStormxXx 10 лет назад +2

    This is so good!! Thanks!

  • @VincentStevenson
    @VincentStevenson 7 лет назад +1

    You saved my life

  • @anchaljaiswal718
    @anchaljaiswal718 4 года назад

    Thanku so much for uploading these vedios and these are very helpful to me

  • @CantoErgoSum11
    @CantoErgoSum11 9 лет назад +8

    I am getting confused at 15:05. shouldn't the velocity expectation value be negative like eqn. 1.31 in the book? It's a small detail, but I keep trying this on my own and I wind up with a minus sign in front of the i*h_bar/m.

    • @saybrook_7283
      @saybrook_7283 9 лет назад

      +CantoErgoSum11 He forgot a minus sign at about 12:24

    • @executorarktanis2323
      @executorarktanis2323 4 года назад

      i know i also paniced like bruh we missed it

  • @joannalada5815
    @joannalada5815 10 лет назад +4

    Thank you so so so much for all of these videos! They are the only videos I have found on QM that really explain each step, so thank you :)
    I just have a quick question- for the 'integrative by parts' at 11:50, why do we keep the integral limits for the v term? Thank you!

  • @katgirl3000
    @katgirl3000 2 года назад

    These are fantastic! Many thanks! When you went from to a minus sign appeared and I wondered where that came from (in front of ih_bar/m). I've done the derivation a couple of times and just don't see it. Thanks!

    • @daddy_myers
      @daddy_myers 2 года назад

      That comes from the IBP (Integration by Parts) bit; as the boundary terms go to zero, you're left with a negative integral - that negative is pretty much the negative that's on that integral, just factored out.

    • @kontiimanalatit8987
      @kontiimanalatit8987 Год назад

      It is carried over from the first integration by parts is done. He forgot to carry it over I guess

  • @andrewstallard6927
    @andrewstallard6927 6 лет назад +8

    4/3*h*A^2/m

  • @camdenfitzgerald2557
    @camdenfitzgerald2557 8 лет назад

    i am a senior in high school, i just found these videos today. and by far they are some of the best explanations to qm, qed,qcd that I have ever seen. You talk about a book, is there a download i could use to read it? Thank you

    • @II-op5vv
      @II-op5vv 8 лет назад

      The book is by Daniel Griffith, you can get it for 13 dollars on amazon(paperback though), I paid $50 for the hardcover.

    • @camdenfitzgerald2557
      @camdenfitzgerald2557 8 лет назад +2

      Overlord Master omg thank you so much. Peace!

    • @the-fantabulous-g
      @the-fantabulous-g 4 года назад +1

      David J Griffiths, Introduction to Quantum Mechanics. Sorry I'm a bit late

    • @speedspeed121
      @speedspeed121 3 года назад

      @@camdenfitzgerald2557 It's been four years. Did you study physics in University? You should be done or a senior by now.

    • @camdenfitzgerald2557
      @camdenfitzgerald2557 3 года назад +1

      @@speedspeed121 yes! I actually went through with it and studied physics and got my bachelors in it!
      I will admit most of the time I was a little over my head learning this stuff, but I’m super glad I did.

  • @felonhighman5146
    @felonhighman5146 2 года назад

    What does it mean when an expectation of the momentum operator is a complex number ?

  • @UcranianoUKR
    @UcranianoUKR 11 лет назад

    which textbook do you use in your class? is it the griffiths 2nd ed?

  • @zenojimneuromansah8665
    @zenojimneuromansah8665 11 лет назад

    Can you assume that A is real and positive? Can A ever be negative? Intuitively it seems like this (real and positive A) should be true because the infinite integral of A times Psi(x) must be 1 (by normalization). I'm only asking this because in almost every problem/example I've done A has been positive.

  • @Warriorpend2
    @Warriorpend2 6 лет назад

    At 12:15, why is the uv portion of the integration-by-parts technique being evaluated at infinity and minus infinity? Should you just be multiplying u and v together and then subtracting from that the integral of yada yada?

  • @abbiravindhran8424
    @abbiravindhran8424 6 лет назад

    how do you find the values of kinetic energy from a linear combination wavefunction?

  • @123string4
    @123string4 9 лет назад

    At 12:55 when applying integration by parts, why couldn't you apply it to the first integral and get an a partial derivative on psi star? The operator would then act on the complex conjugate of the wave function. What's wrong with this?

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

    Dude this is awesome, but couldn't you use the formula that says the time derivative of the exp. Value (times i hbar) is equal to the exp. Value of the commutator of that operator with the Hamiltonian ?
    Would have saved you the 2 i.b.p.

  • @azoooz86
    @azoooz86 7 лет назад +1

    at 13:40 why (du = d(psi)/dx)? shouldn't be du = d(psi)?

  • @SplashEazy
    @SplashEazy 7 лет назад

    So at 26 minutes you mention how you can't just move the operator around in the integrand. Yet when you were deriving the momentum operator that is precisely what you did in your first step when you pulled the x out front.

    • @officiallyaninja
      @officiallyaninja 2 года назад +2

      that's because x commutes with psi and psi conjugate. but most operators In general don't.

  • @marcosette6995
    @marcosette6995 11 лет назад

    Is there an index of the lessons?

