Quantum harmonic oscillator via power series
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- เผยแพร่เมื่อ 20 ก.ค. 2024
- This video describes the solution to the time independent Schrodinger equation for the quantum harmonic oscillator with power series, including change of variables, removal of the asymptotic solution, the resulting recurrence relation, how the requirement of normalizability leads to quantization, and the structure of the solutions in terms of Hermite polynomials. (This lecture is part of a series for a course based on Griffiths' Introduction to Quantum Mechanics. The Full playlist is at th-cam.com/users/playlist?list=...)
Brant, you have a talent for breaking this down manageable bite sized pieces. Griffiths is also good at explaining the topic in a plain and accessible way. Betwixt the two of you I finally understand QM.
Give this guy some gold.
That's right. The power series must terminate or the solutions go to infinity (not physical), but the power series only terminates for special values of K, which correspond to special values of the energy. Those special energies will make the series terminate either for even powers or for odd powers, but never both, so physics requires us to choose.
We lose a lot of freedom that way, but that's what physics requires, and that's one of the really strange things about quantum mechanics.
Thank you.
At 26:16 you meant to write (K-1) and not (K+1).
You explained termination part very well. Thank you!
You're a lifesaver. Thanks a lot. This really really helped. God bless you
I don't get why at 31.40 the function "h(ξ)" approach to e^ξ^2. Help!
Very clever. Much better than the generic power series method that I was taught (in a time when, in Spain, mathematicians fled from applications as if they were the Plague). As a matter of fact, before I saw this video, I tried myself to get power series solutions to the infinite square well QM problem, running into poorly behaved power series.
Thanks for the video! This was a really helpful breakdown of the analytic method.
I have something I've struggled to understand. Usually when the textbooks explain how the polynomial h approaches e^(g^2) (g is that greek letter, but I dont have it on my keyboard), they first explain that the recursion formula becomes approximately a_(j+2)=(2/j)•a_j for large values of j. Then the say that this implies that a_j is approximately C/(j/2)!, where C is a constant. But the problem I have with this is that using that formula, you could easily obtain an expression relating the (a_j)'s for odd values of j and those for even values of j, which breaks the argument that a_0 and a_1 are arbitrary constants with no relation to each other. The only solution that I can come up with is that there are two constants, C_1 and C_2, in the equation for a_j (and not just one constant C), one for the odd values of j and one for the even values of j. But if those two constants have different signs (one is negative and one positive), then the terms in the series will alternate in signs, which means that the series might not blow up at g=infinity (the terms "cancel each other out"), and most importantly, the series certainly won't become e^(g^2).
This video is amazing. Thank you.
Why is your power (e^2)/2 ? the characteristic polynomial is r=+- xi, so shouldn't it be e^xi *psi
I am missing something.
Thank you so much for this video. Got back on track now.
Look like the ray series (ansatz) in seismology but definitely the ray series does NOT terminate and we have to stay with asymptotic solutions. Will think a bit more..
You are really amazing!! thank for this video
very nice explanation, thank you SO much
nice explanation.. really helpful
thank u so much sir
thanks for really informative video.... God bless u
THANK YOU SO MUCH THIS HELPED ME A LOT!!
thank you so much, really good video!
Old video but imma leave a comment anyway
For h_3 I get the answer: ( ξ - (2/3) ξ^3 ) a_1
However, given the Hermite polynomial for H_3 at the end, it sort of seems like a_1 should be -12 for some reason
Is it due to normalization? Or have I missed something
Got the same answer 😃 and same doubt 😬
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You are a life saver
I'm confused. What's the relationship between h(xi) and the Hermite polynomials? The latter sort of just popped up out of nowhere
Hermite polynomials solve that very same differential equation for h(xi). h(xi) results to be a Hermite polynomial series, normalized with a_0 = 1. The last expression of psi(x) is the solution for each energy eigenstate already normalized for every value of n bringing in the Hermitian polynomials (that considers a_0 = 1).
Helps a lot .thnku so much
2:56 just out of pure curiosity, could you have just substituted the 'xsi' representation of 'x' into the bottom 'x' of the derivative, and 'cancelled' all the coefficient of 'xsi' by multiplying the derivative by 'wm/h-bar', in order to describe the partial derivative in terms of 'xsi'. Since, the resulting expression is identical to the one shown in 6:50 ! If not..why? Is it because it's mathematically unorthodox???
Yes i also feel that we can do it like what you said:
bascically what i feel is that he is non-dimensionalizing the differntial equation since we often do it when we want to solve differntial equation in computer....
