Born's Statistical (Probabilistic) Interpretation of Quantum Mechanics

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  • Опубліковано 28 вер 2024

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  • @FortheLoveofPhysics
    @FortheLoveofPhysics  8 місяців тому

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  • @dheeraj_matics25
    @dheeraj_matics25 5 місяців тому +4

    If there will be unlimited like option in UA-cam, I would not restrict my self to it for this video. Thanks a lot❤

  • @deepikadiwan4228
    @deepikadiwan4228 7 місяців тому +3

    Throughout the playlist you have been patiently and compassionately unfolding the mysterious layers of quantum word, it's more thrilling than any movie fir me. Thankyou so much Sir......❤

  • @arjunsinha4015
    @arjunsinha4015 8 місяців тому +3

    Sir it's time to explain Born-Oppenheimer approximation (Just a small request)

  • @zzzoldik8749
    @zzzoldik8749 8 місяців тому +2

    You said the function tell about probability to find particle but I have heard for interpretating double slit that particle through both slit together in one time. This source from UNSW, they said that even single particle can get constructive interference with it self.

    • @lowersaxon
      @lowersaxon 8 місяців тому

      Yes, it is said so.

    • @lepidoptera9337
      @lepidoptera9337 5 місяців тому

      That's complete nonsense. There are no particles in quantum mechanics to begin with. A quantum is a small amount of energy. It's the energy that the measurement system absorbs irreversibly from the free field. There simply is no photon until that absorption process takes place. This is no different from classical probability theory. There is no outcome of a dice throw until all of the kinetic energy of the rolling dice has been dissipated in the table and the system has come to rest, again. The main difference is that in quantum mechanics the equivalent of "the table", which we call "the measurement system" does not have to absorb all of the energy of the quantum system. It has a spectral function and that spectral function can be zero for some or even most energies. The measurement system can therefor only have outcomes for quanta for which it is absorptive. Amounts of energy which it can't absorb can not be measured and are therefor not allowed in the projection of the ensemble on the final state after measurement.

  • @KipIngram
    @KipIngram 3 місяці тому

    I want to give you some major kudos for really underscoring the proper interpretation of what the wave function is telling us about FUTURE measurements. Earlier on in the video I was worried you were going to do the oh so common "particle is in many places at once" thing, which I absolutely hate, but you dodged that bullet quite nicely. I love how you said very clearly that it's not THE PARTICLE that's "partially here, partially there, etc." but rather the probability of later finding the particle here or there. Very nice. I adhere to Feynman's wisdom on this - that we really should not talk about the values of things we have not yet measured as actually existing before we do so.

    • @lepidoptera9337
      @lepidoptera9337 3 місяці тому

      It doesn't tell us anything about future measurements. At most the wave function tells us something about the averages of future measurements on many identical (and identically prepared) systems. If we don't have identical preparation, then we need the density matrix. The wave function alone won't do. :-)

    • @KipIngram
      @KipIngram 3 місяці тому

      @@lepidoptera9337 Sure - it doesn't tell us what individual future measurements will be - it only tells us some information about their statistics. Given the fundamental non-determinism of things that's the best we can do.

    • @lepidoptera9337
      @lepidoptera9337 3 місяці тому

      @@KipIngram Exactly. That's the major misunderstanding about quantum mechanics. It's an ensemble theory. It is NOT a theory of individual system evolution. It can't be any other way. Relativity won't allow that.

    • @KipIngram
      @KipIngram 3 місяці тому

      @@lepidoptera9337 Two thumbs up - two thumbs up.

  • @adrinfpv
    @adrinfpv 8 місяців тому +1

    You are an awesome teacher, love your videos. I am still waiting for the explanation, WHY does the square of a wave function represent the probability? I understand Einstein inspired Born, but the derivation is missing. Can you help enlighten us?

    • @FortheLoveofPhysics
      @FortheLoveofPhysics  8 місяців тому +1

      It's accepted as one of the many postulates, in the foundations of quantum mechanics.

    • @adrinfpv
      @adrinfpv 8 місяців тому

      @@FortheLoveofPhysics okay understood, just like the Plancks' constant

    • @lepidoptera9337
      @lepidoptera9337 5 місяців тому

      @@FortheLoveofPhysics That's not an answer. That's a cop-out. It's NOT a postulate, to begin with, and it's only true for systems for which individual quantum measurements are independent. You can probably make it work for photon detection on a thermal source, but you definitely can't make it work on a laser within the coherence time scale. The wave function formalism can be derived more or less trivially from Kolmogorov's axioms. There are several (rather boring) theory papers about that. Please inform yourself.

  • @ELECTRONICS_TIMES
    @ELECTRONICS_TIMES 5 місяців тому

    The best and clear explanation sir 👌
    9:08 What if only modulus instead of modulus square? Without considering right hand terms complex and its conjugate.
    16:23 How einstein got that idea (square the amplitudes of light gives probability of occurrence of photon)?

    • @lepidoptera9337
      @lepidoptera9337 5 місяців тому +1

      The square of the amplitude is the intensity. That photon count and intensity are proportional follows directly from independence of individual photon detections and energy conservation.