    • @sphericalchicken
      @sphericalchicken  11 лет назад +2

      The closest I have is the overall playlist: ruclips.net/p/PL65jGfVh1ilueHVVsuCxNXoxrLI3OZAPI

    • @marcosette6995
      @marcosette6995 11 лет назад

      Brant Carlson
      Thanks

  • @Salmanul_
    @Salmanul_ 4 года назад

    can psi and psi* be switched in momentum and velocity operators?

    • @Makazakariah
      @Makazakariah 4 года назад

      no im pretty sure its always psi* first

    • @Salmanul_
      @Salmanul_ 4 года назад

      @@Makazakariah ohh I see, do you know why it is like that?

    • @κπυα
      @κπυα 2 года назад

      ​@@Salmanul_ I think that is related to the definition of inner product adopted.
      We can write:
      ⟨Q⟩ = ⟨ψ|Qψ⟩ = ⟨Qψ|ψ⟩ = ∫Q*ψ* ψ dx
      Operators Q with this property are called hermitian operators, and observables are represented by that kind of operator.

  • @Harry-ub2fv
    @Harry-ub2fv 4 года назад +1

    KE=p^2/2m hence
    (-ih*d/dx)^2/2m
    But please tell how (d/dx)^2 = d^2/dx^2

    • @Jehannum2000
      @Jehannum2000 4 года назад +2

      Squaring an operator means making the operator act twice.

    • @Harry-ub2fv
      @Harry-ub2fv 4 года назад

      @@Jehannum2000 Oh thanks, I was reading that wrongly.:)

  • @Dyslexic-Artist-Theory-on-Time
    @Dyslexic-Artist-Theory-on-Time 6 лет назад +1

    Do you think it is logical that if the future is unfolding relative to the atoms, if we look down at the individual atoms we will find probability? This is an invitation to see a theory on the nature of time! In this theory we have an emergent uncertain future continuously coming into existence relative to the spontaneous absorption and emission of photon energy. Within such a process the wave particle duality of light and matter in the form of electrons is forming a blank canvas that we can interact with forming the possible into the actual! The future is unfolding with each photon electron coupling or dipole moment relative to the atoms of the periodic table and the individual wavelengths of the electromagnetic spectrum. As part of a universal process of energy exchange that forms the ever changing world of our everyday life the ‘past’ has gone forever. At the smallest scale of this process the ‘past’ is represented by anti-matter annihilation with the symmetry between matter and anti-matter representing the symmetry between the future and the past as the future unfolds photon by photon. In such a theory the mathematics of quantum mechanics represents the physics of ‘time’ with the classical physics of Newton representing processes over a period of time, as in Newton’s differential equations. In my videos I explain how this process is relative to temperature and the phase changes of matter.

  • @nisarqasim9965
    @nisarqasim9965 8 лет назад +1

    Sir please upload the mathematical method of physics(Authors Arfken) lectures and also classical mechanic(J B Marion) please sir that is sympathetic petition.Also upload the Griffth Electrodynamics lectures.

  • @FlintPet
    @FlintPet 4 месяца назад

    Anybody got the answer to the Understaning Check?

  • @muhammadziaulislamarsalan1392
    @muhammadziaulislamarsalan1392 8 лет назад

    sir.. value is negative what does this mean??physical significance please!!

  • @BLVGamingY
    @BLVGamingY 9 месяцев назад

    YAYY the static wavey has no momentum YUPPIIEEE

  • @wulphstein
    @wulphstein 3 года назад

    String theory doesn't seem to be a contender for a quantum gravity theory because a quantum gravity theory has to answer the question: what is spacetime made of? Twistor theory is the closest to answering "what is spacetime made of". But it falls short because "twisters" cannot be experimentally detected or isolated. It is the same situation with Loop Quantum Gravity. Those loops cannot be isolated either. But it is worse than that. Neither of those theories are compatible with big bang expansion. You would expect that, whatever spacetime is made of, that such "atoms of spacetime" require an "expansion" parameter because the big bang "expanded" from a point.
    Someone needs to come up with a quantum gravity theory that answers the question: what is spacetime made of/what are the atoms of spacetime. I would suggest calling it, the Expanding Graviton. The Expanding Graviton has three primary characteristics. 1) Expanding Gravitons are the carriers of the physics constants. 2) Expanding Gravitons are equivalent to wave functions; thus wave functions and their operators are real things; and now they are a manifestation of gravitons. 3) Expanding gravitons expand spherically, at the speed of light, with radius r = ct. Where in physics have we seen something that looks like an expanding sphere? Two places. First, the derivation of time dilation. Second, two slit interference pattern.
    A proper quantum gravity theory should have an experiment associated with it. I would suggest that quantum entanglements between photons, are (Captured) Gravitons; it's not really an Expanding Graviton if it is captured between two photons whose positions can be controlled with optical fiber, lenses, mirrors, etc. I would also comment that it takes a tiny amount of energy to capture a graviton; if you add up all of the gravitons in the universe, then that number of gravitons multiplied by the capture energy of a graviton should equal DARK ENERGY. More importantly, if a quantum entanglement IS a graviton, then we can do experiments on it.
    Someone needs to create a quantum gravity theory that looks like the outline above.

  • @ВладимирКоролёв-х1м

    Damn this is so nonrigorous it hurts.

    • @turboleggy
      @turboleggy 3 года назад +3

      I think they start making it rigorous in ch
      3 this is just building upto it