Finite square well bound states

[@leetingfung]

video less than 30min still way better than lectures in whole semester

[@changbadinesh]

I need exact analytical solution to the energies ..can YOu do it for me?

[@annakiseleva7418]

THANK YOU SO MUCH YOU'RE LITERALLY SAVING MY INTEREST IN QUANTUM HERE

[@BLVGamingY]

at around 6:00 you assume the function only contains sines and cosines instead of letting it be complex exponentials, it ends up the same because once you assume those complex exponentials form an even function, then you arrive at sine anyways

[@c4esium137]

21:52 n should only take odd integer values no?

[@sharairamirez9274]

hi, great video c: I was wondering, what happens if the potential is odd?

[@datsmydab-minecraft-and-mo5666]

Shouldnt it be the second derivative of psi for when you wrote the time independant schrodinger equation?

[@ifrazali3052]

Yes

[@ifrazali3052]

Yes

[@eliaskaroui5665]

please help why E is greater then the potential V in Region 2 ?2:35 why is V greter then E in Region 1 and 3 ?

[@puikihung5882]

see the graph at 1:57. E(x) is always in between o and -V0. Hence E(x) is a negative number. In region 1 and 3, by definition, V(x) is 0, so EV.

[@kemalaziz9696]

Look at the bottom of 3:26, there is an error.

[@NiflheimMists]

Yeah, leftmost term should have d2ψ/dx2 instead of just ψ

[@pranavgeorge992]

In the 2nd region E>V, does this mean the particle is free in that region?

[@account1307]

No a particle is only free if its energy is greater than the potential everywhere, in the second region the energy of the particle is greater than V but it is still less than the maximum value of the potential globally :)

[@BLVGamingY]

​@account1307 bruh he meant in the region and, in the region, the answer is yes, but since the region is finite there are some limitations

[@himangshuchakraborty1760]

Thank you so much.. On 31st may'2021 I've presentation on "Finite square well potential". Nd I found this☺

[@niamphmotley]

where does the 4 come from ?

[@emillytabara9410]

Hi! great explanation . Which software do you use to write?

[@davidhand9721]

Do you have a course on QFT, too? Can anyone recommend one that is as clear and good as this?

[@onlineearning8323]

I need a pdf solution of chapter 11 problems. quantum scattering.. can I get it from you?

[@Domenzain31]

The S.E. is missing the second derivative with respect to x in minute 3:00

[@kokori100]

SOS!I have a question @ 04:30 . for x

[@kartikaloria8256]

|E|

[@rubenlauwaert6673]

I think it is just an assumption that (E

[@puikihung5882]

see the graph at 1:57. E(x) is always in between o and -V0. Hence E(x) is a negative number. In region 1 and 3, by definition, V(x) is 0, so EV.

[@aniruddhavichare5425]

best explanation

[@Chemidan92]

At 9:30 "if we want to have an even function for psi we cannot have a sin term". Can someone explain please

[@LimosRock1]

sin(x) is an odd function so even if you had any even terms, the function for psi will always be odd if there's a sin

[@ashwith]

Would be incorrect to not use the fact that \Psi can be odd or even and do the problem the hard way?

[@gyalofabundance]

Thanks for the explanation! Where did you graph the functions?

[@the-fantabulous-g]

Program's called SAGE, you should be able to find it by searching sage graphing or something similar

[@monicapym8948]

This is exactly what I was looking for, amazing!!

[@Paradox586]

Why only even solution boundary conditions? I don’t get why sine is not considered for even solutions

[@NiflheimMists]

Sine is an odd function, because it is antisymmetric about the origin. Cosine is an even function because it is symmetric about the origin. Both symmetric (even) and antisymmetric (odd) wavefunctions are both, in general, solutions for symmetric potentials, because the magnitude squared of either a symmetric or antisymmetric function is symmetric.

[@HankGussman]

Just plot a graph of sin function from -pi/2 to +pi/2. You will see sin function is not symmetric about vertical axis.

[@xichen9674]

Why is E negative? Is it always negative?

[@monku1521]

The outside of the well is 0 joules. It's when the potential is zero. Since we can't escape the potential (bound states), the Energy is less than zero.

[@vivekpanchal3338]

Thanks for explainning Better than everyone. 👌

[@SampleroftheMultiverse]

Your video is well done!

[@user-rg1nt9lf4s]

very good content sir. thank you .. sir kindly make a video on Bound States for Potential Wells with no rigid walls.

[@arupmarik]

What about last slide (no9)

[@rukwoo4418]

What about cos( la)=0?

[@YourAverageHater]

What happens if the potential is positive? Solution without imaginary roots?

[@zeenaligog]

+Дејан Гујић yes if the V is +, inside the well ,the solutions are exponential without imaginary

[@JohnDavid-iq9rz]

If E = Potential Energy(V) + Kinetic Energy(K), then, how E could be lesser than V ? For E< V, K will have to be negative, which is impossible. Please comment.

[@NiflheimMists]

E cannot be less than V. But it can be negative, if V is. V < 0 E < 0 V < E < 0 is an allowed energy

[@frede1905]

@@NiflheimMists Yes, E CAN be less than V(x) for several points (so for several values of x). It's just that E can't be less than V(x) for ALL points (so for all values of x), because then the wave function can't be normalized (it will either be 0 everywhere, or it will blow up at x=-inf. and/or x=inf.).

[@aniruddhavichare5425]

E is negative because its the binding energy

[@shivanandashekhar6580]

Easily understood

[@beedeelovesyouall]

Thank you so muccchhhhhh. Ur GODDDDDDDD

[@dlmoney11]

cos()/sin() = cot()