  • @hg5332
    @hg5332 8 місяців тому +1

    Good day, professor. Could you please let me know the book name you've shown in minute 16 of this video?

    • @doutormanhattan5680
      @doutormanhattan5680 5 місяців тому

      It is not a book. I will send the link below, but YouYube may exclude it.

  • @lowersaxon
    @lowersaxon 8 місяців тому

    Your lectures are excellent, imho. The Schroedinger equation has to be called the Schroedinger equation, nevertheless. Greetings to sunny India.

  • @sauravneogi7024
    @sauravneogi7024 8 місяців тому

    Sir next video on general theory of relativity .... Derivation and solution

  • @lowersaxon
    @lowersaxon 8 місяців тому

    How can a energy ( conservation) equation, i.e the Schroedinger equation, have no physical significance??

    • @lepidoptera9337
      @lepidoptera9337 5 місяців тому

      The SE is NOT energy conserving. The SE is unitary, i.e. it conserves the (infinite) number of systems in the quantum mechanical ensemble. This is often misrepresented in lectures and textbooks on QM.

  • @hazimahmed8713
    @hazimahmed8713 8 місяців тому

    I have a doubt sir. It has nothing to do with this video but anyways, do atoms have temperature? I got the answer from Google that temperature is a macroscopic property and atoms are microscopic objects so they should not have a temperature. But it didn't satisfy me because temperature is defined as avg kinetic energy of particles and atoms have electrons within and electrons have kinetic energy. So atom must have a temperature right. But when I thought deeper I stumbled upon an insight that maybe it could be because that atoms cannot exchange heat with one another. And therefore we can never say two atoms are in thermal equilibrium if a meaningful concept of heat exchange is not even defined for them. And without thermal equilibrium we cannot define temperature simply by zeroth law of thermodynamics. But then I got the idea that maybe atoms can exchange heat that maybe there is something like a quantum mechanics analog of heat like there is for angular momentum. So maybe their temperature is quantized? What do u think sir?

    • @lepidoptera9337
      @lepidoptera9337 5 місяців тому

      Atoms in thermal equilibrium have temperature. It's the requirement of "being in thermal equilibrium" that requires that a thermal bath is present. That, however, is already so in classical thermodynamics.

  • @salimburki5972
    @salimburki5972 8 місяців тому

    Sir probability current density and born s interpretation are same thing

    • @lepidoptera9337
      @lepidoptera9337 5 місяців тому

      No, they are not. Born requires an external measurement system, whereas a probability current density does not. Born is physical for systems that exchange energy irreversibly, the probability current density is not. It doesn't give you results except for the case of scattering problems in the free field, i.e. for quantum field theory, which is far less general than Born.

  • @Theories_stuff
    @Theories_stuff 8 днів тому

    Sir,can we say that integration of mod psi square is a distribution of energy through out the x axis in a given region as generally amplitude square represents the energy.

  • @salimburki5972
    @salimburki5972 8 місяців тому

    Excellent.

  • @victorvilchis
    @victorvilchis 8 місяців тому

    Wow man amazing!

  • @GUNALANM-k5n
    @GUNALANM-k5n 8 місяців тому

    thank

  • @KipIngram
    @KipIngram 3 місяці тому

    I'm insanely jealous of your chalkboard handwriting. 😞

  • @lepidoptera9337
    @lepidoptera9337 5 місяців тому

    It's not a statistical interpretation. It's an ensemble interpretation.

  • @lowersaxon
    @lowersaxon 8 місяців тому

    The Schroedinger equation. The Das equation. The Eiffel tower. The paper clip. The Euler equation. The Galilei transformation. Never ‚The Eiffel‘s tower’, wrong. ‚The paper’s clip’, wrong. ‚The Euler’s equation’, wrong. Wrong, wrong, wrong. Sorry.

  • @luciferbikkuangmin9528
    @luciferbikkuangmin9528 8 місяців тому

    Excellent.

  • @ajittiwari9766
    @ajittiwari9766 Місяць тому

    Is there any paid lectures on Quantum Mechanics Where Divyajyoti Sir completes all the lectures on Quantum Mechanics 😮😍

  • @KipIngram
    @KipIngram 3 місяці тому

    Actually it contains information about what those quantities WILL BE - because they don't exist until you measure them. For example, position doesn't exist until the system is in a position eigenstate, which it will be as soon as you do a position measurement. As far as the position before you measure? Don't do that. Don't ask about it. It doesn't exist.

  • @harshsingh7384
    @harshsingh7384 8 місяців тому +1

    Respected sir, I love this series, to a point where I've made notes of every single lecture. As this is the only series which I could understand. It's my humble request to just be a little more consistent with it, perhaps a lecture every week. Thank you!

  • @owaisfarooq939
    @owaisfarooq939 8 місяців тому +1

    How can we proof max born statistical interpretation mathematically

    • @lowersaxon
      @lowersaxon 8 місяців тому

      It works mathematically, they say thats the proof.

