Harmonic oscillator: Differential equation
Вставка
- Опубліковано 20 сер 2024
- MIT 8.04 Quantum Physics I, Spring 2016
View the complete course: ocw.mit.edu/8-0...
Instructor: Barton Zwiebach
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu
No where literally nowhere anybody explained HO in this way.
He gave reasons for every single step which I always wondered why we do like that!
Pure beauty.
I often wonder if physics at top universities are easier because I find myself constantly coming here for a better explaination. Even the lecture notes online help me with my tutorials and assignments
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our professors have no excuse, this guys first language is undoubtedly not English and elucidates these concepts better than a native speaker!
Vaya, en verdad es una inspiración este hombre, pensar que estudio en la uni, ¡orgullo peruano!
dang, didn't realize until now he actually wrote the schrodinger's equation in a much cleaner fashion that's mathematically this much easier to read, and I can see the concepts this much more easily! Not the most important when it comes to solving the problem but in terms of arithmetics things get cleaned up very nicely!
Much better than the Griffiths explaination! Thanks
Very different way of introducing quantum mechanics- I liked it.
Talk about clear and concise.
El buen Barton :")
What are the mechanisms responsible for energy loss of a harmonic ascillator
Can anyone expand upon the statement around 3:40 that the first derivative in the Taylor series going to zero at a minimum will leave the Taylor approximation of the potential looking like a quadratic? Is this just assuming that cubed and higher terms fall off fast? Why can you assume that?
We can assume that when we are considering a very small interval
Can someone please explain why a^2=h/mw is actually used? It's obtained by equating the units right? So the correct form must be a^2 = r. h/mw right, where r is some number?
your 'r' is actually the unitless 'u' which is extracted out. considering x now equals a·u.
[E]=[h^2/ma^2] = [mw^2x^2]. Equate the two expressions for [E]. You will get h^2/m^2w^2 = x^2.a^2 . As we talking dimensionally, x and a have the same dimensions of length. Therefore they can be treated as equal and the RHS becomes a^4, and therefore a^2 = h/mw.
sir how do you got h cross / m omega
ua-cam.com/video/XQIbn27dOjE/v-deo.html 💐
[E]=[h^2/ma^2] = [mw^2x^2]. Equate the two expressions for [E]. You will get h^2/m^2w^2 = x^2.a^2 . As we talking dimensionally, x and a have the same dimensions of length. Therefore they can be treated as equal and the RHS becomes a^4, and therefore a^2 = h/mw.
@@ShubhamSingh-lz1mb whats the tradeoff from equating [h^2/ma^2] = [mw^2x^2], we know the dimension is energy but certainly h^2 posses a value since its a constant, and we know dimension shouldnt have a definite value (its like a variable)
👍🏼
What about this: ua-cam.com/video/FSlEqYo9MDA/v-deo.html
This MIT professor vs that indian professor.