Не послушавшись рукописи, наш текст продолжил свой путь.
I think the plot done at around 39:30 curved against the wrong axis
Thank you sir💓
Plzz tell me ..
To satisfy the graph in which equation we have to put the values of E and (xi).
Can anyone explain me why we want to remove the asymptotic behavior, and what effect h(xi) has on the asymptotic solution? Thank you
To answer your first QUESTION why we remove the asymptotic behaviour I'm gonna say it's because of pure mathematical reasons
In the assumed solution of si we Set B=0 because that'd give us infinite value of the solution and the wave function at large values of c .
Same way if c is large the polynomial becomes zero at large values and that's unacceptable from mathematical point of view but we still got in a better position than the last equation so we remove and continue solving .
Watch from 10:00 min you'll understand Mr Brant explains it there.
Thank you
Basically when for large values of c the power series goes to zero that in general isn't a good representation of power series that's why we need to kick out the asymptotic behaviour
Psi ----> as si
C---------> is the sign , the variable that replaces x and psi is a function of this ...
How did the physicists know that writing x in that way makes the differential equation simpler?
We can change the variables x = αξ and see if there is any useful choice of α.
Please can you provide the answer for n = 3 so that we can check our answer. Also why do our worked out solutions not exactly match the provided ones in the tables ( e.g. For n=2)
Check the video at 45:07 , it gives you a rough structure of the solution of the problem
10:10 should be (ξ^2 -1) rather than (ξ^2 -2)
i was wondering the same thing.Thanks
you're a real g
I did not get it. If we get the solution as e-z^2/2 for psi, why do we multiply by h? You said that the reason was it is hard to approximate using the power series but why dk we nedto approximate it with power series?
Thank you...
Потом мы покатались на аттракционе вислоухие горки, которые почему-то назывались Волшебные портянки.
Superb
Now it is clear why k=2n +1 thanks
THANKSSS MAN
Tq
damn your handwriting bro! but thanks for the drop!
Один другому говорит: «что-то не так - воздух и все остальное кругом экологически чистое, а все что мы едим - натуральное, органическое, но почему-то никто не живет дольше тридцати».
(а у него был вес 6 тонн и скорость 3 км в час - страшный зверь!) и протаранил теремок.
What's the name of the book?
Introduction to Quantum Mechanics - David J. Griffiths
Игроку предстоит победить это громадное чудовище, но для этого необходимо подготовиться и набрать команду.
Но если сим подарит букет пожилому, то последний возненавидит его (отношения между ними испортятся) и.
"All of the energy and matter that existed still exists. Matter does not create energy of itself. The actions of matter enable energy to become manifest".
Gerry Nightingale why don't you go back to third grade before you run your mouth
Overlord Master Which aspect of what I wrote can you disprove?
Or do you even understand any of it?
Gerry Nightingale I'm not saying your comment is wrong. My point is that the statement is quite well known to children. There is no real point of commenting that here, it's like I commented, "Gerry Nightingale is an idiot." It's a common fact that most children know already.
Overlord Master Is that all you have? No rebuttal at all? So, you have no comprehension at all of what your objecting to? Why? It's very simple in construct and readily understood.
What sort of parent names a child "Overlord Master"...or are you using that 'name' to 'hide and troll' with?
I think that is the reasonable assumption.
(why not complain to your 'Masters' on the various physics Forums? Perhaps they can exert greater influence than a pretentious troll with delusions of grandeur)
Overlord Master I wasn't aware 'trolls' could do anything other than be 'trolls'...I believe the 'hunting' aspect only exists in your mind.
(you are not the 'Noel Coward' you believe yourself to be...nor even literate)
Why would I want to analyze a 'troll?'
There isn't sufficient motivation to do so, as the knowledge would amount to nothing of any value.
You cannot rebut a single word of the concepts I wrote...and that is sufficient unto itself as a 'successful hunt' on my part.
Yes...take your child-boy fantasy 'name' and run away to 'snipe' at others with your facile 'name-calling'...it's all you have.
(although I'm curious why an obvious 'physics troll poseur' would want to 'comment' on such a 'thread' involving complex issues of the nature of the relationships of energy and matter)
Oh...enjoy your {BLOCK}
at 42:55 i dont understand how -2(n-j) came T_T pls halp
you sub in k=2n+1 in your original recursion relation. Hopefully that clears it up!
В Симс 4 появилась и смерть от эмоций.
Мой парень гуляет просто так.
x times kkkksii
Моя девушка гуляет просто так.
Еще мне хочется отдохнуть в Кремле.