    • @owaisfarooq939
      @owaisfarooq939 8 місяців тому

      😂

    • @PioneersOppenheimer
      @PioneersOppenheimer 8 місяців тому

      @@lowersaxon lol

    • @lepidoptera9337
      @lepidoptera9337 5 місяців тому

      You start with Kolmogorov's axioms, which will give you both conventional probability theory as well as quantum mechanics, then you work in that your quantum system is connected to an irreversible measurement system with a spectral function. Then you calculate the most likely states of the two connected systems using the density matrix formalism.

  • @Wonderful_life354
    @Wonderful_life354 8 місяців тому +1

    SIR ❤

  • @hanzalamalik7236
    @hanzalamalik7236 Місяць тому

    Sir you are brilliant

  • @saibaMasuku
    @saibaMasuku 6 місяців тому

    Isn't the whole idea that these particles are so small that the slightest disturbances introduced from taking a measurement is enough to move it to a new location? So its still pretty much like classical physics except its hard to prove that it is just because its a bitch to measure?

    • @lepidoptera9337
      @lepidoptera9337 5 місяців тому

      No, that is not the idea. There are no particles of light. A photon is a quantum (small amount) of energy. Planck got that right. Einstein got that right, with the exception that he stipulated that these quanta should have corpuscular character by having location and path. We know that the latter assumption is not true (and Einstein didn't have to make it and he had no physical evidence for it, either, he just made a very unfortunate and false guess).

  • @physicistkarthikeyan5289
    @physicistkarthikeyan5289 Місяць тому

  • @岡安一壽-g2y
    @岡安一壽-g2y 3 місяці тому

    The important things about quantum mechanics are explained in an easy-to-understand way.
    The notes on the board are easy to read. And the diagrams are beautiful.
    By the way, surprisingly and unfortunately, most physicists believe that when two essentially indistinguishable coins are tossed, the probability of getting
    ①Both heads
    ②One head and one tail
    ③Both tails
    are all 1/3.
    Let's think about next question.
    [Question]
    Find the probabilities that the following events will occur when two dice are rolled.
    ①Both show odd numbers
    ②One shows odd number and another shows evenn number
    ③Both show even number
    P(E):the probability that event E occurs.
    【Distinguishable dice A and B】
    (the number that dice A shows,the number dice B shows):event
    If we only judge whether the numbers indicated by the dice are odd or even, the following four events will occur.
      (odd,odd) (odd,even)
      (even,odd) (even,even)
    If each event occurs with equal probability, the probability is 1/4.
    Therefore, P(①)=1/4, P(②)=1/2, P(③)=1/4 (1)
    If we judge the number(1,2,···,6) indicated by the dice ,the following 36 events will occur.
    (1,1), (1,3),(1,5) (1,2),(1,4),(1,6)
    (3,1), (3,3),(3,5) (3,2),(3,4),(3,6)
    (5,1), (5,3),(5,5) (5,2),(5,4),(5,6)
    (2,1), (2,3),(2,5) (2,2),(2,4),(2,6)
    (4,1), (4,3),(4,5) (4,2),(4,4),(4,6)
    (6,1), (6,3),(6,5) (6,2),(6,4),(6,6)
    If each event occurs with equal probability, the probability is 1/36.
    Therfore
    P(①)=9×(1/36)=1/4, P(②)=18×(1/36)=1/2, P(③)=9×(1/36)=1/4 (2)
    (2) matches (1).
    【Indistinguishable two dice】
    (odd,even)is the same event as(even,odd).
    Thererore, if we only judge whether the numbers indicated by the dice are odd or even, the following three events will occur.
      (odd,odd)
      (even,odd) (even,even)
    If each event occurs with equal probability, the probability is 1/3.
    Therefore
    P(①)=1/3, P(②)=1/3, P(③)=1/3 (3)
    (1,3) is the same event as (3,1).
    If we judge the number(1,2,···,6) indicated by the dice ,the following 21 events will occur.
    (1,1)
    (3,1), (3,3)
    (5,1), (5,3),(5,5)
    (2,1), (2,3),(2,5) (2,2)
    (4,1), (4,3),(4,5) (4,2),(4,4)
    (6,1), (6,3),(6,5) (6,2),(6,4),(6,6)
    If each event occurs with equal probability, the probability is 1/21.
    Therefore
    P(①)=6×(1/21)=2/7, P(②)=9×(1/21)=3/7, P(③)=6×(1/21)=2/7 (4)
    (4) contradicts (3).

  • @HM-pb9kd
    @HM-pb9kd 8 місяців тому

    How much quantum mechanics do you need for IPhO?
    The focus of the exam mainly lies on classical mechanics, but the quantum physics portion is not properly drfined for the exam.

  • @riyanshraj1265
    @riyanshraj1265 8 місяців тому

    I would love to see you teach general relativity

  • @mazharali5348
    @mazharali5348 8 місяців тому

    Sir, love this series of lectures ❤

  • @muneerahmed2793
    @muneerahmed2793 8 місяців тому

    Thank you sir

  • @imadyounas3454
    @imadyounas3454 8 місяців тому