TISE in 3d radial behavior
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- เผยแพร่เมื่อ 13 พ.ค. 2013
- The radial equation resulting from separation of variables in the time-independent Schrodinger equation has solutions involving special functions. The infinite spherical well is a good example, giving spherical Bessel functions as solutions. (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=...)
This is want I've wanted to know since learning about s, p and d orbitals, and special functions. Thanks so much!
Cheers Brant, this has been useful for my dissertation topic, keep up the good work!
This was extremely useful for me, as someone who has only taken a 200 level undergrad quantum course!
Thank you so much for your wonderful job!
thank you so much for these vids
very nice sir
in slide 3 shouldn't there be a + sign with l(l+1) term
can someone please do an explanation or tell me where in the video i can find the answers to the CHECK YOUR UNDERSTADNING questions please
I don't understand, from this derivation, why l must be less than n. Plotting the radial part of the wave function here there are still solutions for l greater than or equal to n, what gives?
Good question, actually for an infinite potential well, n can go up to infinity and hence you have unlimited allowed angular momenta "l", while for finite square wells, like of atoms ex:Hydrogen atom, .... There you will be solving for a slightly different TISE where the potential operator does have a value for V on the boundaries...therefore you will have a limitation on the energy levels driven by the quantum number "n"... hence the recurrence of the power series solution has to terminate... subsequently you have a limitation on "l" = n - j(max) - 1... hence finally "l" has an upper limit equals n-1... If you still don't get it, review the hydrogen atom radial part solution.
The result at 6:26 was the cause of much frustration for me. You tried to do too many steps at once, and ended up with V(r) - hbar^2/2m ... blah blah blah. The result should be V(r) PLUS hbar^2/2m blah blah blah. Try taking your radial equation (with the u/r substituted in) and from there, isolate Eu. You'll see. Sorry if I sound grumpy btw, I just wrestled with your result for about 20 minutes before checking with the book and discovering the right answer.
What about l
l cannot be negative