TISE in 3d radial behavior

[@roshan8853]

This is want I've wanted to know since learning about s, p and d orbitals, and special functions. Thanks so much!

[@ninjaQK]

Cheers Brant, this has been useful for my dissertation topic, keep up the good work!

[@sshamsi]

This was extremely useful for me, as someone who has only taken a 200 level undergrad quantum course!

[@haoqinggenius4335]

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

[@fornasm]

is this a course in mathematics of physics???????

[@ThePacmanamcap]

Thank you so much for your wonderful job!

[@mankalememartin1371]

thank you so much for these vids

[@iqranasir9224]

in slide 3 shouldn't there be a + sign with l(l+1) term

[@balabhadraharipal287]

very nice sir

[@alexanderdavidsonbryan7264]

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?

[@mohameda.444]

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.

[@KingCrocoduck]

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.

[@Salmanul_]

What about l

[@patasefios]

l cannot be negative